Pearson Advanced Chemistry Series
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Inorganic Chemistry
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Physical Chemistry: Thermodynamics, Statistical Mechanics, and Kinetics
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Physical Chemistry
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Physical Chemistry: Principles and Applications in Biological Sciences
by Ignacio Tinoco Jr. University of California Berkeley, Kenneth Sauer University of
California Berkeley, James C. Wang Harvard University, Joseph D. Puglisi
Stanford University, Gerard Harbison University of Nebraska Lincoln,
David Rovnyak Bucknell University
Quantum Chemistry
by Ira N. Levine Brooklyn College, City College of New York
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F I F TH EDITION
Physical Chemistry
Principles and Applications in Biological Sciences
Ignacio Tinoco, Jr.
University of California, Berkeley
Kenneth Sauer
University of California, Berkeley
James C. Wang
Harvard University
Joseph D. Puglisi
Stanford University
Gerard Harbison
University of Nebraska, Lincoln
David Rovnyak
Bucknell University
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Data obtained from the RCSB Protein Data Bank (www.pdb.org). H.M. Berman, J. Westbrook, Z.
Feng, G. Gilliland, T.N. Bhat, H. Weissig, I.N. Shindyalov, P.E. Bourne (2000) The Protein Data Bank.
Nucleic Acids Research, 28: 235–242. Data files contained in the PDB archive include:
Top Structure: PDB ID 3CLN
Y. S. Babu, C. E. Bugg, W. J. Cook (1988) Structure of calmodulin refined at 2.2 A resolution. J. Mol.
Biol. 204: 191–204.
Middle Structure: PDB ID 1CM1
M.E. Wall, J. B. Clarage, G. N. Phillips Jr. (1997) Motions of calmodulin characterized using both
Bragg and diffuse X-ray scattering. Structure 5: 1599–1612.
Bottom Structure: PDB ID 1CFD
H. Kuboniwa, N. Tjandra, S. Grzesiek, H. Ren, C. B. Klee, C.B., A. Bax (1995) Solution structure of
calcium-free calmodulin. Nat. Struct. Biol. 2: 768–776.
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ISBN–10: 0-13-605606-7; ISBN–13: 978-0-13-605606-5
Brief Contents
Chapter 1
Introduction
1
Chapter 2
The First Law: Energy Is Conserved
Chapter 3
The Second Law: The Entropy of the Universe Increases
Chapter 4
Free Energy and Chemical Equilibria
Chapter 5
The Statistical Foundations of Biophysical Chemistry 151
Chapter 6
Physical Equilibria
Chapter 7
Electrochemistry
Chapter 8
The Motions of Biological Molecules
Chapter 9
Kinetics: Rates of Chemical Reactions 305
Chapter 10
Enzyme Kinetics
Chapter 11
Molecular Structures and Interactions: Theory
Chapter 12
Molecular Structures and Interactions: Biomolecules
Chapter 13
Optical Spectroscopy
489
Chapter 14
Magnetic Resonance
539
Chapter 15
Macromolecular Structure and X-Ray Diffraction
Appendix
Mathematics
Appendix
Tables
13
55
101
196
238
264
378
408
453
574
618
628
iii
Contents
Preface xv
New to This Edition xvii
About the Authors xx
Chapter 1
Introduction
1
Neuroscience 2
The Human Genome and Beyond 3
Transcription and Translation 5
Ion Channels 9
Single-molecule Methods 10
Reference 11 • Suggested Reading
Chapter 2
12 •
Problems
The First Law: Energy Is Conserved
Concepts 13
Applications 14
Energy Conversion and Conservation
Systems and Surroundings 15
Energy Exchanges 15
Work
16
Heat
20
Internal Energy
12
13
14
21
Constant Volume Heat Capacity
24
Constant Volume Heat Capacity of Diatomic Gases 24
Constant Volume Heat Capacity of Monatomic Solids 25
Heat Capacity of Molecular Solids and Liquids
State and Path Variables
26
Reversible Paths and Reversible Processes
Equations of State
The Enthalpy
25
27
28
30
The Constant Pressure Heat Capacity of an Ideal Gas 31
Dependence of the Energy and Enthalpy of a Pure Substance on p, V, and T
Liquids or Solids
Gases
32
35
Phase Changes 36
Chemical Reactions
39
Heat Effects of Chemical Reactions
Temperature Dependence of DrH
Standard Enthalpies of Formation
iv
40
42
42
31
Contents | v
The Energy Change DE for a Reaction
44
Computing Reaction Energies from First Principles
Quantum Chemical Calculations
Bond Energies
44
44
44
Molecular Interpretations of Energy and Enthalpy
Summary 47 • References 50
Suggested Reading 50 • Problems 50
Chapter 3
46
The Second Law: The Entropy of the Universe Increases
Concepts 55
Applications 55
Toward the Second Law: The Carnot Cycle
A New State Function, Entropy 58
The Second Law of Thermodynamics 60
Molecular Interpretation of Entropy 62
Fluctuations
56
64
Measurement of Entropy 65
Chemical Reactions 65
Third Law of Thermodynamics
66
Temperature Dependence of Entropy
66
Temperature Dependence of the Entropy Change for a Chemical Reaction
Entropy Change for a Phase Transition
Pressure Dependence of Entropy
70
Spontaneous Chemical Reactions
71
Gibbs Free Energy
70
72
DG and a System’s Capacity to Do Nonexpansion Work
Spontaneous Processes at Constant T and p
Calculation of Gibbs Free Energy
73
73
Temperature Dependence of Gibbs Free Energy
Pressure Dependence of Gibbs Free Energy
Phase Changes
75
77
80
Helmholtz Free Energy 80
Noncovalent Reactions 80
Hydrophobic Interactions
Proteins and Nucleic Acids
82
83
Use of Partial Derivatives in Thermodynamics
Relations Among Partial Derivatives
87
The Thermodynamic Square 91
The Gibbs-Helmholtz Equation 92
Summary 93 • References 95
Suggested Reading 95 • Problems
96
87
72
68
55
vi | Contents
Chapter 4
Free Energy and Chemical Equilibria
Concepts 101
Applications 102
Partial Molar Gibbs Energy
Chemical Potential
101
102
102
The Sum Rule for Partial Molar Quantities
Directionality of Chemical Reaction
Reactions of Ideal Gases
102
103
104
Dependence of Chemical Potential on Partial Pressures 104
Equilibrium Constant
Nonideal Systems
Activity
105
108
108
Standard States
109
Activity Coefficients of Ions
117
Equilibrium and the Standard Gibbs Free Energy
119
Calculation of Equilibrium Concentrations: Ideal Solutions
Temperature Dependence of the Equilibrium Constant
Biochemical Applications of Thermodynamics
Thermodynamics of Metabolism
Isothermal Titration Calorimetry
121
126
129
135
139
Double Strand Formation in Nucleic Acids 140
Ionic Effect on Protein–Nucleic Acid Interactions
Summary 143 • References 146
Suggested Reading 146 • Problems
Chapter 5
143
146
The Statistical Foundations of Biophysical Chemistry
Concepts 151
Applications 151
Maxwell Boltzmann Statistics
152
The Boltzmann Distribution
152
The Maxwell-Boltzmann Distribution
153
The Maxwell-Boltzmann Distribution and the Speed 155
Statistical Thermodynamics
156
Statistical Mechanical Internal Energy
Work
Heat
156
157
158
Most Probable (Boltzmann) Distribution 158
Statistical Mechanical Entropy
162
Examples of Entropy and Probability 163
Partition Function: Applications
164
151
Contents | vii
The Random Walk
166
Calculation of Some Mean Values for the Random-Walk Problem 168
Diffusion
170
Average Dimension of a Linear Polymer
Helix–Coil Transitions
171
173
Helix–Coil Transition in a Polypeptide
173
Helix–Coil Transition in a Double-Stranded Nucleic Acid 176
Binding of Small Molecules by a Polymer 180
Identical-and-Independent-Sites Model
Langmuir Adsorption Isotherm
181
183
Nearest-Neighbor Interactions and Statistical Weights
184
Cooperative Binding, Anticooperative Binding, and Excluded-Site Binding
N Identical Sites in a Linear Array with Nearest-Neighbor Interactions 188
Identical Sites in Nonlinear Arrays with Nearest-Neighbor Interactions 188
Summary 190 • References 193
Suggested Reading 193 • Problems
Chapter 6
Physical Equilibria
193
196
Concepts 196
Applications 196
Membranes and Transport
Ligand Binding
Colligative Properties
Phase Equilibria
196
197
197
197
One component systems
197
Solutions of Two or More Components
Membranes
202
214
Lipid Molecules
Lipid Bilayers
214
214
Phase Transitions in Lipids, Bilayers, and Membranes
Surface Tension
218
Surface Free Energy
221
Vapor Pressure and Surface Tension
Biological Membranes
Colligative Properties
222
223
225
Boiling-Point Elevation and Freezing-Point Depression
Osmotic Pressure
216
227
Molecular-Weight Determination
229
Summary 230 • References 231
Suggested Reading 232 • Problems
232
225
186
viii | Contents
Chapter 7
Electrochemistry
238
Concepts 238
Applications 238
Basic Electricity 239
Capacitance and Electrical Neutrality
Ground and the Reference Potential
The Electrochemical Cell
240
241
241
Reversibility in the Electrochemical Cell
241
Electrical Work, Electrochemical Potential, and Free Energy
Standard Electrochemical Potentials
Concentration Dependence of e
Transmembrane Equilibria
242
243
245
247
Donnan Effect and Donnan Potential 248
Plasma Membrane Potentials and the Na+, K+ ATPase
250
Biological Redox Reactions and Membranes 254
Oxidative Phosphorylation
255
NADH-Q Reductase (Complex I)
255
Succinate Dehydrogenase (Complex II)
256
Coenzyme Q – Cytochrome c Oxidoreductase (Complex III)
Cytochrome c Oxidase (Complex IV)
257
Mitochondrial Oxidation of NAD + 258
ATP Synthase
258
Summary 259 • References 260
Suggested Reading 260 • Problems
Chapter 8
260
The Motions of Biological Molecules
Concepts 264
Applications 265
Molecular Motion and Molecular Collisions
The Collision Tube
264
265
266
Random Walks in a Gas 268
Diffusion
269
The Diffusion Coefficient and Fick’s First Law 269
Fick’s Second Law
270
The Einstein-Smoluchowski Relation
270
Determination of the Diffusion Coefficient 272
Values of the Diffusion and Self-Diffusion Coefficient
The Frictional Coefficient
f and D
274
275
Shape Factor
276
Diffusion Coefficients of Random Coils
Sedimentation
278
Analytical Centrifugation
Sedimentation Equilibrium
279
282
277
274
257
Contents | ix
Molecular Weights from Sedimentation and Diffusion
Density-Gradient Centrifugation
Viscosity
283
284
285
Measurement of Viscosity
Viscosities of Solutions
Electrophoresis
285
286
286
Gel Electrophoresis
287
Conformations of Nucleic Acids
290
Pulsed-Field Gel Electrophoresis
Protein Molecular Weights
Protein Charge
290
292
293
Macromolecular Interactions
294
Size and Shape of Macromolecules 294
Summary 295 • References 298
Suggested Reading 298 • Problems 299
Chapter 9
Kinetics: Rates of Chemical Reactions
305
Concepts 305
Applications 306
Kinetics 306
Rate Law
308
Order of a Reaction
308
Experimental Rate Data
310
Zero-Order Reactions
310
First-Order Reactions
311
Second-Order Reactions
317
Renaturation of DNA as an Example of a Second-Order Reaction
Reactions of Other Orders
324
Determining the Order and Rate Coefficient of a Reaction 325
Reaction Mechanisms and Rate Laws
Parallel Reactions
329
Series Reactions (First Order)
Equilibrium and Kinetics
Complex Reactions
327
330
333
335
Deducing a Mechanism from Kinetic Data
337
Temperature Dependence 338
Transition-State Theory 341
Electron Transfer Reactions: Marcus Theory 344
Ionic Reactions and Salt Effects 345
Isotopes and Stereochemical Properties 347
Very Fast Reactions 348
Relaxation Methods
348
Relaxation Kinetics
349
321
x | Contents
Diffusion-Controlled Reactions 354
Single-Molecule Kinetics 356
Photochemistry and Photobiology 358
Vision
361
Photosynthesis
361
Summary 364 • References 369
Suggested Reading 369 • Problems 370
Chapter 10
Enzyme Kinetics
378
Concepts 378
Applications 378
Catalytic Antibodies and RNA Enzymes—Ribozymes
Enzyme Kinetics 380
Michaelis–Menten Kinetics
Kinetic Data Analysis
382
385
Two Intermediate Complexes
Competition and Inhibition
Competition
389
391
391
Competitive Inhibition
391
Noncompetitive Inhibition
Allosteric Effects
393
394
Single-Molecule Kinetics
396
Summary 398 • References 400
Suggested Reading 400 • Problems
Chapter 11
378
401
Molecular Structures and Interactions: Theory
Concepts 408
Application to Vision 410
Origins of Quantum Theory
412
Origins: Blackbody Radiation
Origins: Hydrogen Emission
412
413
Origins: Photoelectric Effect
414
Origins: Electrons as Waves
414
Origins: Heisenberg Uncertainty Principle
415
Origins: Classical Waves and Quantization
415
Quantum Mechanical Calculations 416
Wave Mechanics and Wavefunctions 417
The Schrödinger Equation
419
Solving Wave Mechanical Problems
421
Outline of Wave Mechanical Procedures
Particle in a Box
Example of a Particle-in-a-Box Calculation
Tunneling
429
421
423
427
408
Contents | xi
Simple Harmonic Oscillator
Rigid Rotator 433
Hydrogen Atom 434
Electron Distribution 436
430
Electron Distribution in a Hydrogen Atom
Many-Electron Atoms
Hybridization
442
Origins: Postulates 444
Summary 445 • References 449
Suggested Reading 449 • Problems
Chapter 12
436
440
449
Molecular Structures and Interactions: Biomolecules
Concepts
453
Molecular Orbitals
453
Delocalized Orbitals
458
Molecular Structure and Molecular Orbitals
Geometry and Stereochemistry
Transition Metal Ligation
459
460
461
Charge Distributions and Dipole Moments 463
Intermolecular and Intramolecular Forces
463
Bond Stretching and Bond Angle Bending 464
Rotation Around Bonds
Noncovalent Interactions
464
466
Electrostatic Energy and Coulomb’s Law
466
Net Atomic Charges and Dipole Moments
Dipole–Dipole Interactions
London Attraction
472
473
van der Waals Repulsion
474
London–van der Waals Interaction
475
The Lowest-Energy Conformation
476
Hydrogen Bonds
469
477
Hydrophobic and Hydrophilic Environments
Molecular Dynamics Simulation
Monte Carlo Method
480
481
Molecular Dynamics Method
481
Outlook 483
Summary 484 • References 485
Suggested Reading 485 • Problems
Chapter 13
Optical Spectroscopy
Concepts 489
Applications 490
Electromagnetic Spectrum
490
489
485
479
453
xii | Contents
Color and Refractive Index 490
Absorption and Emission of Radiation
Radiation-Induced Transitions
Classical Oscillators
492
492
494
Quantum Mechanical Description
Lifetimes and Line Width
494
496
Role of Environment in Electronic Absorption Spectra
Beer–Lambert Law
497
498
Proteins and Nucleic Acids: Ultraviolet Absorption Spectra 502
Amino Acid Spectra
503
Polypeptide Spectra
504
Secondary Structure
505
Nucleic Acids
505
Rhodopsin: A Chromophoric Protein
Fluorescence
506
507
Simple Theory
508
Excited-State Properties
509
Fluorescence Quenching
512
Excitation Transfer
Molecular Rulers
514
516
Fluorescence Polarization
Phosphorescence
517
518
Single-Molecule Fluorescence Spectroscopy
519
Optical Rotatory Dispersion and Circular Dichroism
Polarized Light
520
Optical Rotation
522
Circular Dichroism
524
Circular Dichroism of Nucleic Acids and Proteins
Vibrational Spectroscopy
Infrared Absorption
Raman Scattering
526
527
528
Summary 530 • References 532
Suggested Reading 532 • Problems
Chapter 14
Magnetic Resonance
Concepts 539
Applications 540
Nuclear Magnetic Resonance
533
539
541
Nuclear Spin Energy Levels
The Spectrum
520
541
543
A Pulse in the Rotating Frame
545
Interactions in Nuclear Magnetic Resonance 547
524
Contents | xiii
Chemical Shifts
547
Spin–Spin Coupling, Scalar Coupling, or J-Coupling
Relaxation Mechanisms
552
Nuclear Overhauser Effect
555
Multidimensional NMR Spectroscopy
555
Determining Macromolecular Structure by NMR
Electron Paramagnetic Resonance
560
561
Magnetic Field Gradients, Diffusion and Microscopy
Magnetic Resonance Imaging
NMR Hardware Overview
562
565
566
Summary 567 • References 568
Suggested Reading 568 • Problems
Chapter 15
549
569
Macromolecular Structure and X-Ray Diffraction
574
Concepts 574
Applications 575
Lattices 575
Symmetry
576
Symmetry in Three Dimensions
578
Images 579
X-Rays 579
Emission of X-Rays
Image Formation
579
580
Scattering of X-Rays
580
Diffraction of X-Rays by a Crystal
585
Measuring the Diffraction Pattern
587
Bragg Reflection of X-Rays
Intensity of Diffraction
Unit Cell
589
591
592
Determination of Molecular Structure
593
Calculation of Diffracted Intensities from Atomic Coordinates: The Structure Factor
Calculation of Atomic Coordinates from Diffracted Intensities
The Phase Problem
Direct Methods
594
595
595
Isomorphous Replacement
597
Multiwavelength Anomalous Diffraction
Determination of a Crystal Structure
Accessing Crystal Structures
599
599
602
Scattering of X-Rays by Noncrystalline Materials
Absorption of X-Rays
602
604
Extended Fine Structure of Edge Absorption (EXAFS)
604
X-Rays from Synchrotron Radiation and Free-Electron Lasers 605
Electron Diffraction
606
593
xiv | Contents
Neutron Diffraction 607
Electron Microscopy 608
Resolution, Contrast, and Radiation Damage
608
Transmission and Scanning Electron Microscopes
Image Enhancement and Reconstruction
609
Scanning Tunneling and Atomic Force Microscopy
Summary 611 • References 614
Suggested Reading 614 • Problems
Appendix
Mathematics
609
610
615
618
Derivatives 618
Integration 621
Improper Integrals 622
Exponents and Logarithms 623
Series and Approximations 624
Mathematics for Quantum Mechanics: Hilbert Space 625
References 627
Tables
628
Selected Answers 639
Index 645
Preface
There is a deep sense of pleasure to be experienced when the patterns and symmetry of
nature are revealed. Physical chemistry provides the methods to discover and understand
these patterns. We think that not only is it important to learn and apply physical chemistry
to biological problems, it may even be fun. In this book, we have tried to capture some of
the excitement of making new discoveries and finding answers to fundamental questions.
This is not an encyclopedia of physical chemistry. Rather, we have written this text
specifically with the life-science student in mind. We present a streamlined treatment that
covers the core aspects of biophysical chemistry (thermodynamics and kinetics as well as
quantum mechanics, spectroscopy, and X-ray diffraction), which are of great importance
to students of biology and biochemistry. Essentially all applications of the concepts are
systems of interest to life-science students; nearly all the problems apply to life-science
examples.
For the fifth edition, we have extensively revised and updated the treatment of
biophysical chemistry, bringing in theoretical approaches earlier and also updating the
text to current IUPAC conventions. We have added a new chapter on electrochemistry and
expanded our treatment of single molecule methods, quantum mechanics, and magnetic
resonance.
Chapter 1 introduces representative areas of active current research in biophysical
chemistry and molecular biology: the human genome, the transfer of genetic information
from DNA to RNA to protein, ion channels, and cell-to-cell communication. We encourage
students to read the current literature to see how the vocabulary and concepts of physical
chemistry are used in solving biological problems.
Chapters 2 through 4 cover the laws of thermodynamics and their applications to
chemical reactions and physical processes. Essentially all of the examples and problems
deal with biochemical and biological systems. For example, after defining work as a
force multiplied by the distance moved (the displacement), we discuss the experimental
measurement of the work necessary to stretch a single DNA molecule from its randomcoiled form to an extended rod, introducing the intuitive and accessible concept of
molecular force microscopy. We also include a new and more comprehensive treatment
of heat capacities, beginning with the kinetic theory of gases, which is now treated much
earlier in chapter 2, and moving systematically to a consideration of what affects the heat
capacity of a protein. Molecular interpretations of energies and entropies are emphasized
in each of chapters 2 through 4. We also introduce isothermal titration calorimetry in
chapter 3. Despite this new content, the length of chapters 2 through 4 as been reduced by
over 30 pages, largely by eliminating redundant material.
In chapter 5, we show how the thermodynamic laws discussed in chapters 2 through 4
can be explained by a statistical treatment of molecular motion and interactions, and apply
these statistical methods to the conformation of proteins and DNA and the binding of
ligands. This section has been combined with the conceptually-related statistical treatment
of Maxwell-Boltzmann gases and appears much earlier. In chapter 6, we immediately use
these statistical insights to explain physical phenomena such as phase transitions, ligand
binding, and surface and membrane effects.
In chapter 7, we present a new and integrated treatment of electrical and electrontransfer phenomena in biophysics, starting with classical electrochemistry, and considering
how the chemical processes of electron transfer are linked to the physical processes of ion
translocation to explain most of biological energy transduction.
xv
xvi | Preface
Chapters 8 through 10 cover molecular motion and chemical kinetics. Chapter 8 starts
with a discussion of molecular collisions, random walks, and brownian motion. Fluorescence
microscopic tracking of single protein molecules diffusing in membranes is shown to
beautifully corroborate Einstein’s equation relating average distance traveled by a single
molecule to its bulk diffusion coefficient. Following this direct experimental demonstration
of thermal motion of a molecule, we consider the bulk transport of molecules by diffusion,
sedimentation, viscous flow, and electrophoresis. The next two chapters deal with general
chemical kinetics and enzyme kinetics, including single-molecule enzyme kinetics.
In the 5th edition, we reflect the rapidly expanding importance of quantum mechanics
and diverse powerful spectroscopies in understanding molecular biological phenomena
by presenting these subjects in four more focused and augmented chapters. Chapter 11,
“Molecular Structures and Interactions: Theory,” now focuses solely on the origins and key
introductory results of quantum theory, including a review of the postulates. Chapter 12,
“Molecular Structures and Interactions: Biomolecules,” presents molecular orbital theory,
interactions, and an overview of computational methods applied to macromolecules.
Similarly, the treatment of spectroscopy is now more focused in two separate chapters
on optical (chapter 13) and magnetic (chapter 14) methods, respectively. Chapter 13
increases emphasis on absorption and fluorescence, and includes new material on protein
IR spectroscopy. Chapter 14 introduces the classical framework for NMR in more detail
and covers new methods in multidimensional and diffusion NMR.
Chapter 15 discusses X-ray diffraction, electron microscopy, and scanning microscopies
(such as atomic force microscopy), and emphasizes how structures are determined
experimentally. We added a new section on crystal lattices and symmetry, and expanded
the discussion of modern methods such as X-ray imaging and free-electron lasers.
A new appendix in the fifth edition is an accessible, self-contained, and pragmatic
review of the mathematics expectations in this text. We hope the carefully defined scope
of the mathematics (a characteristic of previous editions) will be reassuring in preparing
to study this text.
We are gratified by the number of faculty who have elected to use this book over
the many years since it was first published. We are also grateful for the many students
and faculty who have given us their thoughts and impressions. Such feedback has helped
improve the book from edition to edition. We are particularly grateful to those of our
colleagues who commented on the fifth edition:
Noah W. Allen, III—University of North Carolina, Asheville
Jason Benedict—University of Buffalo
Tim Keiderling—University of Illinois, Chicago
Ruth Ann Murphy—University of Mary Hardin Baylor
Tatyana Smirnova—North Carolina State University
Keith J. Stine—University of Missouri-St. Louis
Gianluigi Veglia—University of Minnesota
Jeff Woodford—Missouri Western State University
Danny Yeager—Texas A&M University
Kazushige Yokoyama—State University of New York, Geneseo
We welcome your comments.
Ignacio Tinoco, Jr.
Kenneth Sauer
James C. Wang
Joseph D. Puglisi
David Rovnyak
Gerard Harbison
New to This Edition
The major theme of this revision is to update and reorder a classic text to reflect changes
in scientific knowledge and student learning styles, while retaining the time-tested
central core.
• Over 200 new and revised figures to help students visualize and understand the
concepts discussed within each chapter.
• At least 20-25% new and revised end-of-chapter problems
• New treatment of the most modern methods, including free-electron laser X-ray
imaging, single-molecule microscopy, and isothermal titration calorimetry.
• Quantum mechanics has been split into two chapters covering basic theory
and molecular properties; spectroscopy likewise has been split into optical
spectroscopy and NMR.
• New chapter on electrical phenomena (chapter 7), which integrates
electrobiochemistry, redox biology, and electrophysiology in a single location.
• New focus on single molecule microscopy, dynamics, spectroscopy, and kinetics.
• New molecular based development of heat capacities from ideal gases to proteins,
along with new sections on calorimetric measurements.
• Better integration of statistical theories of molecular conformation and binding,
which now appear earlier in the text and directly precede experimental treatment
of phase transitions, binding, and membranes.
• Addition of MasteringChemistry for Physical Chemistry. The
MasteringChemistry platform is the most widely used and effective online
homework, tutorial, and assessment system for the sciences. It delivers self-paced
tutorials that focus on your course objectives, provide individualized coaching,
and respond to each student’s progress. The Mastering system helps instructors
maximize class time with easy-to-assign, customizable, and automatically graded
assessments that motivate students to learn outside of class and arrive prepared
for lecture or lab.
Chapter-by-Chapter Changes
Chapter 1
• new sections on neuroscience and on single-molecule methods
• the Human Genome Initiative section was brought up to date
Chapter 2
• new treatment of molecular-force microscopy
• new development of the concept of the heat capacity at a molecular level,
beginning with the kinetic theory of monatomic gases, and extending to proteins
• new and clearer discussion of paths, states and phase transitions
Chapter 3
• expanded introduction and illustrations of Carnot cycle and associated principles
• the thermodynamic square helps students see thermodynamics holistically
xvii
xviii |
New to this Edition
Chapter 4
• expanded introduction and illustrations of chemical potential
• isothermal titration calorimetry is included to reflect its broad importance in drug
discovery and characterizing substrate interactions
Chapter 5
• reorganized treatment of statistical thermodynamics, beginning with the MaxwellBoltzmann distribution of gases, and proceeding to statistics of discrete and
quantized systems, helix-coil statistics of macromolecules, and ligand binding
Chapter 6
• the Clapeyron equations for phase transitions are now developed from the visual
concept of three-dimensional free energy surfaces
Chapter 7
• brand new chapter on electrical phenomena in biophysics
• begins with electrochemical cells and biophysical applications of the Nernst equation
• moves on to transmembrane equilibria, the Donnan effect, ion pumps, and
neuroelectrophysiology
• ends with biological redox reactions and a full, up-to-date exploration of oxidative
phosphorylation
Chapter 8
• new treatment of molecular collisions and their effect on mean square
displacements
• new development of the diffusion equations in one, two and three dimensions
using Einstein’s original arguments
• new discussion of single molecule microscopic measurements of diffusion in twoand three-dimensions
• new treatment of effect of shape on diffusion and sedimentation coefficients
• improved and rewritten section on analytical ultracentrifugation
Chapter 9
• sections on differential and integrated rate equations rewritten and clarified
• new section on single-molecule kinetics
Chapter 10
• new discussion of Michaelis-Menten kinetics with an emphasis on direct leastsquares fitting of data, compared to older, statistically inferior linearization methods
Chapter 11
• now pedagogically focused solely on introducing quantum mechanical origins
and key applications
• stronger connection between classical and quantum mechanics is forged
• postulates are now included which can help students build their understanding of
the logical structure of quantum mechanics
New to this Edition | xix
Chapter 12
• streamlined for focus solely on molecular orbitals, intermolecular and
intramolecular interactions
• focus on analyzing coefficients in molecular orbital wavefunctions through
introductory material and problem sets
Chapter 13
• introducing optical principles of spectroscopy solely will help students and
instructors better organize and build their approach to spectroscopy in biophysics;
• focus on visualizing spectroscopic concepts such as waves, the transition dipole,
and the Franck-Condon principle
• protein infra-read spectroscopy gives valuable information on structure,
complementary to circular dichroism
Chapter 14
• introduces the vector model for pulsed magnetic resonance
• characterizing molecular rotation and size via relaxation, connecting to StokesEinstein principles of chapter 8
• protein 3D-NMR methods and gradient diffusion methods for particle size and
microscopy
Chapter 15
• new and highly visual treatment of lattices and symmetry
• new treatment of ‘frontier’ research techniques such as x-ray imaging using
free-electron lasers
About the Authors
Ignacio Tinoco was an undergraduate at the University of New Mexico, a graduate
student at the University of Wisconsin, Madison, and a postdoctoral fellow at Yale. He
then went to the University of California, Berkeley, where he has remained. His research
interest has been on the structures of nucleic acids, particularly RNA. He was chairman
of the Department of Energy committee that recommended in 1987 a major initiative to
sequence the human genome. He is a member of the National Academy of Sciences and
of the American Academy of Arts and Sciences. His present research is using singlemolecule methods to determine how the ribosome synthesizes proteins.
Kenneth Sauer grew up in Cleveland, Ohio, and received his A.B. in chemistry from
Oberlin College. Following his Ph.D. studies in gas-phase physical chemistry at Harvard, he
spent three years teaching at the American University of Beirut, Lebanon. A postdoctoral
opportunity to learn from Melvin Calvin about photosynthesis in plants led him to the
University of California, Berkeley, where he has been since 1960. Teaching general
chemistry and biophysical chemistry in the Chemistry Department has complemented
research in the Physical Biosciences Division of the Lawrence Berkeley National Lab
involving spectroscopic studies of photosynthetic light reactions and their role in water
oxidation. His other activities include reading, renaissance and baroque choral music,
canoeing, and exploring the Sierra Nevada with his family and friends.
James C. Wang was on the faculty of the University of California, Berkeley, from
1966 to 1977. He then joined the faculty of Harvard University, where he is presently
Mallinckrodt Professor of Biochemistry and Molecular Biology. His research focuses
on DNA and enzymes that act on DNA, especially a class of enzymes known as DNA
topoisomerases. He has taught courses in biophysical chemistry and molecular biology
and has published over 200 research articles. He is a member of Academia Sinica, the
American Academy of Arts and Sciences, and the U.S. National Academy of Sciences.
Joseph Puglisi was born and raised in New Jersey. He received his B.A. in chemistry
from The Johns Hopkins University in 1984 and his Ph.D. from the University of California,
Berkeley, in 1989. He has studied and taught in Strasbourg, Boston, and Santa Cruz, and
is currently professor of structural biology at Stanford University. His research interests
are in the structure and mechanism of the ribosome and the use of NMR spectroscopy to
study RNA structure. He has been a Dreyfus Scholar, Sloan Scholar, and Packard Fellow.
Gerard Harbison was born in the United Kingdom and raised there and in Ireland. He
received his B.A. in biochemistry from Trinity College, Dublin, and his Ph.D. in biophysics
from Harvard University. After a brief postdoctoral sojourn at the Max-Planck Institute for
Polymer Research in Mainz, Germany, he joined the faculty of Stony Brook University, and
then moved to the University of Nebraska Lincoln. He is a Dreyfus Teacher-Scholar, Lilly
Foundation Teacher-Scholar and Presidential Young Investigator. His research interests
are in nuclear magnetic resonance and electronic structure theory.
David Rovnyak, a native of Charlottesville, Virginia, earned his B.S. in Chemistry at the
University of Richmond and Ph.D. in physical chemistry from the Massachusetts Institute
of Technology. After performing post-doctoral study at the Harvard Medical School under
an NIH-NRSA fellowship, he joined Bucknell University where he has been recognized
with the Bucknell Presidential Teaching Award for Excellence. His research focuses on new
methods for NMR spectroscopy and physico-chemical behavior of bile acids.
xx
Chapter 1
Introduction
Physical chemistry is everywhere. Physical chemical principles are basic to the methods
used to determine the sequence of the human genome, obtain atomic resolution structures
of proteins and nucleic acids, and learn how biochemicals react and interact to make a cell
function. Once you learn physical chemistry, you will subconsciously apply your knowledge to each scientific paper you read and to each explanation of an experiment you hear
or propose yourself. “What provides the energy for the reaction? What about the Second
Law? The proposed mechanism seems to violate microscopic reversibility. An intermediate
is proposed; I can identify it by fluorescence, or nuclear magnetic resonance (NMR), or
Raman scattering.” Most importantly, you will realize that to study any biological process
you need to use the methods of physical chemistry. Understanding how the brain works has
become among the most active areas of biological research. The neurons in the brain communicate with each other by electrical signals and the transfer of neurotransmitters. The
structures of all the players (neurotransmitters, neurotransporters, receptors, ion channels,
synapsis proteins, and so forth) need to be determined by X-ray diffraction or by NMR.
These molecules move, change conformation, and interact with other molecules in response
to action potentials. Their positions, interactions, rates of motion, shapes, and sizes are
measured by high resolution optical microscopy, atomic force microscopy, or electron
microscopy. Functional magnetic resonance imaging, and infrared absorption and scattering reveal which parts of the brain are most active while you are thinking about whatever
you are doing. In the next fifty years new methods and new applications of old methods will
reveal new answers and completely new questions to ask about every biological process.
Physical chemistry is a set of principles and experimental methods for exploring
chemical and biological systems. The power of physical chemistry lies in its generality.
The principles described in this book can be applied to systems as large as the cosmos
and as small as an individual atom. Physical chemistry has been especially powerful in
understanding fundamental biological processes. In the following chapters, we will present
the principles of thermodynamics, transport properties, kinetics, quantum mechanics and
molecular interactions, spectroscopy, and scattering and diffraction. We will also discuss
various experimentally measurable properties such as enthalpy, electrophoretic mobility,
light absorption, and X-ray diffraction. All these experimental and theoretical methods give
useful information about whatever problem you want to solve. We emphasize the molecular
interpretation of these methods and stress biochemical and biological applications. By learning the principles behind the methods, you will be able to judge the conclusions obtained
from them. This is the first step in inventing new methods or discovering new concepts.
First, a quick tour of the book. Chapters 2 through 4 cover the fundamentals of
thermodynamics and their applications to chemical reactions and physical processes.
Because these chapters review material usually covered in beginning chemistry courses,
1
2 Chapter 1 |
Introduction
we emphasize the applications to biological macromolecules. Chapter 5 covers the statistical basis of thermodynamics; it provides a molecular interpretation of thermodynamics.
Cooperative binding of ligands to macromolecules, plus helix-coil transitions in nucleic
acids and proteins are described. Chapter 6 covers physical equilibria, including osmotic
pressure, equilibrium dialysis, and membrane equilibria. Chapter 7 deals with electrochemistry, including Galvanic cells, Donnan equilibria, and transmembrane potentials.
The effect of sizes and shapes of molecules on their translational and rotational motions
in gases, liquids, and gels are discussed in chapter 8. The driving forces for molecular
motion are either random thermal forces that cause diffusion or the directed forces in sedimentation, flow, and electrophoresis. Chapter 9 describes general kinetics, and chapter 10
concentrates on the kinetics of enzyme-catalyzed reactions. Chapter 11 introduces the
quantum mechanical principles necessary for understanding bonding and spectroscopy,
and chapter 12 describes calculations of protein and nucleic acid conformations using
classical force fields (Coulomb’s Law, van der Waals’ potential). Chapter 13 includes the
main spectroscopic methods used for studying molecules in solution: ultraviolet, visible,
and infrared absorption; fluorescence emission; circular dichroism; and optical rotatory
dispersion. Chapter 14 is devoted to nuclear magnetic resonance (NMR); it discusses the
fundamentals of the method for determining structures of proteins and nucleic acids.
Chapter 15 starts with the scattering of electromagnetic radiation from one electron and
proceeds through the diffraction of X-rays by crystals. Scanning microscope methods are
introduced. The appendix contains numerical data used throughout the book, unit conversion tables, and the structures of many of the biological molecules mentioned in the text.
We encourage you to consult other books for background information and greater depth
of coverage. Standard physical chemistry texts offer applications of physical chemistry to
other areas. Biochemistry and molecular biology texts can provide specific information
about such areas as protein and nucleic acid structures, enzyme mechanisms, and metabolic pathways. Finally, a good physics textbook is useful for learning or reviewing the
fundamentals of forces, charges, electromagnetic fields, and energy. A list of such books
is given at the end of this chapter.
In the following sections, we highlight several important biological problems that
physical chemistry can address. These examples are meant to give you an overview of
how physical chemistry is applied in the biological sciences. Read them for pleasure, without trying to memorize them. Our aim is simply to illustrate some current research from
the scientific literature and to point out the principles and methods that are used. We hope
to motivate you to learn the material discussed in the following chapters. Many articles in
journals such as Nature or Science apply the methods and concepts described in this book.
Read such articles to learn how the book will improve your understanding.
Neuroscience
Eric Kandel, a winner of the 2000 Nobel Prize in Physiology or Medicine, states that
the last frontier of the biological sciences is to understand consciousness and the mental
processes by which we experience our surroundings. An adult human brain has about
80 billion neurons, and each is connected to about 10,000 other neurons. The interactions
and communication among the neurons is who we are. How this all works is left as an
exercise to the reader because nobody else has explained it yet. However, in this book you
will learn some of the ideas and methods that have been used up to now to begin to find
answers, and more importantly, you may learn how to discover the next ‘last frontier’.
Magnetic resonance imaging (MRI) uses nuclear magnetic resonance in the presence
of a magnetic field gradient (the strength of the magnetic field varies with position) to
produce an image. The resonance frequency of a nucleus, such as a proton in water,
The Human Genome and Beyond | 3
depends on the magnetic field strength and thus on its position in the sample. Therefore,
the intensity at each frequency provides the distribution of water in the sample. For humans
the image is analogous to an X-ray picture, but it is much more sensitive to soft tissue. A
special application of MRI called functional MRI (fMRI) is used to learn which areas of
the brain are most active when you see, hear, feel, smell, or think about different things.
Brain activity uses more oxygen than usual, which means more blood flow and a higher
concentration of oxyhemoglobin from the lungs. This produces a signal that is different
from the slower blood flow with more deoxygenated hemoglobin in the rest of the brain.
An fMRI image of a brain of a person listening to music will show activity in a region
different from that of a person looking at a picture. These types of images are beginning
to teach us which parts of the brain are involved in specific inputs, thoughts, and outputs.
The molecular basis of brain activity depends on the electrical and chemical interaction among the neurons. Neurons are specialized cells with a cell body containing the
nucleus with its DNA, an axon that is much longer than the body, and many dendrites for
communicating to other neurons. The communication between neurons is done at synapses, where the cells touch. Ion channels open, neurotransmitters are released by one cell
and absorbed by the other, and somehow an image is seen, a thought is formed, an idea
occurs. Reading this paragraph should trigger hundreds of questions in your mind. I can
answer nearly all of them: “Nobody knows”. The obvious questions include: What are the
structures of all the proteins, nucleic acids, and small molecules involved? How do these
molecules get to their sites of action, and how fast do they do it? What are the interactions
that trigger and control all the effects? Finally, how do these molecular effects lead us to
think, to remember, to dream?
The Human Genome and Beyond
The instructions for making all the molecules that occur in the brain and all the other
organs in your body are stored in sequences of base pairs in your DNA. The structure of
DNA, determined by X-ray diffraction to be two interwound strands, started the molecular
understanding of how genes were stored and replicated (Watson and Crick, 1953). The
tenth anniversary of the Human Genome Project was celebrated in 2011; it had determined
the sequence of a human genome of 3 billion 13 * 109 2 base pairs (see Science, 2011
and Lander, 2011). Genes are sequences of base pairs in double-stranded DNA. In human
sperm the DNA is packaged in 23 chromosomes: 22 autosomes plus a male Y chromosome
or a female X chromosome. In human eggs the DNA is packaged in 22 autosomes plus a
female X chromosome. Thus, each of us acquires 23 pairs of chromosomes; the XX pair
makes us female, the XY male. When the human genome project started it was thought
that about 100,000 genes coded human proteins. Now the number is estimated to be 20,000
to 25,000, not very different from fruit files. So what is going on? Most of us think we are
smarter than a fruit fly.
A DNA sequence of base pairs does not directly code for a sequence of amino acids
in a protein; it codes for a sequence of bases in RNA. Some of the RNAs are part of the
translation machinery that produces proteins (ribosomal RNAs, transfer RNAs), and some
are messenger RNAs that are translated into a sequence of protein amino acids. However,
the messenger RNAs in humans are often modified before they are translated; pieces called
introns are removed and the remaining exons are spliced together before the messenger
RNA is translated. Alternative splicing can thus produce more than one protein from one
DNA sequence—one gene. Furthermore, an increasing number of DNA sequences have
been found to code for regulatory RNAs, not for proteins; they are RNA genes. These RNAs
are not translated into proteins, but they control which proteins are made and when they are
made. The 20,000 to 25,000 genes mentioned earlier are protein-coding genes; there may
4 Chapter 1 |
Introduction
be an equal or larger number of RNA-coding genes that help define you. Interactions not
yet discovered will eventually explain some of the differences between you and a fruit fly.
The Human Genome Project is thus typical of nearly all scientific projects. The completion
of one project reveals many new projects to investigate.
Determining the precise sequence of 3 billion nucleotides is a heroic task, which has
been made possible by the application of many biophysical techniques to separate and
characterize molecules. We emphasize the human genome, but the genomes of all kinds of
organisms from bacteria to plants to extinct animals are being sequenced, and are revealing new insight into how organisms evolve, differentiate, and exist. The key method used
to determine DNA sequence is to measure the fluorescence of fluorescently-labeled DNA
bases. A different fluorophore is added to each of the four bases in DNA; this allows the
sequence to be read using automated equipment. A newer method of sequencing uses an
electric field to pull a single-stranded DNA (or RNA) through a narrow pore in a membrane
(Pennisi, 2012). The four bases are different sizes and decrease the current passing through
the pore by different amounts. A trace of current vs. time as the nucleic acid strand passes
through the pore provides the sequence. The speed of sequencing has increased and the
cost has decreased so much that soon we can all be sequenced routinely at birth. Also,
DNA sequences have been placed on silicon chips and glass slides. Using the principles
of Watson–Crick base pairing, scientists can rapidly identify changes in DNA sequences.
These “genes on a chip” are revolutionizing the way that genes and gene expression are
analyzed. Many times in science, fundamental advances are allowed by improvements in
instrumentation to measure physical properties.
Once the sequence of an organism’s genome has been determined, a difficult task
begins. What does the string of the four different letters of the DNA alphabet (A, C, G, T)
mean? The DNA sequence is first transcribed into the RNA alphabet (A, C, G, U). A
messenger RNA is then translated into a protein sequence of 20 amino acids in a threeletter code.* As there are 43 (64) three-letter words with an alphabet of four letters, the
code must be redundant. In the genetic dictionary, most amino acids are coded by two
or four different words. Three amino acids have six words each (arginine, leucine, and
serine), and two have only one word (tryptophan, methionine). Three of the words do
not code for amino acids but instead signal for protein synthesis to stop: UAA, UAG,
UGA. One word, AUG, codes for the start of protein synthesis (it also codes for methionine). Sequences before the starting AUG and after the terminating UAA, UAG, or
UGA control and regulate the synthesis of the protein. Using the known genetic code,
scientists can predict the sequence for a protein that is coded by a given gene. What does
this protein do? What does it look like?
Physical chemistry provides the principles that allow bioinformatics scientists to make
sense of the vast DNA sequence data of a genome. The protein sequence predicted from
a gene is first compared to known protein sequences. If the protein is an essential part of
some biochemical process that is common to many or all organisms, related proteins have
likely been studied. Computer sequence comparisons establish the relationship between
a novel protein and known proteins. The sequences of two proteins of similar biological
function from different organisms are almost never identical. However, different protein
sequences can adopt similar three-dimensional structures to perform similar functions.
Different amino acids can have similar physical properties. For example, both isoleucine
and valine have greasy aliphatic side chains and can often be exchanged for each other in a
protein with little effect on its activity. Likewise, negatively charged amino acids (aspartic
*The
structures and names of the nucleic acid bases and the protein amino acids are given in table A.9 in the
appendix.
Transcription and Translation | 5
acid or glutamic acid) can often be swapped, and so on. Using this type of logic, computer
programs can sometimes predict the function of the unknown gene by its relation to a
known protein.
The weakness of this approach is obvious. You require the sequence of a known protein
with which to compare the new gene. In addition, what truly determines the function of
a protein is not the sequence of amino acids but rather how these amino acids fold into a
three-dimensional structure that can perform a specific function—for example, catalysis
of a reaction. Biophysical chemists can determine the three-dimensional structures of
biological macromolecules, using methods described in this book. Unfortunately, the rate
at which structures can be determined lags behind that of sequencing a gene. Nonetheless,
comparisons of protein structures often reveal similarities that simple protein-sequence
comparisons miss. Triosephosphate isomerase is a protein involved in metabolism, and it
has a barrel-like three-dimensional structure. This structure is a rather common motif in
proteins, but sequence comparisons by computer can rarely identify its presence.
Determining the three-dimensional structure of a protein would be easy if it could
be predicted from its sequence. A protein’s amino acid sequence contains the physical
characteristics that determine the most stable three-dimensional fold. Biophysical chemists
have shown that proteins almost always adopt the most stable three-dimensional structure
as determined by the principles of thermodynamics. Thus, physical chemistry provides
the framework to predict protein structure. However, predicting the most stable threedimensional structure of a protein is a very difficult task because a large number of relatively
weak interactions stabilize its structure (chapter 12). Precise treatment of these interactions
is impossible, so biophysicists and computation biologists use a number of approximations
to calculate a protein structure from its sequence (for an example, see Das & Baker, 2008).
This is a valid approach to many complex biological problems. How can a scientist know
whether a computer program is actually working? Well, she could try it on a sequence of
a protein of known structure. But this is of course biased, for our scientist already knows
the answer. Scientists in this field in fact resort to friendly competitions. They are asked
to predict the structures of proteins whose structures are not known at the beginning of
the competition but will be revealed by the end. This provides an unbiased test of various
algorithms. This example shows a glimpse of the human side of the scientific process.
Although current algorithms cannot predict the structures of protein to the same precision
as experimental methods, they are improving. Computer prediction of protein folding and
RNA folding is now a highly active area of biophysical research.
In addition to predicting structure and function of a protein from the sequence, you can
try to improve the function. For example, will the catalytic activity increase or the specificity change if a crucial aspartic acid is changed to a glutamic acid? Changing one amino
acid at a time is slow and tedious; but instead, by randomly changing the RNA sequence
that codes for the protein, many mutants of the protein can be made. A selection process is
then used to find the one with the desired optimized function. Furthermore, in producing
better functions, or new functions, you need not be limited to naturally-occurring amino
acids. The translation machinery can be tricked into producing proteins containing amino
acids not found in nature (see Brustad and Arnold, 2011).
Transcription and Translation
Genetic information must be faithfully transmitted from DNA to messenger RNA to
protein. Copying DNA to RNA is called transcription; reading RNA to produce a protein
is called translation. Two central macromolecular machines are responsible for these processes: RNA polymerases transcribe RNA from DNA, and the ribosome translates RNA
6 Chapter 1 |
Introduction
into protein. In both systems a series of repetitive tasks must be performed with high
fidelity. These machines must be directional, because they copy information in only one
direction. The machines are processive, in that once they start the process of transcription
or translation, they continue through hundreds or even thousands of steps of the process.
Finally, these biological processes are highly regulated. Associated factors determine
when, where, and how rapidly these processes begin and end. Physical methods have
provided important insights into how transcription and translation occur.
The process of transcription was first investigated in simple organisms such as bacteria.
The protein that catalyzes transcription consists of only one or a few polypeptide chains. In
contrast, in eukaryotic organisms such as humans, the RNA polymerase enzyme consists of
ten or more polypeptide chains, reflecting the higher degree of regulation in higher organisms. Transcription begins at specific signals in the DNA called promoters. These DNA
sequences bind specific transcription factors that enhance or prevent transcription. This is
an essential feature in the regulation of gene expression. The activity of these transcription
factors can be affected by attaching a phosphate group to a protein or by binding of a small
molecule cofactor. The classic example is a protein that binds both to small sugars and to
DNA, like the lac repressor (Bell and Lewis 2001). These DNA-binding proteins recognize
specific promoter sequences, which control the expression of genes for sugar metabolism
enzymes. When the lactose concentration reaches a certain level, the sugar binds to specific
sites on the protein and changes its conformation, such that it binds tighter to its DNA site,
thus turning off transcription of genes that would produce more sugars. This is an example
of feedback inhibition. This example of biological regulation can be explained by the laws
of chemical equilibrium and thermodynamics, discussed in chapters 2 through 5.
The high fidelity of transcription is ensured by an elegant kinetic mechanism, determined using the methods of enzyme kinetics described in chapter 10. During a round
of polymerization, a nucleoside triphosphate enters the active site of RNA polymerase
and pairs with the single-stranded DNA, which has been opened from its double helical
form (figure 1.1). The three-dimensional structure of this essential enzyme from bacteria
(E. coli, see Opalka et al., 2010) and higher organisms (yeast, see Cramer et al., 2000) has
been solved. The shape of the active site is such that only the correct Watson–Crick base
FIGURE 1.1 Three-dimensional
structure of the RNA polymerase
from E. coli, the enzyme responsible for transcribing the RNAs.
The enzyme breaks base pairs
in the double-stranded DNA (in
black) to produce an open loop.
The new RNA strand (in blue)
is synthesized complementary
to one of the DNA strands in
the loop. This structure is a
combination of X-ray diffraction
and cryo-electron-microscopy
data plus computer modeling
(Opalka et al., 2010). The
coordinates were obtained from
the Protein Data Base. (Courtesy
of Troy Lionberger, University of
California Berkeley.)
Transcription and Translation | 7
pair is tolerated; the wrong nucleoside triphosphate does not make a good fit into the active
site and is more rapidly ejected. For DNA polymerase, the enzyme that copies DNA during
cell division, the push for fidelity is so strong that the enzyme contains an editing function.
If a wrong nucleotide is incorporated into the DNA, it is snipped out, and the correct
nucleotide is incorporated. This drive for fidelity is understandable, considering the drastic
effects mutations can have on protein function. On the other hand, the polymerases have to
perform their functions rapidly, so they have evolved a trade-off between high fidelity and
reasonable rates of polymerization. Such trade-offs are a hallmark of biological chemistry.
The regulation of transcription is a central process in biology; the requirement for
a complex macromolecular assembly to perform RNA transcription in higher organisms
derives from the need for regulation. Cells must sense outside stimuli and respond, usually
by rapidly synthesizing or degrading a protein or proteins. Recent biochemical experiments
have revealed elaborate signal transduction pathways. A protein on the surface of a cell,
called a receptor, will bind to an external signal, which may be a specific hormone or
another extracellular-signaling molecule. The receptor molecule spans the cell membrane,
and the binding of the hormone causes a change in its three-dimensional structure, activating an enzymatic activity (a kinase) that adds a phosphate group to a protein. When a
phosphate group modifies a protein, the protein’s shape and activity can change. Often, a
cascade of kinase events occurs, where protein 1 phosphorylates protein 2, which, in turn,
phosphorylates protein 3, and so on. The final targets of these cascades are often transcription factors which can turn transcription on or off depending on the desired result of a
signaling event. Certain human cancers occur when these signaling pathways—and thus
the ability of a cell to respond to external stimulus—are disrupted. Signaling pathways
are very complex and biologists are still identifying their many components. Physical
methods and reasoning, however, will be required to unravel the mechanisms of these
signaling pathways.
The ribosome (figure 1.2), where translation occurs, is more complex than RNA polymerases. The ribosome in bacteria consists of two subunits which weigh 0.80 * 106 and
1.4 * 106 daltons. These enormous subunits each consist of at least one RNA chain and
20 to 30 proteins. The adaptors between the genetic code of RNA and the protein amino
acid, first proposed by Crick, are called transfer RNAs (tRNAs). A single loop of the tRNA
contains three nucleotides—the anticodon—that can form Watson–Crick base pairs with
a given codon; the amino acid that corresponds to that codon is attached at the 3⬘-end of
the tRNA. The three-dimensional structure of tRNA shows that these two parts are located
7.5 nm apart. The ribosome is able to select the correct tRNA that binds to the appropriate
codon. The messenger RNA (mRNA) runs through a cleft between the subunits, and the
anticodon portion of the tRNA interacts with the smaller (30S) subunit. Once the correct
tRNA is selected at the A-site, the 3⬘-end of the tRNA sits within the larger subunit,
where peptide bond formation is catalyzed between the amino acid and a peptide-chain
containing tRNA (which is bound at the adjacent codon at the P-site). The ribosome then
must move by three nucleotides in the mRNA to the next codon; this precise directional
movement is called translocation.
The basic mechanism of translation was delineated over 40 years ago, but molecular
details of how the ribosome performs the task of protein synthesis have been revealed only
recently (reviewed in Schmeing and Ramakrishnan, 2009, and Moore, 2012). The structure of the ribosome (Ban et al., 2000; Schuwirth et al., 2005) showed that the biological
functions of the ribosome are dominated by the RNA components; RNA catalyzes the
formation of the peptide bond, making it an RNA enzyme—a ribozyme. Kinetic studies,
similar to those done on polymerase enzymes, have revealed the origins of translational
fidelity. The strategy used by the ribosome is somewhat similar to that used by polymerases.
In the case of the ribosome, the base pairing between codon and anticodon occurs about
8 Chapter 1 |
Introduction
FIGURE 1.2 (top) The
architecture of the ribosome;
the large, 50S, subunit is on top
and the small, 30S, subunit is
on the bottom. Three transfer
RNAs are shown reading the
sequence in the messenger
RNA. The structure was solved
by X-ray crystallographic
methods (Zhang et al., 2009).
(bottom) A close-up view of
the Watson-Crick base pairing
between each codon on the
messenger RNA with the
anticodon of each transfer RNA
as it occurs on the amino acid
site (A-site), peptide site (P-site),
and exit site (E-site) of the ribosome. The coordinates are from
the Protein Data Base. (Courtesy
of Shannon Yan, University of
California Berkeley.)
P-site
E-site
A-site
tRNAs
3’
5’
mRNA
5’
3’
7.5 nm from the site of peptide bond formation; the ribosome couples this base pairing to
another chemical reaction: hydrolysis of guanosine triphosphate, GTP, which is bound to a
protein factor that escorts the tRNA to the ribosome. Rate constants for tRNA dissociation
and ribosomal conformational changes are modulated by whether the correct or the incorrect tRNA is present. Structural biologists have obtained detailed views of the ribosomal
particles. The two subunits of the ribosome interact through an interface that is entirely
RNA. Adjustments of this interface allow the ribosome to translocate down the mRNA. As
biochemical experiments have predicted, the structures show that RNA forms the critical
Ion Channels | 9
active sites for tRNA binding and peptidyl transfer. The RNA folds into a complex threedimensional structure, which the protein components of the ribosome (many of which
bind to ribosomal RNA) stabilize. The molecular rationale for how the ribosome performs
translation will only be revealed by physical chemical investigations.
Ion Channels
Cells perform spectacular feats of chemistry. Ion channels are proteins that span the lipid
membrane of a cell and specifically allow one ion type to traverse the channel. Ion channels
are critical for many biological processes, including signaling by neurons. Ion channels
can be remarkably selective. Potassium ion 1K+ 2 channels are about 10,000-fold more
selective for K+ than for Na+ (sodium) even though their ionic radii are 1.33 and 0.95 Å,
respectively. Also, the ion channels must allow a large number of ions to pass across a
membrane in a directional manner in a short time period. Finally, many ion channels are
controlled by external conditions. They are opened or closed to ion passage by factors
such as the voltage difference across the membrane. The methods of physical chemistry
have been invaluable in determining how ion channels work (Doyle et al. 1998). When ion
channels do not function properly, the results can be disastrous. Many human diseases are
linked to impairment in these molecular highways. For example, cystic fibrosis, one of
the most common genetic diseases, is caused by mutations in a Cl- (chloride ion) channel.
The disrupted function for this channel leads to a buildup of thick, fibrous mucus in the
lungs, which impairs breathing.
Determining the three-dimensional structure of a K+ channel to atomic resolution
was a significant breakthrough in understanding how ion channels work. The protein is a
tetramer of identical subunits. Long rods of alpha helix span the membrane. The protein
is not merely a tube through which potassium flows. The overall shape of the protein is
like a flower, with the petals opening toward the outside of the membrane and narrowing
at the inside of the membrane (figure 1.3). The ions pass through a channel in the center
of the tetrameric protein. How are potassium ions specifically selected and transported?
cell exterior
cell interior
FIGURE 1.3 The threedimensional structure of a
K+ channel, showing schematically the position of the channel
within a cellular membrane.
Intracellular (in the cell interior)
and intercellular (on the cell
surface) domains are indicated.
Potassium ions are shown as
spheres and are transported
directionally from the exterior
to the interior of the cell.
(By permission of Roderick
Mackinnon, MD; Professor,
The Rockefeller University;
Investigator, Howard Hughes
Medical Institute.)
10 Chapter 1 |
Introduction
The top of the flower-shaped tetramer is a selectivity filter. This region of the protein is
rich in negatively charged amino acids, which specifically select for cation binding. The
shape and chemical properties of this filter region are such that potassium ions bind most
favorably, while smaller or larger cations are excluded. The presence of a second potassium
ion helps nudge the first further down the channel. Beyond this selectivity filter, the channel widens and is lined by mainly hydrophobic side chains. This may seem surprising, as
one might expect the whole pore to be lined by negatively charged amino acids. However,
the function of the pore is to transport a large number of cations through the channel. If
the pore were too negatively charged, the laws of electrostatics (chapter 12) predict that the
cation would stick within the channel. The three-dimensional structure—combined with
the principles of physical chemistry—takes the mystery out of ion channels.
What turns an ion channel on or off? Voltage-gated ion channels are required for
nerve signaling and control the flow of ions in response to transmembrane voltages. Biophysicists have studied voltage-gated potassium channels that have sequence similarity to
the ion channel whose structure the preceding paragraph describes (Glauner et al. 1999;
Cha et al., 1999). Based on this similarity, a model for what the related protein might
look like in three dimensions was created. This low-resolution model allows the design
of experiments to test how the structure of the protein responds to voltage gating. We will
see in chapters 13 and 14 that spectroscopy allows the investigation of protein structure as
well as its time-dependent changes. With the technique of fluorescence energy transfer,
in which fluorescent dyes are attached to the ion channel at defined places, biophysicists
use spectroscopy to measure the distances between the dyes and to determine changes
in distance caused by conformational alterations. For instance, an alpha helix rich in
positively charged amino acids senses the change in voltage and twists and changes its
position. This twisting movement changes the shape of the pore and allows ions to pass.
This example shows the synergy of various physical measurements. Structural studies
lead to models of activity that spectroscopic and kinetic measurements can test. Physical
studies can transform what was originally just a gene sequence for the ion channel into a
drama of molecular forms and movements.
Single-molecule Methods
Nearly all of the physical methods that have been used to determine structures and properties
of biological molecules have been applied to ensembles of molecules. Even experiments on a
micro scale, such as 1 μL of solution containing 1 μM concentration has ~6 * 1011 molecules.
Previously we needed that many molecules to get a detectable signal. However, recently it has
become possible to measure properties of individual molecules in solution, or on membranes
or surfaces. Of course, electron microscopes could see individual molecules on films in a
high vacuum, and atomic force microscopes could see proteins and similar size molecules on
surfaces, but single molecules in biological cells or solutions were not accessible. Now individual fluorescent molecules, or any molecule with an attached fluorophore can be studied.
Neurotransporters with their cargo can be watched as they move down an axon in a neuron.
Messenger RNAs can be seen moving from the nucleus, where they are synthesized, to the
cytoplasm, where they are translated. Fluorescence resonance energy transfer, FRET, can
reveal distances in the nm range between two fluorophores. The theory of FRET was derived
by quantitatively considering the electromagnetic interaction between two oscillating dipoles;
the efficiency of energy transfer was found to depend on their inverse distance raised to the
sixth power, (1/r)6 (see chapter 13). This large dependence on distance allows the interactions
between proteins or RNAs to be monitored as a biological function occurs. Measurement of
fluorescence is highly sensitive, but any spectroscopy can in principle be applied to a single
molecule, such as absorption, scattering, or nuclear magnetic resonance.
Physics | 11
Individual molecules can be manipulated using laser tweezers. A micron-sized polystyrene bead is attached to the molecule, and a laser beam focused to a diffraction-limited
spot can trap the bead. Moving the laser spot moves the bead and molecule. If two beads
are attached to a protein or nucleic acid, the molecule can be unfolded from its compact
form to an extended polypeptide or polynucleotide. The work required to unfold it is
measured as a force times distance, and the free energy of unfolding is obtained from the
reversible work (chapter 2).
The unique advantage of single-molecule methods over ensemble methods is that you
are not measuring an average over ˜1010 molecules. You are observing one molecule at a
time. This is most important in kinetics where there may be many intermediates present in
solution on the way from reactants to product. The average over many intermediates may
be very difficult to sort out. However, by following a single molecule through the reaction
every intermediate in the process is seen. Combination of X-ray diffraction on crystals with
single-molecule methods in solution is providing a motion picture of molecular machines
such as the ribosome and RNA polymerases. (See reviews by Bustamante et al., 2012, and
Tinoco and Gonzalez, 2011.)
Reference
The following textbooks can be useful for the entire course.
Physical Chemistry
Atkins, P. W. and J. DePaulo. 2009. Physical Chemistry. 9th ed.
New York: Freeman.
Levine, I. N. 2008. Physical Chemistry. 6th ed. New York:
McGraw-Hill.
Engel, T., and P. Reid. 2010. Physical Chemistry. 3rd ed.
New York: Pearson.
Silbey, R. J., R. A. Alberty, and M. G. Bawendi. 2004. Physical
Chemistry, 4th ed. New York: Wiley.
Biophysical Chemistry
Cantor, C. R., and P. R. Schimmel. 1980. Biophysical Chemistry.
Pts. 1–3. San Francisco: Freeman.
Kuriyan, J., B. Konforti, and D. Wemmer. The Molecules of Life.
2012. Hamden, CT: Garland Science.
Van Holde, K. E., W. C. Johnson, and P. S. Ho. 2005. Principles
of Physical Biochemistry. 2nd ed. Upper Saddle River, NJ:
Prentice Hall.
Biochemistry
Mathews, C. K., K. E. van Holde and K. G. Ahern. 2000.
Biochemistry. 3rd ed. Reading, MA: Addison-Wesley.
Voet, D., and J. G. Voet. 2010. Biochemistry. 4th ed. New
York: Wiley.
Berg, J. M., J. L. Tymoczko, and L. Stryer. 2010. Biochemistry.
7th ed. San Francisco: Freeman.
Molecular Biology
Alberts, B., D. Bray, J. Lewis, M. Raff, K. Roberts, and
J. D. Watson. 2007. Molecular Biology of the Cell. 5th ed.
New York: Garland.
Watson, J. D., T. A. Baker, S. T. Bell, and A. Gann. 2007.
Molecular Biology of the Gene. 6th ed. Redwood City,
CA: Benjamin/Cummings.
Krebs, J. E., E. S. Goldstein, S.T. Kilpatrick. 2009. Lewin’s
Genes X. New York: Oxford University Press.
Physics
Giancoli, D., 2004. Physics. 6th ed. Upper Saddle River, NJ:
Prentice Hall.
Halliday, D., R. Resnick, and J. Walker. 2010. Fundamentals of
Physics. 9th ed. New York: Wiley.
12 Chapter 1 |
Introduction
Suggested Reading
Ban, N., P. Nissen, J. Hansen, P. B. Moore, and T. A. Steitz. 2000.
The Complete Atomic Structure of the Large Ribosomal
Subunit at 2.4Å Resolution. Science 289:905–920.
Bell, C. E., and M. Lewis. 2001. The lac repressor: a second generation of structural and functional studies. Current Opinion
Structural Biology.11:19–25.
Brustad, E. M., and F. H. Arnold. 2011. Optimising non-natural
protein function with directed evolution. Current Opinion
Chemical Biology. 15:201–210.
Cha, A., G. E. Snyder, P. R. Selvin, and F. Bezanilla. 1999.
Atomic Scale Movement of the Voltage-Sensing Region in
a Potassium Channel Measured via Spectroscopy. Nature
402:809–813.
Bustamante, C., W. Cheng, and Y. Mejia. 2011. Revisiting the
Central Dogma One Molecule at a Time. Cell 144:480–491.
Cramer, P., D. A. Bushnell, J. Fu, A. L. Gnatt, B. Maier-Davis,
N. E. Thompson, R. R. Burgess, A. M. Edwards, P. R. David,
and R. D. Kornberg. 2000. Structure of RNA Polymerase II
and Implications for the Transcription Mechanism. Science
288:640–649.
Das, R. and D. Baker. 2008. Macromolecular modelling with
rosetta. Annual Review of Biochemistry. 77:363–382.
Doyle, D. A., C. J. Morais, R. A. Pfuetzner, A. Kuo, J. M. Gulbis,
S. L. Cohen, B. T. Chait, and R. MacKinnon. 1998. The
Structure of the Potassium Channel: Molecular Basis of
K+ Conduction and Selectivity. Science 280:69–77.
Glauner, K. S., L. M. Mannuzzu, C. S. Gandhi, and E. Y. Isacoff.
1999. Spectroscopic Mapping of Voltage Sensor Movement
in the Shaker Potassium Channel. Nature 402:813–817.
Lander, E. 2011. Initial impact of the sequencing of the human
genome. Nature 470: 187–197.
Moore, P. B. 2012. How should we think about the ribsome?
Annual Review of Biophysics. 41:18.1–18.9.
Opalka, N., J. Brown, W. J. Lane, K.-A. F. Twist, R. Landick,
F. J. Asturias, and S. A. Darst. 2010. Complete Structural
Model of Escherichia coli RNA Polymerase from a Hybrid
Approach. PLoS Biol 8(9): e1000483.
Pennisi, E. 2012. Genome sequencing. Search for pore-fection.
Science 336:534–537.
Schmeing, T. M. and V. Ramakrishnan. 2009. What recent
ribosome structures have revealed about the mechanism of
translation. Nature. 461:1234–1242.
Science, 2011. The tenth anniversary of the human genome
sequence is discussed in the February 21, 2011 Vol. 331, 6020
issue of Science. Its impact on human genetics, biomedical
research, evolution, and biology in general are discussed.
See Human Genome 10th Anniversary (pp. 1005–1009) and
Genome Sequencing Annivesary (pp. 1024–1027).
Schuwirth, B. S., M. A. Borovinskaya, C. W. Hau, W. Zhang,
A. Vila-Sanjurjo, J. M. Holton., and J. H. Cate. 2005.
Structures of the bacterial ribosome at 3.5 A resolution.
Science 310:827–834.
Tinoco, I., and R. Gonzalez, Jr. 2011. Biological mechanisms, one
molecule at a time. Genes and Development 25:1205–1231.
Watson, J. D. and F. H. Crick. 1953. Molecular structure of
nucleic acids; a structure for deoxyribose nucleic acid.
Science 171:737–738. Genetical implications of the structure
of deoxyribonucleic acid. Science 171: 964–967.
Zhang, W., J. A. Dunkle, J. H. Cate. 2009. Structures of the
ribosome in intermediate states of ratcheting. Science
325:1014–1017.
Problems
1. Choose a subject in this chapter that you are not familiar
with. Read about it in Google or Wikipedia and give a
one-page summary in your own words.
2. Read a paper in the scientific literature that sounds interesting
to you.
a. Record the complete reference to it: authors, title,
journal, volume, first and last pages, and year.
b. Summarize the purpose of the paper and why it is
worthwhile.
c. List the methods used and state how the measurements
are related to the results.
d. What further experiments could be done to learn more
about the problem being studied?
Chapter 2
The First Law: Energy Is
Conserved
The First Law: you can’t win.
The Second Law: you can’t break even.
The Third Law: you can’t come close to breaking even, unless it’s really cold.
— A common scientific joke, with many variations.
Concepts
A scientific law is a regularity observed in nature and formulated after a large number of
observations. Because they are solidly grounded in experimental observations, scientific
laws may be modified after further experience, but they are rarely refuted.
The branch of science we call thermodynamics deals with the exchange of energy.
Scientific observations of how the exchange of energy happens ultimately lead to laws that
govern the direction of all natural processes.
The First Law of Thermodynamics states that energy is conserved; that is, in any
process, energy can neither be created nor destroyed. The law is based on experiments
carried out in the early 19th century. In one such experiment, a falling weight, via a pulley,
turned a paddle that stirred a bucket of water. As the weight fell and the paddle turned, the
temperature of the water increased. This experiment showed that the potential energy of the
weight could be converted into internal energy manifested in the increased temperature of
the water, just as could be achieved by applying heat to the water. The mechanical motion
of the apparatus churning the water is described as work. Classical thermodynamics
centers on work and heat, but other forms of energy such as electrical energy have since
been incorporated into the First Law.
Since Einstein proposed his famous equation E = mc 2 , we have known that the
individual laws requiring the conservation of mass and energy are incomplete, and that
what is really conserved is a combined quantity called mass-energy. However, absent
nuclear reactions and radioactivity, the conversion of energy into mass, or mass into energy,
is not observed, and so in most of everyday life, the original formulation of the First Law
is perfectly valid.
The Second Law of Thermodynamics, which was formulated at about the same time,
deals not with the quantity of energy but with its flow. Scientists observing steam engines,
in which wood or coal are burned to generate heat that is converted into work, noticed
that the conversion of heat into work was very inefficient. Pursuing the theory of these
heat engines, they showed that no engine, no matter how well designed, can convert heat
into work with perfect efficiency. Some of the heat must be discharged into a colder
133
14 Chapter 2 |
The First Law: Energy Is Conserved
environment in order to drive the conversion of the rest into work. In fact, the observation
that heat always spontaneously flows from objects at high temperature into objects at lower
temperature has been shown to be equivalent to the Second Law. This law therefore dictates
the direction of all spontaneous processes and not simply the flow of heat. Being able to
determine the direction of any process that occurs spontaneously is incredibly useful.
As we will see in chapter 3, the Second Law introduces the concept of entropy. Entropy
is the quantity that increases in any spontaneous process.
The Third Law of Thermodynamics is the most recent. It was only clearly stated in
the 1920s. It shows, within limitations, that the entropy of systems goes to zero as the
temperature approaches zero Kelvin. The Third Law is important in the biological sciences
because it gives us an absolute measure of the entropy. But, of course, biological systems
never naturally approach zero Kelvin. For that reason, we will give it limited attention in
this book.
Everything in biology, from nearly instantaneous chemical reactions to the incredibly
slow process of evolution, is governed by the three laws of thermodynamics. The laws
explain the hybridization of DNA, the folding of proteins, the pumping of ions across
cell membranes, and the metabolism of foodstuffs. Understanding these laws is crucial to
understanding life itself.
Applications
Thermodynamics applies to everything from black holes, so massive that even light cannot
escape from them, to massless neutrino particles. Thermodynamics can answer these and
many other questions:
• How do you calculate the work done when a muscle contracts or stretches?
• How can chemical reactions be used to do work or to produce heat?
• How much heat can be generated by burning 1 gram (g) of sugar or eating and digesting 1 g of sugar?
Many examples of the applications of thermodynamics will be illustrated in this and the
subsequent three chapters.
Energy Conversion and Conservation
A large set of experiments, done over a period of many years, has shown that energy can be
converted from one form to another, but that the total amount of energy remains constant. We
will eventually discuss this quantitatively, but a couple of examples will make the idea clearer.
Consider the cellphone that you have carelessly left on the window ledge of the
fifth floor of your apartment building. Perched precariously, high above the sidewalk, it
possesses gravitational potential energy. Knocked off the ledge, it accelerates towards its
untimely end. First the potential energy is converted to kinetic energy, and then the friction
of the air begins to convert the kinetic energy into heat. Once it hits the hard concrete,
some of the kinetic energy is retained in the splintered fragments of the ‘phone as they
fly in every direction. Some of the kinetic energy is transferred to planet Earth, which is
jogged an imperceptibly tiny amount in its orbit by the impact. But many new forms of
energy also appear. Some heat is produced. There may be some light energy in the form
of sparks. Some energy is used to break the chemical bonds in the metal, glass, and plastic
components as they shatter. Some sound energy is certainly produced.
A more interesting question: how much energy arrives from the Sun and how is it
transformed? Sunlight hitting the desert or a solar collector is mainly transformed into heat.
However, some of the light energy can be converted into electrical energy using solar cells,
and sunlight striking a green leaf is partly transformed into useful chemical energy through
photosynthesis. It is vitally important that we know and understand the various kinds of
energy involved in biochemical processes and what limits, if at all, their interconversion.
Energy Exchanges | 15
Systems and Surroundings
Clear definitions are perhaps more important in thermodynamics than in any other branch
of science. But our first pair of definitions is easy. The system is defined as the specific
part of the universe on which we choose to focus; in other words, the system is whatever
we say it is. It might be the Sun, Earth, a person, a mammalian liver, a single living cell, or
a mole of liquid water. Everything else in the universe we call the surroundings, and what
separates the system from the surroundings is termed the boundary (figure 2.1).
Thermodynamics deals with three idealized kinds of systems, illustrated in figure 2.1.
• Isolated systems have no exchange of any kind — neither matter nor energy — with
the surroundings.
• Closed systems can exchange energy but not matter with the surroundings.
• Open systems exchange both matter and energy with the surroundings.
Isolated systems are the most difficult to construct, because it is hard to completely cut
off energy transfer to a system. However, the contents of a sealed and thermally insulated
flask come very close to an isolated system, especially over a short period of time with negligible heat flow in and out. An isolated system is really an idealized concept, which can only
be approached in the limit. Such concepts are common in thermodynamics. They are often
easy to theoretically analyze but can usually only be realized in practice by extrapolation.
A closed system can simply be one within a physical enclosure. Many chemical
reactions are performed in a closed system, a sealed flask being an example. The chemicals
stay inside the flask, but heat can come in or out.
The most difficult type of system to consider (and also the most common) is the open
system. An open system can exchange both matter and energy with the surroundings. A
fertilized egg being hatched by a hen is an example: oxygen goes in and carbon dioxide
comes out of the egg; heat is also exchanged between the egg and its surroundings.
We should emphasize that it is entirely up to us to specify our system and to define
real or imaginary boundaries that separate it from its surroundings. If we specify as our
system 10 mol of H2O (180 g) poured into a beaker and left to evaporate, then the liquid
water remaining in the beaker [(180 - x) g] plus the water that has evaporated from it
(x g), constitutes a closed system.
As another example, if a solution containing the enzyme catalase is added to an open
beaker containing a hydrogen peroxide solution, the enzyme will accelerate the reaction
2 H2 O2 S 2 H2 O + O2
and oxygen gas will come out of the beaker. If we choose the liquid content in the beaker
as our system, we have an open system. But if we choose the liquid content plus the oxygen
evolved as our system, we have a closed system.
We can choose the system according to our interests, objectives, and convenience. We
must, however, always define the chosen system clearly, to avoid confusion.
Energy Exchanges
Now that we have defined a system, we can focus on how its energy can be changed. In
thermodynamics, we are interested in changes in energy, more than the absolute value of
energy, which is in any case difficult to define. We can add energy to a system in several
ways. Adding matter to an open system, for example, increases the chemical energy of
the system because the matter can undergo various physical and/or chemical reactions.
We do not have to think about the large amount of energy potentially available from
nuclear reactions unless such reactions are actually occurring in the system; that is, usually
we do not have to include the E = mc2 energy term because this term does not change
significantly in an ordinary reaction.
It is convenient to divide energy exchange between system and surroundings into
different types. Two of the most common types of energy exchanges are work and heat.
Open system
heat
matter
Closed system
heat
Isolated system
FIGURE 2.1 The three kinds
of thermodynamic systems.
The black box is the system
boundary. In an open system,
matter and energy can pass
through the boundary between
system and surroundings. In a
closed system, only energy can
pass through the boundary. In
an isolated system, nothing can
pass through the boundary.
16 Chapter 2 |
The First Law: Energy Is Conserved
Work
(a)
x0
In classical mechanics, work is defined as the product of a force times a distance. In thermodynamics, the system can do work on the surroundings, or the surroundings can do
work on the system. To calculate the mechanical work done by the system or on the system,
multiply the external force on the system by the distance moved — the displacement:
work = external force * displacement
x
(b)
Fex
(c)
x
We must be consistent about the sign of the work to keep proper accounting of the energy
exchanges between the system and the surroundings. Physical chemists and biochemists
generally follow the convention that the work is positive if the surroundings are doing
work on the system, and negative if the system is doing work on the surroundings. So
when we use Eq. 2.1, we need to keep a watchful eye to make sure that the sign of work,
which depends on the proper choice of signs for the force and the displacement, is always
consistent with this convention.
Some physicists and engineers define work done by the system as positive, and so one
has to be careful not to blindly apply formulas from textbooks in those disciplines without
checking the sign convention used. Sign conventions can be a nuisance, but as long we
specify these clearly, the nuisance can be kept to a minimum.
Work of Extending a Spring
Any spring has an equilibrium length along its axis equal to x 0 . This is its length when
not subject to an external force. We can apply an external force Fex either to extend or
to compress the spring to a new length x, as shown in figure 2.2. At this new length, the
stretched or compressed spring itself exerts a spring force Fsp equal in magnitude, and
opposite in sign, to the external force.
According to Hooke’s Law, the spring force Fsp that balances Fex exactly is directly
proportional to the change in the length of the spring, and has a sign such that it opposes
the change in length away from equilibrium (i.e., opposite to x - x 0)
Fsp = -Fex = -k(x - x 0),
Fex
FIGURE 2.2 (a) A spring
under no external force, with
an equilibrium length x0.
(b) a spring subjected to a
compressive (negative) external
force Fex; x 6 x0 (c) a spring subjected to a extensive (positive)
external force Fex; x 7 x0.
(2.1)
(2.2)
where k is a constant for a given spring (if we do not extend it beyond its elastic limits).
The negative sign on the right-hand side reflects the spring force acting in the opposite
direction to the change in length. The magnitude of k will be different for different springs.
To calculate the work done in changing the length of the spring, we choose an x-axis
with one end of the spring fixed at x = 0. The other end is free to move along this axis;
its position when there are no forces acting on it is x 0; in general, its position is at x.
As we change the length of the spring, the spring force changes. This means we cannot
simply multiply the force by the change in length; we must integrate the force multiplied
by the differential of the displacement.
w = 3 Fex(x)dx = 3 k(x - x0)dx
(2.3)
In the preceding equation w is positive if the external force Fex and the displacement dx are
of the same sign, and negative if they are of opposite sign. This is consistent with our sign
convention for w: if the direction of the external force is the same as the direction of the
displacement, the external force is doing work on the system (the spring), and w is positive.
The work done on the spring when its length is changed from x 1 to x 2 is:
x2
w = 3 k(x - x0)dx
x1
Changing variables: x⬘ = x - x 0 and so dx⬘ = dx .
(2.4)
Energy Exchanges | 17
x2 - x0
w = 3
kx⬘ dx⬘ = 兩 1> 2 kx⬘ 2 兩 xx21 -- xx00
x1 - x0
(2.5)
= 1>2 k c ( x 2 - x 0)2 - ( x 1 - x 0 ) 2 d
x0
This integration is depicted in figure 2.3. The green line graphs the external force as a
function of displacement x. The area under the green line — in other words, its integral —
is the work done. The right angled triangle whose apex is at x 0 and whose right edge is the
vertical line at x = x 1 has width equal to x 1 - x 0 and vertical height equal to k(x1 − x0).
Its area is therefore 1>2 k(x 1 - x 0)2 . Likewise, the triangle bounded by a vertex at x 0 and
the vertical line at x = x 2 has an area 1>2 k(x2 - x0)2 . The difference between these two
triangles is an area of the light green polygon between values x = x 1 and x = x 2 . This is
the work done stretching the spring from x 1 to x 2 . A graph of this sort — with force on
the y axis and displacement on the x axis — is called a force-extension curve.
We have thus calculated the work done on the system by extending the spring. If the
spring was originally extended, the system would do work on the surroundings when the
spring returned to its equilibrium position. A muscle fiber does work in a similar fashion.
It can do work by stretching or contracting against an external force.
According to the Système International (SI), the unit of work, and indeed of any form of
energy, is the joule (J). One joule is 1 newton * meter. The unit of force is the newton (N),
and the spring constant k must therefore have units of N/m or N m−1. One newton is the
magnitude of the force that will cause an acceleration of 1 m s−2 when applied to a mass of
1 kilogram (kg): 1 N = 1 kg m s−2. Older units for energy are the erg: 1 erg = 10 −7 J; and
the calorie: 1 cal = 4.184 J. Additional energy conversion tables can be found in table A.3
in the appendix. Most international scientific organizations now insist on SI units which we
will use throughout this book. The basic SI units of length, mass, and time are meter (m),
kilogram (kg), and second (s). While other units still survive, particularly in older published
works, their use is discouraged.
Work Done Against a Constant Force
The force due to the gravitation attraction of the Earth on mass m is given by Fex = mg,
where g is the acceleration due to gravity on the Earth’s surface. This force is directed
down towards the Earth’s center, and it depends on the inverse square of the distance. On
the Earth’s surface, approximately 6370 km from the center of the planet, the acceleration due to gravity is approximately 44 times that on a TV satellite in geosynchronous
orbit, 42,000 km from the center. However, since almost all of us live and work within
10 km of the Earth’s surface, the variation in our distance from the center of the Earth
is negligible.
Even at the Earth’s surface, the acceleration due to gravity varies from a sea level
value of 9.832 m s−2 at the poles to 9.780 m s−2 at the equator, mostly because the centrifugal acceleration caused by the Earth’s rotation reduces the pull to the center to
the Earth at the equator. Other effects such as altitude and variations in density of the
Earth’s crust also affect local gravity. For this reason, scientists have adopted a standard
gravity; approximately the mean gravity at a latitude of 45°N. Standard g is called
g0: g0 = 9.80665 m s−2.
If a force does not change significantly with distance, it can be treated as a constant
for the purpose of the integration in Eq. 2.2
w = 3 Fex (x) dx = 3 mg dx ⬇ mg 3 dx.
Fex
(2.6)
x1
x2
x
FIGURE 2.3 The force-extension
curve for a spring obeying Hooke’s
Law. On extending the spring
from x1 to x2, the work done on
the system is the area of the light
green polygon.
18 Chapter 2 |
The First Law: Energy Is Conserved
Integrating this between two values of x, x 1 and x 2, gives
θ laser
x2
r
tip
FIGURE 2.4 Molecular force
microscopy; one end of a
biomolecule is attached to a
surface and the other to an
atomic force microscope tip.
As the tip is pulled back from
the surface, the force bends
the cantilever.
Force (pN)
(2.7)
x1
surface
A
B
C
Extension (nm)
FIGURE 2.5 Force-extension
curve for single stranded DNA.
At low levels of extension, the
apparatus straightens the DNA,
which takes little force (A).
However, once the DNA is fully
extended (C), the force required
to further stretch it is large.
Force (pN)
w = mg 3 dx = mg ( x2 - x1 ) = mgh ,
cantilever
A
B
C
65 pN
Extension (nm)
FIGURE 2.6 Force-extension
curve for double stranded DNA.
At low levels of extension, the
apparatus straightens the DNA
with little force (A). Once the
DNA is fully extended (B), the
force increases to 65 pN, and
begins to unzip the DNA (C),
converting a gently pitched
Watson-Crick double helix to an
extended single-stranded conformation. When this process is
complete the DNA behaves as if
single stranded.
where h = x 2 - x 1.
Molecular Force Microscopy
Using technology that became available in the last twenty years, starting with the
invention of the scanning tunneling microscope, we can reproduce Robert Hooke’s
experiments on springs, but now using biological molecules. The essence of these
experiments is simple; we tether one end of a biological molecule to a surface, using
a chemical cross-linking agent or a very tight noncovalent binder, and the other end to a
movable object, such as a polymer bead or an atomic force microscope (AFM) tip. We
then move the object away from the surface, either the bead with a pair of tightly focused
counter-propagating laser beams called ‘laser tweezers’ or the AFM tip mechanically. This
applies an external force, stretching the molecule.
Figure 2.4 shows the basics of a molecular force microscope. A linear biopolymer is
attached chemically to a surface and also to the tip of an atomic force microscope; the
tip is glued on the end of a flexible cantilever — a very narrow, flexible wafer of silicon
oxide or nitride. As the cantilever is withdrawn from the surface, the distance by which
it is displaced can be measured very precisely using a laser interferometer. The force on
the tip also bends the cantilever, and the angle by which it bends can be measured by the
deflection in the direction of the laser and directly converted into a force. So, measuring
these two quantities, the distance the cantilever moves from the surface and the bending
angle, allows us to construct a force-extension curve for the biomolecule.
Few, if any, biomolecules obey Hooke’s Law, but some, like single-stranded DNA
(ss-DNA), have a comparatively simple force extension curve. Unstretched ss-DNA can be
considered a random coil (figure 2.5A), or worm-like chain, and as you might expect, as
you stretch it, it uncoils rather easily (figure 2.5B). However, once the coils are gone and
the chain becomes nearly straight, further stretching requires elongating covalent bonds
beyond their ‘natural length’, or opening bond angles more than the nearly-tetrahedral angle
most would like to have, or moving torsion angles into sterically unfavorable ‘eclipsed’
conformations. This requires considerable force, and so the force-extension curve bends
dramatically upwards (point C).
However, with double-stranded DNA, one can convert it to a stretched form by
unwinding the strands and breaking the hydrogen bonds that hold together the base pairs
in the double helix; this takes much less force than needed to snap a covalent bond —
65 pN, in fact, for DNA from the double-stranded virus phage l. At that force, the double
helix begins to ‘unwind’ (figure 2.6C), and the stretched chain increases in length by
70%, as shown by the flat plateau in figure 2.6. The detailed structure of the stretched
form of DNA is not known. Different structures can form depending on whether one or
both DNA strands are attached to the pulling surfaces, and whether ‘nicks’ occur in the
strands. Reducing the force on the ends of the DNA allows the DNA strands to re-anneal
and re-form the coiled form of B-DNA.
E X A M P L E 2 .1
B-DNA (ds-DNA) has a length of 0.34 nm per base pair. The stretching of l phage
DNA occurs at a force of 65 pN, while that of the artificial DNA poly(dAdT) occurs
at 35 pN. Compute the work done in kJ/mol in stretching a base-pair length of DNA
by 70%.
Energy Exchanges | 19
SOLUTION
The change in length of a nucleotide length of DNA on stretching is ( 0.70 * 0.34 ) nm =
0.24 nm or 2.4 * 10−10 m . For l DNA, the force required is 65 pN = 6.5 * 10−11 N . The
work done in Joules is therefore 2.4 * 10−10 m * 6.5 * 10−11 N = 1.6 * 10−20 J . For one
mole of base pairs, the work done is that for a single base pair, multiplied by Avogadro’s
number: w = 1.6 * 10−20 J * 6.022 * 1023 mol−1 = 1.0 * 104 J mol−1, or 10 kJ mol−1 . For
poly(dAdT), the work done is w = 2.4 * 10−10 m × 3.5 * 10−11 N * 6.022 * 1023 mol−1
* 10−3 kJ−1 J = 5.1 kJ mol−1.
In practice, integrating curves such as those shown in figure 2.5 and figure 2.6 requires
either fitting the curve to a function (based on theory or a curve-fitting procedure), or else
using computer numerical integration.
Work Increasing or Decreasing the Volume of a Gas
Consider a system such as a gas or liquid enclosed in a container with a movable wall or
piston (figure 2.7). The system can expand and do work on the surroundings if the external
pressure pex is less than the pressure of the system p, or the surroundings can do work on
the system by compressing it if pex is greater than p.
By definition, pressure is just force per unit area, and so the external pressure pex is
related to the external force Fex by pex = Fex >A , where A is the cross-sectional area of
the piston. We can then write
Fex dx = Fex >A * A dx = -pex dV ,
(2.8)
since the product A dx = -dV is the change in volume of the gas container. Why the negative sign? If the change in the x coordinate has the same direction and therefore the same
sign as the external force, the volume of the container decreases, and so dV is negative if
dx is positive. Thus, from the equation above, force times displacement is equivalent to
pressure times the negative of the change of volume.
The work done on the system is therefore:
w = 3 Fex dx = - 3 pex dV
(2.9)
Work associated with a change in the volume of the system is often termed pressure–volume
work, or pV work. The SI unit of pressure is the pascal, or Pa; 1 Pa = 1 N m - 2 . The SI
unit of volume is of course the cubic meter, or m3. 1 m3 is a larger volume than most
chemists and biochemists use in practice, and so we commonly substitute the liter(L);
1 L = 10 - 3 m3. Similarly, 1 pascal is a much lower pressure than ambient, and so we
often work in bars; 1 bar = 105 Pa. A bar is approximately the pressure exerted by the
atmosphere at sea level. However, we have to be careful with non-SI units, and we advise
you to convert all units to SI before doing any calculations.
SI carries with it a sort of warranty; if you use the correct formula, and all units are
SI units, the result will be in an SI unit. Thus, for example, if we multiply the SI unit of
pressure (1 Pa) by the SI unit of volume (1 m3), the result is the SI unit of energy (1 J).
Fex
x
pex
A
dx
dV=–Adx
EXAMPLE 2.2
Calculate the pV work done when a system containing a gas expands from 1.0 L to 2.0 L
against a constant external pressure of 10 bar.
SOLUTION
First convert to SI units!
pex = 10 bar * 105 Pa bar - 1 = 106 Pa; V 1 = 1 L * 10 - 3 m3 L - 1 = 10 - 3 m3;
V 2 = 2 * 10 - 3 m3
FIGURE 2.7 Relationship
between force and pressure. As
the gas is compressed under an
external force acting on a piston
of area A, it creates a pressure
pex. The change in displacement
along the direction of the force,
dx, reduces the volume of the
system by an amount dV = ⫺Adx.
20 Chapter 2 |
The First Law: Energy Is Conserved
The external pressure is constant, and so it can be taken outside the integral:
V2
V2
w = - 3 pex dV = -pex 3 dV = -pex ( V2 -V1 )
V1
V1
= -10 Pa * (2 * 10
6
-3
m3 -10 - 3 m3) = -103 J
The system does work; the sign of w is negative.
Friction
Friction — specifically, kinetic friction — is the force that acts on a surface in motion
and in contact with a solid, liquid, or gas to retard the motion. The frictional force is
approximately independent of the velocity and the contact area but proportional to the load
on the surface. Because friction involves a force applied over a distance, an object moving
against a frictional force does work on the surroundings. This work is quite different from
the work done compressing a piston or extending a spring. When we extend the spring
or compress the piston we store potential energy that can be released when the spring is
allowed to contract or the gas to expand. The work we do on the system can be returned
to us as work done by the system on the surroundings. These processes are considered
conservative processes.
In contrast, frictional work is converted to heat, not to potential energy, and cannot
easily be reconverted back to work. As we will see in chapter 3, a heat engine is needed to
convert heat energy back into work. This is an intrinsically inefficient process. Therefore,
frictional work is called a dissipative process rather than a conservative process. It is essential to understand, however, that, whether the process is conservative or dissipative, energy
is always conserved. A conservative process stores work as potential energy whereas a
dissipative process converts it to heat.
Heat
When two bodies are in contact with each other, their temperatures tend to become equal.
Energy is being exchanged. The hot body will lose energy and cool down; the cold body
will gain energy and warm up. This energy exchange is said to occur by heat transfer, and
the energy that passes through the system-surroundings boundaries because of a temperature difference between the two is called heat. The sign convention for heat is the same
as that for work: heat is positive if it flows into the system and negative if it flows out of
the system.
For a closed system, the transfer of an infinitesimal quantity of heat dq will result in
a change dT in its temperature. The ratio dq/dT is called the heat capacity of the system
and is given the symbol C. The heat capacity for every material is the quantity of heat that
is needed to raise its temperature by 1⬚C or 1 K.
dq
= C or dq = CdT
dt
(2.10)
C is a distinct property of a given material, and will also tend to vary with temperature.
Thus, the heat q gained when the temperature of the system changes from T1 to T2 should
be evaluated from the integral
T2
q = 3 C dT .
(2.10a)
T1
If C is constant in the temperature range T1 to T2, then it can be removed from the integral, and
q = C( T2 - T1 ) .
(2.10b)
Energy Exchanges | 21
For a hot body in contact with a cold body, the heat capacity of both bodies must be known,
and Eqs. 2.10 –2.10b must be applied to each body individually.
The heat capacity C is an extensive quantity; that is, it increases with the amount of
material in the body. An intensive quantity is one that remains unchanged when a system
is subdivided. If 100 g of water at a uniform temperature is divided into two portions, the
temperature of each part is unchanged. Thus, temperature is an intensive property. Similarly,
the pressure of a system is also an intensive property. For a pure chemical substance, we
can convert C into an intensive quantity — one that depends only on the properties of the
substance, and not its amount — by dividing by the number of moles of substance present:
Cm = C>n, where n is the number of moles and Cm is the molar heat capacity. In general,
when any property has an equivalent molar property (such as how volume has molar volume)
the property is extensive and the molar property is intensive. Energy is extensive, but energy
mol−1 is intensive; mass is extensive, but density or mass volume−1 is intensive.
A heat capacity is usually determined by measuring the change in temperature when
a known quantity of electrical energy is dissipated in the material. The SI unit of heat
capacity is J K−1; that of molar heat capacity is J mol−1 K−1 . In older books, heat capacities
are often given in cal K−1 . Until it was discovered that the heat capacity of liquid water
depends slightly on temperature, one calorie or cal was the quantity of heat needed to raise
the temperature of 1 g of water by 1 K. Later it was simply defined as 1 cal = 4.184 J.
We also occasionally encounter the term specific heat capacity, or c (lower case),
which is the heat capacity divided by the mass: c = C>m . It has the SI unit J kg−1 K−1,
but it is often given in J g−1 K−1 .
EXAMPLE 2.3
Calculate the heat, in joules, necessary to increase the temperature of 100.0 g of liquid
water from 25.0°C to 75.0°C at ambient (1 bar) pressure. The molar heat capacity of
liquid water is 75.4 J mol−1 K−1 and is nearly independent of temperature.
SOLUTION
( 100.0 g ) > ( 18.015 g>mol ) = 5.551 mol
( 75.0⬚C - 25.0⬚C ) = 50 K
5.551 mol * 50.0 K * 75.4 J mol - 1K-1 = 2.09 * 104 J.
Internal Energy
You are probably already familiar with the idea that when we transfer heat energy to
a body, raising its temperature, the energy goes into motion of its constituent atoms or
molecules. The field that studies how that energy is distributed among the countless modes
of motion possible in a substance is called statistical mechanics. We will consider statistical
mechanics in more detail in chapter 5; here we are only concerned with what happens when
we transfer heat to a substance. The simplest instance of statistical mechanics is the kinetic
theory of gases, which was the first successful attempt to explain the properties of a class
of materials (ideal gases, which obey the ideal gas equation of state pV = nRT ) in terms
to the motion of atoms and molecules. The theory was constructed by the German physicist
Rudolf Clausius, on two premises:
• Gases are made up of molecules in ceaseless random motion, colliding with each other
and with the container.
• Gases have no attractive forces between them, and their collisions are elastic (they
neither gain or lose energy).
Our analysis starts with a rectangular-sided container bounded by edges of length a, b, and
c, corresponding to the x, y, and z directions. Let’s consider a single gas molecule in that
22 Chapter 2 |
vy
-vz
The First Law: Energy Is Conserved
vy
vz
c
b
FIGURE 2.8 A particle
collides with the z wall of a
box, reversing the sign of the
z component of its momentum,
while leaving the x and y
components unchanged.
container, of mass m, moving with a velocity vector v. The momentum of that molecule
is given by:
p = mv
(2.11)
It is unfortunate that the symbol for momentum, p, is similar to that for pressure, p.
However, momentum is a vector, and so will either be in bold font, or, if we are discussing
one component of the momentum, it will have a subscript corresponding to a Cartesian
direction (e.g., px). Pressure, in contrast, is always a scalar.
For the moment, let’s assume the molecule is a point particle, and therefore, even if
there are other molecules present, it will not collide with them. Since we have forbidden
collisions except with the wall of the container, the particle will simply bounce ceaselessly
off the walls, like a frictionless billiard ball. We show this behavior for a two-dimensional
box in figure 2.8.
Each time the ball hits a wall, its momentum in the direction normal to the wall
will change sign. Let’s consider a collision with the z wall (the wall perpendicular to the
z direction). If the component of the momentum along the z direction is pz,1 = mv z before
the collision, it is pz,2 = -mv z after the collision. The momentum along the other two
directions is unchanged by the collision (see figure 2.8).
The change in the z component of the momentum per collision with the z wall is
therefore
⌬pz = pz, 2 - pz, 1 = -2mv z .
(2.12)
Momentum, like energy, is a conserved quantity. If momentum is lost by the particle, it
must be gained by the wall, which recoils very slightly every time a particle collides with
it. So the gain in momentum in the wall is:
⌬pz, wall = - ⌬pz = 2mv z
(2.13)
The particle typically collides with the z wall hundreds of times per second. The exact time
interval t between collisions is obtained by dividing the distance traveled in the z direction
between collisions, which is twice the length of the box in that direction or 2c, divided by
the magnitude of the velocity in the z direction, vz.
t = 2c>v z
(2.14)
If there are t seconds between collisions, in 1 second there will be 1/t collisions. The
number of collisions per second is therefore 1/t = v z /2c. In a time ⌬t there are v z ⌬t/2c
collisions, and in this increment of time ⌬t the total momentum transferred to the wall by
this single particle is
⌬pz, total = vz ⌬t/2c * ⌬pz, wall ,
(2.15)
which from Eq. 2.13 becomes
⌬pz, total = v z ⌬t/2c * 2mv z = mv 2z ⌬t/c.
(2.16)
This momentum transfer occurs in tiny bursts. However, if we ‘smooth out’ the collisions,
we can treat the transfer of momentum as a continuous process. Think of it this way: in
a building with a corrugated iron roof, at the beginning of a rain shower, we can hear the
raindrops collide individually with the roof. However, as the rain gets heavier, and there
are hundreds of drops hitting the roof per second, the individual impacts merge into a
continuous roar. The mathematical operation of going from the discrete to the continuous
is accomplished by replacing the differences in Eq. 2.16 by differentials.
⌬pz, total/⌬t _ dpz, total/dt = mv 2z /c
(2.17)
Energy Exchanges | 23
The derivative of momentum is force. So the time-averaged force on the z wall due to this
single particle is:
Fz = mv2z >c
(2.18)
Pressure is force per unit area. The area of the z wall is ab. So the pressure on the z wall is
p = Fz >ab = mv2z >abc = mv2z >V,
(2.19)
since the product abc is the volume V of the container.
We’ve gone as far as we possibly can with a single particle. A real gas is composed of
a large number of individual particles, each with its own z velocity. We’ll assume the gas
is pure and that therefore all of the masses of the particles are the same. If we label the
particles 1,2,3… the total pressure is therefore the sum of the pressure of particles 1, 2, 3…
p =
mv2z,1
V
+
mv2z,2
V
+
mv2z,3
V
N
Á = a m b v2z, i
V ia
=1
(2.20)
The mean square velocity in the z direction is just the sum of all squared velocities, divided
by the number of particles N. It is therefore:
1
N
8 v 2z 9 = a b a v 2z, i
N
(2.21)
i= 1
But there is nothing special about the z direction. By symmetry, the mean square velocities
in the x, y and z directions are the same. Furthermore, by Pythagoras’ theorem, the total
velocity squared is v 2 = v 2x + v 2y + v 2z = 3v 2z . So combining this with Eqs. 2.20 and 2.21
we get:
mN
p = a
b 8v2 9
(2.22)
3V
Multiplying both sides by V:
pV = 1>3 mN 8 v 2 9
(2.23)
N, the total number of molecules, is related to the total number of moles n, by n = N/N A,
where NA is Avogadro’s constant, the number of particles in a mole. The best current
value of Avogadro’s constant is (6.0221418 { 0.0000003) * 1023. The mass of a mole
M is equal to the mass m of the particle times the same constant: M = m * N A. So the
product mN is the same as the product nM. Replacing this in Eq. 2.18, and then comparing
the result with the Ideal Gas Law, we get:
pV = 1>3 nM 8 v 2 9 = nRT
(2.24)
Canceling the number of moles, and multiplying across by 3/2, we get
> 2 M 8 v 2 9 = 3>2 RT .
1
(2.25)
But the left-hand side of this equation is the total kinetic energy of all the gas particles.
We call this total energy the molar internal energy Um = 1> 2 M 6 v 2 7, an intensive
quantity. Similarly, if we multiply both sides of Eq. 2.25 by n, we get U, the internal
energy, an extensive quantity.
This equation is remarkable for what it is missing: p and V! It says that the internal
energy of an ideal gas depends only on the temperature, and is entirely independent of
the volume and the pressure. This makes intuitive sense; since the theory assumes no
interactions between the gas molecules, reducing or increasing the distance between them
should make no difference.
24 Chapter 2 |
The First Law: Energy Is Conserved
Constant Volume Heat Capacity
What happens when we heat the container? In classical thermodynamics, we express the
First Law by the equation
⌬U = w + q.
(2.26)
That is, there are two ways we can add energy to a system; we can heat it, or do work on
it. As long as we keep the volume fixed, we do no work: by Eq. 2.9, doing work requires
changing the volume. Therefore, at constant volume,
⌬U = qV .
(2.27)
In thermodynamics, a subscripted variable means we keep that variable constant. If we
now take the partial differential with respect to temperature, and include Eq. 2.25, we get
a
0q
0U
b = a b = 3>2 nR .
0T V
0T V
(2.28)
But dq/dT is the heat capacity (Eq. 2.10) so we call (0q/0T)V the constant volume heat capacity.
We can once again divide across by the number of moles, n, to make everything intensive.
CV,m = a
Gas
CV,m/R
Helium
Neon
Argon
Krypton
Xenon
1.5000
1.5000
1.5000
1.5000
1.5000
0qm
0Um
b = a
b = 3>2 R
0T V
0T V
(2.29)
This is a remarkable result; it says that the heat capacities of all ideal gases that can be
regarded as point particles should be the same! How well does it work?
Quite excellently, it appears, as we can see by the table on the right! Amazingly, 127 g
of xenon has the same heat capacity and the same internal energy as 4 g of helium. Theory
predicts CV,m /R for an ideal gas to be 3/2; experiment shows it is 1.5000; and, therefore,
agrees with theory to five significant digits. In fact, modern theoretical chemistry can
predict most thermodynamic properties of simple gases more accurately than they can
be measured.
Constant Volume Heat Capacity of Diatomic Gases
Gas
CV,m/R
y (cm−1)
H2
2.467
4395
N2
2.503
2360
CO
2.506
2170
O2
2.535
1580
F2
2.770
892
Since we’ve succeeded with the heat capacities of monatomic gases, let’s see if we can
equally elegantly explain the molar constant volume heat capacities of diatomic gases at
1 bar and 298 K.
While they’re nowhere nearly as similar to each other as the monatomics, it is clear
the first four diatomic gases in the table are very close to a value of CV,m/R = 5/2, with
fluorine a little bit larger. Why is this? A diatomic gas, in addition to moving translationally in three dimensions like monatomic gases, can also rotate. However, only rotations
about two dimensions are meaningful; rotation about its long axis leaves the molecule in an
identical state. These two additional degrees of freedom or modes can also absorb energy,
just as the three translational modes absorb energy. Because the molecule now has five
active modes, CV,m/R has a value of approximately 5/2. The principle that energy will be
shared out equally among all active modes is called the equipartition theorem.
What about fluorine? One additional mode available to a diatomic is the vibrational
mode along the bond axis. Why is this mode inactive in the first four diatomics and only
partly active in fluorine? The answer lies in quantum mechanics, which we will treat in
more detail in chapter 11. But briefly, the energy required to raise a diatomic molecule
into its first vibrational excited state is often quite high. Vibrational energies are typically
given in wavenumbers, and are listed in the table. It can be seen that fluorine, and to a
lesser extent oxygen, have comparatively low vibrational frequencies, and CV,m/R values
slightly higher than 5/2, because a small but significant fraction of the molecules can be
thermally excited to a higher vibrational state at room temperature.
Constant Volume Heat Capacity of Monatomic Solids | 25
Moving away from gases and towards ‘condensed matter’ — liquids and solids — let’s
now consider the heat capacities of the simplest solids: metals, which are composed of
crystals with isolated atoms in a simple lattice. The molar heat capacities of metals at room
temperature are all close to 3R; this phenomenon is so general it has been given the name
Dulong and Petit’s Law.
Why do atoms in a crystal lattice have molar heat capacities of 3R, while those in an
ideal gas have CV,m values of (3/2)R? The answer, again, lies in the number of modes available. An atom in a crystal lattice interacts with its neighbors, and as it moves away from
its equilibrium position in the lattice, its potential energy increases. Atoms in an ideal gas
do not interact with each other at all, and so have no potential energy. So atoms in a lattice
can put RT/2 units of internal energy into kinetic energy along a particular direction, and
RT/2 units of energy into potential energy; thus, per mole, their total internal energy is
double that of the same atom in an ideal gas at the same temperature.
Vibrational modes always have kinetic and potential energy components, and so
can always absorb RT units of energy per mode, when thermally accessible. Rotational
modes, like translational modes, can absorb RT units of energy in a crystal lattice,
but only RT/2 units in an ideal gas, where there is no potential energy associated with
rotation.
We can gain additional insight from the molar heat capacities of a couple of metals as a
function of temperature. These are shown in figure 2.9. The heat capacities of both sodium
and aluminum drop to zero at 0 K; this is a consequence of quantum mechanics and is true
of all substances. They rise rapidly above a temperature of about 10 K and eventually reach
a plateau near 3R = 24.94 J mol - 1K - 1; as can be seen, aluminum, a hard metal where the
lattice vibrations are spaced quite far apart, rises more slowly and has not quite reached
the plateau at room temperature. In sodium, a softer metal, it is easier to displace atoms
from their equilibrium positions, and so the heat capacity increases faster as a function of
temperature, but still reaches approximately the same plateau.
CV, m (J mol–1 K–1)
Constant Volume Heat Capacity of Monatomic Solids
Metal
CV,m/R
Mg
Al
Fe
Pt
Pb
2.91
2.78
2.96
2.97
2.98
25
20
Na
15
Al
10
5
50 100 150 200 250 300
T (K)
FIGURE 2.9 The constant
volume heat capacities of
sodium and aluminum, as a
function of temperature.
Heat Capacity of Molecular Solids and Liquids
125
Cp, m (J mol–1 K–1)
In figure 2.10, we show the temperature dependence of the constant pressure molar heat
capacities of glycine and alanine. We use these because constant volume heat capacities
are very difficult to obtain for most solids.
At low temperature, the two curves resemble each other and those of metals, except
that in addition to translations, restricted rotations of the molecules within the lattice
(called librations) also contribute 3R to the heat capacities. At higher temperatures,
instead of reaching a plateau at 6R, as the amino acids seem to be trying to do at
100 K, the heat capacities continue to increase, as more and more vibrational modes
become thermally accessible and contribute to Cp,m. Alanine, a larger molecule with
more low-frequency vibrational modes, diverges appreciably from glycine at these higher
temperatures.
Finally, we consider the heat capacities of more complex molecular solids and liquids;
a variety of these are given in table 2.1. All of them are at 298 K except where specified.
There are some general trends:
• With the exception of water ice, all of the heat capacities exceed 6R, which is the heat
capacity due to lattice vibrations and hindered rotations alone. Vibrations play an
important role in the heat capacities of most molecules at ambient temperatures.
• As can be seen from the last three rows, the molar heat capacities increase with
molecular mass M, because increasing the number of atoms increases the number of
low-energy vibrations that absorb heat.
100
75
50
ala
gly
25
50 100 150 200 250 300
T (K)
FIGURE 2.10 The constant
pressure heat capacities of the
amino acids glycine and alanine,
as a function of temperature.
26 Chapter 2 |
The First Law: Energy Is Conserved
TABLE 2.1 Heat capacities of molecules
Compound
H2O (l, 273 K)
Cp. m
J mol−1K−1
M
g mol−1
75.4
18.0
H2O (s, 273 K)
38.1
18.0
potassium iodide
52.8
166.0
a-D-glucose
219.2
180.2
L-tyrosine
216.4
181.2
anthracene
211.7
178.2
decanoic acid
375.6
172.3
dodecanoic acid
404.3
200.3
tetradecanoic acid
432.0
228.4
• For molecules of similar mass (rows 3–8), those that are more flexible (which usually
means softer solids, such as decanoic acid) have more low frequency vibrations
and therefore higher heat capacities than harder materials composed of more rigid
molecules, molecules linked together by hydrogen bonds, or ionic solids.
• Liquids have higher heat capacities at the melting points than the corresponding solids;
an example is water and water ice.
For the same reasons, we find that macromolecules in random coils, such as single-stranded
nucleic acids and denatured proteins, have higher heat capacities than the corresponding
folded proteins and double-stranded DNA and RNA. We shall see later that this difference
in heat capacities is important in understanding the stability of native proteins, and to a
lesser extent, nucleic acids.
State and Path Variables
Consider a closed system in the absence of all external fields. This statement describes a
system that we can only approximate on Earth. We cannot turn off gravity, for example.
However, for many practical purposes we can ignore the effects of gravity.
If the system consists of a pure liquid, specifying the pressure p, volume V, and temperature T of the liquid is sufficient to specify many other properties of the liquid, such
as its density, surface tension, refractive index, and so forth. These other properties of the
system, and p, V, and T, are called variables of state (or state variables). The variables of
state depend only on the state of the system, not on how the system arrived at that state.
The useful characteristic of variables of state is that when a few ( p, V, T, chemical composition, etc.) are used to specify a system, all other variables of state are determined implicitly.
Any property of the system that depends only on the variables of state must itself be
a variable of state. As we have seen, the internal energy U of an ideal monatomic gas is
a function only of the temperature, a state variable, and so it is itself a state variable. As
we have seen, we can raise the temperature and thus increase the internal energy of an
ideal, monatomic gas by heating it. However, we could also raise its temperature by doing
frictional work. This might seem like an odd idea — gases don’t exert much friction on
objects moving through them — but recall that a space vehicle or a meteorite entering the
Earth’s atmosphere glows red hot from friction. Move a surface fast enough through a gas,
and you will certainly heat it up!
So while the internal energy of a system is called a state variable, depending only on
the state of the system and not how it arrived where it is, heat and work are path variables,
State and Path Variables | 27
which depend on the route taken to the final state of the system. Properties of the system
that we can measure, such as pressure, volume, and temperature, are state variables. When
we measure the temperature of a beaker of water, we don’t care whether it warmed up by
sitting in strong sunlight or on a hot plate.
Once we specify the path, however, a path variable becomes a state variable. For
example, the heat capacity, dq/dT is clearly a path variable, because it depends on q, itself
a path variable. But if we specify a constant volume path, then CV , the heat capacity at
constant volume, is a state variable, equal to dU/dT.
Because state variables do not depend on the path taken, but only on the state of the
system, the change in a state variable as a result of a physical process depends only on the
initial and final state, and not how we get there. For this reason, state variables are intimately
linked to conservation laws. If we take the system from state 1 to state 2, and then back to
state 1 — a cyclic path — all the state variables must return to their initial values. This means
that internal energy cannot be reduced or increased over a cyclic path; if it could, we could
generate internal energy in a system by taking it around a cyclic path which would violate
the First Law. Because work and heat are not state variables, we can create or destroy them
in a cyclic path, and in fact we will see in chapter 3 how we can use a cyclic path to turn heat
into work. However, the internal energy, the sum of heat and work, must be unchanged, and
therefore whatever heat we consume must be turned into work.
Reversible Paths and Reversible Processes
The external pressure pex and the pressure of the system p are not equal when the system is
undergoing compression or expansion; if they were, nothing would happen. Unfortunately,
if pex is very different from p, then the work will not simply change the volume of the gas,
but impart velocity to it, and thence generate frictional heat. If pex is kept nearly equal
to p during the process (to be precise, in the limit where pex - p goes to zero), the process
is said to be reversible. A very small change in pressure will reverse the process from an
expansion to a compression, or vice versa. The sign of the work will change from positive
to negative, depending on an infinitesimal change in the pressure. If pex and p differ
significantly, irreversible expansion or compression will occur, depending on whether
pex is smaller or greater than p.
To visualize a reversible compression or expansion, consider the expansion of an ideal
gas in a cylinder with a frictionless and weightless piston, as illustrated in figure 2.11a.
Suppose that the pressure outside the cylinder is always at 1 bar and the pressure inside the
cylinder is initially at 2 bar. If we remove the stops that hold the piston in position, the gas
will expand irreversibly until a final state is reached at which its pressure becomes 1 bar.
During the course of expansion, the pressure of the system is always significantly greater
than that of the surroundings (that is why the expansion is called irreversible). The two
become the same only at the end of the expansion.
We can carry out the expansion in a different way. Instead of holding the piston in
position with stops at the beginning, we put many small weights on top of the piston to
make up for the pressure difference, as shown in figure 2.11(b). The expansion is then carried out in a stepwise manner by removing one weight at a time. If the number of weights is
very large (and the weight of each very small), the pressure of the system is always almost
the same as that of the surroundings during the course of expansion. Furthermore, if we
add rather than remove a weight, the process will be reversed.
A reversible path is one in which the process can be reversed by an infinitesimal
change of the variable that controls the process. The path illustrated in figure 2.11(a) is
not a reversible one; the path illustrated in figure 2.11(b) becomes a reversible one when
the weights are infinitesimally small and their number approaches infinity.
28 Chapter 2 |
The First Law: Energy Is Conserved
FIGURE 2.11 Comparison of
(a) an irreversible expansion
and (b) a reversible expansion.
In a reversible expansion, the
internal pressure is always
nearly equal to the external
pressure.
(a)
Irreversible
expansion by
removal of stops
pex
stop
p
pex
stop
p
(b)
Reversible
expansion by
one-at-a-time
removal of
weights
pex
pex
p
p
In the special case when a system is expanding irreversibly against a vacuum, pex is
zero and no work is done.
Reversible heating or cooling is similar to reversible expansion or compression. Here,
the temperature difference between the system and the surroundings must be very small—
infinitesimal. Heat flows in or out, depending on very small changes in temperature. If the
temperatures of two bodies differ significantly, irreversible heat exchange occurs when the
two come into contact.
Most processes considered by classical thermodynamics assume reversibility.
Equations of State
A few variables of state are usually sufficient to specify all others. This means that
equations exist that can relate the variables of state. Such equations are termed equations
of state. The simplest and most frequently used equations of state link p, V, and T.
State and Path Variables | 29
V = m/r
(2.30)
At a higher level of accuracy, the volume of a solid or liquid does change somewhat
with T and p. Experimental data for 1 mol of liquid water are shown in figures 2.12 and
2.13. The molar volume is plotted versus temperature at constant pressure in figure 2.12,
and versus pressure at constant temperature in figure 2.13. Equations can be obtained for
V as a function of p and T for liquid water from these data.
If we want to substantially improve on the primitive equation of state V = V ⬚, we
can introduce a linear dependence on temperature and pressure. This gives the following
intensive equation of state, which can be made extensive by multiplying both sides by n.
V m = V m⬚[1 + a(T - T⬚)][1 - k(p - p⬚)]
(2.31)
Vm (m3 × 10–6)
19.0
18.8
18.6
18.4
18.2
18.0
18.2
18.0
17.8
17.6
17.4
In this expression, a is the coefficient of volume expansion, defined by
1 0V
a = a b
V 0T p
k = -
1 0V
a b
V 0p T
200 400 600 8001000
p (bar)
(2.32)
and k (the Greek letter kappa) is the isothermal compressibility…
(2.33)
20 40 60 80 100
T (°C)
FIGURE 2.12 The molar
volume of liquid water as a
function of temperature at a
constant pressure of 1 bar.
Vm (m3 × 10–6)
Solids or Liquids
The volume of a solid or a liquid does not change very much with either pressure or temperature. For the time being, we will not consider changes from solid to liquid or any other
change except p, V, and T. Therefore, a first approximation for the equation of state of a
solid or liquid is that the volume is constant, or V = V ⬚, where V ⬚ is the volume under
standard conditions (298 K, 1 bar). This means that calculating the volume of a solid or
liquid simply requires finding the density or specific volume at one temperature and using
that value for any temperature and pressure. The volume is related to the density r by
FIGURE 2.13 The molar
volume of liquid water as
a function of pressure at a
constant temperature of 298 K.
a is the fraction by which the volume of the system increases per degree temperature
increase, while k is the fraction by which it decreases per Pa of increased pressure. T⬚ and
p⬚ are the standard temperature and pressure, respectively. Isothermal compressibilities and
coefficients of expansion are tabulated in handbooks of chemical and physical data. The
dependence of the volume of liquid water on temperature (figure 2.12) is more complicated.
The plot is not a straight line, and the molar volume of water actually has a minimum value
at 277 K (4°C). Most of the time, we use the simplest approximation, that V is independent
of T and p for a solid or liquid.
30
20
Vm (L)
Gases
The volume of a gas varies greatly with T and p but is nearly independent of the type of
gas. The simplest approximate equation of state for gases is the Ideal Gas Law, ( pV = nRT ),
which has already been introduced in Eq. 2.24.
A plot of how the volume of 1 mol of an ideal gas depends on pressure is shown in
figure 2.14. Note that, although 1000 bar is necessary to change the volume of liquid water
by 3%, a change from 1 bar to 1000 bar for an ideal gas will cause a change in volume
by a factor of 1/1000. The ideal gas equation has the great advantage that it contains no
constants applying to individual gases; it applies to all gases if the pressure is low enough.
It is an exact limiting equation for all gases as p approaches zero. For higher pressures, it is
an approximation for real gases. The answers obtained using Eq. 2.24 are usually accurate
within a percent for most gases near room temperature and atmospheric pressure.
500 K
10
273 K
0
2
4 6
p (bar)
8
FIGURE 2.14 Molar volume
of an ideal gas as a function of
temperature and pressure.
10
30 Chapter 2 |
The First Law: Energy Is Conserved
There are several more accurate equations of state for gases; these correct the ideal
gas equation with parameters specific to individual gases. One example is the van der
Waals gas equation:
a p+
n2a
b (V - nb) = nRT ,
V2
(2.34)
where a and b are parameters specific for each individual gas. The van der Waals a constant
is an approximate measure of the attractive forces between molecules, and the b constant is
an approximate measure of the intrinsic volume of the gas molecules. When both of these
are zero, Eq. 2.34 simplifies to the ideal gas equation.
The equations of state that we have been discussing have all been applied to systems
containing only one component. For mixtures, the number of grams or moles of each
component must be specified, and the equation of state will depend on the amounts of the
components. For gases that can be approximated by the ideal gas equation, the results are
particularly simple. The ideal gas equation can be applied to each gas in the mixture as
if the others were not there. This is often called Dalton’s Law of Partial Pressures. The
partial pressure of each gas can be calculated as follows:
pi =
niRT
V
(2.35)
The total pressure is just the sum of the partial pressures:
p = a pi
(2.36)
i
The partial pressures can also be obtained from the total pressure and the mole fractions
xi of each component:
pi = x i p
xi =
ni
a i ni
(2.37)
The equations for partial pressures are useful because often we are interested in only one
component of a gas mixture. For example, for living organisms, the partial pressures of
oxygen and carbon dioxide in the air they breathe are more important than the total pressure.
The Enthalpy
As we have seen, heat is not a state variable. However, over many years, physical chemists
collected heats for various processes; heats of solution, reaction, freezing, etc., and found
that in practice they didn’t depend very much on exact conditions. It was obvious, therefore,
that there exists a state variable similar but not identical to heat. That state variable is called
the enthalpy, and it is defined as:
H = U + pV
(2.38)
We can see why the enthalpy H is so similar to the heat q by differentiating:
dH = dU + pdV + V dp
(2.39)
Differentiating Eq. 2.26 and substituting in Eq. 2.9, we obtain:
dU = -pexdV + dq
(2.40)
dH = dq - pexdV + pdV + Vdp
(2.41)
Inserting this in Eq. 2.39:
Dependence of the Energy and Enthalpy of a Pure Substance on p, V, and T | 31
If we now assume a reversible process, the second and third terms on the right-hand side
cancel, because pex = p. What remains is:
dH = dqrev + V dp
(2.42)
The second term on the right is zero if we keep the pressure constant, so we conclude that
the differential of the enthalpy, as we have defined it, is equal to the differential of the heat
at constant pressure and in a reversible process.
dH = dqp, rev
(2.43)
It is important to remember, though, that the enthalpy is defined by Eq. 2.38, not Eq. 2.43,
and the identity of enthalpy change and heat is confined to special, albeit quite common,
circumstances. The enthalpy H is also known as the heat content, because the enthalpy change
⌬H at constant pressure is equal to the heat absorbed or given off. We emphasize, however,
that the derivation of Eq. 2.40 and 2.43 requires that only pressure–volume work occurs. If
the system does other types of work (electrical, for example) during a process, these equations
do not apply.
For an ideal gas, pV = nRT. Inserting this in Eq. 2.38, we get:
H = U + nRT
(2.44)
The Constant Pressure Heat Capacity of an Ideal Gas
If we differentiate Eq. 2.43 with respect to temperature, we obtain the constant pressure
heat capacity Cp.
a
0q
0H
b
= a b
= Cp
0T p, rev
0T p, rev
(2.45)
We can also differentiate Eq. 2.44 with respect to temperature at constant pressure:
a
0H
0U
b
= a b
+ nR
0T p, rev
0T p, rev
(2.46)
But as we have seen, the internal energy of an ideal gas is independent of pressure or
volume, so we can remove the constant pressure condition. It is also a state function, and
so does not care if the path is reversible:
a
dU
0U
0U
b
=
= a b = CV
0T p, rev
dT
0T V
(2.47)
Combining 2.45, 2.46, and 2.47 gives the important ideal gas relationship:
Cp = CV + nR
(2.48)
So, just as the derivative of the energy with respect to temperature at constant volume
is the constant-volume heat capacity, so the derivative of the enthalpy with respect to
temperature at constant pressure is the constant pressure heat capacity. Enthalpy bears the
same significance for constant pressure processes that internal energy does for constant
volume processes. In biochemistry, most processes happen at constant or nearly constant
pressure, and so enthalpy will be far more significant to us than energy.
Dependence of the Energy and Enthalpy of a Pure
Substance on p, V, and T
It is often important to know how temperature and pressure affect phase changes (the melting of a solid, the condensation of a gas, etc.), chemical reactions, or other chemical and
physical processes. We first consider the effects of T and p on the energy and enthalpy of
32 Chapter 2 |
The First Law: Energy Is Conserved
TABLE 2.2 Physical Properties* of Water, H2O, M = 18.0153 g mol−1
Solid H2O (ice)
(at 0 C = 273.15 K, and 1 bar)
Density (r) = 0.9167 g mL−1 = 916.7 kg m−3
Molar volume Vm = 19.65 mL mol−1 = 1.965 × 10−5 m3 mol−1
Vapor pressure = 0.00612 bar = 612 Pa
Molar enthalpy of fusion (melting) ΔfusHm = 6007 J mol−1
Absolute molar entropy Sm = 41.3 J mol−1 K−1
Molar heat capacity at 1 bar (Cp,m) = 38.0 J mol−1 K−1
Liquid H2O
Temperature
(°C)
Density
kg m−3
Surface
tension
mN m−1
Vapor
pressure
(Pa)
Molar heat
capacity
Cp,m
Enthalpy of
vaporization,
kJ mol−1
Viscosity
(mPa s)
0
999.8
75.65
612
76.01
44.91
1.792
20
998.2
72.74
2536
75.38
44.08
1.002
40
992.2
69.60
7370
75.29
43.27
0.653
60
983.2
66.24
19900
75.39
42.44
0.466
80
971.8
62.67
46120
75.61
41.56
0.354
100
958.3
58.91
101320
75.95
40.66
0.282
Absolute molar entropy Sm = 63.3 J
= 86.9 J
mol−1
K−1
mol−1
K−1
at 273.15 K
at 373.15 K
Boiling point at 1 bar pressure: 372.76 K
Gaseous H2O (steam)
(at 99.6°C)
Absolute molar entropy Sm = 195.8 J mol−1 K−1
Molar enthalpy of vaporization (boiling) ΔvapHm = 40657 J mol−1
Molar heat capacity at 1 bar (Cp,m) = 36.5 J mol−1 K−1
*Some of the properties listed will be defined and discussed in later chapters.
pure solids, liquids, and gases. The results can then be applied directly to the effects of
T and p on the enthalpy and energy changes of phase transitions and of chemical reactions.
The methods we use in the next several sections are similar to those we will need in later
chapters to calculate the effects of T and p on other thermodynamic properties.
Liquids or Solids
Consider a one-component system that undergoes a change in p, V, and T only. External
fields, friction, and surface effects are ignored. Only two of the three variables p, V, and T
need be specified because the equation of state allows calculation of the third. The number of moles of the component does have to be specified (V is an extensive property). Our
goal here is to calculate the energy and enthalpy changes for a variety of processes. We
can get from the initial state to the final state by many paths. It is usually easiest to use a
series of reversible steps, where either p, V, or T is held constant in each one; the energy or
enthalpy contributions from the individual steps are then calculated and summed up to give
the overall energy or enthalpy change. For illustration we use one of the most important
biological molecules, water. Some properties of water are given in table 2.2.
Dependence of the Energy and Enthalpy of a Pure Substance on p, V, and T | 33
First, let’s consider heating or cooling some liquid water in an open flask at a constant
pressure provided by the atmosphere. The reaction is
n mol H2O(l) at T 1, p1, V 1 S n mol H2O(l) at T 2, p2, V 2 .
The heat is calculated from Eq. 2.45; at constant p:
T2
qp = 3 Cp dT
(2.49)
T1
The subscript p on the heat q and the heat capacity C remind us that the heat depends on
the path. We have already shown (Eq. 2.43) that for a constant pressure, reversible path,
the heat is equal to the enthalpy change.
In general, Cp will depend on the pressure and the temperature. However, we can
usually neglect the effects of p and often neglect the effect of T, that is, Cp, m for liquid water
is 75.4 { 1% J mol - 1 K - 1 from 0⬚ to 100⬚C and for pressures less than a few hundred bar.
Therefore, for a temperature change at constant p, if Cp is independent of T,
T2
⌬H = qp = Cp 3 dT = Cp ( T2 - T1 ) .
(2.50)
T1
Because values of heat capacity are often given per mole, we can write
⌬H = qp = nCp, m ( T 2 - T 1 ) .
(2.50a)
Heat capacities are always positive, so if T2 is greater than T1, we know that heat is
absorbed, and qp is likewise positive.
The work is calculated from Eq. (2.6). At constant p, for a reversible process
V2
wp = -p 3 dV = -p ( V2 - V1 ) .
(2.51)
V1
We can compute the change in volume; it equals n(V m,2 - V m,1), where Vm, the molar
volume, equals M/r. Heating water between 80° and 100°C, for example, changes the
molar volume by only 2.6 * 10 −7 m3 mol−1, and so between these two temperatures, at
1 bar, the expansion work is −105 Pa * 2.6 * 10 −7 m3 mol−1 = −2.6 * 10 −2 J mol−1. This
is insignificant compared with the heat. In general, for a liquid or solid at pressures close
to ambient:
wp _ 0
(2.51a)
and it follows therefore that ⌬Hp _ ⌬Up.
If the water is heated in a closed and very strong container that keeps the volume
constant at T 1, p1, V 1 S T 2, p2, V 1, the heat and work are, at constant V,
qV = nCV, m ( T 2 - T 1 )
w V = 0.
(2.52)
(2.53)
CV is the heat capacity at constant volume; for a solid or liquid, it is often not very different
in magnitude from Cp. The pV work for a constant volume process is obviously zero.
If the water is kept at constant temperature while the pressure is changed, there is
only a negligible volume change, and there is no appreciable work done or heat transferred.
Thus, for an isothermal process involving liquid or solid.
qT _ 0
(2.54a)
w T _ 0.
(2.54b)
34 Chapter 2 |
The First Law: Energy Is Conserved
To obtain ⌬U for the changes discussed above, we just sum up q and w:
⌬U = qp + w p
(2.55a)
= nCp,m ( T 2 - T 1 ) - p ( V 2 - V 1 ) ,
(2.55b)
where p = p1 = p2 because the pressure remains unchanged for an isobaric (constantpressure) process. ⌬U can also be expressed if an isochoric (constant-volume) process
is used:
⌬U = qV + w V = qV + 0 = nCV, m ( T 2 - T 1 )
(2.56)
To obtain ⌬H we use the definition of enthalpy:
⌬H = ⌬U + ⌬ ( pV )
(2.57)
Substituting Eq. 2.55b into 2.57 at constant pressure, we see that the p⌬V terms cancel
to give
⌬H = nCp, m ( T 2 - T 1 ) ,
(2.58)
which is also what we would get by integrating Eq. 2.45. Thus, for temperature or pressure
changes of a solid or a liquid, to a good approximation, ⌬H _ ⌬U.
The exact relation between ⌬U and ⌬H depends on the equation of state. The general
relationship between Cp and CV , for any equation of state, is
Cp, m = CV, m +
a2VmT
,
k
(2.59)
with a and k defined in Eqs. 2.32 and 2.33, respectively. It is a useful exercise to show that
Eq. 2.59 reduces to Eq. 2.48 for an ideal gas.
To summarize, for any change from (T1, p1, V1) to (T2, p2, V2) we can obtain ⌬U and
⌬H by choosing a convenient path and then combining the q’s and w’s. For example, we
could use an isothermal plus an isobaric path or an isothermal plus an isochoric path.
EXAMPLE 2.4
Calculate ⌬U and ⌬H in joules for heating 1 mol of liquid water from 0°C and 1 bar to
100°C and 10 bar. The molar volume of water is almost independent of pressure; it can
be calculated from the average density of water, 985 kg m−3, given in table 2.2.
SOLUTION
Choose a path such as an isothermal path plus an isobaric path. First, the pressure is
raised from 1 bar to 10 bar at 0°C. Then, the temperature is raised from 0° to 100°C at
10 bar. The overall energy change ⌬U is the sum of energy changes for the two steps,
⌬U (isothermal) for the constant-temperature path and ⌬U (isobaric) for the constantpressure path ⌬U = ⌬UT + ⌬Up = qT + w T + qp + w p.
For a liquid, qT _ 0 and w T _ 0 (Eq. 2.54) and w p _ 0 because the volume change
is small.
Thus, ⌬U _ qp + w p = nCp, m ( T 2 - T 1) = 1 mol * 75.4 J mol - 1 K-1 * 100 K = 7540 J.
To calculate ⌬H, we use Eq. 2.38: ⌬H = ⌬U + ⌬(pV ) = ⌬U + p2V 2 - p1V 1 .
Because the volume of a liquid is essentially independent of pressure, V 2 _ V 1 and
p2V2 - p1V1 _ (p2 - p1)V1. V1, moreover, is the molar volume, V m = M/r.
Thus, ⌬H = ⌬U + (p2 - p1)r/M
= 7540 J + 1 mol * (10 bar − 1 bar) * 105 Pa/bar * /(0.018015 kg mol−1)/985 kg m−3
= 7540 J + 1 mol * 9 * 105 Pa * 985 kg m−3/(0.018015 kg mol−1)
= 7540 J + 16.45 J = 7557 J
Dependence of the Energy and Enthalpy of a Pure Substance on p, V, and T | 35
Gases
10
⌬U = nCV,m(T 2 - T 1)
(2.56)
⌬H = nCp,m(T2 - T1) = nCV,m(T2 - T1) + nR(T2 - T1).
(2.53)
Of course, the heat capacities for gaseous H2O must be used here.
As we saw in the previous section, in general, when we heat solids or liquids, we can
ignore the work, but for gases, because the volume change is large, the expansion work
is significant. The equation for calculating the constant-pressure work under reversible
conditions on heating an ideal gas is the usual one:
w p = -p(V 2 - V 1) = -nR(T 2 - T 1)
(2.60)
For an ideal gas at constant pressure p, of course, p(V 2 - V 1) = nR(T 2 - T 1). From
Eq. 2.55a and Eq. 2.50a we therefore get
qp = ⌬U - wp = nCV,m ( T2 - T1 ) + nR ( T2 - T1 ) = ⌬H ,
8
p (bar)
⌬U and ⌬H for ideal gases are, as we have shown, dependent only on temperature. When
gaseous H2O changes from T1, p1, V1 to T2, p2, V2 , therefore, the change in energy and
enthalpy are just (assuming temperature independent heat capacities)
6
4
500 K
2
273 K
0
10
20
Vm (L)
30
FIGURE 2.15 The expansion
work at a constant external
pressure of 6 bar and reversible
conditions (p = pex) is the area
of the shaded rectangle of
height p and width V2 - V1 on
a p versus V plot. The curves
represent paths of isothermal
expansion.
(2.61)
which is what we showed for all substances, not merely ideal gases, in Eq. 2.43.
Figure 2.15 illustrates that work done when heating a gas with a constant external
pressure pex = p can be thought of as the area of a rectangle of height p and width
(V 2 - V 1) in a p versus V plot. In a real piston, of course, pex 6 p and the work is less.
V2
w = - 3 pex dV
V1
V2
w = - 3 pdV
V1
V2
reversible
10
V
2
nRT
nT
dV = -R 3
dV
V1 V
V1 V
w = -3
ideal gas
p (bar)
V2
T
w = -nR 3
dV
V
V1
pVm = RT
8
closed system
6
p1
4
2 p2
V2
V2
dV
= -nRT ln a b
w = -nRT 3
V1
V1 V
isothermal process
0
At constant temperature, ⌬U and ⌬H are zero for an ideal gas. The work at constant
temperature is simply the area under the isotherm, and is the shaded area in figure 2.16.
For an isothermal, reversible, ideal gas expansion or compression,
wT = -nRT ln
V2
.
V1
(2.62a)
Because volume is inversely proportional to pressure for an ideal gas at constant temperature,
wT = -nRT ln
p1
.
p2
(2.62b)
Since ⌬U is zero for isothermal expansion of an ideal gas, then the First Law requires that
qT = -w T. Thus for an isothermal, reversible expansion or compression of an ideal gas:
qT = nRT ln
p1
V2
= nRT ln
p2
V1
(2.63)
10
20
Vm (L)
30
FIGURE 2.16 Reversible isothermal expansion for an ideal
gas. The shaded area represents
the work done. Note that if the
gas undergoes an irreversible
isothermal expansion against a
constant pressure (p2), the final
pressure, the work done will
correspond to the crosshatched
area under the dashed line.
The work done by the system
on the surroundings, from a
given initial state to a given final
state of the system, is always a
maximum for a reversible path.
The work done on the system by
the surroundings is a minimum
for the reversible process.
36 Chapter 2 |
The First Law: Energy Is Conserved
For an isochoric process, wV = 0. The heat is just the energy change:
qV = ⌬UV = nCV,m(T 2 - T 1)
(2.64)
For any change of p, V, and T for an ideal gas, we can compute ⌬U and ⌬H from
Eqs. 2.60 and 2.53.
Phase Changes
The preceding section has dealt with changes of p, V, and T only. Now we want to consider changes in phase. The names of some of the phase changes (or phase transitions)
are as follows:
Phase change
Name
Symbol for the enthalpy
Gas S liquid or solid
Condensation
- ⌬ vapH, - ⌬ subH
Solid S liquid
Fusion, melting
⌬ fusH
Liquid S solid
Freezing
- ⌬ fusH
Liquid S gas
Vaporization
⌬ vapH
Solid S gas
Sublimation
⌬ subH
There can also be phase changes between different solid phases (such as diamond
changing to graphite) and between different liquid phases (such as a solute partitioning
between oil and water). We are particularly interested in the thermodynamics of reversible changes that occur at constant T and p. If water in a beaker is heated on a hot plate
at a pressure of 1 bar, the temperature of the water increases until a temperature of
99.6°C is reached. It then undergoes a phase transition (vaporization): the temperature
stays at 99.6°C until all the water is boiled off. A phase change is usually a good way to
store energy. It takes about 75 J to heat 1 mol of water from 98.6° to 99.6°C, but it takes
40,667 J to change 1 mol of water from liquid to gas at 99.6°C.
A phase transition is reversible if, by an infinitesimal change in the system, we can
reverse its direction. Ice melting slowly to water at 0°C is reversible. If the surroundings are at an infinitesimally higher temperature than the system, then heat will flow
from surroundings to system and the ice will melt. If we lower the temperature of the
surroundings slightly, then heat will flow from system to surroundings and water will
freeze. Thus, the direction of the phase transition is reversed by a tiny change in the
surroundings. In general, phase changes at any given pressure have a characteristic temperature, at which the transition is reversible. At any other temperature, the transition
is irreversible. So, for example supercooled water freezing at −10°C is an irreversible
phase transition, as is ‘bumping’ of superheated water when heated in a flask without
boiling chips. On the other hand, vaporization of water at its normal, 1 bar boiling point
of 99.6°C is a reversible transition, as long as the partial pressure of water in the vapor
phase is 1 bar.
The work involved in a reversible phase transition from phase a to phase b is
wp = -p⌬V,
(2.65)
where ⌬V = Vb - Va . The heat absorbed by the system at constant p is qp. If the phase
transition is a reversible one, and only pV work is involved, qp is equal to ⌬H = Hb - Ha
according to Eq. 2.43.
Phase Changes | 37
Values of ⌬H have been tabulated for various reversible phase changes, such as the
enthalpy of vaporization or fusion. Because most of these measurements are carried out at
constant pressure under reversible conditions, we often read, particularly in older works,
about a heat of vaporization, or a heat of fusion.
The energy change of a phase transition at constant p is
⌬ fU = ⌬ fH - p⌬ fV.
(2.66)
Often a ⌬ fH or a ⌬ fU value is known at one T and p but is needed at another. Suppose we want to calculate the amount of heat removed when liquid water evaporates at
human skin temperature (around 35°C). We want to know how effectively vaporization of
sweat can cool us. If we know ⌬Hm for vaporization of water only at 99.6°C — ⌬ vapH⬚,
at the standard pressure of 1 bar — we can use an indirect path to calculate the ⌬Hm for
the desired reaction at 35°C:
H2O (l), T2 = 35°C, p = 1 bar
H2O (g), T2 = 35°C, p = 1 bar
step 1
step 3
H2O (l), T1 = 99.6°C, p = 1 bar
step 2
H2O (g), T1 = 99.6°C, p = 1 bar
We can calculate ⌬H for each step in the chosen path and add the ≤H’s to obtain the
overall ⌬H
Step 1:
⌬Hm = Cp,m(l)(T 1 - T 2)
Step 2:
⌬Hm = ⌬ vapH⬚
Step 3:
⌬Hm = Cp,m(g)(T2 - T1)
Total:
⌬Hm(T 2) = ⌬ vapH⬚ + [Cp,m(g) - Cp,m(l)](T 2 - T 1)
The generalization for the temperature dependence of enthalpy of a phase change is
⌬ fHm(T2) = ⌬ fHm(T1) + ⌬ fCp,m ⌬T ,
(2.67)
where ⌬ fHm = Hm(phase b) - Hm(phase a) and ⌬ fCp,m = Cp,m(phase b) - Cp,m(phase a).
Note that, strictly speaking, the partial pressure of water at 35°C will not be 1 bar; however,
since neither the enthalpy of a gas nor that of a liquid depends significantly on pressure,
we can use the 1 bar value for the enthalpy change at any lower pressure.
EXAMPLE 2.5
Check the value for ⌬ vapHm of vaporization of water at 20°C given in table 2.2 by using
Eq. 2.66 and ⌬ vapHm (100°C) from the table.
SOLUTION
⌬ vapHm(20⬚C) = ⌬ vapHm(100⬚C) + ⌬ vapCp,m(100⬚C - 20⬚C)
= 40657 J mol−1 - (36.5 J mol−1 K−1 - 75.4 J mol−1 K−1)(80 K)
= 43769 J mol−1
The table gives 44080 J mol−1. The temperature dependence of the heat capacities,
which we assumed to be zero but is not, accounts for the small discrepancy.
38 Chapter 2 |
The First Law: Energy Is Conserved
The energy of a phase change at constant p can be obtained from Eq. 2.65. If one of
the phases in the phase change is a gas (condensation, vaporization, or sublimation), the
volume of the solid or liquid is so much smaller than the volume of the gas that it can
be ignored. Furthermore, the gas phase can be approximated as an ideal gas. As an example,
for vaporization
⌬ vap Um = ⌬ vapHm - p⌬ vapV m = ⌬ vapHm - p[V m(g) - V m(l)]
_ ⌬ vapHm - pVm(g) = ⌬ vapHm - RT .
EXAMPLE 2.6
Calculate (a) the change of energy and enthalpy per kg on freezing liquid water at 0°C
and 1 bar, and (b) the change of energy and enthalpy per kg on vaporizing liquid water
at 0°C and 1 bar.
SOLUTION
a. The enthalpy of freezing is - ⌬ fusH 6.007 kJ mol−1/0.018015 kg mol−1 = −333.4 kJ kg−1 .
The energy of freezing is - ⌬ fusU = -(⌬ fusH - np⌬ fus Vm).
The molar volume change on freezing is
- ⌬ fusV m = V m(s) - V m(l) = M[r - 1(s) - r - 1(l)]
- ⌬ fusU = -(⌬ fusH - npM[r - 1(l)]) = -(⌬ fusH - mp[r - 1(s) - r - 1(l)])
= −333.4 kJ kg−1 + 9 J kg−1 = -333.4 kJ kg−1 .
b. The enthalpy of vaporization is ⌬ vapH = 44.91 kJ mol - 1/0.018015 kg mol - 1
= 2493 kJ kg−1 .
The energy of vaporization is ⌬ vapU = ⌬ vapH - nRT
= 2493 kJ kg−1 - (1.00 kg/0.018015 kg mol−1) * 8.31447 J mol−1 K−1 * 273.15 K
= 2493 kJ kg−1 - 126 kJ kg−1 = 2367 kJ kg−1 .
There is a significant difference between ⌬ fH and ⌬ fU when one of the phases is a
gas [part (b)], but the difference is insignificant when neither phase is a gas [part (a)].
Now that we know how to calculate the enthalpy both for changes in temperature and
for phase changes, we can plot the enthalpy of a pure substance, such as water, as a function
of temperature. Because there is no absolute value of the enthalpy, we do this with respect
to the enthalpy at some standard temperature, typically 298.15 K, which we arbitrarily set
to zero. Such a plot is shown in figure 2.17.
In the range Tfus = 273.15 K S Tb = 372.76 K (the stable range for liquid water at
1 bar pressure), Hm is given by:
T
Hm - H⬚m = 3 Cp,m(water)dT ,
T⬚
which can be approximated closely by
Hm - H⬚m = Cp,m(T - T⬚).
Chemical Reactions | 39
50
Hm – Hm° (kJ mol–1)
40
FIGURE 2.17 The molar
enthalpy of water at 1 bar
pressure and temperature T,
relative to that at 298.15 K,
computed as described in
the text.
Tb
H2O, 1 bar
30
ΔvapHm
20
10
0
250
–10
ΔfusHm
300
350
400
T (K)
Tfus
Below Tf = 273.15 K , the enthalpy becomes
Tf
T
Hm - H⬚m = 3 Cp,m(water)dT - ⌬ fusHm + 3 Cp,m(ice)dT
T⬚
Tf
⬵ Cp,m(water) 3 Tf - T⬚ 4 - ⌬ fusHm + Cp,m(ice) 3 T - Tf 4 .
Above Tb = 372.76 K , the enthalpy is
Tb
T
Hm - H⬚m = 3 Cp,m(water)dT + ⌬ vapHm + 3 Cp,m(steam)dT
T⬚
Tb
⬵ Cp,m(water) 3 Tb - T⬚ 4 + ⌬ vapHm + Cp,m(steam) 3 T - Tb 4 .
A similar plot for the energy looks nearly identical up to Tb, but the step up at the boiling
point is smaller, and the slope in the vapor region, being CV,m and not Cp,m, is considerably shallower.
Chemical Reactions
A chemical reaction is the most important way the energy and enthalpy of a system can
be changed. We can write a general chemical reaction as
nAA + nBB + c S nPP + nQQ + c
That is, nA mol of A reacts with nB mol of B to give nP mol of P and nQ mol of Q. More
terms can be included if there are more reactants and/or products. The conditions of p, V,
and T must be specified for both the products and the reactants. The change in a property of
the state for the chemical reaction, such as ⌬ H, can be expressed in terms of the algebraic
sum of this property for the reactants and products:
⌬H = H(products) - H(reactants)
⌬ rH = nPHm,P + nQHm,Q - nAHm,A - nBHm,B
(2.68)
Similar equations can be written for other variables of state. Remember, for any change
in a variable of state, it does not matter how the reaction actually takes place; only the
initial and final states are important.
40 Chapter 2 |
The First Law: Energy Is Conserved
Heat Effects of Chemical Reactions
We have already shown that when there is no work other than the pV type,
⌬U = qV
(2.27)
⌬H = qP .
(2.43)
By measurement of the heat of chemical reactions at constant V or p, values of ⌬ rU and
⌬ rH for the reactions are obtained. The heat is usually measured by surrounding the
reaction vessel with a known amount of a liquid (water, for example) and measuring the
temperature rise of the liquid. (The heat capacity of the liquid can be calibrated by measuring the temperature rise when a known amount of electrical energy is passed through a
heating coil immersed in it.) If the reaction is carried out in a vessel with strong walls that
keeps the volume constant (a “bomb calorimeter”), the measured heat of a reaction gives
⌬ rU; if the reaction is carried out at constant pressure, the measured heat of reaction gives
⌬ rH. Note that if either ⌬ rU or ⌬ rH is measured, the other can be obtained from Eq. 2.68.
⌬ rH = ⌬ rU + ⌬ r(pV )
(2.69)
In the equation above, ⌬ r(pV ) is equal to pV of the products minus pV of the reactants. If
gases are involved in the reaction, we can ignore the volumes of the solids or liquids and
use the ideal gas equation for the gases.
⌬ r(pV ) = ⌬ r(nRT)
(2.70a)
If the reaction is carried out without a change in temperature, this reduces to
⌬ r(nRT) = RT⌬ rn.
(2.70b)
⌬ rn is just the number of moles of gaseous products minus the number of moles of gaseous
reactants.
Reactions that give off heat (q negative, ⌬ rU negative at constant volume, and ⌬ rH
negative at constant pressure) are said to be exothermic, and reactions that absorb heat
(q positive, ⌬ rU positive at constant volume, and ⌬ rH positive at constant pressure) are
termed endothermic. The amount of heat that can be obtained from a chemical reaction
is of great practical importance.
The biochemical reactions necessary to sustain life in an adult person produce about
6000 kJ day−1 of heat, which must be replenished in the form of chemical energy from food.
This basal metabolic rate is about the same as the power requirement of a 70 W lightbulb
(1 W = 1 J s−1) . If one does more than resting in bed, more than this basal level of energy is
needed, typically in the range of 8000–12,000 kJ day−1 per person. Each gram of protein or
carbohydrate provides about 15 kJ, and each gram of fat provides about 35 kJ. (Confusingly,
the energy unit used by nutritionists in the US is Cal, 1 Cal = 1 kcal = 4.184 kJ. Elsewhere
in the world, kJ are used).
The heat at constant pressure, and thus the enthalpy changes for many reactions, has
been measured. These reactions and their enthalpies can be combined to calculate the
enthalpies for many other reactions. For example, we know ⌬ rH1 for the oxidation of solid
glycine (NH3 + CH2COO - ) at 25°C to form CO2, ammonia, and liquid water:
(1) 3O2(g, 1 bar) + 2NH3 + CH2COO - (s) S 4CO2(g, 1 bar) + 2H2O(l) + 2NH3(g, 1 bar)
⌬ rH1 = -1163.5 kJ mol - 1
The ⌬ rH2 for the hydrolysis of solid urea is also known:
(2) H2O(l) + NH2CONH2(s) S CO2(g, 1 bar) + 2NH3(g, 1 bar)
⌬ rH2 = 133.3 kJ mol - 1
If we subtract these two reactions, treating the chemicals and their ⌬ rH values as algebraic
quantities, we get
Chemical Reactions | 41
3O2(g, 1 bar) + 2NH3 + CH2COO - (s) - H2O(l) - NH2CONH2(s) h
4CO2(g, 1 bar) + 2H2O(l) + 2NH2(g, 1 bar) - CO2(g, 1 bar) - 2NH3(g, 1 bar).
Rearranging and canceling, we get
(3) 3O2(g, 1 bar) + 2NH3 + CH2COO - (s) S 3CO2(g, 1 bar) + 3H2O(l) + NH2CONH2(s)
⌬ rH3 = -1296.8 kJ mol - 1 .
This equation is of more biochemical interest because urea, rather than ammonia, is the
main oxidative metabolic product of amino acids.* However, the biological reaction does
not involve solid glycine and solid urea but rather aqueous solutions. Therefore, we use the
reactions and heats for the dissolution of 1 mol of urea and of glycine:
(4) NH3 + CH2COO - (s) h NH3 + CH2COO - (aq)
⌬ solnH4 = -15.69 kJ mol - 1
(5) NH2CONH2(s) h NH2CONH2(aq)
⌬ solnH5 = -13.93 kJ mol - 1
Although the enthalpies of solution depend somewhat on concentration, here we will use
enthalpies for very dilute solutions (designated aq) and assume that they do not depend on
concentration. From reaction (3) we now subtract two times reaction (4) and add reaction (5):
(6) 3O2(g, 1 bar) + 2NH3 + CH2COO - (s) - 2NH3 + CH2COO - (s) + NH2CONH2(s) h
3CO2(g, 1 bar) + 3H2O(l) + NH2CONH2(s) - 2 NH3 + CH2COO - (aq) + NH2CONH2(aq)
(7) 3O2(g, 1 bar) + 2NH3 + CH2COO - (aq) S 3CO2(g, 1 bar) + 3H2O(l) + NH2CONH2(aq)
⌬ rH7 = ⌬ rH3 - 2⌬ solnH4 + ⌬ solnH5 = -1314.2 kJ mol - 1
We now know the enthalpy change for the reaction of a dilute aqueous solution of glycine
with O2 gas to form a dilute aqueous solution of urea plus CO2 gas plus 3 mol of liquid H2O.
The enthalpy for reaction (7) will be quite close to that of the naturally occurring
reaction in the human body. However, it should be clear that, by adding or subtracting the
enthalpies of other reactions, we can calculate ⌬H for any reaction we like. That is, if we
think it is important, we can find the heat of solution of glycine in a defined buffer solution
instead of pure water. We can specify that instead of CO2 (g, 1 bar) as a product, we have
a carbonate solution of a certain pH. It may be very difficult to measure directly the ⌬H
of the reaction we want, but we can always find the ⌬H by using a convenient alternative
path. In other words, if we want the ⌬H for reaction A h B, we may use a path that
is a sum of many other reactions:
∆rH
A
∆rH1
B
∆rH2
C
∆rH3
D
P
∆rH4
E
∆rH5
F
∆rH6
⌬ rH = ⌬ rH1 + ⌬ rH2 + ⌬ rH3 + ⌬ rH4 + ⌬ rH5 + ⌬ rH6
It is also convenient to remember that if
⌬ rH1
is the enthalpy for
then
- ⌬ rH1 is the enthalpy for
and
-n⌬ rH1 is the enthalpy for
A h P
P h A
nA h nP .
are often confused by the units of ⌬ H for an equation like reaction (3). The value of ⌬ rH3 given is
the enthalpy change associated with the formation of 1 mol of urea, but at the same time it forms 3 mol of CO2
and 3 mol of H2O . To deal with this problem, we adopt the convention of calculating thermodynamic quantities
(⌬ U, ⌬ H, etc.) per mole of reaction as it is written. In the present case, a mole of reaction is defined for reaction
(3) as the amount of reaction that produces 1 mol of urea, 3 mol of CO2, and so on. Obviously, it is just as valid
to describe the reaction as follows:
13a2 O2(g, 1 bar) + (2/3)NH3 + CH2COO - (s) S CO2(g, 1 bar) + H2O(l) + (1/3)NH2CONH2(s)
For this reaction, ⌬ rH3a = - 1296.8/3 kJ mol - 1 = - 432.3 kJ mol - 1 .
*Students
42 Chapter 2 |
The First Law: Energy Is Conserved
Temperature Dependence of ⌬ rH
Similar to what we showed earlier for phase changes, if ⌬ rH is known at one temperature
T1, it can be calculated at any other temperature T2:
A(T2)
ΔrH(T2)
P(T2)
Cp,P(ΔT)
–Cp,A(ΔT)
A(T1)
⌬ rH = H(products) − H(reactants)
⌬ rCp = Cp(products) − Cp(reactants)
⌬ T = T2 − T1
P(T1)
ΔrH(T1)
⌬ rH(T 2) = ⌬ rH(T 1) + ⌬ rCp ⌬T
(2.71)
and we assume that Cp is not a function of temperature.
Equation 2.71 can also be derived by a different method. Take the reaction
nAA + nBB h nPP + nQQ
and express ⌬ H in terms of the molar enthalpies of the reactants and products:
⌬ rH = nPHm,P + nQHm,Q - nAHm,A - nBHm,B
(2.68)
The derivative of ⌬ H with respect to temperature is
dHm,Q
dHm,P
dHm,A
dHm,B
d⌬ rH
= nP a
b + nQ a
b - nA a
b - nB a
b
dT
dT
dT
dT
dT
(2.72)
= Cp,P + Cp,Q - Cp,A - Cp,B = ⌬ rCp .
(2.73)
Thus,
⌬rH(T2)
3
⌬rH(T1)
or
T2
d⌬ rH = 3 ⌬ rCpdT
(2.74a)
T1
T2
⌬ rH(T2) - ⌬ rH(T1) = 3 ⌬ rCp dT .
If ⌬ rCp is independent of temperature,
(2.74b)
T1
⌬ r H(T2) - ⌬ r H(T1) = ⌬ rCp(T2 - T1),
(2.75)
which is the same as Eq. 2.71.
Standard Enthalpies of Formation
Thermodynamicists are interested in the change of energy of a system rather than its
absolute value. In principle, we can calculate the absolute energy of a system from its mass
using E = mc2. In practice, c2 is such an enormous number that we cannot measure the
masses of reactants and products accurately enough to calculate the relatively tiny energy
change of a chemical reaction. Since the enthalpy is just the energy plus pV , the error in
energy carries over to it and to other energy functions.
In calculating the enthalpy change ⌬ rH for a reaction, we need to know the molar
enthalpies of the reactants and products, but only relative to each other. We can therefore
arbitrarily choose a zero of enthalpy, just as we arbitrarily choose mean sea level as the
reference point for measuring elevations on the surface of Earth. For example, in the reaction to form water vapor from its elements:
Chemical Reactions | 43
H2(g, 1 bar) + 1>2O2(g, 1 bar) h H2O(g, 1 bar)
⌬ rH = 1Hm(H2O, g, 1 bar) - 1Hm(H2, g, 1 bar) - 1>2Hm(O2, g, 1 bar)
If we choose Hm for the elements O2 and H2 equal to zero, the enthalpy of H2O is equal
to the enthalpy of the reaction:
⌬ rH = 1Hm(H2O, g, 1 bar) - 1 * 0 - 1>2 * 0
Thermodynamicists have adopted the convention of assigning zero enthalpy to all
elements in their most stable states at 1 bar pressure. These are called standard states
and are designated by a superscripted circle. The standard enthalpy H⬚ of a compound is
defined to be the enthalpy of formation of 1 mol of the compound at 1 bar pressure from
its elements in their standard states.
H⬚ = ⌬ f H⬚,
(2.76)
where ⌬ f Hm⬚ is the enthalpy of formation of 1 mol from elements in their most stable form
at 1 bar. It should be clear that the standard enthalpy is a defined quantity that depends on
the choice of the standard state.
The standard enthalpies of thousands of substances have been determined, often by
measuring qV for combustion reactions at some known temperature, adding ⌬(pV ), and
then subtracting the enthalpies of formation of the combustion products. Tables A.5–7 in
the appendix give a few standard enthalpies at 25°C. The enthalpies at 25°C and 1 bar
pressure for many reactions can be calculated from such tabulated H⬚ data. Note that 1 bar
pressure is part of the definition of the standard state, but the temperature is not specified
in the definition. However, nearly all the tables available are for 25°C. For our general
reaction at 25°C,
⌬ rH⬚ = nPH⬚P + nQH⬚Q - nAH⬚A - nBH⬚B .
(2.77)
EXAMPLE 2.7
Use the Tables A.5–7 in the appendix to calculate the value of ⌬ H for reacting 1 g
of solid glycylglycine (NH3 + CH2CONHCH2COO - (s)) with O2 to form solid urea
(NH2CONH2(s)), CO2 gas, and liquid H2O at 25°C, 1 bar.
SOLUTION
We must begin with a balanced stoichiometric equation for the reaction:
3O2(g, 1 bar) + NH3 + CH2CONHCH2COO - (s) S 3CO2(g, 1 bar) + 2H2O(l) + NH2CONH2(s)
⌬ rH⬚ = H⬚urea + 2H⬚H2O + 3H⬚CO2 - H⬚glycylglycine - 3H⬚O2
= −333.17 + 3(−393.51) + 2(−285.83) − (−745.25) − 3(0) = −1340.11 kJ
The enthalpy change calculated is for 1 mol of glycylglycine. To find ⌬ rH per gram, we
must divide by the molecular mass, 132.12 g mol−1:
⌬ rH = -10.14 kJ g - 1
The negative sign means that heat is given off in the reaction.
What about other temperatures and pressures? The enthalpy at any temperature can
be obtained from Eq. 2.71. The temperature dependence of the standard chemical reaction
is written as
⌬ rH⬚(T) - ⌬ rH⬚(298 K) = ⌬ rCp⬚(T - 298 K),
(2.78)
44 Chapter 2 |
The First Law: Energy Is Conserved
where ⌬ rCp⬚ = nPC⬚p,P + nQC⬚p,Q - nAC⬚p,A - nBC⬚p,B The value of H° for elements is
chosen as zero, but C⬚p is not zero.
The pressure dependence of ⌬ rH is negligible near ambient pressure. For ideal gases,
H is independent of p; for solids or liquids, H is not very dependent on p. For geologic
processes or biology in the deep ocean, where the pressures may become very large,
this approximation is not valid. We will also generally ignore the effect of concentration
on ⌬ rH.
It should be clear that a heat of reaction and a ⌬ rH can be measured for any condition
of concentration, solvent, pH, pressure, and so on. We make these approximations because
they are accurate enough for most purposes and because they simplify the calculations.
The Energy Change ⌬U for a Reaction
For a reaction at constant pressure, ⌬ rU can be calculated from ⌬ rH by combining
Eqs. 2.69 –2.70:
⌬ rU = ⌬ rH - RT⌬ rn,
(2.79)
where, once again, ⌬ rn is the number of moles of gaseous products minus the number of
moles of gaseous reactants.
Computing Reaction Energies from First Principles
Quantum Chemical Calculations
The advent of modern quantum chemistry programs allows us to calculate gas-phase thermodynamics of chemical reactions involving small molecules (say, < 20 atoms) to accuracies of perhaps a few kJ mol−1. For very small molecules (perhaps up to 5 atoms), advances
in statistical and quantum physics have made computations of thermodynamic properties
in the gas phase more accurate than experiment, and in fact the data for diatomic gases
given in the appendix are computed, not measured. As molecules get bigger, we have to
trade off accuracy for speed of computation, and as a result quantum computations are less
accurate; they are also only formally correct in the gas phase, and corrections for solvation,
particularly if there are charged species involved, introduce considerable more inaccuracy.
Nevertheless, quantum calculations are often a useful alternative when enthalpies of formation are unavailable and reaction enthalpies hard to measure.
Bond Energies
A simpler but less accurate method of approximating enthalpies of reaction uses bond
dissociation energies D. This essentially involves finding an alternative path for the reaction where the individual steps involve breaking and making chemical bonds. The bond
dissociation energy is the enthalpy at 25°C and 1 bar for the reaction
A -B(g) S A(g) + B(g).
The bond dissociation energy should be called a bond dissociation enthalpy, but the
tradition for energy is strong. In many cases, the amount of energy necessary to break a
particular type of bond in a molecule is not too dependent on the molecule. For example,
the average energy necessary to break a C i H bond is 411 kJ mol-1 { 3% in a wide range
of organic compounds.
Some average bond dissociation energies are given in table 2.3. We expect the values for ⌬ H estimated from the table to be reasonable. One notable exception is for molecules that are stabilized greatly by electron delocalization. For example, the data that
were used to obtain values for C i C and C “ C bonds in table 2.3 came from molecules
Computing Reaction Energies from First Principles | 45
TABLE 2.3 Average Bond Dissociation Energies at 25°C
Bond
D (kJ mol−1)
CiC
359
C“C
611
C‚C
827
CiH
411
CiN
303
CiO
361
C“O
709
CiS
294
NiH
383
OiH
452
SiH
359
HiH
436.0
N‚N
945.4
O“O
498.3
C (graphite)
716.7
S (rhombic sulfur)
277.0
Data computed by the authors using a database of 27 small molecule (C1−C4) enthalpies of formation.
Data collected by Gerard Harbison.
with single bonds and isolated double bonds. Molecules with conjugated double bonds
(C “ C bonds separated by only one C i C bond) are more stable than those containing
the same number of isolated single and double bonds. This difference in energy is called the
resonance energy.
Resonance energy is important for aromatic molecules such as benzene or phenylalanine, for the carboxyl groups in carboxylic acids, for amino acids, porphyrins, carotenoids,
and so on. For such molecules, the difference in heat of formation calculated from bond
energies and that experimentally measured is an estimate of the resonance energy. In
other cases, ring strain produces energy effects that need to be added to the average bond
dissociation energies.
The easiest way to show how table 2.3 can be used is with some examples.
EXAMPLE 2.8
Calculate the heat of formation for gaseous cyclohexane using table 2.3 and compare
with the measured value in the appendix (table A.6).
SOLUTION
The reaction for the formation of cyclohexane is
6C(graphite) + 6H2(g) S C6H12(g).
We can write it as a sum of bond-breaking and bond-forming reactions:
(1) 6C(graphite) S 6C(g)
⌬H1 = 6D(graphite) = 6 * 716.7 = 4300 kJ
46 Chapter 2 |
The First Law: Energy Is Conserved
This is the enthalpy required to remove 6 mol of carbon atoms from a crystalline lattice
of graphite, which is the standard state for elemental carbon.
(2) 6H2(g) S 12H(g)
⌬H2 = 6D(H2) = 6 * 436.0 = 2616 kJ
(3) 6C(g) + 12H(g) S (6 C i C + 12 C i H) = C6H12(g)
⌬H3 = -6D(C i C) - 12D(C i H) = -6 * 359 - 12 * 411
= -7086 kJ
Note that bond formation energies are just the negative of bond dissociation energies:
⌬ fHm = ⌬H1 + ⌬H2 + ⌬H3 = -170 kJ
The value in table A.6 in the appendix is −123.15 kJ/mol for cyclohexane, which is in
pretty good agreement. This is because cyclohexane is a molecule that contains normal
bonds; also, there is no significant bond angle strain in the six-membered ring.
EXAMPLE 2.9
Calculate the heat of formation for gaseous benzene using table 2.3 and compare with
the measured value in table A.6 in the appendix.
SOLUTION
The reaction for benzene is
6C(graphite) + 3H2(g) S C6H6(g).
(1) 6C(graphite) S 6C(g)
⌬H1 = 6D(graphite) = 6 * 716.7 = 4300 kJ
(2) 3H2(g) S 6H(g)
⌬H2 = 3D(H2) = 3 * 436.0 = 1308 kJ
(3) 6C(g) + 6H(g) S (3 C i C + 3 C “ C + 6 C i H) = C6H6(g).
⌬H3 = -3D(C i C) - 3D(C “ C) - 6D(C i H)
= −3 × 359 − 3 × 611 − 6 × 411
= −5376 kJ
⌬ fHm = ⌬H1 + ⌬H2 + ⌬H3 = 232 kJ
The value in table A.6 in the appendix is 82.93 kJ for C6H6(g)! Benzene is about
150 kJ lower in enthalpy than would be expected for a molecule made up of three
C i C bonds, three C “ C bonds, and six C i H bonds. The measured value for
benzene differs significantly from that calculated for the classical structure of
benzene. An explanation of the large thermodynamic stability of benzene was an
important goal for chemists interested in chemical bonding. This energy is what we
call the resonance energy.
Molecular Interpretations of Energy and Enthalpy
We have considered various ways of changing the energy and enthalpy of a system. They
include:
• Heat and work exchanges between the system and surroundings
• Chemical reactions or phase changes within the system
Summary | 47
Let’s consider what the molecules are doing when the energy and enthalpy change. For
example, if the system consists of an ideal gas, we find that its internal energy is a function
of temperature but not of pressure or volume. We can increase the internal energy (and the
temperature) by adding heat to the system or doing work on it. The individual ideal gas
molecules have higher translational, rotational, vibrational, and (at high enough temperatures) electronic energies at the higher temperature. The internal energy of an ideal gas is
just the sum of the energies of all the individual gas molecules. There are no interactions
between ideal gas molecules. The heat capacity is the amount of energy required to raise
the temperature by 1 degree. Its magnitude is a measure of how many ways the molecules
have of storing energy.
For nonideal gases and all liquids and solids, interactions between molecules become
significant. For these systems, the energy and enthalpy can change at constant temperature.
Potential energy can be stored in intermolecular interactions. Thus, compressing the system
at constant temperature can increase its energy by increasing the repulsion between molecules as the distances between them decrease. The energy of a system of liquids, solids,
or real gases is the sum of the energies of all the molecules plus the energy of interactions
among all the molecules. Raising the temperature of a liquid, for example, raises the energy
of individual molecules and also changes the intermolecular interactions. The heat capacity
of a substance in the liquid phase is thus greater than that in the gas phase.
Chemical reactions and phase changes cause abrupt changes in energy and enthalpy.
For chemical reactions, the changes are a consequence of making and breaking bonds.
For phase changes, the changes result from the different intermolecular interactions in
solids, liquids, and gases. The energy and enthalpy changes in a chemical reaction can
be hundreds of kilojoules per mole; in a phase change, tens of kilojoules can be involved.
These are both large compared to the effects of temperature alone; a temperature change
of 100 degrees may result in no more than a few kJ mol−1 energy change.
Chemical bonds represent one of our greatest energy resources. We run automobiles
on the energy derived from the chemical combustion (oxidation) of the hydrocarbons
of gasoline. A vehicle would not run very far on the energy released by the cooling of
20 gallons of water from 100°C, even if all that thermal energy could be converted into
mechanical work. It is important we keep the relative magnitudes of these quantities in
mind, especially when we make approximations in thermodynamic calculations.
Summary
State Variables (SI units in bold)
Symbol
Units
Definition
Volume
Name
V
mL, L, m3
(length)3
Pressure
p
bar, MPa, GPa, Pa
also atm, Torr (deprecated)
force/area
Temperature
T
K, °C
also °F (deprecated)
Energy
U
J, kJ
also erg, cal, Cal (deprecated)
Enthalpy
H
J, kJ
also erg, cal, Cal (deprecated)
H = U + pV
48 Chapter 2 |
The First Law: Energy Is Conserved
Unit Conversions
Volume
106 mL = 103 L = 1 m3
Pressure
1 atm = 760 torr = 101325 Pa = 1.01325 bar
Temperature
K = °C + 273.15
°C = (°F - 32)/1.8
Energy and enthalpy
1 Cal = 103 cal = 4184 J
1 J = 107 erg
General Equations
Energy, U
(In a closed system, heat and work being the only forms of energy the system exchanges with surroundings.)
Enthalpy, H
⌬U = q + w
(2.26)
H = U + pV
(2.38)
⌬H = ⌬U + ⌬(pV )
(2.57)
dq
= C
dt
(2.10)
Heat, q
Heat absorbed by system is positive.
T2
q = 3 CdT
(2.10a)
T1
Work, w
Work done on system is positive.
Stretching or compressing a spring:
w = 1>2 k 3 1x2 - x0 2 2 - 1x1 - x0 2 2 4 ,
(2.5)
where k is the Hooke’s Law constant: k = -F(x - x 0)
x 0 is the length of the spring in the absence of an external force
x1, x2 are the respective initial and final lengths of the spring
Expansion or compression of a gas
w = 3 Fex dx = - 3 pex dV ,
(2.9)
where pex is the external pressure.
wp = - pex 1V2 - V1 2 ,
(2.51)
where the external pressure is constant.
Pressure–Volume Work Only
T2
⌬UV = U2 - U1 = qV = 3 CVdT ⬵ CV 1T2 - T1 2
T1
(2.27, 2.52)
Summary | 49
T2
⌬Hp = H2 - H1 = qp = 3 Cp dT _ Cp ( T2 - T1 )
(2.56)
T1
Solids and Liquids
We assume in these equations that for a solid or liquid V is independent of T and p, that n is the number of moles, and
that Cp,m and CV,m are the respective constant-pressure and constant-volume molar heat capacities.
⌬U = nCV,m(T 2 - T 1) (any change of p and T)
(2.56)
⌬H = nCp,m(T 2 - T 1) (any change of p and T)
(2.58)
Gases
We assume that gas properties can be approximated by the ideal gas equation, that n is the number of moles, and that
Cp,m and CV,m are the respective constant-pressure and constant-volume molar heat capacities, which are temperature
independent.
⌬U = nCV,m ( T 2 - T 1 ) (any change of p, V , T)
(2.50a)
⌬H = nCp,m ( T 2 - T 1 ) (any change of p, V , T)
(2.53)
Cp,m = CV,m + R
(2.48)
w p = - nR(T 2 - T 1)
(2.60a)
w T = - nRT ln
p1
V2
= - nRT ln
(reversible)
p2
V1
(2.62)
qT = - nRT ln
p1
V2
= nRT ln
(reversible)
p2
V1
(2.63)
Phase Changes
For a phase change, phase a S phase b, which occurs at constant T and p,
⌬ fH = Hb - Ha = qp
(2.56)
⌬ fU = Ub - Ua = ⌬ fH - p⌬ fV = Hb - Ha - p(V b - V a)
(2.66)
⌬ fH(T 2) = ⌬ fH(T 1) + ⌬ fCp ⌬T = ⌬ fH(T 1) + (Cp,b - Cp,a)(T 2 - T 1)
(2.67)
⌬ fH(p2) _ ⌬ fH(p1)
wp = - p⌬ fV .
Chemical Reactions
For a chemical reaction
nAA + nBB + c S npP + nQQ + c
that occurs at constant T and P,
⌬ rH = npHm,P + nQHm,Q - nAHm,A - nBHm,B
(2.67)
H⬚ = ⌬ fH⬚ ,
(2.76)
where ⌬ fHm⬚ is the enthalpy of formation of 1 mol from elements in their most stable form at 1 bar.
⌬ rU = ⌬ rH - RT⌬ rn,
(2.79)
where Δrn is (number of moles of gaseous products−the number of moles of gaseous reactants).
⌬ rH(T2) - ⌬ rH(T1) = ⌬ rCp(T2 - T1) ,
where ⌬ rCp = Cp,P + Cp,Q - Cp,A - Cp,B.
⌬ rH(p2) _ ⌬ rH(p1)
(2.75)
50 Chapter 2 |
The First Law: Energy Is Conserved
References
Freshman chemistry texts are good for reviewing basicthermodynamics.
1. Munowitz, M. 1999. Principles of Chemistry. New York:
Norton.
2. Petrucci, R. H., F. P. Herring, J. D. Madura, and C.
Bissonnette. 2011. General Chemistry: Principles and Modern
Applications. 10th ed. Upper Saddle River, NJ: Pearson.
The following textbooks on thermodynamics can be useful as
supplements to chapters 2 through 5.
3. DeVoe, H. 2001. Thermodynamics and Chemistry. Upper
Saddle River, NJ: Prentice Hall.
4. McQuarrie, D. A., and J. D. Simon. 1997. Physical chemistry:
a molecular approach. Herndon, VA: University Science
Books.
5. Hammes, G. G. 2007. Physical chemistry for the biological
sciences. New York: Wiley.
6. Klotz, I. M., and R. M. Rosenberg. 2008. Chemical
thermodynamics: basic concepts and methods. 7th ed. New
York: Wiley.
7. Alberty, R. A. 2003. Thermodynamics of biochemical
reactions. New York: Wiley.
Thermodynamic data for inorganic and organic chemicals can be
found on the internet.
The Committee on Data for Science and Technology (CODATA)
has a web page (http://www.codata.org/) that has links to thermodynamic databases and publications.
The National Institute of Science and Technology (NIST) web
page (http://webbook.nist.gov/chemistry/) has a very extensive
database of chemical and thermodynamic properties of organic
compounds.
The International Union of Pure and Applied Chemistry’s
Gold Book (http://goldbook.iupac.org/) is an invaluable
reference for defitions, symbols, etc.
Suggested Reading
Janshoff, A., M. Neitzert, Y. Oberdörfer, and H. Fuchs. 2000.
Force Spectroscopy of Molecular Systems—Single Molecule
Spectroscopy of Polymers and Biomolecules. Angew. Chem.,
Int. Ed. 39:3212–3237.
Neuman, K. C. and A. Nagy. 2008 Single-molecule force
spectroscopy: optical tweezers, magnetic tweezers and
atomic force microscopy. Nature Methods 5: 491-505.
Problems
1.
a. A hiker caught in a rainstorm absorbs 1.00 L of water in
her clothing. If it is windy so that the water evaporates
quickly at 20°C, how much heat is required for
this process?
b. If all this heat is removed from the hiker (no significant heat was generated by metabolism during this
time), what drop in body temperature would the hiker
experience? The clothed hiker weighs 60 kg, and you
can approximate the heat capacity of hiker and clothes
as equal to that of water. (Moral: stay out of the wind if
you get your clothes wet.)
c. How many grams of sucrose would the hiker have to
metabolize (quickly) to replace the heat of evaporating
1.00 L of water so that her temperature would not
change? You can use the enthalpy of reaction at 25°C;
the sucrose (s) reacts with oxygen (g) to give carbon
dioxide (g) and water (l).
2. Photosynthesis by land plants leads to the fixation each
year of about 1 kg of carbon on the average for each square
meter of an actively growing forest. The atmosphere is
approximately 20% O2 and 80% N2 but contains 0.0390%
CO2 by volume.
a. What volume of air (25°C, 1 bar) is needed to provide
this 1 kg of carbon?
b. How much carbon is present in the entire atmosphere
lying above each square meter of Earth’s surface? (Hint:
Remember that atmospheric pressure is the consequence
of the force exerted by all the air above the surface; 1 bar
is equivalent to 1.020 × 104 kg m−2.)
c. At the current rate of utilization, how long would it take
to use all the CO2 in the entire atmosphere above the
forest? (This assumes that atmospheric circulation and
replenishment from the oceans, rocks, combustion of
fuels, respiration of animals, and decay of biological
materials are cut off.)
3. Calculate the work (in J) done on the system for each of the
following examples (the system is given in italics). Specify
the sign of the work.
a. A box of groceries weighing 10 kg is carried up three
flights of stairs (10 m total).
b. A muscle of 1 cm2 cross section and 10.0 cm length is
stretched to 11.0 cm by hanging a mass on it. The muscle
behaves like a spring that follows Hooke’s Law. The
Hooke’s Law constant for the muscle was determined
Problems | 51
by finding that the muscle exerts a force of 5.00 N when
it is stretched from 10.0 cm to 10.5 cm.
c. The volume of an ideal gas changes from 1.00 L to 3.00 L
at an initial temperature of 25°C and a constant pressure
of 1 bar.
d. The volume of an ideal gas changes from 1.00 L to 3.00 L
at an initial temperature of 25°C and a constant pressure
of 0.1 Pa.
e. The volume of an ideal gas changes from 1.00 L to 3.00 L
at a constant temperature of 25°C, and the expansion is
done reversibly from an initial pressure of 1 bar.
4. The membrane protein bacteriorhodpsin, which contains
seven transmembrane alpha helices, was attached by its
N-terminal to a glass slide, and the C-terminal chemically
linked to a magnetic force microscope. The microscope tip
was then pulled away from the slide. As it withdraws, a
helices pull out of the membrane, two at a time, as shown in
the figure, giving a force-extension curve with four ‘teeth’.
6.
7.
8.
f(pN)
400
0
20
40
x (nm)
60
a. If the force required to extract the first pair of
alpha helices in the example above is given by
(f/pN) = 20(x/nm) + (x/nm)2, up to an extension of
10 nm, what is the work needed to extract the first pair
of helices?
b. What would be the work in J needed to extract 1 mol of
pairs of helices?
5. Recently, biological organisms have been discovered living
at great depths at the bottom of the ocean. The properties
of common substances are greatly altered there because of
enormous hydrostatic pressures caused by the weight of the
ocean lying above. The organisms need to equilibrate their
internal pressures with this external hydrostatic pressure to
prevent being crushed.
a. Estimate the hydrostatic pressure (in bar) in the ocean
at depth 2500 m. Assume the density of sea water is
1025 kg m−3. The hydrostatic pressure is given by rgh,
where r is the density and h the depth.
b. Except near sources of heat, such as vents of hot gases
coming from Earth’s mantle, the temperature is close
to 4°C. What is the percentage increase in density of
9.
10.
liquid water under these conditions compared with that
at the surface? The isothermal compressibility k of liquid
water at 4°C is 4.9458 × 10−10 Pa−1 .
c. Consider a balloon filled with oxygen O2 at 1 bar pressure
and occupying a 10 L volume at 293 K. Assuming ideal
gas behavior, what would be its volume at a depth of
2500 m under the ocean surface at 277 K? What is the
percentage increase in density of the O2?
Calculate the heat (in joules) absorbed by the system for each
of the following examples (the system is given in italics).
Specify the sign of the heat.
a. 100 g of liquid water is heated from 0°C to 100°C at 1 bar.
b. 100 g of liquid water is frozen to ice at 0°C at 0.01 bar.
c. 100 g of liquid water is evaporated to steam at 100°C
at 1 bar.
One mole of an ideal gas initially at 27°C and 1 bar pressure
is heated and allowed to expand reversibly at constant
pressure until the final temperature is 327°C. For this gas,
CV,m = 2.5R, constant over the temperature range.
a. Calculate the work w done on the gas in this expansion.
b. What are ⌬U and ⌬H for the process?
c. What is the amount of heat q absorbed by the gas?
Staphylococcus aureus is a spherical cell approximately
1.0 μm in diameter. Assuming it is almost entirely water with
a density of 1 g/mL, compute
a. The mass of the bacterium in kg
b. The mass of the bacterium in kD
c. The number of water molecules it contains
d. How many S. aureus in a row would it take to circle the
Earth? (Earth radius at the equator is 6378.1 km.)
e. The surface area in m2
f. The number of phospholipid molecules in its plasma
membrane, assuming each lipid molecule has an area
of 0.50 nm2
For the following processes, calculate the indicated quantities for a system consisting of 1 mol of N2 gas. Assume ideal
gas behavior.
a. The gas, initially at 10 bar, is expanded tenfold in volume
against a constant external pressure of 1 bar, all at 298 K.
Calculate q, w, ⌬U and ⌬H for the gas.
b. After the expansion in part (a), the volume is fixed,
and heat is added until the temperature reaches 373 K.
Calculate q, w, ⌬U and ⌬H.
c. What will be the pressure of the gas at the end of the
process in part (b)?
d. If the gas at 373 K is next allowed to expand against
1 bar pressure under adiabatic conditions (no heat
transferred), will the final temperature be higher,
remain unchanged, or be lower than 373 K? Explain
your answer.
For the following processes, state whether each of the
thermodynamic quantities q, w, ⌬U , and ⌬H is greater than,
equal to, or less than zero for the system described. Explain
your answers briefly.
52 Chapter 2 |
11.
12.
13.
14.
15.
The First Law: Energy Is Conserved
a. An ideal gas expands adiabatically against an external
pressure of 1 bar.
b. An ideal gas expands isothermally against an external
pressure of 1 bar.
c. An ideal gas expands adiabatically into a vacuum.
d. A liquid at its boiling point is converted reversibly into
its vapor, at constant temperature and 1 bar pressure.
e. H2 gas and O2 gas react in a closed bomb at 25°C, and
the product water is brought back to 25°C.
One mole of liquid water at 100°C is heated until the liquid
is converted entirely to vapor at 100°C and 1 bar pressure.
Calculate q, w, ⌬U, and ⌬H for each of the following:
a. The vaporization is carried out in a cylinder where the
external pressure on the piston is maintained at 1 bar
throughout.
b. The cylinder is first expanded against a vacuum (pex = 0)
to the same volume as in part (a), and then sufficient
heat is added to vaporize the liquid completely to 1 bar
pressure.
An ice cube weighing 18 g is removed from a freezer, where
it has been at −20°C.
a. How much heat is required to warm it to 0°C without
melting it?
b. How much additional heat is required to melt it to liquid
water at 0°C?
c. Suppose the ice cube was placed initially in a 180 g
sample of liquid water at +20°C in an insulated (thermally isolated) container. Describe the final state when
the system has reached equilibrium.
Describe what happens to the temperature, the volume, and
the phase composition (number of moles of liquid or vapor
present) following the removal of 400 J of heat from 1 mol of
pure water initially in each of the following states. Assume
that the pressure is maintained constant at 1 bar throughout
and that the gas behaves ideally.
a. Water vapor initially at 125°C
b. Liquid water initially at 100°C
c. Water vapor initially at 100°C
d. Which of the three processes above involves the largest
amount of work done on or by the system? What is the
sign of w? Explain your answer.
The volume of the average human lung is given by the
equation V = A - B exp[k(p0 - p)], where A = 5.8 L,
B = 2.0 L, and k = 100 bar−1, p0 is the atmospheric
pressure which we can assume is equal to 1 bar, and p is
the pressure in the lung. Calculate the volume of the lung if
the pressure p in the lung is 0.998 bar, and the number of
moles of gas inhaled into the lung as the pressure drops from
1.005 bar to 0.998 bar.
For the following processes, state whether each of the four
thermodynamic quantities q, w, ⌬U, and ⌬H is greater than,
equal to, or less than zero for the system described. Consider
all gases to behave ideally. State explicitly any reasonable
assumptions that you may need to make.
16.
17.
18.
19.
a. Two copper bars, one initially at 80°C and the other initially at 20°C, are brought into contact with each other
in a thermally insulated compartment and then allowed
to come to equilibrium.
b. A sample of liquid in a thermally insulated container
(a calorimeter) is stirred for 1 h by a mechanical linkage
to a motor in the surroundings.
c. A sample of H2 gas is mixed with an equimolar
amount of N2 gas at the same temperature and pressure
under conditions where no chemical reaction occurs
between them.
Some thermodynamic equations or relations discussed in
this chapter are true in general; others are true only under
restricted conditions, such as at constant volume, for an adiabatic process only, for an ideal gas, for pV work only, and
so on. For each of the following equations, state what are
the minimum conditions sufficient to make them true in the
framework of chemical thermodynamics. (Some subscripts
normally present have been omitted. You may assume that
the system is uniform in properties: T, p, U, and so on are
everywhere the same.)
a. ⌬U = q + w
b. ⌬H = q
⌬H
c. Cp =
⌬T
d. ⌬H = ⌬U + RT⌬n
n2a
e. ap + 2 b(V - nb) = nRT
V
f. w = - pex ⌬V
If you set out to explore the surface of the Moon, you would
wear a thermally insulated space suit. In such activity, you
might expect to generate roughly 4 kJ of heat per kilogram
of mass per hour. If all this heat were retained by your body,
how much would your body temperature increase per hour
owing to this rate of heat production? (Assume that your heat
capacity is roughly that of water) What time limit would you
recommend for a moon walk under these conditions?
If a breath of air, with a volume of 0.5 L, is drawn into
the lungs and comes to thermal equilibrium with the body
at 37°C while the pressure remains constant, calculate the
increase in enthalpy of the air if the initial air temperature
is 20°C and the pressure is 1 bar. At a breathing rate of
12 per minute, how much heat is lost in this fashion in 1 day?
Compare your answer (and that of problem 17, assuming a
body weight of 80 kg) with a typical daily intake of 12,000 kJ
of food energy. What problems might you foresee in arctic
climates, where the air temperature can reach −40°C and
below? Assume Cp,m for air is 7R/2.
Human beings expend energy during expansion and contraction
of their lungs in breathing. Each exhalation from the lungs of
an adult involves pushing out about 0.5 L of gas against 1 bar
of pressure. This occurs about 15,000 times in a 24 h day.
a. Estimate the amount of work in breathing done by each
person in the course of 24 h.
Problems | 53
b. To get a feeling for how much work this represents,
imagine using it to raise a mass to the top of a 30-story
building (about 100 m). First, make a guess at how large
a weight could be raised by this amount of work: 1 kg
(2.2 lb), 10 kg, 100 kg, 1000 kg (about 1 ton). Now do
the calculation and draw some conclusions about how
much your body is working even when you are “resting.”
20. One hundred grams of liquid H2O at 55°C is mixed with
10 g of ice at −10°C. The pressure is kept constant, and no
heat is allowed to leave the system. The process is adiabatic.
Calculate the final temperature for the system.
21. A reaction that is representative of those in the glycolytic
pathway is the catabolism of glucose by complete oxidation
to carbon dioxide and water:
C6H12O6(s) + 6O2(g) S 6CO2(g) + 6H2O(l)
Calculate ⌬ rH⬚298 for the glucose oxidation.
22. Alcoholic fermentation by microorganisms involves the
breakdown of glucose into ethanol and carbon dioxide by
the reaction
C6H12O6(s) S 2 C2H5OH (l) + 2 CO2(g) .
a. Calculate the amount of heat liberated in a yeast brew
upon fermentation of 1 mol of glucose at 25°C, 1 bar.
b. What fraction is the heat calculated in part (a) of the
amount of heat liberated by the complete combustion
(reaction with O2) of glucose to carbon dioxide and liquid water at 298 K, calculated in problem 21?
23. The enzyme catalase catalyzes the decomposition of hydrogen peroxide by the exothermic reaction
H2O2(aq) S H2O (l) + 1>2O2(g) .
Estimate the minimum detectable concentration of H2O2 if
a small amount of catalase (solid) is added to a hydrogen
peroxide solution in a calorimeter. Assume that a temperature
rise of 0.02°C can be distinguished. You can use a heat capacity of 4.184 kJ kg−1 K−1 for the hydrogen peroxide solution.
24. Consider the reaction
CH3OH (l) S CH4(g) + 1>2O2(g) .
a. Calculate ⌬ rH⬚298 .
b. Estimate ⌬ rU⬚298 .
c. Write an equation that would allow you to obtain ⌬H at
500°C and 1 bar.
25. Use the standard enthalpies of formation in tables A.5–7 in
the appendix to verify the value given in the text of this
chapter for the oxidation of glycine to give urea and CO2:
3O2(g, 1 bar) + 2NH3 + CH2COO - (s) S
NH2CONH2(s) + 3CO2(g, 1 bar) + 2H2O(l)
26. Yeasts and other organisms can convert glucose
(C6H12O6) to ethanol or acetic acid. Calculate the reaction enthalpy ⌬ rH when 1 mol of glucose is oxidized
to (a) ethanol or (b) acetic acid by the following path at
298 K:
glucose
glucose-6-phosphate
fructose-6-phosphate
glyceraldehyde-3-phosphate
ethanol
acetaldehyde
acetic acid
You can ignore all heats of solution of products or
reactants. The overall reactions are
C6H12O6(s) S 2 C2H5OH (l) + 2 CO2(g)
2O2(g) + C6H12O6(s) S
2 CH3COOH (l) + 2 CO2(g) + 2H2O(l) .
a. Calculate the ⌬ rH for the complete combustion of
glucose to CO2(g) and H2O(l).
27. Estimate the change in ⌬ rH if each reaction in problems
26(a) and 26(b) is carried out by a thermophilic bacterium
at 80°C. Cp for ethanol (l) = 111.5 J mol−1 K−1; for acetic
acid(l) = 123.5 J mol−1 K−1; and for glucose(s) = 219.2 J
mol−1 K−1.
28. A good yield of photosynthesis for agricultural crops in bright
sunlight is 20 kg of carbohydrate (for example, sucrose) per
hectare per hour. (1 hectare = 104 m2). The net reaction for
sucrose formation in photosynthesis can be written
12CO2(g) + 11H2O(l) S C12H22O11(s) + 12O2(g) .
a. Use standard enthalpies of formation to calculate DrH°
for the production of 1 mol of sucrose at 25°C by the
reaction above.
b. Calculate the energy equivalent of photosynthesis that
yields 20 kg of sucrose (hectare−1 h−1) .
c. Bright sunlight corresponds to radiation incident on the
surface of Earth at about 1 kW m−2. What percentage
of this energy can be “stored” as carbohydrate in
photosynthesis?
29. a. Calculate the enthalpy change on burning 1 g of H2(g)
to H2O(l) at 25°C and 1 bar.
b. Calculate the enthalpy change on burning 1 g of n-octane
(g) to CO2(g) and H2O(l) at 25°C and 1 bar.
c. Compare H2O(g) and n-octane(g) in terms of joules of
heat available per gram.
30. Consider the reaction in which 1 mol of aspartic acid(s) is
converted to alanine(s) and CO2(g) at 25°C and 1(CH3) bar
pressure. The balanced reaction is
NH3 + CH(CH2COOH)COO - (s) S
2NH3 + CHCH3COO - (s) + CO2(g) .
a. How much heat is evolved or absorbed?
b. Write a cycle you could use to calculate the enthalpy for
the reaction at 50°C. State what properties of molecules
you would need to know and what equations you would
use to calculate the answer.
54 Chapter 2 |
The First Law: Energy Is Conserved
31. What experiments would you have to do to measure the (a)
energy and (b) enthalpy change for the following reaction at
25°C and 1 atm?
ATP4 - + H2O S ADP3 - + HPO4 2 - + H +
The reaction takes place in aqueous solution using the
sodium salts of each compound. Give as much detail as
possible and show the equations you would need to use.
ATP4 - is adenosine triphosphate; ADP3 - is adenosine
diphosphate.
32. A household uses 22 kWh day−1 of electricity.
a. How many kJ day−1 are used?
b. About 1 kW m−2 of energy from the Sun hits the surface
of Earth. With a 10% efficient solar battery, what area
(in m2) of solar battery would be needed to provide
enough solar energy to supply the household? Assume
an average of 5 h day−1 of sunshine. (A suitable energystorage device would of course be needed to provide
electricity at night.)
33. The enzyme catalase efficiently catalyzes the decomposition
of hydrogen peroxide to give water and oxygen. At room
temperature, the reaction goes essentially to completion.
a. Using standard enthalpies, calculate ⌬ rH⬚298 for the
reaction
2H2O2(g) S 2H2O(g) + O2(g) .
H⬚298 for gaseous H2O2 is −133.18 kJ mol−1.
b. Calculate the bond dissociation energy for the O–O
single bond.
c. The enzyme normally acts on an aqueous solution of
hydrogen peroxide, for which the equation is
2H2O2(aq) S 2H2O(l) + O2(g) .
What is ⌬ rH⬚298 for this process?
d. A solution initially 0.01 M in H2O2 and at 25.00°C is
treated with a small amount of the enzyme. If all the heat
liberated in the reaction is retained by the solution, what
would be the final temperature? (Take the heat capacity
of the solution to be 4.184 kJ kg−1 K−1) .
34. Use bond-energy data to calculate the enthalpy of formation
for each of the following compounds at 25°C.
a. n-Octane, C8H18(g)
b. Naphthalene, C10H8(g)
c. Formaldehyde, H2CO(g)
d. Formic acid, HCOOH(g)
Compare your answers with enthalpies of formation
given in tables A.5–A.7 in the appendix. (The enthalpy of
vaporization of formic acid is 46.15 kJ at 25°C.) Give the
most likely reasons for the largest discrepancies between
your calculated values and the ones in the tables.
35. The standard enthalpies at 25°C of gaseous trans-2-butene
and cis-2-butene are -11.1 and -7.0 kJ mol−1, respectively.
In the cis compound
H3C
CH3
C
C
H
H
the relatively bulky methyl groups interact repulsively with
one another because they are both constrained to remain on
the same side of the molecule. (There is no rotation about
the C i C bond.) Calculate the thermodynamic enthalpy
attributable to this steric repulsion in the cis compound
relative to the trans
H
H3C
C
H
C
CH3
where the methyl groups are on the opposite sides of the
molecule. This is an example of how detailed molecular
information can be deduced from thermochemical data.
Chapter 3
The Second Law: The Entropy
of the Universe Increases
Concepts
Water does not flow spontaneously up a waterfall to enlarge a pool at the top, and a candle
never assembles a column of wax out of thin air by burning in reverse. Expressing this
intuition of the direction of natural processes in objective and precise language is the
domain of the Second Law of Thermodynamics.
The Second Law plays a crucial role in the discovery of a new variable of state, the
entropy S, which is recognized as a measure of the disorder of the system, although the
terms disorder and order do not capture the true meaning of entropy that was discovered by Ludwig Boltzmann (see sidebar). In an isolated system (which exchanges neither
energy nor matter with the surroundings), every change of the system increases its entropy.
If the system is not isolated from the surroundings, then the entropy of the system can
decrease provided there is a larger increase in entropy of the surroundings. Thus, an elegant
restatement of the Second Law emerges: the entropy of the universe—the system plus the
surroundings—always increases.
An additional variable of state, the Gibbs free energy G, can be defined as the enthalpy
H minus the product TS. The Gibbs energy always decreases in a process that occurs
spontaneously at constant temperature and pressure, important conditions for biochemical
processes. By calculating a free-energy change for a reaction, we can tell if the reaction
proceeds spontaneously at constant temperature and pressure.
S = k lnW
Inscribed on Ludwig Boltzmann’s
tombstone in Vienna is the equation S = k ln W, meaning entropy
is proportional to the logarithm
of the number of different ways
W a system can be rearranged
without perceptibly changing it.
Entropy had already been defined
as a thermodynamic variable of
state, but Boltzmann gave it a
statistical meaning.
Applications
The First Law of Thermodynamics does not tell us whether a particular reaction will
spontaneously occur; it just says that whichever way a reaction goes, the energy must be
conserved. The Second Law of Thermodynamics specifies a criterion that predicts the
direction of spontaneous change for a system.
Living organisms maintain homeostasis—the stable and coordinated operation of all
organismal processes such as respiration, etc.—by creating conditions that control the
directions of useful chemical reactions. Consider that a voltage is established across a
cell membrane to drive the synthesis of ATP. The dephosphorylation of this ATP may
be coupled to a reaction that requires a certain minimum concentration of ATP to occur
spontaneously in the cell. A knowledge of thermodynamics allows us to learn how the
conditions can be changed to make impossible reactions possible.
A critical limitation is that thermodynamics tells us a process is spontaneous, but not
how fast it will actually occur, or how the rate will depend on the reaction conditions. For
example, diamond at atmospheric pressure will convert spontaneously to graphite, but this
555
56 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
(a)
p1 , V1
I
p2 , V2
Thot
q1, w1
Thot
IV
q4, w4
q2, w2 II
p4 , V4
q3, w3
p3 , V3
Tcold
III
Tcold
(b)
Thot
2.0
p1, V1
I
p (bar)
p2, V 2
IV
Tcold
Toward the Second Law: The Carnot Cycle
In the early 1800s, steam engines were common but poorly understood. In 1824, Sadi Carnot
published an analysis of an idealized heat engine that underwent a particular cycle of four
steps to return to its original state. Unlike the operation of a real engine, such as a steam
or automobile engine, all steps in this idealized engine are reversible.
We will apply the Carnot cycle to an engine composed of an ideal gas in a piston
cylinder (see figures 3.1 and 3.2). In step I, a hot ideal gas at Thot expands isothermally
and reversibly, with work w1 and heat q1. The gas then expands adiabatically and reversibly in step II (q2 = 0). Recognize that in step II the engine does work (w2 is negative)
without heat input, and so its energy drops and the temperature of the gas is reduced to
Tcold. In step III, the gas is compressed isothermally and reversibly (w3, q3 ). In the last step,
the gas is compressed adiabatically and reversibly (w4, q4 = 0) to return to its original
condition. For one cycle, the total work is w = (w1 + w2 + w3 + w4 ), and the total heat
absorbed is q = (q1 + q2 + q3 + q4 ). Since the initial and final states are the same for
the cyclic process, we know that U = q + w = 0 and so q = -w. The First Law can
be employed to analyze the heat and work for each step of the Carnot cycle. For the first
step (an isothermal reversible expansion),
V2
V2
.
w 1 = - 3 p dV = -nRT hot ln
V1
V1
II
An ideal gas at constant temperature must have constant U (chapter 2). So the First Law
for step I is q1 + w1 = 0, giving
p4, V 4
1.0
process is so kinetically limited that it cannot be observed on human time scales. The rates
of reactions are discussed in the chapters on kinetics.
The application of thermodynamics is universal and is not limited to chemical
reactions. For example, the Second Law is often framed as a constraint on the design of
engines: no engine can be constructed, no matter how ingenious its mechanical design,
which can convert heat into work without also discharging some heat to the surroundings.
III
q1 = -w 1 = nRT hot ln
p3, V3
12
16
V (L)
20
FIGURE 3.1 (a) The Carnot cycle
of an idealized heat engine. Each
step is carried out reversibly.
Steps I and III are carried out
isothermally (step I at Thot and
step III at Tcold) and steps II and
IV adiabatically (q2 = q4 = 0). (b)
A pV diagram for a Carnot cycle,
illustrating the four reversible
steps taken in moving between
the states in (a).
V2
.
V1
The second step is an adiabatic reversible expansion, so q2 = 0 and the First Law gives
w 2 + 0 = U = CV (T cold - T hot ),
using the fact that the energy of an ideal gas depends only on temperature (Eq. 2.25). We
will take a moment to find some additional helpful relationships for reversible adiabatic
processes of ideal gases such as steps II and IV. Since dq = 0 and the differential of work
is -pdV , we have dU = -pdV . Given that dU = CV dT is universally true, then
C V dT = -pdV
= -
nRT
dV .
V
Dividing both sides by T and integrating, we have for step II,
Tcold
CV 3
Thot
CV ln
V
3
dT
dV
= -nR 3
,
T
V2 V
T cold
V3
V2
= -nR ln
= nR ln
.
T hot
V2
V3
(3.1a)
Toward the Second Law: The Carnot Cycle | 57
p2, V2, Thot
p3, V3, Tcold
p3, V3, Tcold
p2, V2, Thot
p4, V4, Tcold
p1 , V1, Thot
p1, V1, Thot
Thot
Tcold
Thot
Tcold
I. Isothermal reversible expansion
q1 is positive
w1 is negative
p4, V4, Tcold
II. Adiabatic reversible expansion
q2 = 0
w2 is negative
III. Isothermal reversible compression
q3 is negative
w3 is positive
FIGURE 3.2 A sketch of the four steps of the Carnot-cycle heat engine. The engine is in thermal contact
with a hot heat reservoir at Thot (step I), a cold heat reservoir at Tcold (step III), or is thermally isolated (steps II
and IV). Steps I and II are expansions in which work is done by the engine. Steps III and IV are compressions
in which work is done on the engine.
Thus, we have a relation between the temperature and volume changes for an adiabatic
reversible expansion (or compression) of an ideal gas that applies for steps II and IV.
The third and fourth steps are similar to the first and second. For the isothermal
step III,
V4
.
q3 = -w 3 = nRT cold ln
V3
For the adiabatic step IV, we have q4 = 0 and w 4 = C V (Thot - Tcold ) and also
CV ln
T hot
V4
= nR ln
.
T cold
V1
(3.1b)
The total heat absorbed is the sum over all steps
qcycle = q1 + q2 + q3 + q4 = nRT hot ln
V2
V4
+ 0 + nRT cold ln
+ 0,
V1
V3
and the total work done by the engine is also summed to give
-wcycle = - (w1 + w2 + w3 + w4 ) = nRThot ln
V2
V4
+ nRTcold ln
,
V1
V3
since w2 and w4 cancel. It is a good check to see that the cycle conforms to the First Law
(e.g., Ucycle = qcycle + w cycle = 0). Notice that unlike U, neither q nor w is zero for this
cyclic path, a reminder that heat and work are not variables of state (chapter 2).
Carnot made an important finding that although the sum of qi for all the steps of
this reversible cycle is not zero, the sum of the quantities qrev >T over the path is zero.
The subscript “rev” is used to emphasize that the heat exchange occurs reversibly. In
general, qrev >T can be evaluated by integrating dqrev >T , with dqrev representing an infinitesimal heat change along a reversible path and T being the temperature at which dqrev is
exchanged.* For the isothermal steps I and III,
*Although we use the symbol dq to express a differential change in heat, we should keep in mind that there is
an important difference between dq and the differential of a state function, such as dU. For any cyclic path from
state 1 back to state 1, the integral dU is always zero. But dq is dependent on the particular path, and its integral
is generally not zero for a cyclic path. Mathematically, we say that dU is an exact differential, but dq (and dw as
well) is not.
IV. Adiabatic reversible compression
q4 = 0
w4 is positive
58 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
dqrev
q1
1
3 T = T hot 3 dqrev = T hot ,
(Step I)
q3
dqrev
1
3 T = T cold 3 dqrev = T cold .
(Step III)
For steps II and IV, the temperature is not constant, but dqrev is zero (adiabatic), giving
dqrev
3 T = 0.
(Steps II and IV)
Then 3 dqrev >T for the cyclic path reduces to the contributions from steps I and III,
q3
q1
V2
V4
V 2V 4
+
= nR ln
+ nR ln
= nR ln
.
T hot
T cold
V1
V3
V 1V 3
Now add Eqs. 3.1a and 3.1b, to find a useful identity for the Carnot cycle,
nR ln
V 2V 4
= 0,
V 3V 1
and therefore,
q1
q3
+
= 0.
T hot
T cold
(3.2)
Equation 3.2 is an important finding. Our system (an ideal gas in a cylinder) has gone through
a cycle and returned to its original state. Carnot recognized that the sum of the quantities
3 dqrev >T for this cyclic path is zero and could represent a state function, just as we have
learned that state functions such as U and H exhibit no change for a cyclic path. But before
we get too excited about this realization, we must ask two questions. First, is Eq. 3.2 just
a special result of choosing an ideal gas or can it apply to all matter? Second, is Eq. 3.2
only an outcome of the Carnot cycle, or is it true for any reversible cycle? The First Law of
Thermodynamics cannot answer these questions. In deducing that Eq. 3.2 is general for all
reversible cycles, Carnot was the first to clearly specify the Second Law of Thermodynamics.
A New State Function, Entropy
Carnot had a superb intuition for his model steam engine, but there is some mystery about
how much he knew as he withheld equations. In modern terms, we can tackle the preceding
questions about Eq. 3.2 by analyzing the efficiency of the Carnot cycle as the total work
done by the engine in one complete cycle -w divided by the heat absorbed at the hot heat
reservoir qhot. For the Carnot engine,
-w
-w
efficiency =
=
.
(3.3)
qhot
q1
The rest of the energy is discharged as heat qcold at the cooler heat reservoir (qcold = q3
for the Carnot engine) and is wasted, as far as the efficiency for converting heat into work
is concerned. With a little algebra, we can solve for efficiency in terms of the temperatures
of the heat reservoirs. From Eq. 3.2,
-q3
q1
=
.
(3.4)
T hot
T cold
From the First Law, the total work done by the engine in one complete cycle is
-w = q1 + q3 .
A New State Function, Entropy | 59
Thus,
q1 + q3
-w
-w
=
=
,
qhot
q1
q1
q3
= 1 +
.
q1
efficiency =
(3.3a)
It follows from Eq. (3.4) that
q3
T cold
= ,
q1
T hot
giving the final result in terms of temperature,
efficiency = 1-
T cold
.
T hot
(3.5)
For any heat engine that operates reversibly at all steps, the engine can also be operated
in the reverse direction as a refrigerator or heat pump. In the forward direction as a heat
engine, there is a net transfer of heat from the hot reservoir to the cold one, and the system
does work on the surroundings. In the reverse direction, as a heat pump, the surroundings
perform work on the system, and there is a net flow of heat from the cold reservoir to the
hot one.
Carnot recognized the impossibility of a certain kind of perpetual motion, and he
concluded that all heat engines operating with reversible cycles between Thot and Tcold must
have the same efficiency. We will illustrate the argument in modern terms with a specific
example (figure 3.3). Suppose Thot and Tcold are 1200 and 300 K, respectively. The efficiency
of a Carnot engine is therefore 1 - 300>1200, or 0.75 (75%). If the engine takes up 100 kJ of
heat in step I (qhot = 100 kJ), 25 kJ would be discharged in step III (qcold = -25 kJ),
and the engine does 75 kJ of work (w = -75 kJ) in each cycle.
Suppose another hypothetical engine also operates on a cycle of reversible steps but has
a lower efficiency of 50%. To get 75 kJ of work out of it, 150 kJ would have to be absorbed
at Thot and 75 kJ would be discharged at Tcold. Although this second engine is an inferior
heat engine compared with the Carnot engine, it becomes a superior heat pump when operated in reverse: an input of 75 kJ of work (w = 75 kJ) would lead to the absorption of 75
kJ of heat at the cooler heat reservoir (qcold = 75 kJ) and the delivery of a total of 150 kJ
of heat to the hotter heat reservoir (qhot = -150 kJ).
If the work output of the Carnot engine is used to drive the hypothetical heat pump,
we can see that the total qhot for the combined engine is 100 kJ - 150 kJ = -50 kJ, and
the total qcold for the combined engine is -25 kJ + 75 kJ = 50 kJ. The combined effect of
the two reversible engines is a net flow of 50 kJ of heat per cycle from the cooler reservoir
to the hotter reservoir, with no change in the surroundings. If any two heat engines operating with reversible cycles between Thot and Tcold differ in their efficiencies, we can always
operate the less efficient heat engine in reverse as a heat pump and use the more efficient
Carnot
Hypothetical
1200 K
1200 K
q=+100 kJ
1200 K
q=-150 kJ
w=-75 kJ
η=75%
+
w=+75 kJ
η=50%
q=-25 kJ
q=-50 kJ
=
Impossible
Engine
q=+75 kJ
300 K
300 K
q=+50 kJ
300 K
“the maximum of [work] resulting from the employment of
steam is also the maximum of
[work] realizable by any means
whatever”
—S. Carnot, 1824*
Carnot published his
extraordinary findings in 1824 in
a popular science format, with
few equations. His work was
largely unnoticed in its time,
but has since been recognized
as the first realization and
application of the Second Law.
*Note that [work] is substituted for
Thurston’s translation of motive
power (Carnot S. 1890. Reflections on
the Motive Power of Heat, transl. R.
H. Thurston. Macmillan and Co.)
FIGURE 3.3 If two reversible
heat engines (grey circles) each
operate between the same
heat baths, but with different
efficiencies, then they form a
new engine whose only result
is the transfer of heat from a
cold bath to a hot bath with no
work on or by the surroundings,
a violation of the Second Law
of Thermodynamics. Thus all
reversible heat engines must
have the same efficiency.
60 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
FIGURE 3.4 Two views are
depicted of the same Carnot
cycle. Since U = 0 for the
cycle, the First Law requires
wcycle = - qcycle. Then the
shaded areas bounded by
the reversible pathways
represented in (a) pV or (b) TS
coordinates must be equal. The
integral symbol with the circular
arrow represents summing the
integral over the whole cycle.
(a)
(b)
Thot
I
p
I
Thot
IV
T
Tcold
IV
II
Tcold
III
II
III
V
S
=
p dV
T dS
one as a heat engine to achieve the result of figure 3.3. The combined actions of these two
engines would contradict vast experience that heat flows spontaneously strictly only from
a hotter reservoir to a cooler reservoir without changing the surroundings. Therefore, the
only way to avoid the impossible engine in figure 3.3 is to conclude that all reversible
engines have the same efficiency.
If all reversible engines have the same efficiency, then Eq. 3.4 and therefore Eq. 3.2
cannot be a consequence of using an ideal gas in our engine or of the particular paths
chosen. In other words, we have defined a new state function S, entropy:
dqrev
S = 3
,
T
which for isothermal processes is then
S =
qrev
.
T
(3.6)
Entropy is an extensive variable of state; S depends on the initial and final states and the
amount of the system. From Eq. 3.6, one sees that the dimensions of entropy are energy/
temperature; we will use units of J K - 1. We have “proven” that entropy is a state function
because we accepted as true that heat flows spontaneously from a hot body to a cooler one.
We can now see an alternative depiction of the four reversible steps of the Carnot cycle by
representing the pathway through T and S coordinates (figure 3.4). It is especially clear in
this depiction that steps II and IV are isentropic.
The Second Law of Thermodynamics
When we calculate the entropy changes of both the system and the surroundings for simple
processes that can occur spontaneously—such as the flow of heat from a hot body to a cooler
one, the expansion of an ideal gas into a vacuum, or the flow of water down a hill—we find
that the sum of the entropy changes is not zero. Unlike energy, entropy is not conserved. The
generalization of a great deal of experience, which is the Second Law of Thermodynamics,
is that the sum of the entropy changes of the system and the surroundings is always positive.
Even zero values can be approached only as a limit, and negative values are never found.
If there is a decrease in entropy in a system, there must be an equal or larger increase in
entropy in the surroundings:
S (system) + S (surroundings) Ú 0 .
(3.7)
For an isolated system, since there is no energy or material exchange between such a
system and the surroundings, there is no change in the surroundings. Therefore,
S (isolated system) Ú 0 .
(3.8)
The Second Law of Thermodynamics | 61
The Ú sign means greater than or equal to; the former applies for irreversible processes
and the latter for reversible ones. There are many different ways of stating the Second Law.
That heat spontaneously flows from a hot body to a cold body, for example, can be taken as a
statement of this law. We express the Second Law by Eqs. 3.7 and 3.8 because these choices
are more convenient for problems in which chemists and bioscientists are interested.
E X A M P L E 3 .1
One mole of an ideal gas initially at p1 = 2 bar, T, and V1 expands to p2 = 1 bar, T,
and 2V1. Consider two paths: (a) irreversible expansion into a vacuum, as shown next,
and (b) the expansion is reversible (not depicted). Calculate qirrev, S (system) and
S (surroundings) for (a) and qrev, S (system), and S (surroundings) for (b).
2 bar
V1, T
Vacuum
Open
1 bar
V1, T
1 bar
V1, T
valve
V2 = 2V1
SOLUTION
Recall that S (system) is independent of path and must be the same for paths (a) and
(b) because the initial and final states of the system are the same. However, q depends
on the path, and the change in the state of the surroundings is different for (a) and (b).
Thus, the values of S (surroundings) will be different.
Path a: Begin with the First Law analysis for this path,
w = 0
(no work done against surroundings)
U = 0
(U for an ideal gas is independent of volume)
qirrev = U - w = 0
Since no heat or work exchange between the system and surroundings occurs, the
system is isolated and so the surroundings undergoes no change at all. Then,
S (surroundings) = 0 .
Since S (system) must be computed over a reversible path, we can choose the path
of (b), in which the gas expands isothermally and reversibly,
V2
V2
w rev = - 3 pdV = -RT 3
V1
V1
V2
dV
= -RT ln
V
V1
= -RT ln 2 .
Since U = 0 for both paths,
S (system) =
qrev
U - w rev
0 - ( -R ln 2)
=
=
= R ln 2 .
T
T
T
So, for the spontaneous, irreversible process in part (a),
S (system) + S (surroundings) = R ln 2 + 0 7 0 ,
consistent with Eqs. 3.7 and 3.8.
62 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Path b: We have already calculated for this path that
qrev = RT ln 2 ,
S (system) = R ln 2 .
For path b, an amount qrev of heat is transferred reversibly into the system from
the surroundings. This could occur if the surrounding temperature is higher than
T by an infinitesimal amount, so that the surrounding temperature is still T. For
the surroundings, the heat input is -qrev and
-qrev
S (surroundings) =
= -R ln 2 .
T
Note that for this reversible process,
S (system) + S (surroundings) = 0 ,
consistent with Eq. 3.7.
Molecular Interpretation of Entropy
The Second Law tells us that the entropy of the universe is increasing. This does not mean
that a small part of the universe (the system) cannot have its entropy decrease. When
we freeze liquid water to solid ice in an ice tray in the freezer, we are decreasing the
entropy inside the ice tray. However, the large amount of heat dissipated outside the freezer
compartment (remember that this is a heat pump) leads to a greater increase of entropy in
the surroundings. The total entropy change of the ice tray and the surroundings is therefore
positive. Many biological processes involve decreases of entropy for the organism that are
always coupled to other processes that increase the entropy by a greater amount elsewhere.
In photosynthesis in green plants, the large decrease of entropy for converting simple gases
like CO2 and H2O into a very complex organism is offset by greater increases in entropy
occurring elsewhere, such as in the Sun, which produces the light that drives photosynthesis. It is a pessimistic, but accurate, view that anything we do will always increase the
entropy of the universe.
Feynman (1963) explained entropy as the number of ways the insides can be arranged
so that from the outside the system looks the same. For example, 1 mol of ice has lower
entropy (informally: is more ordered) than 1 mol of water at the same temperature, because
water molecules in the liquid may have many different arrangements but still have the
properties of liquid water. Water molecules in solid ice can have only the periodic arrangement of the crystal structure of ice to have the properties of ice. The molar entropies of
water at 1 bar are the following:
Entropy mol-1
H2O (s) (273.15 K)
41.3 J K - 1
H2O (l) (298.15 K)
69.9 J K - 1
H2O (g) (298.15 K)
188.9 J K - 1
These values are consistent with the fact that the water molecules in ice are confined to
positions in the crystalline lattice, whereas in water vapor the molecules are free to move
about in a relatively large space. The liquid state is intermediate, but it is more like the
solid than the gas in terms of entropy; consider that liquids and solids have similar densities so that the molecules of a liquid are still confined to about the same volume as in the
corresponding solid. For monatomic elements, the entropy can be qualitatively related to
Molecular Interpretation of Entropy | 63
the hardness of an element; in hard solids the molecules are more rigidly confined to lattice
positions and have lower entropies than in soft solids. Contrast graphitic carbon, which is
arranged in two-dimensional planes that can easily slide between each other, with the more
rigid three-dimensionally network-bonded structure of diamond (see sidebar, 298 K, 1 bar).
All entropies increase as the temperature is raised since increasing molecular motion
increases the possible positions and configurations of the molecules.
Interpreting entropy as a measure of disorder helps us to understand and predict
entropy changes in reactions. In the gas phase, if the number of product molecules is less
than the number of reactant molecules, entropy decreases. If the number of gas molecules
increases, entropy increases. Two examples at 25°C and 1 bar are shown as follows. The
superscript in r S specifies that the values are for reactions in which all reactants and
products are in their standard states, which we will define later.
Phase
Entropy
Graphite 5.74 J K - 1mol - 1 Soft
graphite
r Sm(J K - 1 mol - 1)
Reaction
2H2 (g) + O2 (g) S 2H2 O(g)
-88.8
PCl5 (g) S PCl3 (g) + Cl2 (g)
170.3
c
c
Reaction
r Sm (J K - 1 mol - 1 )
CO32 - (aq) + H + (aq) S HCO3- (aq)
+148.1
NH 4+ (aq) + CO32 - (aq) S NH3 (aq) + HCO3- (aq)
+146.0
NH 4+ (aq) + HCO3- (aq) S CO2 (aq) + H2O(l) + NH3 (aq)
+94.2
+
NH3 (aq) + H (aq) S
HC2O4- (aq)
-
+2.1
NH4+ (aq)
+ OH (aq) S
C2O42 - (aq)
+ H2O(l)
CH4 (aq) S CH4 (CCl4 )
-23.1
+75
For the first four reactions, neutralization of charges occurs. Because a charged species
tends to orient the water molecules around it, charge neutralization results in disorientation
of some of the solvent molecules and an increase in entropy. The next two reactions involve
transfer of charge but no neutralization. The entropy changes are small as a consequence. The
last reaction involves the transfer of a nonpolar methane (CH4) molecule from a polar solvent
(H2O) to a nonpolar solvent (CCl4). The large positive S is primarily the result of the ordering
of water molecules around a nonpolar molecule. Removal of the nonpolar molecule results in
the randomization of the water molecules and hence a positive S. The ordering of water molecules around nonpolar moieties is important for many reactions in aqueous media, such as the
binding of a nonpolar substrate to the nonpolar region of an enzyme or the folding of a protein.
Boltzmann derived the result that the entropy is proportional to the logarithm of the
number of microscopic states of a system:
S = kB lnW ,
(3.9)
c
c
c
+80.8
c
c
c
If the number of molecules is unchanged by the reaction, a small change in entropy—either
positive or negative—is expected. For reactions in solution, it is more difficult to predict
the entropy change because of the potentially large effects of the solvent. Two solute molecules may react to form one product molecule, but unless we know what is happening to
the solvent—whether it is being ordered or disordered—we cannot predict the change in
entropy. However, we can rationalize and interpret the measured results. A few examples
of reactions carried out at 25°C, 1 bar are as follows:
OH - (aq) + H + (aq) S H2O(l)
Property
Diamond 2.38 J K - 1mol - 1 Hard
c
c
diamond
64 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
where kB is Boltzmann’s constant (1.381 * 10 - 23 J K - 1 ) and W is the number of microscopic states of the system. Note that the gas constant R (8.314 J K - 1 mol - 1) is proportional
to kB by Avogadro’s number R = k B N A . The number of microscopic states is the number of
different ways the inside of the system can be rearranged without changing its properties.
For a simple molecular system, we can calculate its number of microstates and hence its
entropy; for a complex system, it becomes harder to calculate the entropy from the structures
of the molecules and their interactions. We can also use the measured entropy to calculate
W, the number of microscopic states for any system. Solving for W in Eq. 3.9, we obtain
W = eS>k B .
(3.10)
It is possible, in principle, to calculate the entropy and other thermodynamic properties
of complex systems from molecular structures and interactions; the methods used are
described in books on statistical mechanics.
Fluctuations
As we saw in example 3.1, if we assume that a system will spontaneously reach uniform
pressure, we arrive at the conclusion that the sum of entropy changes of the system and
surroundings is positive (the Second Law). The quantitative statement of the Second Law
in Eqs. 3.7 and 3.8 was based on a wealth of observations and can now be used to predict
the direction of a spontaneous process. So the Second Law predicts that the system shown
in figure 3.5(a) will spontaneously tend to reach uniform pressure. Similarly, starting with
the Second Law, we predict that the system shown in figure 3.5(b) will tend to reach uniform composition spontaneously. That is, at constant pressure, gases originally separated
in two halves of a system will tend to mix. Both of these processes can take place without
changing the surroundings. The reverse of these reactions—unmixing gases, or going from
a uniform pressure to unequal pressures—can only be done if the surroundings are also
changed and thus will not occur spontaneously.
The Second Law was based on observations of relatively large systems and measurements averaged over time. Consider instead a system of 10 N2 molecules contained within
a rigid box, whose volume is not very important for this thought experiment. It is evident
that there will not be a uniform number of collisions of the gas particles with any wall so
that pressure will be a fluctuating and poorly defined property (as will density).
Similarly, if we watched a system containing 10 N2 molecules and 10 O2 molecules,
we may occasionally notice that molecules segregate spontaneously so that all 10 molecules
of N2 are on one side of the system and all 10 molecules of O2 are on the other side.
We might wonder if this spontaneous decrease of entropy with no effects in the surroundings disproves the Second Law.
FIGURE 3.5 Two flasks at
the same temperature are
connected through a valve
that can be turned to the
“Open” position (right-side
drawings) to permit the passage
of the molecules from one
flask to another. The Second
Law states that (a) pressures
tend to become uniform and
(b) composition tends to
become uniform.
(a)
pA > p2
pB < p2
N2
O2
Open
valve
p2
p2
(b)
Open
valve
N2 + O2
N2 + O2
Chemical Reactions | 65
Fluctuations will be important only for very small systems and very short times. For
large numbers of molecules and for small numbers of molecules observed for a long time,
the Second Law always applies. This discussion on fluctuations is a reminder that thermodynamics is a macroscopic theory; it applies to matter in bulk and does not acknowledge
molecules in any way, although we can and should connect thermodynamic insights to
molecular structures and interactions. We can estimate the probability that a process will
occur that is not consistent with the Second Law. From Eq. 3.9, we can show that the probability of observing a violation of the Second Law depends exponentially on the number
of molecules N in the system. The probability that a change contrary to the Second Law
occurs in a system containing N molecules is e -N . We thus find that for 5 molecules, the
probability of contradicting the Second Law is about 1/100; for 10 molecules, 1/104; for
20 molecules, 1/108; and for 100 molecules, 1/1043. Furthermore, observing five molecules
for a long time is equivalent to observing many molecules for a short time.
Scold =
Measurement of Entropy
qin
Tcold
>
qin
= S hot
Thot
The experimental measurement of entropy changes is based on the definition of entropy,
Eq. 3.6. The entropy change when a system changes from state 1 to state 2 is
state 2
state 2
S = S 2 - S 1 = 3
dS = 3
state
1
state 1
dqrev
.
T
(3.11)
We must be sure to understand entropy and its relation to the reversible heat given by
Eq. 3.11. Remember that entropy depends only on the initial state (1) and the final state
(2). It does not matter how we get from state 1 to state 2. However, the entropy difference
can be measured as in example 3.1 by finding a reversible path between states 1 and 2,
measuring the heat change, and using Eq. 3.11. For an irreversible path, the entropy change
is not equal to the heat absorbed divided by T; furthermore, the entropy change is always
greater than the irreversible heat divided by the temperature:
dqirrev
S = S 2 - S 1 7 3
.
T
S hot
Thot
qin
(3.12)
EX E R C I S E 3 .1
Show that example 3.1(a) is consistent with Eq. 3.12.
Before we explore the calculation of entropy, notice that the inverse dependence of the
entropy change on T in Eq. 3.11 conveys a remarkable insight: a fixed qrev changes the
entropy of colder bodies more than of hotter bodies (figure 3.6). In other words, this property illustrates how the entropy changes of the two nonadiabatic steps of the Carnot cycle
can have the same magnitude even when they involve different amounts of heat transfer.
Chemical Reactions
For a general reaction
n AA + n BB h n CC + n DD
at a chosen T and p, the entropy change is the difference in the entropies of the reactants
and products:
rS = nCS m,C + nDS m,D - nAS m,A - nBS m,B .
Scold
Tcold
qin
FIGURE 3.6 Consider
performing reversible,
isothermal expansions of equal
molar amounts of an ideal gas
at temperatures Tcold and Thot
such that each gas receives
the same qin. (pV is a unique
constant for each process.) Then
the entropy of the cold gas
increases more than that of the
hot gas.
66 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
where S m, A is the entropy per mol of compound A and S m, B is the entropy per mol of
compound B, and so on, at T and p. (A similar expression was given in chapter 2 for r H
of chemical reactions; see Eq. 2.67.) The entropy of each reactant or product depends on
T and p; thus, rS of a given reaction will depend on T and p. A reference state with a
temperature of 25°C (298 K) and a pressure of 1 bar is often picked, and the entropy per
mol of a substance at this reference state is termed the standard molar entropy, S°m. We
use the symbol r Sm for the molar entropy change of a reaction at 1 bar. Standard molar
entropies of compounds can be obtained from tables (see appendices). Unlike the convention for the standard enthalpies of formation, the standard molar entropies of the elements
are not equal to zero at 25°C and 1 bar. Rather, the Third Law of Thermodynamics will be
seen to define an absolute scale for entropy.
EXAMPLE 3.2
Calculate the entropy change at 25°C and 1 bar for the decomposition of 1 mol of
liquid water to H2 and O2 gas.
SOLUTION
The reaction is H2O(l) S H2 (g) + 21O2 (g). The entropy change at 25°C and 1 bar is
1
r S m (25C) = Sm (O2 (g) ) + Sm (H2 (g) ) - Sm (H2O(l) )
2
1
= (205.25) + 130.79 - 69.91
2
= 163.51 J mol - 1K - 1
The entropy change is positive, consistent with our expectations for a reaction that
involves the formation of two gases from one liquid. There is an increase in disorder.
Third Law of Thermodynamics
The Third Law of Thermodynamics states that the entropy of any pure, perfect crystal is
zero at 0 K (absolute zero),
S A (0 K) K 0 ,
(3.13)
where A is any pure, perfect crystal. The Third Law is an experimental one, as the first
two laws are, but we can understand it in terms of the relation of entropy and disorder.
Near 0 K, the disorder of a substance can approach zero; the number of microstates
approaches 1. For this to happen, the substance must be pure; in a mixture, the entropy
could be reduced by separating the components. So, for a perfect crystal of any pure
compound, we know the entropy at absolute zero. We can obtain the entropy at any other
temperature if we know how entropy changes with temperature.
Temperature Dependence of Entropy
The entropy change for heating or cooling a system is easy to calculate. The heating or
cooling can be done essentially reversibly so that the entropy change is
T2
S2 - S1 = 3
T
1
T2
dqrev
CdT
= 3
.
T
T
T1
Third Law of Thermodynamics | 67
At constant p,
T2
S = 3
Cp dT
(3.14)
T
T1
= Cp ln
T2
,
T1
(3.15)
if Cp, the heat capacity at constant p, is independent of temperature and thus can be placed
outside of the integral (figure 3.7). At constant V,
T2
CV dT
T
T1
T2
= CV ln
,
T1
S = 3
(3.16)
(3.17)
where C V, the heat capacity at constant V, is assumed independent of temperature. Because
Cp and CV are always positive, Eqs. 3.15 and 3.17 show that raising the temperature will
always increase the entropy; this is expected from the increase in disorder.
Ideal Gas (p, 298 K)
Ideal Gas (p, 310 K)
0.04
C p,m
T
0.03
0.02
0.01
298
310
Temperature / K
Area = Sm=
310 C p,mdT
298
T
= C p,m ln
T2
T1
FIGURE 3.7 The area under
the curve described by Cp,m/ T
is the change in entropy for
expanding an ideal gas at
constant pressure (from 298 K
to 310 K in this example). For
an ideal gas, Cp,m is independent
of temperature, an assumption
which can also be applied
when the temperature range
is very small. For finding S
over large temperature ranges,
Cp,m/ T must be measured by
calorimetric methods and
the integral evaluated by
curve fitting.
68 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
EXAMPLE 3.3
Calculate the change in entropy at constant p when 1 mol of liquid water at 100°C is
brought in contact with 1 mol of liquid water at 0°C. Assume that the heat capacity of
liquid water is independent of temperature and is equal to 75 J mol - 1K - 1. No heat is
lost to the surroundings.
SOLUTION
As we bring equal amounts of the water in contact, C p,m is constant and no heat is lost;
the First Law tells us that the final temperature of the mixture will be the average of the
two temperatures, 50°C. We use Eq. 3.15 to calculate the change in entropy of the hot
water and of the cold water and add the results:
Hot water:
S 1 = S (50C) - S (100C) = (1 mol)Cp,m ln
323
373
= -10.79 J K - 1 .
Cold water:
S 2 = S (50C) - S (0C) = (1 mol)Cp,m ln
323
273
= 12.61 J K - 1 .
(1 mol) H2O(100C) + (1 mol)H2O(0C) h (2 mol)H2O(50C):
S 1 + S 2 = 1.82 J K-1.
The entropy change is positive, as it must be for a spontaneous process in an
isolated system.
Temperature Dependence of the Entropy Change for a
Chemical Reaction
To calculate the entropy change for a chemical reaction at 1 bar and some temperature other
than 25°C, we can use Eq. 3.15. We consider a cycle in which products and reactants are heated
or cooled to the new temperature and then add the entropy changes for the cycle:
A(T2)
A(25°C)
∆rS°(T2)
∆rS°(25oC)
298
rS(T2) = rS(25C) + 3
T2
B(T2)
B(25°C)
T
Cp(A)
2
dT
dT
+ 3 Cp(B) .
T
T
298
Third Law of Thermodynamics | 69
We can generalize the result into a more compact form by using the identity
b
a
3 = -3 ,
a
b
T2
r S(T2 ) = r S(T1 ) + 3 r Cp
T1
dT
,
T
(3.18)
where r Cp = Cp (products) - Cp (reactants) .
EXAMPLE 3.4
If a spark is applied to a mixture of H2 (g) and O2 (g), an explosion occurs and water
is formed. The gaseous water is cooled to 100°C. Calculate the entropy change when
2 mol of gaseous H2O is formed at 100°C and 1 bar from H2 (g) and O2 (g) at the same
temperature and each at a partial pressure of 1 bar.
SOLUTION
The reaction for 2 mol of H2O is
2H2 (g) + O2 (g) h 2H2O(g) .
The entropy change at 25°C, 1 bar, is
r S(25C) = 2Sp,m,H2O(g) - Sp,m,O2(g) - 2Sp,m,H2(g)
= 2(188.93) - 205.25 - 2(130.79)
= -88.97 J K - 1 mol - 1 .
To find r S (100°C, 1 bar), we need to use Eq. 3.18; therefore, we need to know the heat
capacities of H2 (g), O2 (g), and H2O(g). They can be taken as constants over the temperature range 25°C to 100°C (apply Eq. 2.48 to the table on Pg. 24, and use Table 2.2 also)
rCp = 2Cp,m,H2O(g) - Cp,m,O2(g) - 2Cp,m,H2(g)
= 2(36.5) - 29.4 - 2(28.8)
= -14.0 J K - 1 mol - 1 .
From Eq. 3.18,
373
r S(100C) = r S(25C) + 3
r Cp
298
= r S(25C) + r Cp ln
dT
T
373
298
= -88.97 - (14.0)(0.224)
= -92.11 J K - 1 mol - 1 .
The entropy of the system decreased; the Second Law thus requires that there be a
greater increase in the entropy of the surroundings. The reaction is exothermic, and
heat is lost to the surroundings at constant temperature.
70 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Entropy Change for a Phase Transition
On heating many compounds from 0 K to room temperature, various phase transitions,
such as melting (fusion) and boiling (vaporization), may occur. The reversible heat absorbed
divided by the equilibrium temperature of the transition gives the entropy change for
the phase transition. It is important to stress that entropy should be calculated at the
equilibrium conditions of phase changes. Otherwise, the transition is not reversible, and
the heat absorbed is not the reversible heat. For a phase transition, at constant p and T,
qp = qrev = H
S =
H
T
(2.43)
(3.19)
,
where S is the entropy of the phase transition and H is the enthalpy of the phase
transition at the equilibrium temperature T.
Third-Law entropies or absolute entropies, such as those in tables A.5 -7 in the
appendix, are obtained by adding together (1) the entropy of the substance at 0 K (which
will be zero if it is a perfect crystal), (2) the entropy increase associated with increasing
the temperature, and (3) the entropy changes associated with any phase changes. For
example, to obtain the Third-Law entropy of a liquid compound at 25°C and 1 bar, we
would use the equation
298
fus H
dT
dT
+
+ 3 Cp,m(l)
T
Tfus
T
Tfus
0
where Tfus is the melting temperature at 1 bar and we have assumed that there are no
solid–solid transitions. If there are, we would add ( H >T ) for each transition and use
an appropriate C p,m for each solid phase.
Sm (25C, 1 bar) = Sm(0 K) + 3
Tfus
Cp,m(s)
Pressure Dependence of Entropy
For solids and liquids, we generally ignore the direct effect of pressure on entropy:
S = S (p2 ) - S (p1 ) ⬵ 0 .
(3.20)
For gases, we approximate the effect of pressure at constant temperature by that on an
ideal gas. The change in entropy dS for a small change dp in pressure is
dqrev
dS =
T
dU - dw rev
.
=
T
For an ideal gas that changes its pressure at constant temperature, dU is zero because U is
independent of p at constant temperature. Thus,
pdV
-dw rev
=
.
T
T
Because pV = nRT is a constant at constant temperature,
dS =
d(pV ) = pdV + V dp = 0 ,
(constant temperature)
and so pdV = -V dp for an ideal gas at constant temperature. Therefore,
dS =
-V dp
-nRdp
pdV
=
=
,
p
T
T
Third Law of Thermodynamics | 71
and
p2
S = -nR 3
p1
dp
p2
= -nR ln .
p
p1
(3.21)
We notice that in Eq. 3.21 if either p1 or p2 is zero, the entropy of n mol of gas becomes
infinite. This should not surprise us; zero pressure means infinite volume, and infinite
volume implies infinite disorder of the molecules in the volume. The point to remember is
that as the pressure of a gas decreases at constant temperature, the entropy will increase.
Equations 3.20 and 3.21 can be used to calculate the entropy change of a substance when
the pressure is changed.
A useful application of Eq. 3.21 is in the calculation of the entropy change of mixing two gases. Two gases, nA mol of A and nB mol of B, are initially in two separate
flasks connected by a tube with a valve in it, similar to the arrangement illustrated in
figure 3.5b. It is convenient to solve the special case with the two gases at the same
pressure (pA = pB = p) and temperature T. If the volumes of the flasks are the same
then nA = nB. After opening the valve, the two gases will eventually become completely mixed, as both our experience and the Second Law of Thermodynamics tell us.
Because we start with two gases at the same pressure and we also assume that the two
gases behave like ideal gases and do not react with each other, there will be no change
in the total pressure p after mixing. The partial pressures of A and B after mixing,
however, become
pA = p[nA >(nA + nB )] = x A p, and pB = p[nB >(nA + nB )] = x B p ,
respectively. [x A = nA >(nA + nB ) and x B = nB >(nA + nB ) are the mole fractions of A and
B, respectively.] From Eq. 3.21, the entropy change accompanying a change of the pressure
of A from p to xAp is, using the ideal gas approximation,
S A = -nA R ln a
xA p
b = -nA R ln x A .
p
S B = -nB R ln a
xB p
b = -nB R ln x B .
p
Similarly,
The total entropy change for mixing the two ideal gases is therefore
mixS = SA + SB = -R[nA ln xA + nB ln xB] .
(3.22)
It is interesting that if the two gases are identical, there is no entropy change upon mixing
the two sides. The entropy increases when the valve is opened only if the two gases are
distinguishable. One gas can be an isotope of the other, but as long as we can tell them
apart, we calculate the same mixS . If the two are indistinguishable, mixS is zero. This
conclusion is entirely consistent with the molecular interpretation of entropy. For two
indistinguishable gases, swapping molecules between the two flasks does not alter the total
number of microscopic states.
Spontaneous Chemical Reactions
It is natural to ask if graphite could be converted into diamonds. One way to answer this
question is to learn if the entropy of the universe increases when the reaction occurs. We
can easily learn from tables A.5 –7 in the appendix whether the entropy of the system
increases, but it is not so easy to learn about the change in entropy of the surroundings. For
processes that occur in an isolated system, however, the surroundings do not change, and
72 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
we can limit our attention to the system: the sign of the entropy change tells whether the
reaction can occur. Very few reactions occur at constant energy and volume and thus do not
affect the surroundings. Conversion of an optically active molecule to a racemic mixture
is an example. The two enantiomers have identical energies, volumes, and entropies, but
the mixture has a larger entropy and nearly the same energy and volume. The conversion
reaction is spontaneous. Other examples are the mixing of two ideal gases or the transfer
of heat from a hot object to a cold one.
Gibbs Free Energy
For most chemical reactions, there are large changes in energy and enthalpy, and we are
interested in whether a reaction will occur at constant T and p. These are the conditions
under which most experiments are done in the laboratory: the system is free to exchange
heat with the surroundings to remain at room temperature, and it can expand or contract
in volume to remain at atmospheric pressure. Constant temperature and pressure are also
common conditions for life processes. We need a criterion of spontaneity that applies to
the system for these conditions. A new thermodynamic variable of state, the Gibbs free
energy, is useful in this situation. The Gibbs free energy, G, is defined as a combination
of enthalpy, temperature, and entropy:
G K H - TS .
(3.23)
Because G is specified by the state variables H, T, and S and because both H and TS are
quantities that increase in proportion to the amounts of materials, the Gibbs free energy is
an extensive variable of state and has the same units as enthalpy or energy. We will see in
the next two sections that if G for a process at constant T and p is negative, the process
can occur spontaneously; if G at constant T and p is positive, the process will not occur
spontaneously; and if G at constant T and p is zero, the system is at equilibrium.
G and a System’s Capacity to Do Nonexpansion Work
To see how the change in Gibbs free energy is related to the spontaneity of a process,
we first examine the relation between G and a system’s capacity to do work. From the
definition of G, for a closed system a small change dG in the Gibbs free energy is related
to changes in the other thermodynamic parameters by
dG = dH - TdS - SdT
(Eq. 3.23)
= dU + pdV + V dp - TdS - SdT
(H = U + PV )
= dqrev + dw rev + pdV + V dp - TdS - SdT
(First Law, reversible)
= TdS + dw rev + pdV + V dp - TdS - SdT
(definition of S)
= dw rev + pdV + V dp - SdT.
(3.24)
We express dwrev as the sum of two types of work: the pressure–volume work, -pdV owing
to a change dV in the volume of the system, and all other forms of work, dw other,
dw rev = -pdV + dw other .
(3.25)
dG = V dp - SdT + dw other .
(3.26)
Equation 3.24 then becomes
At constant temperature and pressure, dp = dT = 0 and
dG = dw other
(3.27)
G = w other .
(3.28)
or, upon integration,
Gibbs Free Energy | 73
Equation 3.28 states that G of a process at constant pressure and temperature is equal to
w other or that - G is equal to -w other. In any process, such as in a chemical reaction, a
system must do the necessary pV work associated with the change of the system from its
initial state to its final state. Therefore, the non-pV work that can be obtained from a reversible process at constant temperature and pressure is -w other, also termed the nonexpansion
work. If a reversible process does electrical work, then -w other is the electrical work that the
process is capable of doing. Because the work done by a system is maximal when a process
is carried out reversibly, -w other = - G is the maximum amount of nonexpansion work
a system can do on the surroundings by a process at constant temperature and pressure.
Spontaneous Processes at Constant T and p
Equation 3.28 provides a criterion for processes that can spontaneously occur at constant
pressure and temperature. Our experience tells us that if a system undergoes a spontaneous
process, it is capable of performing useful work: water flowing down a fall can be utilized
to generate electricity, and heat flowing from a hot to a cold reservoir can be utilized to
power a heat engine. Conversely, for a nonspontaneous process, such as water flowing
uphill or heat flowing from cold to hot, the surroundings must do work on the system.
Thus, -w other, the maximum nonexpansion work a system can do on the surroundings, is
positive for a spontaneous process. From Eq. 3.28, it follows that - G = -w other 7 0
for a spontaneous process at constant temperature and pressure, or
G 6 0 (spontaneous process at constant T and p) .
(3.29)
If a process is not spontaneous, w other must be positive and
G 7 0 (nonspontaneous process at constant T and p) .
(3.30)
A system at equilibrium cannot perform any work. Thus, w other = 0 and
G = 0 (system at equilibrium at constant T and p) .
(3.31)
The preceding discussion can help us explore the meaning of the catchphrase “Conserve
energy!”—a thermodynamically misleading statement since the First Law of Thermodynamics
says that energy is always conserved. At constant temperature and pressure, Eq. 3.28 shows that
the decrease in Gibbs free energy corresponds to the maximum non-pV work a system can do.
This is the maximum energy that could be extracted, for example, from a fuel cell. And this is
also the energy that is available to drive nonspontaneous reactions in a cell. Generally, life processes take place at constant T and p, while pV work is often neglected in cells so that G is the
only free energy to be recovered from a given process.† Therefore, we are especially interested
to learn the change in free energy at 1 bar and 310 K for biological reactions. A compromise
catchphrase that has both social and biological appeal might be, “Conserve free energy!”
Calculation of Gibbs Free Energy
The Gibbs free-energy change for a reaction at constant temperature can be obtained from
the enthalpy and entropy changes,
rG = rH - T rS .
(3.32)
Tables A.5–7 in the appendix give values of fH and S for various substances at 25°C
and 1 bar. These values can then be used to calculate rH and rS, and hence rG, for
chemical reactions at 25°C and 1 bar.
†Cell
volumes can certainly change, generally in response to environmental conditions, but cells do attempt to
maintain constant volume. And pV work can be performed by organisms as the collective motions of major organs
(e.g., lungs).
74 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Values of f G , the standard free energy of formation, of various substances are
also given in tables A.5–7 in the appendix. The molar standard free energy of formation
is defined as the free energy of formation of 1 mol of any compound at 1 bar from its
elements in their standard states at 1 bar. As we did with the enthalpy of formation, we
arbitrarily assign the elements in their most stable state at 1 bar to have zero free energy.
EXAMPLE 3.5
Calculate the Gibbs free-energy change for the following reaction at 25°C and 1 bar.
Will the reaction occur spontaneously?
H2O(l) h H2 (g) +
1
O (g)
2 2
SOLUTION
rG(25C) = f Gm,H2(g) +
1
G
- f Gm,H2O(l)
2 f m,O2(g)
= 0 + 0 - ( -237.1)kJ mol - 1
= 237.1 kJ mol - 1 .
The Gibbs free-energy change is the negative of the Gibbs free energy of formation of
liquid water. We can also calculate H and S from table A.5 (appendix):
1
rH = Hm,H2(g) + Hm,O2(g) - Hm,H2O(l)
2
= 0 + 0 - ( -285.83)
= 285.83 kJ mol - 1
rS = Sm,H2(g) +
1
Sm,O2(g) - Sm,H2O(l)
2
1
(205.25) - 69.91
2
=163.51 J K - 1 mol - 1 .
=130.79 +
Now rG can be computed:
rG(25C) = 285.83 kJ - (298.15)(163.51 * 10 - 3 ) kJ
= 237.08 kJ mol - 1 .
The reaction will not occur spontaneously because rG is positive.
EXAMPLE 3.6
We wish to know whether proteins in aqueous solution are unstable with respect to
their constituent amino acids. As an example, let’s calculate the standard free energy of
hydrolysis for the dipeptide glycylglycine at 25°C and 1 bar in dilute aqueous solution.
SOLUTION
The reaction is
+
H3NCH2CONHCH2COO - (aq) + H2O(l) h 2 + H3NCH2COO - (aq)
glycylglycine
glycine
Gibbs Free Energy | 75
The standard free-energy change when solid glycine dissolves has been measured and
is small. We will assume that this is also true for solid glycylglycine. Therefore, the
free-energy values from tables A.5–7 (appendix) for solid glycine and glycylglycine
will be a good approximation to their aqueous values:
rG (25C) = 2 f G m(glycine, s) - f G m(glycylglycine, s) - f G m(H2O, l)
= 2( -377.69) - ( -490.57) - ( -237.13)
= -27.68 kJ mol-1
The reaction is spontaneous, but it normally occurs slowly. Catalysts such as proteolytic enzymes can cause the reaction to occur rapidly (e.g., to break down proteins in
food). There are many such enzymes in living organisms whose myriad tasks include
regulating cellular functions of various proteins and the programmed death of cells.
A proposed method of storing solar energy is to use sunlight to overcome the large
positive free energy to split water into hydrogen and oxygen gases that in turn can be
used as fuel. Notably, such a feat has been achieved by inorganic catalysts by Prof. Daniel
Nocera and coworkers, whose invention has been dubbed an “artificial leaf” (Science 334,
2011: 645–648). The light-harvesting reactions of green plant photosynthesis overcome
a positive free-energy change, which is almost exactly the same as for the production
of oxygen gas by the direct decomposition of water. However, green plants do not make
hydrogen gas; instead, they reduce carbon dioxide to carbohydrates and other products.
Temperature Dependence of Gibbs Free Energy
For reactions carried out at 25°C and 1 bar, the Gibbs free-energy change rG can often
be calculated from tabulations of enthalpy of formation and entropy data. If rG of a reaction is known at one temperature such as 25°C, how do we obtain its value at a different
temperature T? It is clear from the definition of Gibbs free energy that G (and therefore
rG) depends explicitly on T. Here are three common approaches to estimate rG at one
temperature given thermodynamic information for the reaction at some other reference
temperature such as 25°C.
(i) Since the values of rH and rS for reactions do not vary significantly for temperatures
close to 25°C, Eq. 3.32 is a simple way to calculate rG at other temperatures:
rG(T) ⬵ rH(25C) - T rS(25C) .
(3.33)
For example, values of r H and r S at 25°C from tables A.5–7 in the appendix can be
used to calculate the approximate free energy of the reaction at a “physiological temperature” of 37°C. Another useful equation that is easily derived from Eq. 3.32 is
rG(T) - rG(25C) ⬵ - (T - 298) rS(25C) ,
(3.34)
which shows that the sign of rS indicates how rG will change with temperature. If rS
is negative, then rG increases with increasing temperature.
(ii) A more general expression of the temperature dependence of rG at constant pressure can be obtained by considering a reversible path involving only pV work, such that
Eq. 3.26 becomes
dG = V dp - SdT .
(3.35)
Although we derived Eq. 3.35 for a reversible path, notice that all variables in this equation
are state functions that do not depend on a particular path. Therefore, this equation is
76 Chapter 3 |
A(T2)
The Second Law: The Entropy of the Universe Increases
ΔG(T2)
ΔG1
B(T2)
generally applicable to all paths and not just to the particular path chosen for its derivation.
At constant pressure, Eq. 3.35 becomes
dG p = -SdT,
ΔG3
A(T1)
ΔG2=ΔG(T1)
B(T1)
(3.36)
which can be integrated to give G at constant pressure:
T2
Gp = G(T 2 ) - G(T 1 ) = - 3 SdT .
(3.36a)
T1
T1
G1 = - 3 S A dT
T2
Eqs. 3.36 and 3.36a can also be written in terms of the Gibbs free energy Gm and entropy
Sm per mole of a substance:
G 2 = G(T1)
dG p,m = -S mdT
(3.37)
T2
G3 = - 3 S B dT
T2
Gm = Gm (T 2 ) - Gm (T 1 ) = - 3 S mdT
T1
(3.37a)
T1
G(T2) = G 1 + G 2 + G 3
To calculate the temperature dependence of rG of a chemical reaction
nAA + nBB h nCC + nDD
T2
= G(T 1) - 3 (S B - S A ) dT
T1
T2
we use the same approach for the temperature dependence of H (Eq. 2.73). The free-energy
change of the reaction is
= G(T 1) - 3 S dT
rG = nCGm,C + nDGm,D - nAGm,A - nBGm,B ,
T1
(3.38)
where Gm,C, Gm,D are the molar free energies of compounds C, D, and so on. [Here we
consider the reactants and products as pure substances that are present as separate phases
(solid zinc pellets dropped into a dilute hydrochloric acid solution, steam passed over hot
charcoal, etc.). A more general definition of the molar free energies will be presented in
chapter 4.] Taking the derivative of Eq. 3.38 with respect to temperature gives
nC dGm,C
nDdGm,D
nAdGm, A
nBdGm,B
d rG
=
+
.
dT
dT
dT
dT
dT
At constant pressure, then, Eq. 3.37 can be substituted to give
d rG
= -nCSm,C - nDSm,D + nASm,A + nBSm, B = - r S ,
dT
(3.39)
since
r S = nCSm,C + nDSm, D - nASm,A - nBSm,B
is the entropy change of the reaction. Integrating Eq. 3.39 gives
rG(T2)
3
rG(T1)
T2
d rG = - 3 rSdT ,
T1
T2
rG(T 2 ) - rG(T 1 ) = - 3 rSdT .
(3.39a)
T1
If rS is assumed constant between T1 and T2, then Eq. 3.39a becomes
rG(T 2 ) - rG(T 1 ) = - rS (T 2 - T 1 ) ,
(3.39b)
which is just a general form of Eq. 3.34. The preceding discussion is an important preview
of molar free energies. An alternate route to derive Eq. 3.39b is to sum the free-energy
changes over a convenient reversible path, as shown in the sidebar.
Gibbs Free Energy | 77
(iii) Another useful equation for the temperature dependence of rG at constant pressure
when we have knowledge of the enthalpy change is
T2
rG(T 2 )
rG(T 1 )
rH(T )
= -3
dT ,
T2
T1
T2
T1
(3.40)
which is called the Gibbs–Helmholtz equation. If we assume that r H(T) is independent
of the temperature range, then r H(T) can be selected at the midpoint of T1 and T2,
rG(T 2)
rG(T 1)
1
1
= rH c
d.
T2
T1
T2
T1
(3.41)
Depending on available data, one can also use r H(T1 ) in Eq. 3.41 when the temperature
range is very narrow. This equation is derived at the end of this chapter.
EXAMPLE 3.7
What is the free energy of hydrolysis of glycylglycine at 37°C and 1 bar?
SOLUTION
We can use Eq. 3.39b to find rG(37°C) if we assume that rS is independent of
temperature, or we can use Eq. 3.41 if we assume that r H is independent of temperature. Both will be nearly identical, since r H and r S should be constant over the
small temperature range. As in example 3.6, we use the data for the solid compounds:
r S(25C) = 2Sm,(glycine, s) - Sm,(glycyglycine, s) - Sm,(H2O, l)
= 2(103.51) - 190.0 - 69.91
= -52.9 J K-1 mol-1 .
r H (25C) = 2( -537.2) - ( -745.25) - ( -285.83)
= -43.32 kJ mol-1 .
With Eq. 3.39b,
rG(37C) = rG(25C) - (12 K) r S(25C)
= -27.68 kJ mol-1 - (12 K)( -0.0529 kJ K-1 mol-1 )
= -27.05 kJ mol-1 .
With Eq. 3.41,
rG(37C)
rG(25C)
1
1
= a
b rH(25C)
310
298
310
298
rG(37C)
-27.68
=
+ ( -1.299 * 10 - 4 )( -43.32)
310
298
rG(37C) = -27.05 kJ mol - 1 .
Both methods predict the same free energy of hydrolysis, rG(37°C).
Pressure Dependence of Gibbs Free Energy
We have shown several approaches to calculate the Gibbs free-energy change r G at 25°C
and at other temperatures, often based on values in tables A.5–7 in the appendix. However,
the tabulated values of r G are all defined for 1 bar pressure. To calculate free energy
78 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
at some other pressure, we must know its dependence on pressure. From Eq. 3.35, we see
that at constant temperature SdT = 0 and so
dG T = V dp ,
(3.42)
p2
G(p2 ) - G(p1 ) = 3 V dp .
(3.42a)
p1
For a solid or liquid, at moderate pressures the volume is approximately constant independent of pressure, and can be taken out of the integral:
G(p2 ) - G(p1 ) ⬵ V (p2 - p1 ) .
(3.43)
For a gas, we use the equation of state to write V as a function of p. For an ideal gas, and
approximately for any gas, we can substitute the ideal gas equation in Eq. 3.42a:
p2
p2
nRT
dp = nRT ln .
p1
p1 p
G(p2 ) - G(p1 ) = 3
(3.44)
To calculate the effect of pressure on the free energy of a chemical reaction, we just
apply Eqs. 3.43 and 3.44 to each product and reactant. If all products and reactants are
solids or liquids, we use Eq. 3.43:
rG(p2 ) - rG(p1 ) = rV (p2 - p1 ) ,
(3.45)
where rV =V (products) -V (reactants). Equation 3.45 shows that if the volume of products is greater than the volume of reactants ( rV is positive), increasing the pressure will
increase the free energy of the reaction. If at least one gaseous product or reactant is
involved, we can ignore the volume of the solids or liquids compared to that of the gases
and use Eq. 3.44:
rG(p2) - rG(p1) = ( r n) # RT ln
p2
,
p1
(3.46)
where r n is the number of moles of gaseous products minus the number of moles of
gaseous reactants.
EXAMPLE 3.8
Can graphite (pencil lead) convert spontaneously into diamond at 25°C and 1 bar?
SOLUTION
The reaction at 25°C and 1 bar
C(s, graphite) h C(s, diamond)
has a Gibbs free energy of 2.90 kJ mol - 1 from table A.5 in the appendix. Therefore,
the answer is no; graphite will not spontaneously convert into diamond at 25°C and
1 bar. This result also means that diamond will spontaneously convert into graphite.
For those who treasure their diamonds, it is fortunate that the rate of this conversion is
extremely slow under these conditions.
EXAMPLE 3.9
Will increasing the pressure favor the conversion of graphite to diamond? If so, what
is the minimum pressure necessary to make this reaction spontaneous at 25°C? Will
increasing the temperature favor the reaction?
Gibbs Free Energy | 79
SOLUTION
We can use Eq. 3.45 since we are considering solids. Start by considering the volume
change for the reaction. The densities of graphite and diamond at 25°C can be used
to calculate the respective volumes, assuming they are independent of pressure.
Just knowing that diamond is denser than graphite allows us to see that rV for the
reaction is negative. To answer the first question, increasing the pressure will decrease
the free energy and favor the reaction graphite to diamond. To find the minimum
pressure necessary to allow the spontaneous reaction, calculate the pressure that makes
r G = 0. The densities at 25°C and 1 bar are as follows:
Density (g mL-1)
C (graphite)
2.25
C (diamond)
3.51
The molar volumes at 25°C and 1 bar are obtained by dividing the atomic mass of
carbon by the densities:
Vm ( mL mol - 1)
C (graphite)
5.33
C (diamond)
3.42
For the reaction C (graphite) h C (diamond) , the Gibbs free-energy change at 25°C
and 1 bar is, from Eq. 3.45,
rG(p) = G(1 bar) + rV (p -1)
= 2.90 kJ mol-1 + (3.42 - 5.33) mL mol-1 # (p -1) bar .
We convert mL bar to kJ by multiplying by a ratio of gas constants,
R/R = 8.314 * 10 - 3 kJ>83.14 mL bar = 1.000 * 10 - 4 kJ mL - 1 bar - 1:
rG(p) = 2.90 - 1.91 * 10-4 (p - 1) kJ mol-1 .
We want to find the pressure that makes r G(p) = 0:
0 = 2.90 - 1.91 * 10 - 4 (p - 1) ,
p =
2.90
+ 1 = 15,200 bar .
1.91 * 10-4
The assumption that rV is constant is not valid over this large pressure range, and
the actual pressure needed is much higher than 15,200 bar. Nevertheless, the effect
is real, and diamonds have been made in this way for years, mostly for industrial
uses. To decide whether increasing temperature will favor the reaction, we have to
know the sign of r S (see Eqs. 3.39a and 3.39b). From table A.5 in the appendix,
we find that rS = 2.38 - 5.74 = -3.36 J K - 1 mol - 1; therefore, increasing the
temperature will make rG more positive and will not favor the formation of
diamond. Equation 3.39a rather than Eq. 3.39b must be used to calculate at
what low temperature rG becomes zero, as rS changes significantly over the
wide temperature change. The industrial process does use high temperatures,
but for kinetic reasons. The increase in pressure provides the necessary freeenergy change.
80 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Phase Changes
For a phase change that takes place at its equilibrium temperature and pressure, the change
in Gibbs free energy is zero:
,eqG = 0 .
To calculate the Gibbs free energy at other temperatures and pressures, we can use
G = H - T S
(3.32)
if H and S are known for the phase change at arbitrary T and P. Otherwise, we can
use the methods described in earlier sections of this chapter, “Temperature Dependence of
Gibbs Free Energy” and “Pressure Dependence of Gibbs Free Energy.”
Helmholtz Free Energy
For a process that takes place at constant T and V, the sign of the Gibbs free energy is not
a criterion for equilibrium. A new thermodynamic variable of state is needed; it is called
the Helmholtz free energy, A, and is defined as
A = U - TS .
(3.47)
The criterion for equilibrium is, at constant T and V,
A 6 0
(spontaneous reaction),
(3.48)
A = 0
(reaction at equilibrium),
(3.49)
A 7 0
(nonspontaneous reaction).
(3.50)
We do not use the Helmholtz free energy very much in this book, although for reactions
at constant volume it is as useful as the Gibbs free energy is for reactions at constant
pressure. The Helmholtz free energy is used in engineering and geochemical work, where
the pressure may vary widely and the constant-volume restriction is more appropriate.
Noncovalent Reactions
The examples that we have given so far of the changes in H, S, and G for chemical reactions involve the breaking and forming of covalent bonds. Many important biochemical
processes involve reactions in which no covalent bonds are made or broken; only weaker
bonds and interactions are involved. Examples include the reaction of an antigen with its
antibody, the binding of hormones and drugs to nucleic acids and proteins, the reading of
the genetic message (codon–anticodon recognition), denaturation of proteins and nucleic
acids, and so on. Thermodynamics can help us understand these reactions and decide
which proposed mechanisms are reasonable and which ones are not. Table 3.1 gives some
measured enthalpies for simple reactions, which illustrate the forces involved.
Charged species can interact with each other by ionic interactions. In a NaCl crystal,
for example, very strong ionic interactions exist between the positively charged Na+ ions
and the negatively charged Cl - ions. This is reflected in the large positive enthalpy change
when solid NaCl is separated into Na+ and Cl - ions:
NaCl(s) h Na + (g) + Cl - (g)
( rH (298 K) = 785 kJ mol - 1 )
Despite such strong ionic forces, solid NaCl nevertheless dissolves in water easily, with a
small rH of only 4 kJ mol - 1:
NaCl(s) + H2O(l) h Na+ (aq) + Cl - (aq)
( r H(298 K) = 4 kJ mol - 1 )
The reason is that there are strong interactions between the charged ions and the water
molecules, so the net enthalpy change is small.
Noncovalent Reactions | 81
TABLE 3.1 Enthalpies of Noncovalent Bonds and Interactions*
Reaction
Na+ (g) + Cl-(g) S NaCl(s)
+
-
NaCl(s) + H2O(l) S Na (aq) + Cl (aq)
Characteristic
interaction
r H (kj mol-1)
Ionic
- 785
Ionic and ion–dipole
4
Argon(g) S Argon(s)
London
-8
n@Butane(g) S n@Butane(l)
London–van der
Waals
-20
Acetone(g) S Acetone(l)
London–van der
Waals
-30
Hydrogen bond (g)
-20
Hydrogen bond (g)
-15
Hydrogen bond
(benzene)
-15
Hydrogen bond
(aqueous)
-5
C3H6(l) + H2O(l) S C3H6(aq)
Hydrophobic
-10
Benzene(l) + H2O(l) S benzene(aq)
Hydrophobic
0
H3C
CH3
H3C
O . . .H
O
2
H
O
H
( g)
( g)
Methanol
H
H
H N. . .H N
N
2
H
H
H
( g)
H
Ammonia
H
O
O
2
H
C
H
( g)
N
H
C
H
H. . .O
N
(benzene)
C
CH3
N
H3C
CH3
H
N-Methyl formamide
H
H
H
N
H. . .O
O
H
N
H
N
H
H
H
N
H. . .O
O
H
H
(aq)
Urea
(aq)
Urea
H
N
H
N
H. . .O
O
N
H
H
H
N
H
H
C
H
O
H. . .O
H
H (aq)
(aq)
*The enthalpies were obtained near room temperature except for the vaporization of solid argon; the standard state is the dilute
solution extrapolated to 1 M. Data are from various sources and are rounded to the nearest integer. For the precise values, see
G. C. Pimentel and A. L. McClellan, The Hydrogen Bond (San Francisco: Freeman, 1960); W. Kauzmann, Adv. Protein Chem.
14(1) (1959); and J. A. A. Ketallar, Chemical Constitution (Amsterdam: Elsevier, 1958).
In a neutral water molecule the bonding electrons are localized more on the oxygen atom than on the hydrogen atoms. Therefore, the oxygen atom is more negatively
charged, and the hydrogen atoms are more positively charged. Because H2O is bent,
the geometric centers of the positive and negative charges do not coincide. We say that
water has an electric dipole. The extent of charge separation is expressed in terms of
the dipole moment. If two point charges +e and -e are separated by a distance r, the
82 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
dipole moment is er. Experimentally, the dipole moment of H2O is found to be equivalent to an electron and a unit positive charge separated by 0.38 Å (1Å =10 - 10 m). If
H
water is modeled as three charges on the three atoms, the charge on the oxygen atom is
-0.834 electronic charges, and each hydrogen atom is +0.417. The interaction between
a charged ion and a neutral molecule with a dipole moment is called an ion–dipole
interaction, which is a member of a family called van der Waals interactions. Van
dipole-dipole
der Waals interactions also include dipole–dipole interactions, dipole-induced dipole
δ−
δ+
δ−
δ+
attractions, ion-induced dipole interactions, London interactions (London dispersion
interactions), and steric repulsions.
Ions can interact with neutral molecules with zero dipole moment by ion-induced
dipole interactions. Consider a molecule of CCl4, for example. Though the bonding electrons are localized more on the chlorine atoms, the permanent dipole moment of the
London
e- He
He
molecule is zero because the four Cl atoms are symmetrically located at the four corners
edispersion
+2
+2
of a tetrahedron, with the C atom occupying the center of the tetrahedron. However, if
eea charge is placed near a CCl4 molecule, the charge will distort the electronic distribution in CCl4. We say that the CCl4 molecule becomes polarized, and has now an induced
fluctuation dipole
induced dipole
dipole moment. Interactions between a charged ion and polarized molecules are called
FIGURE 3.8 All van-der Waals
ion-induced dipole interactions.
and London intermolecular
Fluctuation dipole-induced dipole interactions exist even between neutral molecules
attractions (or repulsions) are
with no permanent dipole moments due to fluctuations in the electronic distributions in
due to coulombic interactions
among unit or partial charges.
the molecules. A molecule with no permanent dipole moment may acquire an instanThe force of attraction depends
taneous dipole moment because of a fluctuation. Also called a London interaction,
on the sizes of the interaction
the
fluctuation dipole induces a dipole in a neighboring molecule. London interactions
charges, q1 and q2, and on the
are
weak but always present and always attractive. It is the only attractive force acting
distance r separating them:
q 1q 2
between identical noble gas atoms, responsible for the 6.43 kJ mol-1 enthalpy change
.
F
r
necessary to vaporize liquid argon to gaseous argon, for example. The force depends on
the polarizability of the interacting molecules, which is a measure of how easy it is to
distort the electron clouds of the molecule. As the polarizability increases, so does the
boiling point (see sidebar).
The van der Waals forces are often invoked in a very general sense to explain reactions such as the 21.5 kJ mol-1 enthalpy of vaporization of liquid n-butane. However, very
specific van der Waals interactions may be implicated when careful fitting of molecules is
involved, such as binding of a particular substrate to an enzyme, antigen–antibody binding,
and the function of specific membrane lipids.
The hydrogen bond is one of the important bonds that determine the three vapH(TB)
dimensional
structures of proteins and nucleic acids. A hydrogen atom covalently
-1
Element TB (K) kJ mol )
attached to one oxygen or nitrogen can form a weak bond to another oxygen or nitrogen.
Helium
4.19
0.08
For O -H cO, N -H cN, and N -H cO hydrogen bonds, the bond enthalpy of
Neon
27.02
1.71
15-20 kJ mol - 1 can be compared with values of about 400 kJ mol - 1 for covalent bonds.
Argon
87.25
6.43
This weak bond becomes even weaker in aqueous solution because of competition between
solute–solute hydrogen bonds and solute–water and water–water hydrogen bonds.
Krypton 119.76
9.08
Between urea molecules in water the effective hydrogen-bond strength between urea molBased on data from CRC Handbook of
ecules is just 5 kJ mol - 1. Urea can be considered as a model for a peptide bond, so this
Chemistry and Physics, 90th Edition,
Ed. D. Lide. Boca Raton, FL: CRC
is the magnitude of the expected energy for breaking a hydrogen bond in a peptide in
Press, 2009/2010.
aqueous solution.
Na+
ion-dipole
H
δ− O
δ+
Hydrophobic Interactions
One type of phenomenon that is important in aqueous solutions is the hydrophobic (fearof-water) effect. Water molecules have a strong attraction for each other, primarily as a
consequence of hydrogen-bond formation. The oxygen atom of most molecules of liquid
Noncovalent Reactions | 83
water is hydrogen-bonded to two hydrogen atoms of two other H2O molecules, and the
hydrogen atoms of most molecules are hydrogen-bonded to the oxygen atoms of two other
water molecules. Therefore, the molecules of liquid water form a mobile network: most
water molecules are primarily interacting, through hydrogen bonds, with four tetrahedrally
oriented neighbors. The network is not a rigid one, and change of neighbors occurs rapidly
because of thermal motions.
Consider what happens if a molecule such as propane (C3H8) is introduced into this
network. A hole is created; some hydrogen bonds in the original network are broken.
The C3H8 does not interact with water strongly; it does not form hydrogen bonds. The
water molecules around the C3H8 molecule must orient themselves in a way that reforms
the hydrogen bonds that were disrupted by the hydrocarbon molecule. The net result is
that water molecules around the C3H8 actually become more ordered. Since there is little
change in the number of hydrogen bonds, the enthalpy change is small. The ordering of
water molecules around the hydrocarbon molecule, however, is associated with a negative
entropy change. These interpretations are consistent with experimental data:
C3H8(l) i C3H8(aq)
⌬ r H ⬚(298 K) ⬇ -8 kJ mol - 1
⌬ r S⬚(298 K) ⬇ -80 J K - 1 mol - 1
⌬ rG⬚(298 K) ⬇ +16 kJ mol - 1 .
Now consider two such hydrocarbon groups, R. Each separate group, when exposed
to liquid water, will, from the preceding data, cause an unfavorable free-energy change.
If the two groups cluster together, the disruptive effect on the solvent network will be less
than the combined effects of two separate groups. Therefore, the association of the groups
will be thermodynamically favored:
R(aq) + R(aq) h [R 苲 R](aq) .
separate
clustered
The clustering of the groups is not because they “like” each other, but because they
are both “disliked” by water. The clustered arrangement of the hydrocarbons results
in a decrease in the overall free energy of the system in comparison with the isolated
hydrocarbons in water. Such hydrophobic clustering is important in many biological
systems. For example, hydrocarbon groups in a water-soluble protein are usually found
to cluster in the interior of the protein. Similarly, polar lipid molecules form bilayer
sheets or membranes in water, in which the hydrocarbon portions are buried inside and
the polar or charged portions are on the surface, exposed to the water. Such molecules
are called amphiphilic. Interactions among hydrophobic groups are characterized by
low enthalpy changes and are entropy driven. We note that the hydrophobic effect is
a term that we use to describe the combined effects of London, van der Waals, and
hydrogen-bonding interactions in certain processes in aqueous solutions; it is not a
“force” different from the others we have discussed. In particular, there is no such thing
as a hydrophobic bond.
Proteins and Nucleic Acids
Proteins are polypeptides of 20 naturally occurring amino acids, the structures of which
are given in appendix table A.8. The amino acids are linked by amide bonds from the
carboxyl group of one amino acid to the amino group of the next. The sequence of the
amino acid residues in a polypeptide is called the primary structure. The polypeptides fold
84 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
into different secondary structures held together by hydrogen bonds and other noncovalent
interactions. The amide group
O
C
C
N
H
C
is planar (because of partial double bond character of the central C - N bond) and has a
trans conformation as shown. The different folded conformations of a polypeptide depend
on rotation about the two single bonds attached to the amide group, as shown in figure 3.9.
A common secondary structural element in proteins is the α-helix shown in figure 3.10. It
is a right-handed helix with hydrogen bonds between the C = O of each amide to the N -H
of the amide four residue units away (figures 3.9 and 3.10).
Nucleic acids are polynucleotides of four naturally occurring nucleotides (appendix
table A.8). The nucleotides are composed of a sugar group connected to one of four bases
and to a phosphate group. The nucleotides are linked by phosphodiester bonds [see figure 3.11(b)]. The sequence of the bases in the polynucleotide is the primary structure. In
DNA two strands form a double helix held together by Watson–Crick base pairs as shown
in figure 3.11(a). The stacking of the base pairs, one above the other, plus the hydrogen
bonds between bases, provide the stabilizing enthalpy and free energy of the helix. Both
cross-strand and same-strand London–van der Waals interactions among the bases are
important. The charged phosphate groups repel each other by Coulomb’s Law repulsion
between like charges. The balance of these forces as well as interactions with the solvent,
and other solutes such as positively charged ions, determine the conformation of the double
helix. Two different double helices—secondary structures—are shown in figure 3.12. The
B-form helix is the predominant structure found for DNA (deoxyribonucleic acid) in biological systems. The A-form helix is the structure found in double-stranded regions of
RNA (ribonucleic acid).
Table 3.2 lists some measured enthalpies for biochemical reactions involving changes
in shape (conformation) of molecules. By changes in conformation, we mean changes in the
secondary structure of the molecule. (The primary structure involves the covalent bonds.)
Examples include the change in a polypeptide from a rigid helix to a flexible coil. Denaturation of proteins involves this type of change. The corresponding change in nucleic acids
and polynucleotides is the change from a two-strand helix to two single strands. Hydrogen
bonds between the amides, the nucleic acid bases, and the solvent are important, but so
are London–van der Waals types of interactions. The magnitudes of r H and r S can
help us understand the various interactions involved.
O
+
H3N
H
C
C
R2
C
C
N
N
C
H
O
H
O
H
C
R4
C
H
N
N
C
H
O
H
O
C
C
R6
O
C
N
C
H
O
–
R1
H
R3
H
R5
H
FIGURE 3.9 A polypeptide chain. Residues are commonly numbered by starting from the amino
N-terminus. The residues in the polypeptide chain are separated by dotted lines, and the arrows show the
two bonds (on either side of each -carbon) around which rotation can occur. The two long lines joining
a carbonyl oxygen to an amide hydrogen four residue units away on its carboxyl terminal side represent
hydrogen-bond formation in an -helix structure (see figure 3.10).
Noncovalent Reactions | 85
As mentioned earlier, the strength of a hydrogen bond between urea molecules in an
aqueous environment is about 5 kJ mol-1. From table 3.2, we see that this is consistent
with the magnitude of ⌬ rH ⬚ for the helix-coil transition of poly-L-glutamate in aqueous
solution. In the ␣-helix, the peptide groups are hydrogen bonded as shown in figure 3.10;
in the coil, they are hydrogen bonded to water. The enthalpy increase for denaturation
of a protein such as lysozyme depends greatly on the method used for denaturation (pH,
urea, or temperature), but the large value of ⌬ rH ⬚ clearly indicates many peptide–peptide
hydrogen bonds are being broken.
For DNA and RNA double-strand to single-strand transitions, we need to break two
hydrogen bonds per A· T base pair (in DNA) or A·U base pair (in RNA) and three hydrogen
bonds per G·C base pair (see figure 3.11). The ⌬ rH ⬚ for dissociating a G·C base pair is
larger than for A·T or A·U as expected, but the magnitude of ⌬ rH ⬚ depends on the sequence
of bases. The main interactions that stabilize the double strands are the stacking of the
base pairs on each other; therefore, the nearest neighbor sequences affect the magnitude of
changes in thermodynamic quantities. In DNA, the range of ⌬ rH ⬚ values for “melting” a
base pair ranges from 30 kJ mol-1 for an A·T in a sequence of alternating –A–T–A–T–A–T–
A– to 44 kJ mol-1 for a G·C in a sequence of alternating –G–C–G–C–G–C–. A measure
of the thermodynamics of unstacking is illustrated by the data for polyadenylic acid in
table 3.2. Polyadenylic acid unstacking refers to the transition between an ordered helical
molecule with the adenine bases stacked on top of each other and a much more disordered
molecule with the adenine bases not oriented relative to each other. This change in stacking does not involve hydrogen bonds directly, but it does have a ⌬ rH ⬚ of 36 kJ per mole of
(a)
(b)
CH3
-O
Thymine
O
N
O
H
T
NH
N
H2C
Adenine
N
H
to
sugar-phosphate
chain
P
O
N
O
H
H
<
N
N
O
P
to sugar-phosphate chain
Base
H
H
O
H
H
H
C
to
sugar-phosphate
chain
O
H2C
H
N
O
O
H
N
Cytosine
Base
H
H
A
O
O
H
-O
O
N
P O
Guanine
O
H
N
N
O
G
H
N
H
N
N
O
H2C
Base
H
H
O
H
H
H
to sugar-phosphate chain
FIGURE 3.11 (a) The Watson–Crick base pairs found in DNA. Adenine is complementary to thymine (A•T),
and guanine is complementary to cytosine (G∙C). In RNA, uracil substitutes for thymine (A•U). (b) The sugar–
phosphate chain in RNA differs only in that the sugar is ribose instead of deoxyribose. In the structures
shown in (a), a covalent bond, whether it is a single or a double bond, is represented by a solid line; a hydrogen bond is represented by a dotted line.
FIGURE 3.10 The ␣-helix. The
structure shown here is the
right-handed conformation
commonly found in proteins.
86 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
B-form
A-form
FIGURE 3.12 Double-stranded, right-handed nucleic acid helices. The two polynucleotide strands in each
double helix are antiparallel. One strand is as shown in figure 3.11(b); the other runs in the opposite direction.
The base pairs are tilted relative to the helix axis in the A-form; they are nearly perpendicular to the helix axis in
the B-form. DNA is mostly in the B-form in biological cells; double-stranded regions of RNA are in the A-form.
TABLE 3.2
Thermodynamics of Biochemical Conformational Transitions and Noncovalent Reactions
Studied Near Room Temperature and Neutral pH
Transition or noncovalent reaction
r H
(kJ mol-1)
r S
( J K - 1 mol - 1)
Reference*
Poly-l-glutamate helix-coil transition in 1.0 M KCl
4.5/amide
—
a
≈450
—
a
RNA double-strand to single-strand transition in 1 M NaCl
≈40/base pair
≈104/base pair
b
DNA double-strand to single-strand transition in 1 M NaCl
≈35/base pair
≈88/base pair
b
Unstacking of bases in polyadenylic acid in 0.1 M KCl
36/nucleotide
113/nucleotide
a
31
182
c
-9.2
141
b
Lysozyme denaturation
Binding of Mg–ATP by tetrahydrofolate synthetase
Binding of Netropsin by poly dA∙poly dT
*Based on data from (a) Fasman, Handbook of Biochemistry and Molecular Biology, Vol. 1, 3 (CRC Press, 1976); (b) Landolt-Börnstein Num. Data Functional Relationships in
Science and Technology, Group VII, Biophysics, Vol. 1, Nucleic Acids (1990); (c) Curthoys Rabinowitz, Journal of Biological Chemistry (1971).
Use of Partial Derivatives in Thermodynamics | 87
nucleotide. This rH is mainly London–van der Waals interactions among the bases and
between the bases and the solvent.
The entropy changes for the double-strand to single-strand transition in nucleic acids
are also consistent with the preceding conclusions. About 100 J K-1 mol-1 of entropy
is gained for each base pair disrupted in DNA or RNA; this mainly corresponds to the
increased possibility of rotation around single bonds in the nucleotides.
Two examples of binding of small molecules by a macromolecule are given in table 3.2.
The rH values can be easily explained; heat must be added to favor the binding of the
Mg–ATP complex to the enzyme tetrahydrofolate synthetase, but heat is released when the
antibiotic Netropsin binds to double-stranded polydeoxyadenylic acid·polythymidylic acid
(poly dA·poly dT). The entropy increases for both, which might be surprising because of
the loss of translational entropy when two molecules form a complex. The explanation is
that water must be released when the molecules are bound; there is thus a large net gain in
translational entropy.
Use of Partial Derivatives in Thermodynamics
Partial derivatives are presented in the mathematics appendix, with several examples for
ideal gases. In summary, a partial derivative is the rate of change of a multivariate function with respect to a change in the value of just one of the variables. A partial derivative
of particular importance in discussing chemical and physical equilibria (chapters 4 and 5)
is the partial molar Gibbs free energy, which is also termed the chemical potential. The
chemical potential of a compound A in a solution containing nA mol of A, nB of B, and
so on is defined by
0G
mA K ¢
≤
.
0n A T, p, nB, nC, c
(3.51)
Equation 3.51 describes the change in the Gibbs free energy that occurs upon an infinitesimal change dnA in the number of moles of A, while keeping the temperature, pressure,
and the number of moles of all other constituents constant. Component A may be either
a solute or the solvent in the solution. For pure A, the chemical potential or partial molar
Gibbs free energy of A is simply the Gibbs free energy per mole of A, G m,A. For a more
rigorous expression of Eq. 3.38, the parameters G m,C, G m,D, and so on are to be replaced
by the corresponding partial molar quantities mC, mD, and so on. (We will discuss this point
in detail in the next chapter.)
Relations Among Partial Derivatives
For a state function such as the volume V, it is generally true that if V = V(p,T), the total
differential of V is
dV = a
0V
0V
b dp + a b dT .
0p T
0T p
(3.52)
This equation states that for small changes, dV can be treated as the sum of how V changes with
p (at constant T ) and how V changes with T (at constant p). Also, the order of differentiation
is not important:
c
0 0V
0 0V
a b d = c a b d .
0T 0p T p
0p 0T p T
(3.53)
88 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
An example of the use of these equations follows.
E X A M P L E 3 .1 0
Show that
a
0V
0S
b = - a b .
0T p
0p T
SOLUTION
In the notation of partial derivatives, Eqs. 3.36 and 3.42 become
a
0G
0G
b = -S, and a b = V ,
0T p
0p T
respectively. Because the order of differentiation is unimportant for a state function,
c
0 0G
0 0G
a b d = c a b d
0T 0p T p
0p 0T p T
(3.54)
0S
0V
b = -a b .
0T p
0p T
(3.55)
a
Equation 3.58 gives a general expression on how entropy depends on pressure. In the
special case of an ideal gas, since
a
0S
0V
nR
nR
, we have a b = .
b =
p
p
0T p
0p T
Integrating at constant temperature gives
S2
p2
3 dS = - 3
S1
p1
p2
nR
dp = - nR ln .
p
p1
This is Eq. 3.21, which we derived earlier by a different route.
Partial derivatives find many applications in thermodynamics, and we will illustrate some
of the results here. The object is to obtain equations relating the state variables T, p, V, U, H,
G, and S. We consider here a closed system with no external fields (such as gravitational or
electrical fields) and no chemical reactions or phase changes. Therefore, the only reason that
one of the preceding variables changes is because one or more of the other variables changes.
We usually think of T, p, and V as the independent variables and U, H, G, and S as the dependent variables. Furthermore, because the equation of state tells us how p, V, and T are related,
we need consider only how U, H, G, and S depend on two of the three variables p, V, and T.
Dependence of G on p and T. Recall Eq. 3.35 for dG:
dG = V dp - SdT .
(3.35)
If we need to calculate the change of G resulting from a change of V, we can first find the
change of p and T corresponding to this change of V and then use Eq. 3.35. To calculate
G(p2, T2 ) - G(p1, T1 ) for a change from p1, T1 to p2, T2, we must integrate Eq. 3.35:
p2
T2
G(p2, T 2 ) - G(p1, T 1 ) = 3 V dp - 3 SdT .
p1
T1
Use of Partial Derivatives in Thermodynamics | 89
To determine how G for a chemical reaction or phase change depends on p and T, use
p2
T2
G(p2, T2 ) - G(p1, T1 ) = 3 Vdp - 3 SdT ,
p1
T1
where V and S are, respectively, the volume change and entropy change for the chemical
reaction or phase change. To do the integration, we need to know V as a function of p,
and S as a function of T, which are approximately constant for some systems.
Dependence of H on p and T. We can obtain corresponding equations for H as a
function of p and T. We use the total differential
dH = a
0H
0H
b dT + a b dp ,
0T p
0p T
where we wish to find convenient substitutions for the partial differentials. First, the change
of enthalpy with temperature at constant pressure was found to be the heat capacity at
constant pressure (Eq. 2.45); thus, 10H>0T 2 p = C p. To derive (0H>0p) T , first write the
identity G = H - TS as H = G + TS and simplify the differential as usual:
dH = dG + TdS + SdT,
= V dp - SdT + TdS + SdT,
= V dp + TdS.
(3.56)
Now write the derivative form of Eq. 3.56, specifying that T is constant. Thus,
a
0H
0S
b = V + Ta b .
0p T
0p T
We already have an expression for (0S >0p) T in terms of p, V, and T (Eq. 3.55), giving
a
0V
0H
b = V - Ta b ,
0p T
0T p
and
dH = Cp dT + c V - T a
0V
b d dp .
0T p
(3.57)
Integrate Eq. 3.57 to give H(p2, T2 ) - H(p1, T1 ) for a finite change from p1, T1 to p2, T2.
Dependence of S on p and T. For S, we have
dS = a
0S
0S
b dT + a b dp .
0T p
0p T
In Eq. 3.14, we used the integrated form of 10S >0T 2 p = C p >T . Then, using Eq. 3.55,
dS =
Cp
T
dT - a
0V
b dp .
0T p
(3.58)
90 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Dependence of U on p and T. For U, the most useful variables are T and V, so write
dU = a
0U
0U
b dT + a b dV .
0T V
0V T
From Eq. 2.47, 10U>0T 2 V = C V . Then to obtain (0U>0V ) T , we first consider a reversible
path with pV work only:
dU = dqrev + dw rev
(First Law, reversible path)
= TdS - pdV .
(3.59)*
Taking the derivative of Eq. 3.59 with respect to V at constant temperature gives
a
0U
0S
b = T a b -p .
0V T
0V T
We may make use of a relation (see questions 27 and 28 of the Problems)
a
0p
0S
b = a b
0V T
0T V
(3.60)
to yield
dU = C V dT + c T a
0p
b - p d dV .
0T V
(3.61)
Eq. 3.60 is a member of a family of equalities between partial derivatives derived by
James Clerk Maxwell. More of these helpful Maxwell relations are reviewed in the
next section.
We can now calculate the change in U, H, S, and G for a system in which p1, V1, and T1
change to p2, V2, and T2, using the integrated forms of Eqs. 3.61, 3.57, 3.58, and 3.35. We
need only be given two of these three variables p, V, and T because we can calculate the
third from the equation of state. Notice that we were able to derive many useful results
from the following differentials,
dU = -pdV + TdS ,
(3.59)
dH = V dp + TdS ,
(3.56)
dG = V dp - SdT ,
(3.35)
dA = -pdV - SdT ,
(3.62)
where questions 27–28 of this chapter guide you through the use of Eq. 3.62 to find
Eq. 3.60.
*We
note again, as we did in the derivation of Eq. 3.35, that all variables in Eq. 3.59 are path independent; thus,
Eq. 3.59 is also valid for paths other than the special one chosen.
The Thermodynamic Square | 91
The Thermodynamic Square
The preceding section in particular suggests that thermodynamic equations exhibit a strong
degree of algebraic symmetry. Indeed, in work attributed originally to Max Born, ‡ definite
symmetrical relationships among the thermodynamic variables of state can be represented
in graphical form by a square (figure 3.13).
Observe these general features of the thermodynamic square:
• Each corner is a state variable (V, T, p, S).
• The middle of each segment is a state energy variable (A,G,H,U ).
• The two arrows will assist in sorting out signs.
There are many ways to recall the square. Memorizing one side (e.g., VAT ) is often
enough to reconstruct the remainder. Beyond serving as a mnemonic, the square can be
especially helpful in first learning thermodynamics. By efficiently summarizing major
relations such as the energy definitions, and the criteria for equilibria, the square can help
you form a more holistic appreciation for the state variables and their relationships with
each other. For example, we just saw in the preceding section that the total differentials
of the four energies were the principle sources for deriving the p and T dependence of
U, H, S, and G.
First, notice that each energy is flanked by its natural variables:
A (V , T ), G(T, p), H(S, p), and U(V , S ) .
The natural variables are defined such that, if taken to be independent (i.e., if held constant), then a simple condition for equilibrium is given by the respective energy:
A V, T Ú 0, GT, p Ú 0,
HS, p Ú 0, and UV, S Ú 0 .
We have already learned how A and G are tests for equilibrium. In addition, U is a test
for equilibrium (conversely spontaneity) if V and S are constant, while H tests for equilibrium for any process where S and p are constant.
V
A
G
U
S
T
H
p
FIGURE 3.13 This simple graphical picture indicates the key relations among thermodynamic potentials
and variables. The basic form is described here, but variations exist. The square is particularly useful for
recalling natural variables and Maxwell’s equations.
‡“General
Principles of Classical Thermodynamics” by H. B. Callen (John Wiley & Sons, Inc., 1960). Callen
reports that the diagram of a thermodynamic square was first presented in a lecture by Max Born in 1929, and was
later reported in the literature by F.O. Koenig, J. Chem. Phys. 3, 29–35 (1935).
92 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Second, the energy definitions are obtained by comparing parallel diagonals.
For example H and U are connected by a line parallel to the arrow between p and V.
By analogy, all energy definitions are found this way, where the directions of the two
arrows can be used as a memory aid to remember for the correct signs:
H = U + pV = G + TS ,
G = A + pV = H - TS ,
A = G - pV = U - TS ,
U = H - pV = A + TS .
Third, the differential forms of the four state energy variables are prescribed by crossing
diagonals, which can be verified from inspection:
dU = -pdV + Tds,
dH = V dp + TdS,
dG = V dp - SdT, dA = - pdV - SdT .
Finally, the Maxwell relations are useful for changing variables in thermodynamic
problems, as we have seen in Eqs. 3.55 and 3.60. Consider
a
0T
0V
b = a b .
0S p
0p S
Notice that V-S-p forms a vertical L shape on the square and T-p-S forms a mirror-image
L shape. When the L patterns are horizontal, a sign change is needed. All four Maxwell
relations are
a
0V
0T
b = a b ,
0S p
0p S
(3.63)
a
0p
0S
b = a b ,
0T V
0V T
(3.60)
a
0S
0V
b = -a b ,
0p T
0T p
(3.55)
a
0p
0T
b = -a b .
0V S
0S V
(3.64)
The Gibbs-Helmholtz Equation
Earlier, we presented a route to finding the temperature dependence of the Gibbs free-energy
change when we have knowledge of the enthalpy change (H),
T2
G(T 2 )
G(T 1 )
H(T )
= -3
dT .
T2
T1
T2
T1
Begin by taking the differential of (G>T ) with respect to T at constant pressure:
a
0(G>T)
0T
b = Ga
p
= -
0(1>T)
0T
1
0G
b + a ba b ,
T
0T p
p
G
1
0G
+ a ba b .
T
0T p
T2
(3.40)
Summary | 93
From Eq. 3.36, at constant pressure 10G>0T2 p = -S. Thus,
a
0(G>T )
0T
G
S
+ b
T
T2
b = -a
p
= -a
=
H - TS
TS
+ 2b
T2
T
-H
T2
(3.65)
Multiplying both sides of Eq. 3.65 by T 2 and recalling that dT >T 2 = -d(1>T ), we obtain
a
0(G>T)
0(1>T)
b = H.
(3.65a)
p
For a chemical reaction n AA + n BB h n CC + n DD, it follows directly from Eqs. 3.65
and 3.65 that
T2
rG(T 2 )
rG(T 1 )
rH(T )
= -3
dT ,
T2
T1
T2
T1
(3.40)
1>T2
1
rH da b ,
T
1>T1
= 3
(3.40a)
where rH is the enthalpy change of the reaction
rH = n CHm,C + n DHm,D - n AHm,A - n BHm.B .
If the change in H is small over the temperature range, Eq. 3.40a becomes
rG(T2)
rG(T1)
1
1
= rH c
d.
T2
T1
T2
T1
(3.41)
Summary
State Variables (SI Units in Bold)
Name
Symbol
Units
Definition
Entropy
S
J K , caldeg , eu
dS K dqrev >T
Gibbs free energy
G
J, erg, cal
G K H - TS
Helmholtz free energy
A
J, erg, cal
A K E - TS
-1
-1
94 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Unit Conversions
Entropy
1 cal deg - 1 = 1 eu = 4.184 J K - 1
Energy
1 cal = 4.184 J = 4.184 * 107 erg
General Equations
Efficiency for a Carnot-Cycle Heat Engine
efficiency = 1 -
Tcold
Thot
(3.5)
Second Law of Thermodynamics
S (system) + S (surroundings) Ú 0
(3.7)
S (isolated system) Ú 0
(3.8)
The equal sign applies for a reversible process and the other sign for an irreversible one.
Third Law of Thermodynamics
S A(0 K ) ‚ 0 (A
a pure, perfect crystal)
(3.13)
Changes in Entropy and Gibbs Free Energy
Capacity to do Nonexpansion Work
G = wother
(3.28)
The decrease in the Gibbs free energy ( - G) is the maximal nonexpansion work ( -wother ) a system can do on the
surroundings at constant pressure and temperature.
Spontaneous Reactions at Constant T and p
G 6 0
(spontaneous process)
(3.29)
G 7 0
(not a spontaneous process)
(3.30)
G = 0
(system at equilibrium)
(3.31)
Pressure or Temperature Changes
Solids and Liquids. We assume that the volume of a solid or liquid is independent of T and p and that Cp and CV are equal and do
not depend on T and p:
S (p2, T 2 ) - S (p1, T 1 ) = nCp, m ln
T2
T2
= nCV, m ln
.
T1
T1
Gases. We assume that gases can be approximated by the ideal gas equation, that Cp, m and CV, m are independent of T, and
Cp, m = CV, m + R .
p2
V2
= nR ln
p1
V1
(constant T)
(3.21)
p2
V2
= - nRT ln
p1
V1
(constant T)
(3.44)
S (p2 ) - S (p1 ) = - nR ln
G(p2 ) - G(p 1 ) = nR ln
S(T2 ) - S(T1 ) = nC p, m ln
T2
T1
(constant p; C p, m independent of T)
(3.15)
Suggested Reading | 95
Phase Changes at Constant T and p
G = 0
S =
(at equilibrium)
H
(at equilibrium)
T
Chemical Reactions—Changes in Entropy
T
2
dT
rS (T2 ) = rS (T1 ) + 3 rCp
T
(3.18)
T1
rC p = n CC m,p, C + n DC m,p, D - n AC m,p, A - n BC m,p, B
Chemical Reactions—Changes in Gibbs Free Energy
rG(T2 ) - rG(T1 ) = - rS(T2 - T1 )
( rS independent of T)
rG(T1)
rG(T2)
1
1
= rH c
d
T2
T1
T2
T1
( rH independent of T)
rG(p2 ) - rG(p1 ) = rV (p2 - p1 ) (solids, liquids only; rV independent of p)
(3.45)
rV = n CVm,C + n DVm,D - n AVm,A - n BVm,B
p2
rG(p2 ) - rG(p1 ) = ( r n) # RT ln ˚(at least one gaseous product or reactant)
p1
r n = ¢
(3.46)
number of moles
number of moles
≤ - ¢
≤
of gaseous products
of gaseous reactants
References
In addition to the resources listed in chapter 2, see the following:
1. Atkins, P. W. 1994. The Second Law, Scientific American
Library. New York: W. H. Freeman.
2. Craig, N. C. 1992. Entropy Analysis: An Introduction to
Chemical Thermodynamics. New York: VCH Publishers.
3. Morowitz, H. J. 1996. Entropy and the Magic Flute.
Oxford: Oxford University Press.
4. Saenger, W. 1984. Principles of Nucleic Acid Structure.
New York: Springer-Verlag.
5. Von Baeyer, H. C. 1998. Maxwell’s Demon: Why Warmth
Disperses and Time Passes. New York: Random House.
Suggested Reading
Feynman, R. P., R. B. Leighton, and M. Sands. 1963. The Feynman
Lectures on Physics. Reading, MA: Addison-Wesley.
Van Ness, H. C., 1969. Understanding Thermodynamics.
New York: McGraw-Hill.
Frautschi, S. 1982. Entropy in an Expanding Universe. Science
217:593–599.
Wilson, S. S. 1981. Sadi Carnot. Sci. Am. 245 (August):134–145.
McIver, R. T., Jr. 1980. Chemical Reactions Without Solvation.
Sci. Am. 243 (Nov):186–196.
96 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
Problems
1. One mole of an ideal monatomic gas is expanded from an
initial state at 3 bar and 450 K to a final state at 2 bar and
250 K.
a. Choose two different paths for this expansion, specify
them carefully, and calculate w and q for each path.
b. Calculate U and S for each path.
2. No example is known of an organism evolving a process or
structure resembling a heat engine. Basing your answer on
the properties of a Carnot cycle, propose reasons why a heat
engine might never arise in natural evolution.
3. The temperature of the heat reservoirs for a Carnot-cycle
(reversible) engine are Thot = 1200 K and Tcold = 300 K. The
efficiency of the engine is calculated to be 0.75, from Eq. 3.5.
a. If w = - 100 kJ, calculate q1 and q3. Explain the meaning of the signs of w, q1, and q3.
b. The same engine can be operated in the reverse order. If
w = + 100 kJ, calculate q1 and q3. Explain the meaning
of the signs.
c. Suppose it were possible to have an engine with a higher
efficiency, say, 0.80. If w = - 100 kJ, calculate q 1 and
q 3 . The superscript denotes quantities for this engine.
d. If we use all the work done by the engine in part (c)
to drive the heat pump in part (b), calculate (q1 + q 1 )
and (q3 + q 3 ) the amount of heat transferred. Explain
the signs. Note that the net effect of the combination of
the two engines is to allow a “spontaneous” transfer of
heat from a cooler reservoir to a hotter reservoir, which
should not happen.
e. Show that if it were possible to have a reversible engine
with a lower efficiency operating between the same two
temperatures, heat could also be transferred spontaneously
from the cooler reservoir to the hotter reservoir. (Hint:
Operate the engine with lower efficiency as a heat pump.)
4. The Second Law of Thermodynamics states that entropy
increases for spontaneous processes and that an increase
in entropy is associated with transitions from ordered to
disordered states. Living organisms, even the simplest
bacteria growing in cultures, appear to violate the Second
Law because they grow and proliferate spontaneously.
They convert simple chemical substances into the highly
organized structure of their descendants. Write a critical
evaluation of the proposition that living organisms contradict the Second Law. Be sure to state your conclusion
clearly and to present detailed, reasoned arguments to
support it.
5. Consider the process where nA mol of gas A initially at 1 bar
pressure mix with nB mol of gas B also at 1 bar to form 1
mol of a uniform mixture of A and B at a final total pressure
of 1 bar, and all at constant temperature T. Assume that all
gases behave ideally.
a. Show that the entropy change, mixS m, for this process
is given by mixS m = - x ARln x A - x BRln x B, where xA
and xB are the mole fractions of A and B, respectively.
(Hint: Entropy changes are additive for components A
and B; that is, mixS m = S m,A + S m,B.)
b. What can you say about the value (especially the sign) of
mixS m and how does this correlate with the Second Law?
c. Derive an expression for mixG m for the conditions of this
problem and comment on values (and signs) of its terms.
6. A system consisting of 0.20 mol of supercooled gaseous
water (95°C, 1 bar) partially condenses to liquid water; the
process is done adiabatically at constant pressure.
a. At equilibrium, what is the temperature of the system?
b. How many moles of gaseous water have condensed?
c. Calculate the entropy change for the process.
7. Explain which of the following are correct statements of the
Second Law of Thermodynamics and which are not.
a. All reversible heat engines have the same efficiency.
b. An engine connected to one heat reservoir can do work
on the surroundings.
c. An engine connected to two heat reservoirs at different
temperatures can do work on the surroundings.
d. The entropy of the universe is conserved.
8. The technology to build microscopic motors is now available. Can microscopic motors take advantage of fluctuations
to beat the Second Law and do work at constant temperature?
Consider the following fluctuation motor:
Ratchet
Flea
Paddle
Assume that the motor is so small that one gas molecule hitting a paddle can make it turn. The motor is placed in a box (at
constant temperature) with a few gas molecules. A possible
explanation of how the motor works is the following: gas
molecules randomly hitting the paddles will sometimes make
the shaft turn clockwise and sometimes make it turn counter
clockwise. The flea will randomly be raised and lowered by
the fluctuations, if we don’t consider the ratchet. However,
the ratchet is held down by a spring, so it allows only clockwise rotations. Therefore, the flea will be raised because the
ratchet has converted the random fluctuations into a directed
rotation. Work is done (the mass of the flea times the acceleration of gravity times the distance it is raised) at constant
temperature. The energy comes from the thermal energy of
the gas, so the First Law is not violated.
a. Will the motor work? Discuss the action of the ratchet in
converting random fluctuations into a clockwise motion.
Problems | 97
9.
10.
11.
12.
b. If the motor will not work as described, what simple change
will allow it to work? Clearly, if the motor will work as
described, you need not answer part (b). This problem is
discussed in The Feynman Lectures on Physics, Vol. 1.
A flask of liquid nutrient solution inoculated with a small
sample of bacteria is maintained for several days in a thermostat
until the bacteria have multiplied 1000-fold. Determine whether
each of ⌬T, w, q, ⌬E, ⌬H, and ⌬G is greater than, equal to, or
less than zero for the system described (the system is in italic
type). Describe your reasoning and state explicitly any reasonable assumptions or approximations that you need to make.
Remarkably, a catalytic system that uses light energy to
hydrolyze water into hydrogen H2(g) and O(g) has been
devised (Science 334, 2011: 645–648) to capture and efficiently store solar energy. Hydrogen is “clean burning” in
that it yields only H2O. In this problem you will contrast the
combustion of 1 kg H2(g) to that of 1 kg of n-octane.
a. Calculate the enthalpy, entropy, and free-energy change
on burning 1 kg of H2 (g) to H2O(l) at 25°C and 1 bar.
b. Calculate the enthalpy, entropy, and free-energy change
on burning 1 kg of n-octane(g) to H2O(l) and CO2 (g)
at 25°C and 1 bar.
Use data from the tables A.5–7 in the appendix to answer the
following questions.
a. A friend wants to sell you a catalyst that allows
benzene(g) to be formed by passing H2 (g) over carbon
(graphite) at 25°C and 1 bar. Should you buy? Why?
b. Is the reaction 2NO(g) + O2 (g) S 2NO2 (g) spontaneous at 25°C and 1 bar?
c. Is the reaction to form solid alanine, CH3CHNH2COOH,
and liquid H2O from CH4 (g), NH3 (g), and O2 (g) spontaneous at 25°C and 1 bar?
Under physiological conditions, the true free-energy change
for the hydrolysis of ATP to ADP is influenced by a number
of factors including the presence of an associated Mg2+ ion
MgATP 2 - + H2O S MgADP - + Pi
and is about -50 kJ mol-1. Suppose an individual maintains
a constant amount of 100 g MgATP2-.
N
O
P
O
O-
H2N
N
O
P
O
N
N
O
OH
O
P
O-
O
O-
c. Evaluate the feasibility for using MgATP2- to store
600 kJ.
13. For the following statements, choose the word or words
inside the parentheses that serve to make a correct statement. Each statement has at least one, and may have more
than one, correct answer.
a. According to the Second Law of Thermodynamics, a
spontaneous process, such as a balloon filled with hot
gas cooling to the surroundings at constant pressure,
will always occur (adiabatically, reversibly, irreversibly, without work done).
b. Associated with such a process, there is always an
increase in entropy of (the system, the surroundings,
the system plus the surroundings, none of these).
c. For the example given, the heat gained by the surroundings is just equal to the negative of the (internal energy
change, enthalpy change, entropy change, Gibbs freeenergy change) of the system.
d. To return a system to its initial state requires from the
surroundings a change of entropy whose magnitude is
(greater than, equal to, less than) that which is gained
during the spontaneous process.
14. Consider the reaction
1
C2H5OH(l) h C2H6 (g) + O2 (g).
2
Calculate ⌬ rH⬚(298 K), ⌬ rS⬚(298 K) and ⌬ rG ⬚(298 K).
Estimate ⌬U ⬚(298 K). State what further data would be
needed to obtain ⌬ rH⬚ at 500°C and 1 bar.
15. The shells of marine organisms contain CaCO3, largely in
the crystalline form known as calcite. There is a second
crystalline form of CaCO3 known as aragonite.
a. Based on the thermodynamic and physical properties
given for these two crystalline forms, would you expect
calcite in nature to convert spontaneously to aragonite
given sufficient time? Justify your answer.
b. Will the conversion proposed in part (a) be favored or
opposed by increasing the pressure? Explain.
c. What pressure should be just sufficient to make this
conversion spontaneous at 25°C?
d. Will increasing the temperature favor the conversion?
Explain.
O-
Mg2+
HO
MgATP2a. How much free energy is represented by this amount of
MgATP2-?
b. ATP is not as much for energy storage as for energy
transduction; ATP is hydrolyzed and restored over and
over. If an individual turns over 6000 kJ in energy via
ATP hydrolysis, how many times per day is this 100 g
of ATP replenished?
Properties at 298 K
CaCO3
(calcite)
⌬ fH ⬚m(kJ mol - 1)
-1206.87
-1207.04
⌬ fG ⬚m(kJ mol - 1)
S ⬚m(J K
-1
C p,m(J K
-1
CaCO3
(aragonite)
-1128.79
-1127.71
-1
mol )
92.88
88.70
-1
81.88
81.25
mol )
-3
Density (g cm )
2.710
2.930
16. Consider the reaction that converts pyruvic acid
(CH3COCOOH) into gaseous acetaldehyde (CH3CHO) and
gaseous CO2, which is catalyzed in aqueous solution by the
enzyme pyruvate decarboxylase. Assume ideal gas behavior
for CO2.
98 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
a. Calculate rG(298 K) for this reaction.
b. Calculate rG(298 K) for this reaction at 100 bar. State
any important assumptions needed in addition to ideal
gas behavior.
17. For the following statements, choose the word or words
inside the parentheses that serve(s) to make a correct statement. More than one answer may be correct. At 100°C, the
equilibrium vapor pressure of water is 1 bar. Consider the
process where 1 mol of water vapor at 1 bar pressure is
reversibly condensed to liquid water at 100°C by slowly
removing heat into the surroundings.
a. In this process, the entropy of the system will (increase,
remain unchanged, decrease).
b. The entropy of the universe will (increase, remain
unchanged, decrease).
c. Since the condensation process occurs at constant temperature and pressure, the accompanying free-energy
change for the system will be (positive, zero, negative).
d. In practice, this process cannot be carried out reversibly.
For the real process, compared with the ideal reversible
one, different values will be observed for the (entropy
change of the system, entropy change of the surroundings, entropy change of the universe, free-energy change
of the system).
At a lower temperature of 90°C, the equilibrium vapor pressure of water is only 0.702 bar.
e. If 1 mol of water is condensed reversibly at this temperature and pressure, the entropy of the system will
(increase, remain unchanged, decrease).
f. The entropy of the universe will (increase, remain
unchanged, decrease).
g. The free energy of the system will (increase, remain
unchanged, decrease).
h. Since the molar heat capacity of water vapor is about
one-half that of the liquid, the entropy change upon condensation at 90°C will be (more negative than, the same
as, more positive than) that at 100°C.
18. Consider a system that undergoes a phase change from
phase a to phase b at equilibrium temperature T and
equilibrium pressure p. At these equilibrium conditions,
the heat absorbed per mole of material undergoing this
transition is q, and there is a molar volume change of
V,m. The molar heat capacities at constant pressure for
the a and b phases are Cp,m,a and Cp,m, b ; they are independent of temperature.
a. Evaluate w, U, H, S , and G for converting
1 mol of the system from phase a to phase b at the
equilibrium T and p. Your answer should be in terms
of q, V,m, etc.
b. Evaluate H and S at temperature T* different from
T, but at the same pressure p.
c. The denaturation transition of a globular protein can
be approximated as a phase change a to b. Calculate G, H , and S for a heat of transition
of q = 638 kJ mol - 1 at T = 70C p = 1 bar;
Cp,m,a - Cp,m,b = 8.37 kJ mol - 1 K - 1.
d. Calculate G, H, and S for the same transition at
37°C and 1 bar. Assume that Cp,m,a - Cp,m,b is independent of temperature.
e. If Vm for the transition is + 3 mL mol - 1, will the
equilibrium transition temperature T be increased or
decreased at a pressure of 1000 bar? Explain.
f. Write a thermodynamic cycle that would allow you to
calculate the equilibrium transition temperature corresponding to a pressure of 1000 bar. Use the cycle to write
an equation that in principle could be solved to calculate
T at 1000 bar.
19. Table 3.2 gives thermodynamic data for transitions from
ordered, helical conformations of polypeptides and polynucleotides to disordered states. The enthalpy changes and
entropy changes are positive as expected; heat is absorbed
and entropy is gained. However, when the synthetic polypeptide polybenzyl-l-glutamate undergoes a transition from an
ordered helix to a disordered coil in an ethylene dichloride–
dichloroacetic acid solvent at 39°C, H = - 4.0 kJ (per
mole of amide) and S = - 12 J K - 1 (per mole of amide).
a. Give a possible explanation for the experimental result
that heat is released and entropy is decreased for this
transition. Does increasing the temperature favor the
helix-coil transition?
b. Is the reaction spontaneous at 39°C? What thermodynamic criterion did you use to reach this conclusion?
c. At what temperature (°C) will the helix-coil reaction be
reversible? This temperature is often called the “melting” temperature of the helix. Assume that H and S
are independent of temperature.
d. Can a reaction occur in an isolated system that leads to a
decrease in the entropy of the system? If it can, give an
example; if it cannot, state why not.
20. The following thermodynamic data at 25°C have been tabulated for the gas-phase reaction shown as follows (dotted
lines represent hydrogen bonds):
O. . . H
O
2 H
C
H
O
O
C
C
H
O
H
H. . . O
f H (kJ mol-1)
Sm
(J K-1 mol-1)
HCOOH(g)
-362.63
251.0
(HCOOH)2(g)
-785.34
347.7
a. Write the equation for the standard heat of formation at
25C of HCOOH(g) (specify the pressure and phase for
each component).
b. Calculate rH , rS, and rG for the gas-phase dimerization at 298 K. Is the formation of dimer from monomers spontaneous under these conditions?
c. Calculate the enthalpy change per hydrogen bond formed
in the gas phase. Why is not a similar calculation useful
to estimate the entropy or free energy of hydrogen-bond
formation?
Problems | 99
21. An electrochemical battery is used to provide 1 milliwatt
(mW) of power for a (small) light. The chemical reaction in
the battery is
22.
23.
24.
25.
1
3
N (g) + H2 (g) h NH3 (g) .
2 2
2
a. What is the free-energy change for the reaction at 25C,
1 bar?
b. Calculate the free-energy change for the reaction at 50C,
1 bar. State any assumptions made in the calculation.
c. The limiting reactant in the battery is 100 g of H2.
Calculate the maximum length of time (seconds) the
light can operate at 25C.
For the combustion of hydrogen gas with oxygen gas to yield
water vapor (note data in appendices and chapter 2 for this
question).
a. calculate the standard entropy change (1 bar) at 298 K.
b. estimate the standard entropy change at 50C and state
any assumptions you need to make.
Consider a fertilized hen egg in an incubator—a constant
temperature and pressure environment. In a few weeks, the
egg will hatch into a chick.
a. The egg is chosen as the system. Is the egg an open, an
isolated, or a closed system? Define an isolated system.
b. In the fertilized egg, hen proteins are formed into a
highly ordered chick. Does the entropy of the system
increase or decrease? Does this violate the Second Law
of Thermodynamics? Explain in two or three sentences
why the development of the chick is or is not consistent
with the Second Law.
c. What happens to the energy of the system as the chick
develops? What forms of energy contribute to the change
in energy (if any) of the system?
d. Does the free energy of the system increase, decrease, or
remain the same? How do you know?
Calculate the entropy change when
a. two moles of H2O(g) are cooled irreversibly at constant
p from 120°C to 100°C.
b. one mole of H2O(g) is expanded at constant pressure of
2 bar from an original volume of 20 L to a final volume
of 25 L. You can consider the gas to be ideal.
c. one hundred grams of H2O(s) at -10°C and 1 bar are
heated to H2O(l) at +10°C and 1 bar.
Critically evaluate the following situations. Some of the proposals or interpretations are reasonable, while others violate
very basic principles of thermodynamics.
a. It is commonly known that one can supercool water and
maintain it as a liquid at temperatures as low as -10°C.
If a sample of supercooled liquid water is isolated in
a closed, thermally insulated container, after a time it
spontaneously changes to a mixture of ice and water at
0°C. Thus, it increases its temperature spontaneously
with no addition of heat from outside; furthermore, some
low-entropy ice is produced.
b. An inventor proposed a new scheme for heating
buildings in the winter in arctic climates. Since freezing
c.
d.
e.
f.
g.
h.
temperatures reach only a few feet down into the soil, he
proposes digging a well and immersing a coil of copper
tubing in the water. He will then connect the ends of the
coil to the radiators in the building and use a heat pump
to transfer heat into the building. What would be your
advice to an attorney who is assigned to evaluate this
patent application? Base your evaluation on the relevant
thermodynamics.
On a hot summer’s day, your laboratory partner proposes
opening the door of the lab refrigerator to cool off the
room.
A sample of air is separated from a large evacuated chamber, and the entire system is isolated from the surroundings.
A small pinhole between the two chambers is opened, and
roughly half the gas is allowed to effuse into the second
chamber before the pinhole is closed. Because nitrogen
effuses faster than oxygen, the gas in the second chamber
is richer in nitrogen, and the gas in the first chamber is
richer in oxygen than the original air. The gases have thus
spontaneously unmixed (at least partially), and this is held
to be a violation of the Second Law of Thermodynamics.
Supercooled water at -10°C has a higher entropy than
does an equal amount of ice at -10°C. Therefore, supercooled water cannot go spontaneously to ice at the same
temperature in an isolated system.
A volume of an aqueous solution of hydrogen peroxide
is placed in a cylinder and covered with a tight-fitting
piston. A small amount of the enzyme catalase is placed
on a probe and inserted through an opening in the base of
the cylinder. The catalase catalyzes the decomposition of
the hydrogen peroxide, and the oxygen gas formed serves
to raise the piston. The catalase is then withdrawn and
the hydrogen peroxide is re-formed, causing the piston
to return to its initial position. The piston is connected
to an engine, and net work is obtained indefinitely by
cycles of simply inserting and withdrawing the catalase.
High temperatures can be achieved in practice by focusing the rays of the Sun on a small object, using a large
parabolic reflector, such as is used in astronomical telescopes. Since the energy gathered increases as the square
of the diameter of the reflector, it should be possible to
produce temperatures higher than those of the Sun by
using a large-enough reflector.
The maximum efficiency of a steam engine can be
calculated using the Second Law of Thermodynamics. If
it operates between the boiling point of water and room
temperature, the maximum efficiency is about
373 - 293
80
=
= 0.215 or 21.5% .
373
373
Photosynthesis by green plants occurs almost entirely at
ambient temperatures, yet publications report theoretical
limits as high as 85% of the fraction of the light energy
absorbed can be converted into chemical energy (free
energy or net work). Clearly, such estimates must be
wrong, or else the Second Law cannot apply.
100 Chapter 3 |
The Second Law: The Entropy of the Universe Increases
26. For the following statements, choose the word or words
inside the parentheses that serve to make a correct statement.
Each statement has at least one and may have more than one
correct answer.
a. For a sample of an ideal gas, the product pV remains
constant as long as the (temperature, pressure, volume,
internal energy) is held constant.
b. The internal energy of an ideal gas is a function of only
the (volume, pressure, temperature).
c. The Second Law of Thermodynamics states that the
entropy of an isolated system always (increases, remains
constant, decreases) during a spontaneous process.
d. When a sample of liquid is converted reversibly to its
vapor at its normal boiling point, (q, w, ⌬p, ⌬V, ⌬T, ⌬U,
⌬H, ⌬S, ⌬G, none of these) is equal to zero for the system.
e. If the liquid is permitted to vaporize isothermally and
completely into a previously evacuated chamber that is
just large enough to hold the vapor at 1 bar pressure, then
(q, w, ⌬U, ⌬H, ⌬S, ⌬G) will be smaller in magnitude
than for the reversible vaporization.
27. Starting with the definition of enthalpy, H = U + pV , and
additional relations based on the First and Second Laws of
Thermodynamics, derive the following:
a. dH as a function of T, S, V, and p
b. Equations for (0H>0p) S and (0H>0S ) p
c. The equation (0T >0p) S ⫽ (0V >0S ) p
d. The result of part (a) starting from Eq. 3.35 and the
definition G K H - TS
28. Starting with the definition of the Helmholtz energy,
A = U@TS, and following the example of question 27,
derive the Maxwell relation in Eq. 3.60.
29. Earth’s atmosphere behaves as if it is approximately isentropic—the molar entropy of air is a constant independent
of altitude up to about 10 km. It is well-known that pressure and temperature vary with altitude. Using the isentropic
model, calculate the temperature of the atmosphere 10 km
above Earth, where the pressure is found to be 210 torr. The
temperature and pressure at the surface of Earth (sea level)
are 25°C and 760 torr, respectively. Assume that air behaves
7
like an ideal gas with C p = R . You may ignore gravitation
2
influences. (The temperature measured during rocket flights
above New Mexico is -50°C at 10 km above Earth.)
30. The temperature of a typical laboratory freezer unit is -20°C.
If liquid water in a completely filled, closed container is
placed in the freezer, estimate the maximum pressure developed in the container at equilibrium. The enthalpy of fusion
of water may be taken as 333.4 kJ kg - 1, independent of
temperature and pressure, and the densities of ice and liquid
water at -20°C are 0.9172 and 1.00 g cm - 3, respectively.
Chapter 4
Free Energy and Chemical
Equilibria
“In every one of us there is a living process of combustion going on very similar to
that of a candle, and I must try to make that plain to you.”
—Michael Faraday, 1860
Constant Natural Variables
S,p
S,V
T,V
Concepts
All of the reactions in cells are governed by changes in the Gibbs free energy. Reactions
that make vital biochemicals, contract muscles, and transport molecules must all satisfy a
balance sheet in which the currency is free energy. In this chapter, you will learn how the
amounts of products and reactants contribute to the Gibbs free energy changes for systems
in equilibrium (the free energy change is zero) and for systems undergoing a spontaneous
change towards equilibrium (the free energy change is negative).
When a very small amount of a substance is added to a system at constant temperature and pressure without changing any of the other composition variables, the change in
G per mole of the substance added is the partial molar Gibbs free energy of the added
substance (Eq. 3.51). We will see that ⌬ rG of a reaction at constant temperature and pressure is determined by the partial molar Gibbs free energies of the reactants and products.
When a reaction achieves equilibrium at constant temperature and pressure, the system
has exactly that composition which minimizes the Gibbs free energy.
The height of a boulder determines its potential energy in Earth’s gravitational field.
It may roll to a lower position with a change in potential energy determined by the height
difference that it traversed. Similarly, the partial molar Gibbs free energy of a species is
its chemical potential ( μ) relative to that of other species which are accessible by reaction
pathways. If the reactants are at a higher overall chemical potential than the products, then
the reaction will proceed spontaneously in the forward direction. The chemical potential
is defined as the partial molar Gibbs free energy for reactions occurring at the common
biological conditions of constant temperature and pressure (figure 4.1).
The chemical potential m of a substance is invariably measured relative to the standard
chemical potential m⬚ of the same substance under a chosen set of conditions, its standard
state. By choosing appropriate standard states, m - m⬚ of any substance can be expressed
as a function of more familiar variables: temperature, partial pressure if it is a gas, and
concentration in molarity or molality if it is a solute are some common examples. We
will see that the equilibrium composition of a reaction is specified by the standard
chemical potentials of the participants in the reaction. Also, quantitative descriptions of
Le Châtelier’s principle will follow naturally by understanding how perturbing a system
changes the chemical potentials of the species.
T,p
⎛ ∂G ⎞
⎟
⎝ ∂nA ⎠ T, p, nj ≠ nA
μA ≡ ⎜
⎛ ∂A ⎞
⎟
⎝ ∂nA ⎠ T, V, nj ≠ nA
μA ≡ ⎜
⎛ ∂U ⎞
⎟
⎝ ∂nA ⎠S, V, nj ≠ nA
μA ≡ ⎜
⎛ ∂H ⎞
⎟
⎝ ∂nA ⎠ S, p, nj ≠ nA
μA ≡ ⎜
FIGURE 4.1 Each of the four
energy variables corresponds
to a set of ‘natural variables’.
(chapter 3). By holding constant
a set of natural variables, the
chemical potential (μ) is defined
as the partial molar quantity of
the corresponding energy variable. For example, for chemical
reactions at constant temperature and volume, the chemical
potential of a substance is given
by the partial molar Helmholtz
free energy, A.
1
101
102 Chapter 4 |
Free Energy and Chemical Equilibria
Applications
Thermodynamic analysis of reactions at equilibrium finds broad applications in chemical
and biological systems. Although it is often difficult to describe a complex biological
system in terms of equilibrium thermodynamics, we can usually separate a process into
its components. Important components that can be analyzed thermodynamically include
(1) metabolic reactions in which chemical bonds are broken and new ones formed,
(2) dissociation of H + from acidic compounds and binding of H + to bases, (3) oxidation–
reduction reactions in which electron transfer occurs, (4) interactions involving the aqueous
medium in which metabolites and ionic species occur in the cytoplasm or other biological
fluids, and (5) the assembly and disassembly of membranes and other multicomponent
cellular structures. Applications of thermodynamic analysis to several cases of biological
interest are illustrated in this chapter.
Mathematics note. We have
introduced partial derivatives
in chapters 2 and 3 to describe
changes in thermodynamic
quantities when one system
parameter is altered. Partial
derivatives and total differentials are also reviewed in the
mathematics appendix.
Partial Molar Gibbs Energy
Chemical Potential
Reactions in biological systems usually occur at constant temperature and pressure. The
chemical potential of substance A,
mA K a
0G
b
,
0nA T, p, nj ⬆ nA
(3.51)
is defined as the partial molar Gibbs free energy of A under conditions of constant T and
p, where the notation nj ⬆ nA specifies that the molar quantities of all other chemicals
present in the system is also constant (figure 4.1). Thus, the chemical potential of A, mA,
describes how much the free energy of a system changes when only the number of moles
of A changes. The chemical potential of a system of pure A is the free energy per mole of
A (sidebar: Gm,A). In general, the chemical potential of one species will depend on what
else is in the system, and we must specify the composition of the system by specifying
the values of nj.
Consider first an open system made of substances A, B, C, and D. The free energy G
of the system is a function of the temperature T, pressure p, and the number of moles of
all species nA, nB, nC, and nD:
G = G(T, p, nA, nB, nC, nD)
(4.1)
The total differential (see Eq. 3.52 also) of Eq. 4.1 is
dG = a
G=
n=nA
G=GA
G
n=0
G=0
GA
nA
μA dn
dG
=
0
GA - 0
=
μA
0
Gm,A = μA
= -SdT + Vdp + mAdnA + mBdnB + mCdnC + mDdnD ,
(4.2)
where the substitutions follow from Eqs. 3.36, 3.42, and 3.51. At constant temperature and
pressure, Eq. 4.2 becomes
dG = mAdnA + mBdnB + mCdnC + mDdnD .
(4.3)
(Eq. 3.51)
nA
The Sum Rule for Partial Molar Quantities
0
Imagine that we start with a system containing very small amounts of substances A, B, C,
and D and gradually add materials to it, at constant temperature and pressure, in infinitesimal amounts (dnA, dnB, dnC, and dnD) that are always in proportion to their amounts in
the mixture. In this way, the composition of the system, and hence the chemical potentials
of the components, remain the same as nA, nB, nC, and nD increase. Because the chemical
GA = μA nA
GA/nA = μA
0G
0G
0G
0G
b
dnA + (...) + a
b
dnD ,
b dT + a b dp + a
0T p,nj
0p T,nj
0nA T, p,nj ⬆ A
0nD T, p, nj ⬆ D
dn
Partial Molar Gibbs Energy | 103
potentials are kept constant during the additions, Eq. 4.3 is readily integrated by assuming
the initial state is negligibly close to zero Gibbs energy.
G
nA
nB
nC
(4.4)
G = nAmA + nBmB + nCmC + nDmD (constant T, p)
(4.5)
0
0
0
C + 4D
d = -0.1
4.0
nD
3 dG = mA 3 dnA + mB 3 dnB + mC 3 dnC + mD 3 dnD
0
A + 3B
D
0
or
B
moles
Equation 4.5 is termed the sum rule for the partial molar Gibbs free energies. From the
way we arrived at this equation, we see that similar sum rules can be obtained for other
partial molar quantities. The total volume V of a system containing components A, B, C,
and D, for example, is equal to nAyA + nByB + nCyC + nDyD, where yA, yB, yC, and yD
are the partial molal volumes of the components.
3.0
2.0
-0.3
+0.4
Directionality of Chemical Reaction
1.0
Now consider a closed system consisting of four components A, B, C, and D that are
undergoing a reversible chemical reaction
-0.1
+0.1
aA + bB N cC + dD .
We still have the total differential
dG = -SdT + Vdp + mAdnA + mBdnB + mCdnC + mDdnD .
0
-dnC
-dnD
dnA
dnB
=
=
=
,
a
c
b
d
allowing us to rewrite Eq. 4.2 as
dG = -SdT + Vdp + (cmC + dmD - amA - bmB) dj .
(4.6)
Here, dj is a measure of the extent of the reaction. Positive dj means dnA and dnB are
negative and dnC and dnD are positive: the amounts of A and B are decreasing, and those
of C and D are increasing and the reaction is proceeding from left to right. Conversely, a
negative dj means the reaction is proceeding from right to left. For reactions occurring at
constant temperature and pressure, dT = 0 and dp = 0, giving
dG = (cmC + dmD - amA - bmB) dj .
(4.7)
If the reaction spontaneously proceeds from left to right at constant temperature and pressure, we know that dG must be negative. Since dj is positive if the reaction goes from left to
right, then 3 (cmC + dmD) - (amA + bmB) 4 must be negative. In other words, the reaction
proceeds in the forward direction if the sum of the chemical potentials of the reactants—
each weighted according to their stoichiometric coefficients—is greater than the sum of
the chemical potentials of the products similarly weighted. Thus, the chemical potentials
are so named because they determine the directionality of chemical reactions at constant
temperature and pressure. If the system is at equilibrium then Eq. 4.7 must equal zero:
3 (cmC + dmD) - (amA + bmB) 4 = 0
Observe from Eq. 4.7 that 3 (cmC + dmD) - (amA + bmB) 4 is equal to 1 0G
0j 2 T, p the
change in the Gibbs free energy per mole of the reaction at constant temperature and pressure, which is denoted ⌬ rG:
⌬ rG K a
0G
b = 3 1cmC + dmD 2 - 1amA + bmB 2 4
0j T, p
Extent of Reaction,
1.0
(4.2)
However, the changes dnA, dnB, dnC, and dnD are now related by the stoichiometry of the
reaction since the system is closed. We can then define
-dj K
C
A
(4.8)
FIGURE 4.2 An example of the
relationship between a change
in the extent of reaction (dj)
and the corresponding changes
in the individual amounts of the
reactants and products.
104 Chapter 4 |
Free Energy and Chemical Equilibria
G
G
products
G
Two points should be clarified. First, any quantity reported per mole of the reaction
would depend on how the reaction is written, as we discussed in chapter 2 (Heat Effects
of Chemical Reactions section, footnote on p. 41). For the oxidation of glucose to carbon
dioxide and water, for example, the reaction could be written
C6H12O6 + 6O2 S 6CO2 + 6H2O ,
reactants
in which “per mole of the reaction” refers to the conversion of 1 mol of glucose and 6 mol
of oxygen to the formation of 6 mol of CO2 and 6 mol of water. If we write
G
T,p
forward
spont.
0
G
<0
G
=0
T,p
equilibrium
1
C H O + O2 S CO2 + H2O ,
6 6 12 6
>0
T,p
reverse
spont.
Extent of Reaction,
1.0
FIGURE 4.3 The sum over the
Gibbs free energy of all reactants
and products reaches a minimum
at equilibrium. The slope at any
composition of the closed system
(Eq. 4.8) is the free energy change
for one mole of reaction under
those conditions.
then “per mole of the reaction” refers to the conversion of 1>6 mol of glucose and 1 mol of
oxygen to the formation of 1 mol of carbon dioxide and water. Notice in Eq. 4.8 that ⌬ rG
is defined per mole of the reaction, as it is dependent on the stoichiometric coefficients.
Second, it is important to discuss the meaning of ⌬ rG in Eq. 4.8 as the change in the
Gibbs free energy per mole of the reaction for an infinitesimal change dj. For an infinitesimal change dj, the chemical composition is unchanged and so the chemical potentials of
the reacting species are constant over dj. However, the chemical composition and therefore
the chemical potentials continuously change as the reaction proceeds through finite steps.
That is, ⌬ rG is an instantaneous value that is true only for a given specific composition. As
the reaction advances to new compositions, then ⌬ rG continually changes and eventually
is zero at equilibrium (figure 4.3). Another way to think about ⌬ rG is to imagine that the
reaction occurs in a very large vessel containing large amounts of the reacting substances.
So in the oxidation of glucose, for example, the net conversion of 1 mol of glucose does
not significantly change the total number of moles of glucose, oxygen, carbon dioxide, or
water in the reaction vessel; ⌬ rG for the reaction as written can then be thought of as the
actual free-energy change when 1 mol of glucose combines with 6 mol of oxygen to form
6 mol of carbon dioxide and 6 mol of water under the particular reaction conditions.
Reactions of Ideal Gases
Dependence of Chemical Potential on Partial Pressures
We start with reactions of gases not just because gases are the simplest to describe thermodynamically, but also because we now know that analyses of ideal gases can lead to
powerful generalizations to all substances (e.g., Carnot Cycle). Also, for common biological conditions such as ambient temperature and pressure, gases tend to behave ideally.
We start with an ideal gas at constant temperature, so that dG = Vdp for a change in
pressure (Eq. 3.35). Then, when the pressure of the gas changes from p1 to p2,
G(p2) - G(p1) = nRT ln a
p2
b.
p1
(3.45)
We will now designate p1 = 1 bar as the standard state and denote G(p1 = 1 bar) as G⬚,
G - G⬚ = nRT ln a
p
b.
1 bar
Dividing both sides by n and remembering that the chemical potential of a pure substance
is its molar free energy, we obtain
m = m⬚ + RT ln a
p
b ,
1 bar
(4.9)
Reactions of Ideal Gases | 105
where m is the chemical potential of the ideal gas and m its chemical potential at the standard state of 1 bar. This relation also holds for an ideal gas A in a mixture of ideal gases.
In a mixture, the pressure of A is the partial pressure pA, thus
mA = mAⴰ + RT ln a
pA
b.
1 bar
(4.10)
E X A M P L E 4 .1
Two gases, nA mol of A and nB mol of B, are initially in two separate flasks at the same
temperature T and pressure p. A valve in the tube connecting the two flasks is then
opened to allow complete mixing of the gases. Derive an equation for the change in
Gibbs free energy for mixing the two gases.
A(p)
Open
valve
SOLUTION
Before mixing, the total Gibbs free energy of the system is (see Eq. 4.5)
G(pure) = nAmA(1) + nBmB(1) = nA c mAⴰ + RT ln a
B(p)
p
p
b d + nB c mBⴰ + RT ln a
bd .
1 bar
1 bar
pA , pB
pA , pB
After mixing, the total Gibbs free energy of the system becomes
G(mix) = nAmA(2) + nBmB(2) = nA c mAⴰ + RT ln a
pA
pB
b d + nB c mBⴰ + RT ln a
bd .
1 bar
1 bar
In the above equations, pA and pB are the partial pressures of A and B, respectively,
after mixing. The change in Gibbs free energy is therefore
mixG = G(mix) - G(pure) = nART ln a
pA
pB
b + nBRT ln a b .
p
p
(4.11)
The partial pressure pA is related to p by
nA
pA =
p = xA p .
nA + nB
Then
pA
pB
= xB .
= x A , and
p
p
Substituting these into Eq. 4.11 gives
mixG = nART ln xA + nB RT ln xB .
We have previously calculated the entropy change for this process (Eq. 3.22) and
obtained
mixS = -R 3 nA ln xA + nB ln xB 4 .
(3.22)
The calculated mixG is equal to -T mixS as we would expect since U and H are
zero when two ideal gases at the same temperature and pressure are mixed.
Equilibrium Constant
We now turn to one of the most important topics in thermodynamics: the relationship
between the equilibrium constant of a chemical reaction and the thermodynamic properties
of the reactants and products in this reaction. We start by considering a reaction involving
only gases, the Haber process of nitrogen fixation in which nitrogen and hydrogen react
Mixing of ideal gases.
106 Chapter 4 |
Free Energy and Chemical Equilibria
to form ammonia. Nitrogen is an essential element for the growth of all living organisms.
Most organisms cannot use N2 directly from the atmosphere but require its reduction to
ammonia or oxidation to nitrite or nitrate first. The nitrogen-fixing bacteria, some of which
live symbiotically with plants in root nodules, use a powerful reductant to convert N2
metabolically to reduced nitrogen compounds which can then be transferred to the plant.
In the synthetic method developed by Fritz Haber, the reaction is
N2(g) + 3H2(g) h 2NH3(g) .
From Eq. 4.5, the Gibbs free-energy change for the reaction is
rG = 2mNH3 - mN2 - 3mH2
ⴰ
= 2c mNH
+ RT ln a
3
pNH3
1 bar
- 3c mHⴰ 2 + RT ln a
b d - c mNⴰ 2 + RT ln a
pH2
1 bar
pN2
1 bar
bd
(4.12)
bd.
We will no longer write the “1 bar” terms since numerically they represent dividing by 1
(sidebar). Equation 4.12 then becomes
ⴰ
- mHⴰ 2 - 3mHⴰ 2) + RT 3 2 ln pNH3 - ln pN2 - 3 ln pH2 4
rG = (2mNH
3
p2NH3
pN2 p3H2
K
pNH
1 1 bar3 2 2
pN
pH
1 1 bar2 2 1 1 bar2 2 3
In the reaction quotients of
gases, as in Eq. 4.13, the “1 bar”
terms are not written, but we
must remember that the “1 bar”
terms compel us to specify all
partial pressures in units of bar
in order to make the argument
to the logarithms unitless.
= rG + RT ln a
p2NH3
pN2 p3H2
(4.13)
b.
ⴰ
In Eq. 4.13, rG is equal to (2mNH
- mHⴰ 2 - 3mHⴰ 2) and is the Gibbs free energy per mole
3
of the reaction at T when all participants of the reaction are in their standard states (1 bar
partial pressure). At 298 K, rG = -33.56 kJ mol - 1 from table A.5.
Equation 4.13 relates the free-energy change for the reaction at any set of partial
pressures, rG to the standard free-energy change if all of the species are at 1 bar partial
pressures, rG. The superscript , which is often spoken as “nought”, is the notation for
indicating standard conditions for all species, which in this case is 1 bar partial pressure
of every gas phase participant in the reaction; the absence of the superscript means the
species in the system are at nonstandard conditions, which must be specified. The standard
state does not specify a temperature.
For every gas phase chemical reaction at any temperature and pressure, there exist
partial pressures of products and reactants for which the system is at equilibrium. We may
start with a mixture of nitrogen and hydrogen, or we may start with ammonia; if we wait
long enough (or use an iron catalyst as is done for this reaction so that the wait is not too
long) we will obtain the same equilibrium mixture of nitrogen, hydrogen, and ammonia.
We now ask, “Are the partial pressures pN2 pH2 and pNH3 related when equilibrium is
reached at constant temperature and total pressure?”
Recall that rG = 0 when a system reaches equilibrium at constant temperature and
pressure. Therefore, Eq. 4.13 becomes at equilibrium
rG = -RT ln
2
1peq
NH3 2
eq 3
1peq
N2 2 1pH 2
,
(4.13a)
2
where the superscript “eq” specifies the quantity at equilibrium. The standard free-energy
change rG for a particular reaction is, however, a constant at any chosen temperature
(because the states of the reactants and products are all specified to be their standard states
Reactions of Ideal Gases | 107
and the Gibbs free energy is a state function). Therefore, at constant temperature and total
eq
eq
eq
pressure, the quotient ( pNH3)2>( pN2)( pH2)3 must be a constant, which we denote K,
K =
2
( peq
NH3)
eq 3
( peq
N2)( pH2)
.
The constant K is called the equilibrium constant of the reaction. When equilibrium is
eq
reached at a constant temperature and total pressure, the partial pressures peq
NH3, pN2, and
eq
pH2 are not independent: they are related by the equation above.
We will now generalize Eqs. 4.13 and 4.13a for any reaction involving only gases that
do not deviate too much from ideal gas behavior in terms of their free-energy dependence
on partial pressure. Consider a generic reaction of ideal gases at temperature T:
aA(g) + bB(g) h cC(g) + dD(g).
The equilibrium constant of the reaction is directly related to ⌬ rG⬚ by
⌬ rG⬚ = -RT ln K .
(4.14)
At constant temperature and pressure the equilibrium constant of a reaction is related to
the standard free-energy change ⌬ rG⬚, and not ⌬ rG itself, which is zero at equilibrium:
⌬ rG = ⌬ rG⬚ + RT ln Q ,
c
Q =
(4.15)
d
( pC ) ( pD) ,
( pA )a ( pB)b
(4.16)
where pA, pB, pC, and pD are the partial pressures of A, B, C, and D, respectively, each
divided by 1 bar. The quotient Q is therefore the ratio of the partial pressures (in bar) of
reactants and products, each raised to the power of its coefficient in the chemical equation.
We see that Q will be large if product pressures are large or reactant pressures are small: a
Q > 1 means a positive (unfavorable) contribution to the free energy. Then Q is small if product pressures are small or reactant pressures are large: a Q < 1 means a negative (favorable)
contribution to the free energy. The difference between standard Gibbs free-energy changes
⌬ rG⬚ when all partial pressures are 1 bar, and actual free-energy changes ⌬ rG when partial
pressures are specified by Q can be very significant (figure 4.4, table 4.1).
At equilibrium, we have Eq. 4.14, ⌬ rG⬚ = -RT ln K with
K =
c eq d
( peq
C ) ( pD )
eq b
a
( peq
A ) ( pB )
,
(4.17)
eq
where peq
A , pB , and so on are equilibrium partial pressures, each divided by 1 bar.
Equations 4.14 and 4.15 can also be combined to give
⌬ rG = -RT ln K + RT ln Q = RT ln
Q
.
K
(4.18)
Equations 4.14 through 4.18 may be the most useful thermodynamic equations a
biochemist learns. They relate the free-energy change for a reaction to experimentally measurable quantities. Determination of the standard Gibbs free energies of various substances
permits calculation of ⌬ rG⬚ values for reactions involving these substances, which in turn
permit calculation of the equilibrium constants; conversely, experimental determination of the
G298
G298 (kJ)
20
less favorable
10
10 -3
10 -2
10 -1
10 1
10 2
Q
10 3
-10
more favorable -20
FIGURE 4.4 Following table
4.1, the line depicts quantitative
values for the change of free
energy with Q at 25°C. Values at
any other temperature T can be
calculated by multiplying those
listed in the table by T/298.
TABLE 4.1
Q
⌬G298 ⴚ ⌬Gⴗ298
kJ mol ⴚ1
100
11.410
10
5.705
1
0
0.1
−5.705
0.01
−11.410
108 Chapter 4 |
Free Energy and Chemical Equilibria
equilibrium constant of a reaction allows us to calculate rG of the reaction. Furthermore,
Eq. 4.17, which defines the equilibrium constant, presents a ratio of equilibrium partial
pressures that is constant for a chemical reaction at a given temperature.
EXAMPLE 4.2
What is the Gibbs free-energy change, relative to that under standard conditions, of
forming 1 mol of NH3 at 298 K if (a) 10.0 bar of N2 and 10.0 bar of H2 are reacted to
give 0.0100 bar of NH3, and (b) 0.0100 bar of N2 and 0.0100 bar of H2 are reacted to
give 10.0 bar of NH3? We assume in (a) and (b) that the reaction vessel has been prepared with the specified partial pressures of NH3. The reaction is
ⴰ
rG298
= -33.56 kJ mol - 1 .
N2(g) + 3H2(g) S 2NH3(g)
SOLUTION
a. From Eq. 4.13,
rG - rG = RT lna
p2NH3
pN2 p3H2
b
= (.008314 kJ K - 1 mol - 1)(298 K)ln c
0.01002
d
(10.0)(10.03)
= -45.6 kJ mol - 1 .
The free energy is decreased and therefore favored relative to the standard free
energy by having high pressures of reactants and low pressures of products.
b. Substituting the partial pressures into Eq. 4.13 gives
rG - rG = RT lna
p2NH3
pN2 p3H2
b = 57.0 kJ mol - 1 .
The free energy is increased by having high pressures of products and low pressures
of reactants; the reaction is less favored relative to the standard conditions.
Nonideal Systems
Activity
It was clear to thermodynamicists that Eqs. 4.14 through 4.18 were simple and easy to
use; it would be convenient if they also applied to real gases, liquids, solids, and solutions.
Thermodynamicists accomplished this by defining a new quantity, the activity, such that
for any substance A its activity aA is related to its chemical potential by
mA = mAⴰ + RT ln aA .
(4.19)
Then for our usual reaction of aA + bB S cC + dD, we follow Eq. 4.8 to arrive at a
familiar equation, but based now on the activities of reacting species,
rG = 3 (cmC + dmD) - (amA + bmB) 4 = rG + RT ln a
acC adD
aaA abB
b.
The Gibbs free energy change of any reaction is
rG = rG + RT ln Q ,
(4.20)
Nonideal Systems | 109
with the reaction quotient Q and the equilibrium constant K defined with activities
Q =
acCadD
aaAabB
eq
K K
K K
(4.21)
,
eq
(aC )c (aD )d
a eq b
(aeq
A ) (aB )
.
(4.22)
These equations are of the same form as Eqs. 4.15, 4.16, and 4.17, but their application is
no longer restricted to ideal gases. We should appreciate now some important properties
of activities:
• Activities are unitless numbers. Notice the activity is the sole argument to the logarithm in Eq. 4.19.
• The activity aA of A is defined with respect to a standard state that we will have to
specify. That is, we will see that the difference in chemical potential between two
states, mA - mAⴰ , is a measurable quantity, and so the activity can be calculated from
ⴰ
this difference: aA = e(mA - m A)>RT .
• If the substance A is in its standard state, then aA must be 1 in order to have mA = mAⴰ
in Eq. 4.19.
• To make full use of Eqs. 4.20–4.22, we need to relate the activities to parameters
that are widely used by chemists and biochemists, such as the partial pressures of
gases or the concentrations of species in solutions.
Standard States
The next step is to carefully specify and adopt a suite of standard states for activities of
all substances. We have performed a highly analogous procedure already with enthalpies
of formation (chapter 2). Specifically, recall that any elemental substance at 1 bar in its
stablest state is assigned an absolute molar enthalpy of 0. Then enthalpies of formation of
molecules are always related to these defined zero enthalpies.
Similarly, any substance at the conditions of its standard state has an activity of 1 and
the chemical potential m. Then the chemical potential of the molecule at another set of
conditions is related to it by Eq. 4.19. That is, the quantity m - m is the Gibbs free-energy
change for taking one mole of a substance from its standard state to its actual state.
We now define standard states that are commonly used for activities. Students are
often bewildered by the many different standard states, and it may help to point out that the
particular choices of standard states not only stress convenience, but also allow us to relate
the activities to well-known experimental parameters (e.g., for ideal solutions, the activity
of a component is given by its concentration). Moreover, it should be appreciated that the
system of standard states for activities needs to be applicable to all forms of matter, which
is a more complicated scenario than for enthalpies in which we only needed to define a
standard state for pure elements.
Ideal gases
The standard state for an ideal gas is the gas with a partial pressure equal to 1 bar. The
notation for the standard pressure is pAⴰ = 1 bar. The activity of an ideal gas is defined as
its actual partial pressure divided by its standard pressure
pA
pA
aA = ⴰ =
,
(4.23)
pA
1 bar
where pA is the partial pressure of the ideal gas. The activity of an ideal gas is numerically
equal to its partial pressure in bar. The standard conditions for a reaction involving only
ideal gases are partial pressures of 1 bar for each product and reactant (chapter 3).
a cC a dD
a aAa bB
It has been helpful for clarity to
include the superscript “eq” in
the expression for K. However,
as long as K is specified, the
“eq” superscripts in Eq. 4.17
and Eq. 4.22 are redundant and
conventionally are not written.
That is, we must recognize the
activity from context. If aA is
written in an expression for K, it
is the equilibrium activity. In a
reaction quotient Q, an activity
aA is determined its activity is
determined by the instantaneous reaction conditions.
110 Chapter 4 |
Free Energy and Chemical Equilibria
=1
1
μ μ ° (kJ/mol)
8
m
ea
su
re
6
4
2
0
0
l
po
ra
t
ex
1.0
e
at
2.0
3.0
|lnpA|
p increasing
FIGURE 4.5 By extrapolating
low pressure data to p = 1 bar
(from right to left), we arrive
at a standard state (the origin,
black circle) where mA = m°A,
the chemical potential the gas
would have at 1 bar if it behaved
ideally.
Real gases
The activity of a real gas is again a function of pressure,
gA pA
gA pA
aA =
=
,
pAⴰ
1 bar
(4.24)
with a new term gA, the activity coefficient, which will account for nonideal behavior and
will depend on the total pressure of the system. We know that all gases become ideal at lowenough pressures, so gA must become 1 as the pressure of the system approaches zero. Near
atmospheric pressure, the activity coefficients of gases are close to 1, and real gases can be
treated as ideal gases to a good approximation. From the definition of activity, for a real gas,
gApA
pA
mA - mAⴰ = RT ln a
b = RT ln gA + RT ln a
b.
1 bar
1 bar
Consider a system containing pure A. If mA - mAⴰ is plotted as a function of ln (pA/1 bar),
at low pressure a straight line is obtained with a slope equal to RT since gA = 1 at
low pressures and therefore ln gA = 0. The straight line can then be extrapolated to
pA = 1 bar, where mA - mAⴰ = 0 and mA = mAⴰ , the chemical potential of the gas at its
standard state (figure 4.5). Thus, the standard state for a real gas is a hypothetical one: it
is the extrapolated state where the partial pressure is 1 bar but the properties are those
extrapolated from low pressure.
Pure solids or liquids
The standard state for a pure solid or liquid is the pure substance (solid or liquid) at 1 bar.
Therefore, the activity is equal to 1 for a pure solid or liquid at 1 bar. The Gibbs free energy
of a solid or liquid changes only slightly with pressure (Eq. 3.43), so we can usually use 1
for the activity of a pure solid or liquid at any pressure:
aA = 1 (pure solid or liquid)
(4.25)
At high pressure, the activity of a pure solid or liquid can be calculated from Eq. 3.42a.
EXAMPLE 4.3
Calculate the activity aH2O of liquid water at 293 K and (a) 10 bar and (b) 1000 bar. Use
the following empirical relation for the molar volume of liquid water as a function of
pressure ( p in bar):
Vm,A = 18.07(1 - 45.9 * 10 - 6 p) mL mol - 1
SOLUTION
At constant temperature, the Gibbs free energy of a substance A is related to the pressure by
p2
G(p2) - G(p1) = 3 Vdp .
(3.42a)
p1
This equation can also be written for 1 mol of the substance,
p2
Gm(p2) - Gm(p1) = 3 Vm,A dp ,
(4.26)
p1
where Gm and Vm,A are the molar Gibbs free energy and molar volume of substance A,
respectively. For a pure substance A, mA = Gm,A, and so Eq. 4.26 becomes
p2
mA(p2) - mA(p1) = 3 Vm,Adp .
p1
(4.27)
Nonideal Systems | 111
We set p1 = 1 bar so mA(p1) = mAⴰ , and let p2 be any pressure p in bar. Thus,
p
mA(p) - mAⴰ = 3 Vm,Adp .
(4.28)
1
Substitute the above equation for molar volume into Eq. 4.28 and integrate:
p
mH2O(I)(p) - mHⴰ 2O(I) = 3 3 18.07(1 - 45.9 * 10 - 6p) 4 dp
1
= c18.07(p - 1) -
18.07 * 45.9 * 10 - 6 2
(p - 12)d (mL bar mol - 1)
2
Thus,
ln aH2O =
mH2O(l) - mH2O(l)
RT
18.07(p - 1) -
=
18.07 * 45.9 * 10 - 6 2
(p - 12)
2
.
RT
a. p = 10 bar. Substituting p = 10 bar, R = 83.14 mL bar mol - 1 K - 1, and
T = 293 K into the equation gives ln aH2O = 0.00676 or aH2O = 1.01, which is
very close to the value 1.000, the activity of liquid water at 1 bar.
b. Similarly, at p = 1000 bar, ln aH2O = 0.732 or aH2O = 2.08. At this high pressure,
the activity deviates significantly from 1.
EXAMPLE 4.4
Calculate the equilibrium constant at 25C for the decarboxylation of liquid pyruvic
acid to form gaseous acetaldehyde and CO2:
O
O
CH3CCOOH(l)
CH3CH( g)
CO2( g)
SOLUTION
From tables A.5–7 in the appendix,
rG = rG ⴰ (acetaldehyde) + rG ⴰ (CO2) - rG ⴰ (pyruvic acid)
= -133.30 kJ mol - 1 + (-394.36 kJ mol - 1) - (-463.38 kJ mol - 1)
= -64.28 kJ mol - 1 .
From Eq. 4.14,
K = 10 - G>2.303RT = e-rG>RT .
The factor 2.303RT = 5.708 kJ mol - 1 for 25C occurs often:
K = 10 - rG>5.708 ( rG in kJ mol - 1)
= 1064.28>5.708
= 1.83 * 1011
Pyruvic acid, a liquid, is very unstable with respect to CO2 and acetaldehyde at room
temperature. The equilibrium constant is K = (aacetaldehyde)(aCO2)>(apyruvic acid), where the
112 Chapter 4 |
Free Energy and Chemical Equilibria
activities are the values at equilibrium. The activity of pyruvic acid is of the order of 1;
the activities of the gaseous products are related to their partial pressures according to
Eq. 4.24. The very large equilibrium constant tells us that thermodynamically the reaction
will go to completion; that is, all pyruvic acid will be converted to gaseous acetaldehyde
and CO2. If the reaction occurs rapidly, the sudden conversion of a condensed material to
gaseous products can cause an explosion. A suitable catalyst is needed at room temperature, however, to accelerate this reaction.
Solutions
A solution is a homogeneous mixture of two or more substances. There can be solid solutions, liquid solutions, or very complicated mixtures of components as might be found in a
living cell. The activity of each substance will depend on its concentration and the concentrations of everything else in the mixture. Following Eq. 4.24, we write the concentration
dependence of the activity in a deceptively simple equation,
activity = (activity coefficient)a
concentration value
b.
standard concentration value
(4.29)
The activity coefficient is not a constant; it incorporates the complicated dependence of the activity of A on the concentrations of A, B, C, and so on. Another
complication is that different concentration units are routinely used in Eq. 4.29. The
commonly used concentration units and their standard states are discussed in the
following section.
Mole fraction and solvent standard state
A general strategy to work with solutions is to make a distinction between components that
are viewed as solutes and components that are viewed as solvents. It is typical to designate
one component (usually H2O) as the solvent.
The standard state using mole fraction (x) as the concentration unit is often called the
solvent standard state. The solvent standard state for species A in a solution is the pure
component A, so that the standard state mole fraction is xAⴰ = 1. Recall that the mole
fraction of species A is the number of moles of A divided by the total number of moles of
all components in the solution
nA
,
(4.30)
xA =
nT
aA K
gAxA
xAⴰ
=
gAxA
1
As we encountered with the
“1 bar” terms for the activities of
gases, there is an implicit step
in Eq. 4.31 of dividing by 1, the
standard mole fraction.
where nA is the number of moles of A and nT = nA + nB + ... is the total number of moles
of all components. Following Eq. 4.29, the activity of A is
aA = gAxA ,
(4.31)
where gA is the activity coefficient of A and xA is the mole fraction of A. In order for
the standard state of pure A to have an activity of 1, we require that gA S 1 as a system
approaches pure A. That is, the activity aA becomes equal to the mole fraction xA
lim aA = xA ,
xAS1
which is read “in the limit as xA approaches 1, aA = xA.” Examples of the use of the solvent
standard state are aqueous solutions that are sufficiently dilute in solutes so that xH2O _ 1
and so to a very close approximation we have aH2O = 1.
We now summarize the useful equations for a solvent in a solution:
Very dilute solution: asolv = 1
(4.32)
Nonideal Systems | 113
Dilute solution or ideal solution: asolv = Xsolv
(4.33)
Real solution: asolv = gXsolv
(4.34)
Equation 4.33 defines an ideal solution: for any concentration, the solvent has the properties of the solvent in a dilute solution. To determine the activity coefficient and therefore
the activity in Eq. 4.34, we have to measure the Gibbs free energy of the solvent in the
solution. This can be done by methods involving measurements of the vapor pressure of
the solvent in a solution, the freezing-point depression, the boiling-point elevation, or the
osmotic pressure. These will be discussed in chapter 6.
Solute standard states
For solutions in which certain components never become very concentrated, such as dilute
aqueous salt solutions, we define a solute standard state whose free energy can be obtained
from measurements in dilute solutions. The solute standard state for a component is defined
as the extrapolated state where the concentration is equal to 1 molar (M) or 1 molal (m),
but the properties are those extrapolated from very dilute solution. The solute standard state
is a hypothetical state: a 1 M or 1 m ideal solution. For real solutions, the ideal behavior is
not approached except for concentrations that are much less than 1 M or 1 m. The definition of the solute standard state should become clearer after reading the following sections.
Molarity
Concentrations are commonly measured in molarity, with units of mol L - 1 = M. (We
use the symbol c for molarities.) The activity of solute B on the molarity scale and with a
solute standard state is, from Eq. 4.29,
aB = gBcB ,
(4.35)
where gB is the activity coefficient of B on the molarity scale and cB is the molarity of
species B. Note that the activity of B, aB, for a molecule B in a solution will be different if
Eq. 4.35 is used rather than Eq. 4.31. The free energy per mole of B in a specified solution
has a definite value, but different choices of standard states produce different values of aB.
In contrast to the solvent standard state, where gA = 1 in the actual standard state
of the pure solvent, in real solutions the activity of a solute at 1 M concentration is most
likely to be less than 1 (that is, gB 6 1 for cB = 1 M). So we appear to have a problem:
the activity must be 1 in the standard state, but the activity of B in a real solution at 1 M
concentration will not be 1 (figure 4.5). An ingenious solution has emerged by recognizing
that the standard chemical potential m does not have to correspond to any actual solution, but just be a reference point that all scientists agree to use. The standard state for
the molarity scale is chosen so that the activity becomes equal to the concentration as the
concentration approaches zero:
lim aB = cB
cB S 0
The molarity scale for activities is used mainly for solutes. The meaning of this standard
state is seen in figure 4.6, in which aB depends linearly on cB for dilute B only, where gB
approaches 1. Thus the standard state for the molarity scale is the point at cB = 1 extrapolated from the linear region of aB. The activity coefficient and therefore the activity of B
can be measured through its effect on the vapor pressure of the solvent in solution or, if
component B is volatile, its vapor pressure can be measured. Electrolytic cells provide easy
methods for measuring certain activities (chapter 7).
To obtain the free energy in the standard state, we measure mB as a function of ln cB
in dilute solutions and extrapolate linearly to ln cB = 0 (cB = 1) to find mBⴰ (figure 4.7).
The standard state answers the question: what would be the chemical potential of the solute
B at 1 M if it were an ideal solution?
gAcB
=
cBⴰ
1M
Once again, there is an implicit
step in Eq. 4.35 of dividing by
the standard state concentration of 1 M.
aB K
gBcB
114 Chapter 4 |
Free Energy and Chemical Equilibria
FIGURE 4.6 Activity of a solute
as a function of molarity for a
real solution (solid curve) compared with that of an ideal solution (dashed line) extrapolated
from very dilute conditions. The
solute standard state lies on this
extrapolated line, as shown.
Ideal
solution
Solute
standard
state
Real
solution
1
aB
0.5
1
1.5
cB
FIGURE 4.7 Chemical potential
of a solute plotted against the
logarithm of molarity for a real
solution (solid curve), compared
with that of an ideal solution
(dashed line) extrapolated from
very dilute conditions.
Ideal
solution
μB
Real
solution
μB
-4
-3
-2
-1
0
1
ln cB
If a solute is a strong electrolyte in aqueous solution it dissociates completely into
its component ions. When we speak of the partial molar Gibbs free energy or chemical
potential of NaCl in aqueous solution, we use the sum rule for the chemical potentials,
mNaCl K mNa + + mCl ⴰ +
ⴰ = mNa
+ mCl
+ RT ln aNa + + RT ln aCl ⴰ +
ⴰ = mNa
+ mCl
+ RT ln (aNa +
#
aCl - ) .
We can also employ the convention of expressing
ⴰ
mNaCl = mNaCl
+ RT ln aNaCl .
Comparison of this expression with Eq. 4.30 shows that
aNaCl = aNa + # aCl - .
Similarly, for Na2SO4, the activity in aqueous solution is
aNa2SO4 = a2Na + # aSO42 - .
gAmB
=
mBⴰ
1m
As we have seen, there is again
an implicit step in Eq. 4.39 of
dividing by the standard state
concentration of 1 m.
aB K
gBmB
(4.36)
(4.37)
(4.38)
Molality
Another concentration unit that is frequently used is molality, m, with units of moles of
solute per kilogram of solvent. The activity of B on the molality scale is
aB = gBmB .
(4.39)
The discussion about molarity applies identically to molality. We have the following:
Dilute solution or ideal solution: aB = mB
(4.40)
Real Solution: aB = gBmB
(4.41)
Nonideal Systems | 115
Again, the standard state is an extrapolated state, where mBⴰ is obtained by linearly extrapolating mB measured in dilute solution to ln mB = 0 (mB = 1). Molality is used instead of
molarity for the most accurate thermodynamic measurements. Since molalities are defined
by weight, not volume, they can be measured quite accurately, and they do not depend on
temperature. For dilute aqueous solutions, 1 L of solution contains about 1 kg of water and
therefore the numerical values of molarity and molality are very close.
Biochemical standard state
It is tempting to think at this stage that we can begin to tackle biochemical reactions and
processes with the above conventions for solvent and solute species. Yet we still require
an additional set of conventions, frequently called the ‘biochemical standard state’, that
specifically apply to biochemical reactions. A broad principle of the biochemical standard
state is to more closely approximate the complex conditions in which life processes occur.
Any quantity expressed in the biochemical standard state is denoted with a prime, as in
K and G.
(i) pH 7. Consider the phosphorylation of glyceraldehyde-3-phosphate (abbreviated G3P)
to 1,3-diphosphoglycerate (1,3-dPG) in the glucose metabolism path
G3P + phosphate + NAD+ S 1,3-dPG + NADH + H+ .
In general, we wish to make thermodynamic measurements of biochemical reactions under
pH buffered conditions so that H+ is a constant throughout the reaction and should not be
viewed as a variable in an equilibrium constant. Further, there is little biological importance to extrapolate biochemical thermodynamics to the extreme acidic conditions of 1 M
H+ concentrations. Thus the biochemical standard state specifies the reaction at pH 7. By
specifying pH 7 we are setting aH + = 1 in the biochemical standard state. All other products and reactants follow the molarity standard state. The chemical (K) and biochemical
(K) equilibrium constants are therefore written, respectively, as
K =
a1,3 - dPGaNADHaH +
a1,3 - dPGaNADH
, and K =
.
a3PGaNAD +
a3PGaNAD +
(ii) Activities are total concentrations. We have assumed so far that we knew the concentration of each species involved in a system. For a molecule that dissociates in solution, this
may be very difficult to determine. For example, suppose a reaction involves H2PO4- ; however, the species actually present in solution include H3PO4, H2PO4- , HPO24 - , and PO34 - .
The distribution of these species will depend markedly on pH, so the concentration of
H2PO4- may be difficult to specify. Therefore, biochemists define the activity of a molecule
as the total concentration of all species of that molecule at pH 7.0:
species
Dilute solution: aA = a ci, A
(at pH 7.0)
(4.42)
i
In our example (sidebar), this sum over all species is the total concentration of phosphate
added—that is, the concentration determined analytically. Knowledge of ionization constants is thus not needed, nor is it necessary to specify the concentration of each of the
actual species involved in the reaction.
Equation 4.42 applies to many components of biochemical reactions. For example,
proteins have many weakly acidic/basic groups that are all in individual acid-base equilibria.
Further, biomolecules can be in equilibrium with multiple conformations. Also, proteins
and small biomolecules such as ATP can bind ions in specific or nonspecific manners. For
example, electrospray mass spectrometry of the protein ubiquitin reveals that the protein
is present with zero, one, or more associated sodium ions (figure 4.8).
For the coupled equilibria in
aqueous solution,
H3PO4 L H2PO4- + H+ ,
H2PO4- L HPO24 - + H+ ,
+
3HPO24 L PO4 + H ,
the activity of phosphate in the
biochemical standard state can
be written from Eq. 4.42 as
api = [H3PO4] + [H2PO4-]
3+ [HPO24 ] + [PO4 ],
which is the total phosphate that
the system was prepared with
and so is known.
116 Chapter 4 |
Free Energy and Chemical Equilibria
(iii) Ionic strength (I) and [Mg2+] are specified. Biochemical reactions generally occur in
the presence of salts that can significantly affect the thermodynamics of the reactions under
study. The ionic strength (I ) reports on the concentration and charge of ions in solutions.
The ionic strength has units of molarity and is computed as a sum over all ions
Ubq
I = 12 a ciZ2i ,
(4.43)
i
Ubq.Na
Ubq.Na2
where ci is the concentration of the i’th ion and Zi is the charge of the i’th ion. Commonly,
biochemical data is reported for I = 0.250 M. Since magnesium ions are frequently involved
in biochemical reactions, an emerging trend is to fix the magnesium concentration at
0.001 M, often expressed as pMg = 3 1pMg = -log 3 Mg2 + 42.
8550
8600
mass (g/mol)
FIGURE 4.8 A protein may
exist in equilibrium with a
number of different bound
ions such as sodium. Here the
protein ubiquitin (Ubq) has
been analyzed with electrospray
mass spectrometry, and shows
a set of peaks that differ by
23 mass units, indicating bound
sodium ions. (Courtesy of
Dr. P. Findeis.)
EXAMPLE 4.5
What concentration of sodium chloride corresponds to an ionic strength of I = 0.250 M?
SOLUTION
Set up Eq. 4.43 for Na+ and Cl- ions. Adding NaCl (s) to an aqueous solution means
that the concentrations of the ions are equivalent (cNaCl = cNa + = cCl - ), so we have
0.250 M = 21(cNa+ (+1)2 + cCl- ( - 1)2)
= 12cNaCl(1 + 1)
= cNaCl .
Thus, a solution with I = 0.250 M has the same ionic strength as a 0.250 M solution
of sodium chloride.
(iv) Activity coefficients are incorporated into the standard Gibbs free energy. We explain
next an important consequence of Eq. 4.42 that defined activities as concentrations. As
we will discuss shortly, when specifying a saline environment with a given ionic strength,
the activity coefficients (g) of the reacting species should not be assumed to be 1, but they
also may not be realistically knowable. By defining activities as concentrations only, the
biochemical standard state incorporates the activity coefficients as additional constants
into the standard Gibbs energy. That is, for our generic reaction of species A and B reacting to form C and D,
rG = rG + RT ln a
= rG + RT ln a
= rG + RT ln a
a Cc a Dd
b = rG + RT ln a
a aA a bB
gCc gDd
cCc cDd
gAa gB
cAacBb
b + RT ln a
b
cCc cDd
cAa cBb
(gC cC)c(gD cD)d
(gAcA)a(gBcB)b
b
b
b.
Notice that there is no notational distinction to indicate that the activity coefficients
have been folded into the standard Gibbs free-energy change. If the ionic strength is not
specified or is zero, then the standard Gibbs energy is for the hypothetical state of 1 M
concentrations but for an ideal solution (all g>s equal 1). If the ionic strength is specified,
then the activity coefficients are incorporated into rG as above, which will now be
dependent upon the ionic strength.
Nonideal Systems | 117
An important difference between the biochemical standard state and all others is that
the equilibrium constant applies only at pH 7. For other pH>s, one needs either to repeat the
experiments to determine the new equilibrium concentrations or to use known ionization
constants to calculate how the concentrations of reactive species depend on pH.
Consider the hydrolysis of ethyl acetate to produce acetic acid and ethanol:
O
O
CH3COCH2CH3 + H2O
EtOAc + H2O
CH3COH + CH3CH2OH
HOAc + EtOH
The equilibrium constant is
K =
(aHOAc)(aEtOH)
.
(aEtOAc)(aH2O)
We can use the dilute solution standard states for ethanol, acetic acid, and ethyl acetate,
meaning their activities are their molarities in dilute solution. For water we use the pure
liquid standard state and replace aH2O by 1 in dilute solution. These standard states allow
us to calculate a standard free energy from the equilibrium constant, rG = -RT ln K.
However, we may be interested in the reaction at other pH values because acetic acid
is a weak acid and can dissociate into a hydrogen ion and an acetate ion. If we consider
the pH as an independent variable, we can write the reaction as
EtOAc + H2O m OAc- + H + + EtOH ,
K =
(aOAc - )(aH + )(aEtOH)
.
(aEtOAc)(aH2O)
To use the biochemical standard state, we measure the equilibrium concentrations at pH 7
and set aH + = 1 and aOAc- = [OAc - ] + [HOAc]; as before aH2O = 1, aEtOH = 3 EtOH 4 ,
and aEtOAc = [EtOAc]. The calculated rG = -RT ln K applies at pH 7.
Activity Coefficients of Ions
To simplify thermodynamic calculations, we often make the approximation of setting
activity coefficients equal to 1. For small uncharged molecules and for low concentrations of univalent ions in dilute aqueous solutions, this approximation may not be too bad.
For multivalent ions such as Mg2 + or PO34 - , however, activity coefficients may be very
different from 1, even at millimolar concentrations. In ionic solutions, the total number
of positive charges is always equal to the total number of negative charges (electrical
neutrality). Therefore, we cannot separately measure the activity coefficient of a positively
or negatively charged ion; we can measure only the mean activity coefficient of an ion and
its counterion. For example, consider NaCl in aqueous solution by summing the chemical
potentials of the individual ions:
mNa + = mNa + + RT ln (gNa + cNa + )
+ mCl - = mCl - + RT ln (gCl - cCl - )
_____________________________________
mNa + + mCl - = mNa + + mCl - + RT ln (gNa + cNa + gCl - cCl - )
Recognizing that charge balance requires cNa + = cCl - , we get
mNaCl = mNaCl + RT ln(gNa + gCl - c2NaCl) .
Now we may define for NaCl or any n:n electrolyte:
g { = (g + g - )1/2 (1:1, 2:2, or n:n electrolyte) ,
(4.44)
118 Chapter 4 |
Free Energy and Chemical Equilibria
which allows us to write the chemical potential of a dissociated electrolyte as
mNaCl(aq) = mNaCl + RT ln (g { cNaCl)2 .
For H2SO4 or any 1:2 electrolyte,
g { = (g2+ g - )1/3 .
(1:2 electrolyte)
(4.45)
(1:3 electrolyte)
(4.46)
For LaCl3 or any 1:3 electrolyte,
g { = (g + g3- )1/4 .
These equations are readily generalized to any type of salt. Figure 4.9 illustrates some
measured activity coefficients for a 1:1 electrolyte (HCl), a 1:2 electrolyte (H2SO4), and
a 2:2 electrolyte (ZnSO4).
To understand why activity coefficients are so different from 1 for electrolytes and why
the ions with the larger charges have the smaller activity coefficients, we need to consider
the interactions between the solutes. Activities can be viewed as effective concentrations.
In an HCl solution, the effective concentration of the H + ions is reduced by the surrounding Cl - ions; the effective concentration of the Cl - ions is reduced by the surrounding H +
ions. Figure 4.9 shows that this decrease in effective concentrations of the ions is greater
in H2SO4 solution, which has a doubly charged ion, SO42 - .
Debye and Hückel (1923) calculated activity coefficients for individual ions using
Coulomb’s Law for the interaction energy between charged particles. Their result applies
only in very dilute solutions (less than 0.01 M), but it can provide a qualitative understanding of activity coefficients. For ions in water at 25C,
log gi = -0.509Z2iI1/2 ,
(4.47)
where Zi = charge on ion ({1, {2, {3, etc.) and I = ionic strength = 21 ici Z 2i . The
numerical value, 0.509, depends on the temperature and the dielectric constant of the
solvent. The dielectric constant is a measure of the strength of an electric field in a solvent
compared to that in a vacuum; it is discussed in chapter 9.
1.0
0.8
FIGURE 4.9 Mean activity
coefficient of various electrolytes
at 25C. The activity coefficients
are much less than 1 even for
0.1 M solutions; at concentrations above 1 M, the activity
coefficients for some electrolytes
increase and become even
greater than 1.
Mean activity coefficient,
±
HCl
0.6
0.4
0.2
H2SO4
ZnSO4
0
0
0.1
0.2
0.3
Molarity (c)
0.4
0.5
Equilibrium and the Standard Gibbs Free Energy | 119
The sum in the ionic-strength equation is over all ions in the solution. Ionic strength has
been found to be a good measure of how different salts affect equilibria and reaction rates.
The ionic strength rather than the concentration of the salt characterizes the effect; thus,
the presence of a salt not directly involved in a reaction may nevertheless exert a significant
effect. Note that 0.01 M NaCl, 0.0033 M MgCl2, and 0.0025 M MgSO4 all have ionic strength
equal to 0.01 M.
Equilibrium and the Standard Gibbs Free Energy
Here we consider several reactions and see how the equilibrium constant and the standard Gibbs free energy depend on the choice of standard states when we evaluate
rG = -RT ln K . Free energies of reactions, particularly complicated biochemical reactions, can be measured via the equilibrium constant for the reaction. Conversely, once
standard free energies of reactants and products are obtained, it is easy to calculate the
equilibrium constant and equilibrium concentrations of reactants and products. For review,
the Gibbs free-energy change at arbitrary concentrations specified by Q is
rG = rG + RT ln Q .
(4.20)
We begin with the dissociation of acetic acid, HOAc, into acetate and hydrogen ions:
HOAc(aq) m H +(aq) + OAc-(aq)
If we choose the molarity scale for all species, a = gc, then
(aH + )(aOAc - )
(cH + )(cOAc - ) g + g g+gg2{
K =
=
= Kc
= Kc
,
gHOAc
gHOAc
(aHOAc)
(cHOAc) gHOAc
(cH + )(cOAc - )
Kc K
.
(cHOAc)
(4.48)
The activity coefficients gHOAc and g { are dependent on the concentrations of the solutes
and approach 1 as the solution becomes very dilute. Therefore, for very dilute solutions,
K and Kc are equal. While technically breaking the rules, the apparent units of K are
often specified as a helpful reminder of the chosen standard states of the reactants and
products. (sidebar)
For more concentrated solutions, we need to know the activity coefficients in order
to calculate Kc from K, or vice versa. The mean ionic activity coefficient g { , which for a
1:1 electrolyte like HOAc is equal to (g + g - )1/2 (Eq. 4.44), can be measured or estimated
by calculation, but it can be more difficult to do so as the concentrations increase. We saw
in the preceding section that ions are very nonideal, and in 0.1 M NaCl the mean ionic
activity coefficient g { is 0.78; in 0.01 M NaCl, it is 0.90. Therefore, 0.01 M solutions might
be the upper concentration limit for ignoring g>s for “dilute” solutions of 1:1 electrolytes.
From the thermodynamic equilibrium constant, we can calculate the standard free
energy, rG (Eq. 4.14):
HOAc(aq) m H +(aq) + OAc-(aq)
ⴰ
ⴰ
- - m
rG = mHⴰ + + mOAc
HOAc
and the standard states of all species are the extrapolated 1 M hypothetical state—that is,
1 M concentration in water but with the properties of a very dilute solution. For the Gibbs
free-energy change at any arbitrary concentration, we use Eq. 4.20.
For example, for
rG
HOAc(10 - 4 M, aq) S H + (10 - 4 M, aq) + OAc - (10 - 4 M, aq) ,
K and Kc are unitless. We know
that K is unitless because all
activities are unitless. Further,
Kc is also unitless because each
concentration term is actually
the concentration in M divided
by 1 M, the chosen standard
state, (e.g., see the discussions
of Eqs. 4.13, 4.35). However, it is
common to express Kc in terms
of the apparent concentration
units that would occur if we did
not divide quantities by their
standard states. For the example
given in Eq. 4.48, Kc would have
the apparent unit of M.
120 Chapter 4 |
Free Energy and Chemical Equilibria
the Gibbs free-energy change can be evaluated at 25 C
(10 - 4)(10 - 4)
10 - 4
= rG + (8.314 J K - 1 mol - 1)(298 K) ln (10 - 4)
rG = rG + RT ln Q = rG + RT ln
= rG - 22,820 J mol - 1.
This calculation tells us that dilution by a factor of 104 decreases the Gibbs free energy by
22,820 J mol - 1 at 25C, making the dissociation reaction that much more favorable. For
comparison, applying Le Châtelier’s principle to lowering the concentration also predicts
that more acetic acid dissociates.
A hydrolysis reaction illustrates the use of other standard states. Consider the hydrolysis of ethyl acetate (EtOAc) to acetic acid (HOAc) and ethanol (EtOH):
H2O + CH3COOCH2CH3 m CH3COOH + CH3CH2OH ,
K=
If we chose the molarity
standard state for water instead,
we would find its concentration
as 55.5 M (1000/18) to give the
activity:
aH2O = 55.5 M>1 M = 55.5
aHOAc aEtOH
aH2O aEtOAc
We can choose the molarity standard state for all solute species, and the mole fraction
for water:
(cHOAc)(cEtOH) gHOAcgEtOH
#
K=
gH2OgEtOAc
(xH2O)(cEtOAc)
In dilute aqueous solution, the mole fraction of H2O approaches 1, and all the activity
coefficients approach 1; hence, only concentrations appear in K. It should be clear that the
numerical value of K will depend on our choice of standard states. Its value will depend
on whether we use 55.5 or 1.0 for the activity of water (sidebar).
Because one of the species, acetic acid, is ionizable, we could choose the biochemical standard state for HOAc. The concentration of the un-ionized species HOAc depends
greatly on the total concentration of HOAc because of its dissociation into OAc - and H+.
The ratio of concentrations at equilibrium,
(cHOAc)(cEtOH)
,
(cEtOAc)
will be difficult to measure, particularly due to the dependence on cHOAc. However, if we
are mainly interested in the reaction near physiological conditions, we can simplify our
problem by choosing the biochemical standard state. We study the hydrolysis in a pH 7
buffer. Now we can use cHOAc equal to the sum of acetic acid and acetate. The pH 7 buffer
ensures that the acetic acid is essentially all in the form of acetate, and the HOAc concentration will be directly proportional to the acetate over a wide range of total concentration.
EXAMPLE 4.6
What is the difference in the standard Gibbs free energies for the hydrolysis of ethyl
acetate for (a) the molarity and (b) mole fractions standard states for H2O? You may
assume a dilute solution.
SOLUTION
Case a: In the molarity convention, we must use 55.5 as the activity of water:
(cHOAc)(cEtOH)
rG = rGaⴰ + RT ln
(cH2O)(cEtOAc)
(cHOAc)(cEtOH)
1
+ RT ln
,
= rGaⴰ + RT ln
55.5
(cEtOAc)
and the apparent units for Q all cancel.
Equilibrium and the Standard Gibbs Free Energy | 121
Case b: With the mole fraction convention for water, we have
rG = rGbⴰ + RT ln
(cHOAc)(cEtOH)
,
(1)(cEtOAc)
and the apparent unit on Q is M.
Since rG is the same regardless of our choice of standard states, we have
rGaⴰ = rGbⴰ + RT ln (55.5) = rGbⴰ + (8.314 J mol-1 K-1)(298 K) ln (55.5)
= rGbⴰ + 9950 J mol - 1 .
There is about a 10 kJ difference in rG values, illustrating the importance of clearly
communicating the choices of standard states for reacting species.
The decision among the possible standard states and equilibrium constants is simply
to choose the most convenient for the system being studied.
If we do not specify the ionic strength (I ), then we will either have to know the activity coefficients explicitly or we will have to assume that the solutions are dilute enough
so that we can set all activity coefficients equal to unity. Even in very dilute solutions, we
should realize that our computations should be regarded as estimates, while in concentrated
solutions our calculations may be in error by factors of 2 or more if we do not specify
activity coefficients.
It is important to remember that the activity coefficient of a species depends on the
concentrations of all species in solution. If one is studying the ionization constant of dilute
acetic acid in water, activity coefficients of the ions are nearly 1; however, the addition of
NaCl to 1 M concentration will have a strong effect on the activity coefficients of the ions
in the dilute acid, even though NaCl does not play a direct role in the dissociation equilibrium. Later we will discuss methods of measuring activity coefficients.
Calculation of Equilibrium Concentrations: Ideal Solutions
Equilibrium constants enable us to calculate concentrations at equilibrium. We may want
to know how to make buffer solutions of any pH, how much product is obtained from a
reaction, how much metal ion is bound in a complex, and so on. Some of these problems
are simple, but others may require a computer to solve. Here we show a general method
for solving equilibrium problems. We will need the equilibrium constants for all of the
equilibria involved, and we assume for now that all solutions are ideal (all g>s =1).
Most scientists solve equilibrium problems by making intuitive approximations to simplify the arithmetic. The “intuition” comes from having solved many similar problems for
example. The method we present here is general and formalizes some of the techniques that
are often regarded as ‘intuition’. The end goal is to learn the concentrations of all species
in a mixture. The problem is essentially an algebraic one of finding simultaneous solutions for a number of equations. So the critical requirement is to have as many equations
as unknowns. The equations are equilibrium expressions, and two types of conservation
equations. They are conservation of mass —the total mass of each element is not altered
by any chemical reaction—and conservation of charge—the number of positive charges
must always equal the number of negative charges in the mixture. Once the number of
equations equals the number of unknowns, a solution can always be obtained, but we will
look for approximations that allow a simple solution.
Suppose a solution is prepared by adding cA moles of acetic acid and cS moles
of sodium acetate to water to a final volume of 1 L at 298 K. The solution will
contain HOAc, OAc - , Na + , H + , and OH - . We have five species in addition to the
122 Chapter 4 |
Free Energy and Chemical Equilibria
solvent (H2O), and so we need five equations. Using brackets to denote molarities, the
five equations are listed below.
3 Na+ 4 = cS = constant
Mass balance:
3 HOAc 4 + 3 OAc- 4 = cA + cS = contant
Charge balance: [Na + ] + 3 H+ 4 = 3 OAc- 4 + 3 OH- 4
Equilibria:
KHOAc =
3 H+ 4 3 OAc- 4
= 1.8 * 10 - 5
3 HOAc 4
KH2O = 3 H+ 4 3 OH- 4 = 1.0 * 10 - 14
We are using the solute (1 M) standard state for H + , OAc - , HOAc, OH - , and Na + and the
mole fraction standard state for H2O. These five equations allow us to solve any problem
involving only these five species. We must be sure that the equilibrium expressions are
independent. Instead of KA = KHOAc and KH2O, we could have used KH2O and KB,
KB =
3 HOAc 4 3 OH- 4
,
3 OAc- 4
or KA and KB. We cannot count KA, KB, and KH2O as three independent equations; only two
are independent since KAKB = KH2O.
EXAMPLE 4.7
What are the concentrations of all species in pure water?
SOLUTION
Mass balance:
3 Na+ 4 = 0, 3 HOAc 4 = 0, 3 OAc- 4 = 0
Charge balance:
3 H+ 4 = 3 OH- 4
Equilibrium:
3 H+ 4 3 OH- 4 = 1.0 * 10 - 14
Therefore,
3 H+ 4 2 = 1.0 * 10 - 14 , 3 H+ 4 = 1.0 * 10 - 7 , 3 OH- 4 = 1.0 * 10 - 7 .
EXAMPLE 4.8
What are the concentrations of all species in a 0.100 M HOAc solution?
SOLUTION
Mass balance:
3 Na+ 4 = 0
3 HOAc 4 + 3 OAc- 4 = 0.100
Charge balance: [H + ] = 3 OAc- 4 + 3 OH- 4
Equilibria:
KHOAc =
3 H+ 4 3 OAc- 4
= 1.80 * 10 - 5
3 HOAc 4
KH2O = 3 H+ 4 3 OH- 4 = 1.00 * 10 - 14
Equilibrium and the Standard Gibbs Free Energy | 123
Except for pure water or solutions close to pH 7, either 3 H+ 4 or 3 OH- 4 will
be negligible in the charge-balance equation. The solutions will be acidic or basic,
so either 3 OH- 4 or 3 H+ 4 can be ignored. Even for a solution at pH 6.5, the 3 H+ 4
concentration is ten times the 3 OH - 4 . We know that an acetic acid solution is acidic,
so we ignore 3 OH- 4 in the charge-balance equation 13 OAc - 4 >> 3 OH - 42 :
Charge balance: [H + ] ⬵ 3 OAc - 4
Remember that we can sometimes ignore species in sums, but we can never ignore
them in products. We can never ignore either 3 H + 4 or 3 OH - 4 in KH2O.
If we set 3 H + 4 = x, we also have 3 OAc - 4 = x and 3 HOAc 4 = 0.100 - x:
KA =
(x)(x)
= 1.80 * 10 - 5
0.100 - x
We can solve for x in the quadratic equation
x 2 + 1.8 * 10 - 5x - 1.8 * 10 - 6 = 0 ,
x =
- 1.8 * 10 - 5 + 23.24 * 10 - 10 + 7.2 * 10 - 6
= 1.33 * 10 - 3 ,
2
3 H + 4 = 3 OAc - 4 = 1.3 * 10 - 3 ,
3 HOAc 4 = 0.100 - 1.33 * 10 - 3 = 9.9 * 10 - 2 ,
3 OH - 4 =
1.0 * 10 - 14
= 7.5 * 10 - 12 .
1.33 * 10 - 3
Our assumption that [OAc - ] W [OH - ] is well verified. Alternatively, we can solve
the problem using successive approximations. The secret in solving by approximations
is to be able to compare x to concentrations of species that are probably much larger.
In this case, we have set x = 3 H + 4 = 3 OAc - 4 . We do not want to set x = 3 HOAc 4
because it is the species that we expect would have the largest concentration. We can
guess that 3 H + 4 and 3 OAc - 4 will have to be much smaller than 0.100 M to satisfy
the equilibrium constant KHOAc = 1.8 * 10 - 5. As a first approximation, assume that
x V 0.1. Then,
KA =
x2
x2
⬵
= 1.80 * 10 - 5 ,
0.1 - x
0.1
x ⬵ 1.34 * 10 - 3 .
Our initial approximation that x V 0.1 is justified. To obtain a more precise answer,
we can use this first solution to generate a second approximation:
x2
= 1.80 * 10 - 5
0.1 - 0.0013
x = 1.33 * 10 - 3
Clearly, no further approximations are needed.
124 Chapter 4 |
Free Energy and Chemical Equilibria
EXAMPLE 4.9
What are the concentrations of all species in a 0.200 M NaOAc solution?
SOLUTION
NaOAc is the salt of a weak acid, and a hydrolysis reaction occurs in water:
OAc - + H2O N HOAc + OH Our general method is listed below.
Mass balance:
Charge balance:
3 Na + 4 = 0.200 M
3 HOAc 4 + 3 OAc - 4 = 0.200 M
3 H + 4 + 3 Na + 4 = 3 OH - 4 + 3 OAc - 4
Equilibria: KHOAc =
3 H + 4 3 OAc - 4
= 1.80 * 10 - 5
3 HOAc 4
KH2O = 3 H + 4 3 OH - 4 = 1.00 * 10 - 14
We recognize from the hydrolysis reaction that OH - is produced. The solution will
therefore be basic, and the term 3 H + 4 in the charge balance equation can be ignored.
Furthermore, we know that 3 Na + 4 = 0.200 M :
Charge balance: 0.200 M = 3 OH - 4 + 3 OAc - 4
Comparing this with the mass-balance equation for 3 HOAc 4 + 3 OAc - 4 , we see that
3 HOAc 4 = 3 OH - 4 .
We want x to be smaller than other species, so set x = 3 HOAc 4 = 3 OH - 4 since
the equilibrium expression for KHOAc tells us that 3 OAc - ]>[HOAc 4 W 1 because
3 H + 4 V 10 - 7 for a basic solution. Thus,
3 OAc - 4 = 0.200 - 3 HOAc 4 = 0.200 - x ,
3H + 4 =
10 - 14 ,
x
and
KHOAc =
(10 - 14/x)(0.200 - x)
10 - 14(0.200 - x)
=
= 1.80 * 10 - 5 ,
x
x2
(0.200 - x)
= 1.80 * 109 .
x2
It is easy to use successive approximation here, so let x V 0.200, giving
0.200
= 1.80 * 109 ,
x2
x = 1.05 * 10 - 5 . (consistent with assumption)
Equilibrium and the Standard Gibbs Free Energy | 125
No further approximations are needed.
3 OH - 4 = 3 HOAc 4 = 1.05 * 10 - 5 M
3H + 4 =
1.00 * 10 - 14
= 9.5 * 10 - 10 M
1.05 * 10 - 5
3 OAc - 4 = 0.20 M
3 Na + 4 = 0.200 M
Note that 3 Na + 4 W 3 H + 4 , which justifies our early approximation in the chargebalance equation.
For buffer problems, the computations are usually simpler because both the acid and
its salt (or the base and its salt) concentrations are large compared with 3 H + 4 or 3 OH - 4 .
For a buffer involving acetic acid and sodium acetate:
3 Na + 4 = cS = constant
[HOAc] + 3 OAc - 4 = cA + cS = constant
Charge balance: 3 Na + 4 + 3 H + 4 = 3 OAc - 4 + 3 OH - 4
Mass balance:
Then
3 Na + 4 = 3 OAc - 4 = cS ,
because 3 H + 4 and 3 OH - 4 V 3 Na + 4 or 3 OAc - 4 . Thus, 3 HOAc 4 = cA and
KHOAc =
3 H + 4 3 OAc - 4
3 H + 4 cS
=
,
cA
3 HOAc 4
which is often rearranged as (see sidebar)
pH = pKA + log
cS
.
cA
(4.49)
Equation 4.49 is sometimes called the Henderson–Hasselbalch equation. (Note that for
real solutions, the proper definition is pH = -log aH + .)
E X A M P L E 4 .1 0
What amount of solid sodium acetate is needed to prepare a buffer at pH 5.00 from 1 L
of 0.10 M acetic acid?
SOLUTION
pH = pKHOAc + log
5.00 = 4.75 + log
cS
cA
cS
= 5.75 + log cS
0.10
cS = 0.18 mol L - 1
mass of NaOAc = (0.18 mol)(82.0 g mol - 1) = 15 g
This answer assumes that the solutions are ideal and that the addition of sodium
acetate has a negligible effect on the volume. In practice, final adjustment of pH can
be done with a pH meter or other electrode system that measures aH + directly.
So far, we have assumed that all activity coefficients of the solutes are close to 1. Similar
calculations can be carried out, however, when the activity coefficients deviate substantially
from 1. Example 4.11 illustrates such a case.
Mathematics note. The letter
p as a prefix has the meaning
of an operator. As we will see
further in quantum mechanics
(Hilbert space in the Mathematic appendix, and chapter 11) an
operator is an instruction that
must be applied to the terms to
the right of the operator. Thus,
the “p operator” is a prescription
to take the negative logarithm
of the quantity that follows,
as in:
pH = - log 3 H+4
pKA = - log KA
pMg = - log 3 Mg2+4 .
126 Chapter 4 |
Free Energy and Chemical Equilibria
E X A M P L E 4 .11
What are the concentrations of all species in a 0.200 M NaOAc solution if we take into
account the mean ionic activity coefficients for the ions in the solution? The mean ionic
activity coefficient for 1:1 electrolytes in this solution is g { = 0.592; we assume the
activity coefficient of the uncharged HOAc is unity.
SOLUTION
We have previously considered this problem in example 4.9 with the assumption that
all activity coefficients are unity. The conservation equations are the same as before.
3 Na+ 4 = 0.200 M
3 HOAc 4 + 3 OAc
-4
(1)
= 0.200 M
3 H+ 4 + 3 Na+ 4 = 3 OH- 4 + 3 OAc- 4
(2)
(3)
The equilibrium constants KHOAc and KH2O are
KHOAc =
=
KH2O
(aH + )(aOAc - )
(cH + )(cOAc - ) g + g =
(aHOAc)
(cHOAc) gHOAc
3 H+ 4 3 OAc- 4 g2{
3 HOAc 4 gHOAc
= 1.80 * 10 - 5 ,
(aH + )(aOH - )
=
= 3 H + 4 3 OH - 4 g2{
(aH2O)
= 1.00 * 10 - 14 .
Substituting g { = 0.755, we obtain
3 H + 4 3 OAc - 4
1.80 * 10 - 5
=
= 3.16 * 10 - 5 ,
3 HOAc 4
(0.755)2
(4)
3 H + 4 3 OH - 4 =
(5)
1.00 * 10 - 14
= 1.75 * 10 - 14 .
(0.755)2
Solve Eqs. (1) through (5) the same way as described in example 4.9.
3 Na + 4 = 0.200 M
3 OAc - 4 = 0.200 M
3 OH - 4 = 3 HOAc 4 = 1.05 * 10 - 5 M
3 H + 4 = 1.67 * 10 - 9 M
The only change introduced by using activity coefficients is in 3 H + 4 ; previously
(example 4.9) we obtained 3 H + 4 = 9.50 * 10 - 10.
Temperature Dependence of the Equilibrium Constant
Once we know the equilibrium constant of a chemical reaction at one temperature, what
can we say about it at another temperature? We show below that the temperature dependence of an equilibrium constant can be readily derived from the temperature dependence
of the change in Gibbs free energy. From Eq. 3.39, at constant pressure:
d⌬ rG
= - ⌬ rS ( p constant)
dT
Equilibrium and the Standard Gibbs Free Energy | 127
When all reactants and products are at their standard states, the above equation becomes
d rG
= - rS. ( p constant)
dT
(4.50)
Substituting rG = -RT ln K into the equation gives
-R ln K - RT
d ln K
= - rS ,
dT
or
-RT
d ln K
= - rS + R ln K .
dT
Multiplying both sides by T, we obtain
-RT 2
d ln K
= -T rS + RT ln K .
dT
Since RT ln K is equal to - rG,
-RT 2
d ln K
= -T rS - rG
dT
= -T rS - ( rH - T rS)
= - rH .
(4.51)
We can write Eq. 4.51 in an alternate form by noting that d(1/T) = -dT/T 2:
- rH
d ln K
=
1
R
d1 > T 2
( p constant)
In partial derivative form, we have
c
- rH
0 (ln K)
d =
.
1
R
01 > T 2 p
(4.52)
Equation 4.52 is called the van’t Hoff equation. Qualitatively, Eq. 4.52 tells us that if a
reaction is exothermic under standard conditions ( rH is negative, - rH is positive),
then as the temperature decreases and 1/T increases, ln K, and therefore K, becomes larger.
An exothermic reaction is favored when the temperature is lowered. Conversely, an endothermic reaction is favored when the temperature is increased.
For example, we know from experience that a large amount of heat is generated upon
mixing a 1 M acid solution (H + ) with 1 M base (OH - ) in the neutralization reaction to
form water. The reverse reaction (the self-ionization of water) therefore absorbs heat:
H2O(l) S H + (aq) + OH - (aq)
ⴰ
rH298
= 55.84 kJ mol-1
Thus, we expect that increasing the temperature of water will result in an increase in Kv,
the equilibrium constant for the self-ionization of water.
While confirming qualitative predictions that can be made from Le Châtelier’s
principle, Eq. 4.52 provides more detailed quantitative information. If rH is known
as a function of temperature, then Eq. 4.52 can be used to calculate K at any temperature if K is known at one temperature. Or we can determine rH from experimentally
measured values of K at different temperatures. A plot of log K versus 1/T, known as
Free Energy and Chemical Equilibria
FIGURE 4.10 A plot of the
logarithm of the ionization
constant of water KW versus
1/T. This is a van’t Hoff plot. The
slope is equal to -⌬H⬚>2.303R
(Eq. 4.52); the standard heat
of ionization is thus found to
be 55.76 kJ mol- 1, in excellent agreement with the value
obtained from calorimetry.
–13
Enthalpy (heat) of ionization = 55.76 kJ/mol
log Kw
128 Chapter 4 |
–14
–15
0.0030
0.0032
0.0034
0.0036
0.0038
1/T (1/K)
a van’t Hoff plot, for experimental values of K w in the temperature range from 0⬚ to
50⬚C is shown in figure 4.10. The plot gives a straight line, indicating that over the
temperature range ⌬ rH⬚ is a constant. From Eq. 4.52, the slope is - ⌬ rH⬚/2.303R. The
best-fit line through the data shown in figure 4.10 gives ⌬ rH⬚ = 55.76 kJ, which agrees
well with the value 55.82 kJ calculated from the calorimetrically measured standard
enthalpies of formation at 298 K.
If ⌬ rH⬚ changes with temperature, a van’t Hoff plot will show curvature and its slope
at any point will equal - ⌬ rH⬚>2.303R at that temperature. When ⌬ rH⬚ is approximately
constant over the temperature range of interest, integration of Eq. 4.52 is
K2
3 d ln K =
K1
ln
- ⌬ rH⬚ T2 1
d ,
R 3T2 T
K2
- ⌬ rH⬚ 1
1
=
a
b.
K1
R
T2
T1
(⌬ rH⬚ = constant)
(4.53)
This equation is often used to calculate an equilibrium constant K2 at T2 when K1 and the
standard enthalpy of the reaction, ⌬ rH⬚, are known. Alternatively, from the equilibrium
constants measured at two different temperatures, one can calculate the enthalpy.
E X A M P L E 4 .12
The equilibrium constant for ionization of 4-aminopyridine is 1.35 * 10 - 10 at 0⬚C and
3.33 * 10 - 9 at 50⬚C. Calculate ⌬ rG⬚ at 0⬚C and 50⬚C as well as ⌬ rH⬚ and ⌬ rS⬚.
SOLUTION
Values of ⌬ rG⬚ at the temperatures are calculated from ⌬ rG⬚ = -RT ln K :
⌬ rG⬚273 = -(8.314 J K - 1mol - 1)(273 K) ln (1.35 * 10 - 10) = +51.58 kJ mol-1
⌬ rG⬚323 = -(8.314 J K - 1mol - 1)(323 K) ln (3.33 * 10 - 9) = +52.42 kJ mol-1
Biochemical Applications of Thermodynamics | 129
Case 1: Find rH first using Eq. 4.53:
ln
K2
1.35
= ln
= -3.205
K1
33.3
rH = -
(8.314 J K - 1mol - 1)( - 3.205)
= 47.0 kJ mol - 1
5.67 * 10 - 4 K - 1
pK2 = 9.83
Now find rS over the same temperature range from
rS =
CH3
Alanine
rH - rG
47,000 J mol - 1 - 51,580 J mol - 1
=
T
273 K
-1
= -16.78 J K mol
-1
C
pK2 = 3.87
COOH
Aspartic acid CH2
(4.50)
rG(T2) - rG(T1) = - rS(T2 - T1)
(4.54)
pK1 = 2.05
pK3 = 9.16
pK2 = 6.00
H+
N
NH3+
CH2
C
-1
= -16.8 J K mol
-1
,
from which rH can be obtained from the calculated rG at either 323 K
or 273 K:
rH = rG - T rS
= 52,420 J mol-1 - 323 K * 16.8 J K-1 mol-1
COOH
pK1 = 1.82
N
H
Histidine
From the calculated rG values at the two temperatures, Eq. 4.54 gives
52.42 - 51.58
rS = kJ K - 1mol - 1
323 - 273
pK2 = 9.11
NH3+
CH2
C
COOH
H
HO
pK1 = 2.20
pK3 = 10.05
Tyrosine
pK2 = 8.95
H3N+
pK3 = 10.53
CH2
C
CH2
H
= 51,580 J mol-1 - 273 K * 16.8 J K-1 mol-1
NH3+
CH2
CH2
= 46.99 kJ mol - 1
= 46.99 kJ mol
COOH
H
3 d rG = - rS 3 dT
= -0.0168 kJ K mol
NH3+ pK3 = 10.3
C
H
Case 2: Find the average value of rS first from rG values at two temperatures by integrating Eq. 4.50, assuming that rS is constant over the
temperature range:
-1
pK1 = 2.35
.
47,000 J mol-1 - 52,420 J mol-1
= -16.78 J K - 1mol - 1
323 K
-1
COOH
H
The same answer is obtained if values of rH and rG at T = 323 K
are used:
rS =
NH3+
COOH
pK1 = 2.18
Lysine
-1
Biochemical Applications of Thermodynamics
Knowledge of equilibrium constants, free energies, and their dependence on
concentration and temperature is important in biochemistry. Here we discuss
a few examples.
Table 4.2 and figure 4.11 show equilibrium constants at 25C and heats
of ionization for various acids. A great deal of information about the acids is
summarized in a small space. For example, some biochemical systems (such
as human beings) work best at 37 C (98.5 F). Table 4.2 allows you to calculate
the pK ’s at 37 C. We see that acetic acid is one of the few acids listed whose
pK is the same at 37 C as at 25 C, because its rH~0. We usually consider
that a neutral aqueous solution has a pH of 7, but this is true only at 25 C.
pK1 = 3.7
NH2
N
H+N
N
N O
OH OH
CH2
O pK = 6.4
2
O P OH
OH
pK3 = 13.1
5-Adenylic acid
FIGURE 4.11 The pK values from table 4.2
of several amino acids and of a nucleotide;
pK = - log K.
130 Chapter 4 |
Free Energy and Chemical Equilibria
TABLE 4.2 lonization constants and enthalpies (heats) of ionization at 25°C
Compound
Ionizing species*
pK†
Acetic acid
HOAc S H+ + OAc
4.76
−0.08
3.5
13.0
+
3.7
17.6
6.4
−7.5
pA S pA2- + H+
13.1
45.6
4.0
15.5
Adenosine
5’-Adenylic acid
(adenosine 5’-phosphate)
+
AH S A = H+
+
pAH S pA = H
-
pA S pA + H
+
-
Adenosine triphosphate (ATP)
+
pppAH S pppA + H
-
pppA S pppA + H
Alanine
+
+
Ammonia
Aspartic acid
Carbonic acid
Fumaric acid
-
+
H3NRCOO S H2NRCOO + H
+
HCO3 S CO23 + H
R(COOH)2 S -OOCRCOOH +
S+
-
+
+
-
+
+
-
H3NRCOO S H2NRCOO + H
+
+
−5.0
2.9
9.87
45.2
9.24
52.2
2.05
7.5
3.87
4.2
7.66
10.24
14.85
3.10
0.4
4.6
−2.9
2.35
9.78
3.92
44.2
H3NRH COOH S H3NRH COO + H
1.80
—
+
H3NRH+COO- S +H3NRCOO- + H+
6.04
29.9
+
H3NRCOO- S +H2NRCOO- + H+
9.33
46.6
9.21
43.5
-
+
+
+
+
38.5
6.36
+
HCN S CN + H
+
H3NRH+COOH S +H3NRH+COO- + H+
+
-
+
-
H3NRH COO S H3NRCOO + H
-
+
-
H3NRCOO S H2NRCOO + H
+
+
ROH S RO- + H+
H3PO4 S
H2PO4-
H2PO4-
S
HPO24
HPO24
+
S PO34 + H
+ H
+
+ H
+
Pyruvic acid
RCOOH S RCOO- + H+
Tyrosine
+
Water
+
OOCRCOO + H
Lysine
Phosphoric acid
H
7.0
2.34
10.0
H3NRCOOH S +H3NRCOO- + H+
Hydrocyanic acid
Phenol
+
NH4+ S NH3 + H+
+
H3NRCOOH S +H3NRHCOO + H+
+
H3NRHCOO- S +H3NR-COO- + H+
+
H3NR-COO- S +H2NR-COO- + H+
H2CO3 S HCO3- + H+
+
Histidine
+
H3NRCOOH S +H3NRCOO + H+
OOCROOH
Glycine
+
rH°(kJ molⴚ1)
+
-
H3NRHCOOH S H3NRHCOO + H
-
H3NRHCOO S +H3NR-COO- + H+
+
H3NR-COO- S H2NR-COO- + H+
H2O S OH + H
+
+
1.26
9.06
53.6
10.53
48.5
9.98
23.6
2.16
−7.95
7.21
+
-
2.16
4.15
12.32
14.7
2.49
12.1
2.20
—
9.11
—
10.1
25.1
14.00
55.82
* The ionizable groups corresponding to the pK’s in the table are shown in figure 4.11. Glycine is like alanine with the methyl group replaced by a hydrogen. Adenosine
is like 5’-adenylic acid except the phosphate is lacking. In ATP the single phosphate (OPO3H2) of adenylic acid is replaced by a triphosphate (OPO2)-(OPO2)-(OPO3H2).
pK = - log K. The K values are thermodynamic equilibrium constants on the molarity scale.
Source: Handbook of Biochemistry and Molecular Biology, 3d ed., Physical & Chemical Data, Vol. 1 (CRC Press,1968, 1976); Dawson’s Data on Biochemical
Research; NBS Tables; Perrin, D. D. pKa prediction for organic acids and bases. (London; New York: Chapman and Hall, 1981); Perrin, D. D. Ionisation constants of
inorganic acids and bases in aqueous solution. (Oxford [Oxfordshire]; New York: Pergamon Press, 1982).
Biochemical Applications of Thermodynamics | 131
E X A M P L E 4 .13
What is the pH of pure water at 37C?
SOLUTION
The temperature dependence of the equilibrium constant is
ln
K2
rH 1
1
= a
b .
K1
R
T2
T1
(4.53)
The ionization constant of water is 1.00 * 10 - 14 at 25C, its rH = 55.82 kJ mol-1
(see table 4.2), K1 = 10 - 14, T1 = 298 K, and T2 = 310 K. Then,
ln
K2
10 - 14
=
-55,820 J mol-1
1
1
a
b
-1
-1 310 K
298 K
8.314 J K mol
= +0.87
K2
10 - 14
= e + 0.87 = 2.39
K 2 = 2.39 * 10 - 14 .
This is the ionization constant of water at 37 C. [H+] and [OH-] are surely low enough
so that the activities of the ions equal the concentrations. At 37 C ,
3 H + 4 3 OH - 4 = 2.4 * 10 - 14
3 H + 4 = 1.55 * 10 - 7 M
pH = -log 3 H + 4
= 6.81 .
E X A M P L E 4 .14
A buffer is prepared by adding 26.8 mL of 0.200 M HCl to 50.0 mL of 0.200 M
tris(hydroxymethyl)aminomethane, a weak base known as Tris and widely used as
a buffer by biochemists. Tris has a pK = 8.3 at 20C: its pK is temperature dependent with a temperature coefficient of pK> T = -0.029 K - 1. The mixture is then
diluted with pure water to a total volume of 200 mL. (a) What is the pH of the
buffer solution at 20C? (b) What is the “buffer capacity” of this solution, expressed
as the change in pH that would occur upon the addition of 1 mmol of strong acid or
base to 1 L of the buffer? (c) What would be the pH of the buffer solution of part
(a) at 37 C?
SOLUTION
The equilibrium involved for Tris buffer is
HOCH2
HOCH2 — C — NH3
HOCH2
Tris·H
HOCH2
HOCH2 — C — NH2 + H+ .
HOCH2
Tris
132 Chapter 4 |
Free Energy and Chemical Equilibria
a. Addition of the HCl solution (26.8 mL of 0.200 M HCl contains 5.36 mmol of HCl)
to the Tris solution (50.0 mL of 0.200 M Tris contains 10.00 mmol of Tris) causes
the conversion of 5.36 mmol of Tris to TrisH + . In the final 200 mL of solution,
3 Tris # H + 4 =
5.36 mmol
= 26.8 * 10 - 3 M ,
200 mL
(10.00 - 5.36) mmol
3 Tris 4 =
= 23.2 * 10 - 3 M .
200 mL
Using Eq. 4.49
pH = 8.3 + log
23.2 * 10 - 3
= 8.3 - 0.063 = 8.2 .
26.8 * 10 - 3
b. The buffer capacity is a measure of the resistance to change in pH upon addition
of H + or OH - . Pure water has no buffer capacity. Addition of 1 mmol of HCl (or
NaOH) to 1 L of water changes the pH from 7 to 3 (or 11). For 1 L of the buffer
solution of part (a), addition of 1.0 mmol of HCl, we would have
3 Tris # H + 4 = (26.8 + 1.0) * 10 - 3 = 27.8 * 10 - 3 M ,
3 Tris 4 = (23.2 - 1.0) * 10 - 3 = 22.2 * 10 - 3 M ,
pH = 8.3 + log
22.2 * 10 - 3
= 8.3 - 0.097 = 8.2 .
27.8 * 10 - 3
The change is only pH = 0.097 - 0.063 = 0.034 pH units. The buffer is still
nominally at pH 8.2. This illustrates the ability of buffer solutions to resist changes
in pH when, for example, H + is produced or consumed in metabolic reactions. Note
that the buffer capacity is a direct function of the buffer concentration. This 50 mM
Tris buffer solution has about ten times the buffer capacity of a 5 mM buffer solution
at pH 8.2.
c. The pK for Tris decreases by 0.029 per degree (K) rise in temperature. At 37 C,
pK37 = pK20 - (0.029)(37 - 20) = 8.3 - 0.49 = 7.8
and the pH of the solution prepared as described above would be
pH37 = pK37 - 0.063 = 7.7.
This is a very significant change with temperature and should be kept in mind in all
biochemical investigations. The temperature coefficient can be related to an enthalpy
change upon dissociation, as tabulated for many weak acids in table 4.2. Beginning
with Eq. 4.53, we can readily derive the equation
rHTⴰ =
2.303RT1T2
K2
2.303RT1T2
log
=
(pK1 - pK2) .
T1 - T2
K1
T2 - T1
For Tris,
(2.303)(8.314 J K - 1 mol - 1)(298 K)(299 K)
(+0.029)
1K
= 49 kJ mol - 1 .
ⴰ
rH298
=
From the pK values given in table 4.2 or those that we calculate at other temperatures, we
can calculate the activities and concentrations of various ionized species present in solution.
For example, consider histidine in solution at 25 C. The equilibria are the following:
Biochemical Applications of Thermodynamics | 133
N H 3
HN
CH2
N H
C
COOH
N H 3
pK1
1.82
CH 2
HN
H
H isH 3
N H
2+
HN
NH pK2
6.00
CH 2
HN
H
N
HN
N
COO − H
H isH
NH 3+
N H2
COO−
C
C
H
H isH 2+
CH2
H
N H 3
COO−
C
COO − H
H isH 2
N H 3+
CH2
C
pK3
9.16
CH 2
HN
H
N
C
COO− H
H
H is −
H isH
We often need to know the concentration of each species at a given pH. Each ionic species
has different chemical reactivity and physical properties; therefore, the properties of the
solution will depend on the concentration of each species.
E X A M P L E 4 .15
Calculate the concentration of each species in a 0.10-M solution of histidine at pH 7.
SOLUTION
K1 = 1.51 * 10 - 2 =
3 H + 4 3 HisH2+ 4
3 HisH23 + 4
K2 = 1.00 * 10 - 6 =
3 H + 4 3 HisH 4
3 HisH2+ 4
K3 = 6.92 * 10 - 10 =
3 H + 4 3 His - 4
3 HisH 4
3 H + 4 = 1 * 10 - 7 M
We assume that activities are equal to concentrations. If necessary, we could estimate
activity coefficients for greater accuracy. The mass-balance equation is
+
3 HisH2+
3 4 + 3 HisH2 4 + 3 HisH 4 + 3 His 4 = 0.100 M .
We now have five equations and five unknowns, so we can solve for the unknowns. It
will be convenient to solve by successive approximations. First, we find the ratios of species from the three equilibrium constants and 3 H + 4 = 1 * 10 - 7 M :
3 HisH2+ 4
3 HisH 4
3 His - 4
5
=
1.51
*
10
,
=
10.0,
= 6.92 * 10 - 3
3 HisH23 + 4
3 HisH2+ 4
3 HisH 4
134 Chapter 4 |
Free Energy and Chemical Equilibria
From the ratios of species, we see that a good first approximation will be to ignore
3 HisH2+
3 4 and 3 His 4 in the mass-balance equation. That is, we need consider only the
two species whose pK’s are closest to the pH of the solution.
So assuming 3 HisH23 + 4 + 3 His - 4 V 3 HisH 4 + 3 HisH2+ 4 , the mass
balance is
3 HisH 4 + 3 HisH2+ 4 ⬵ 0.100 M ,
which now allows all quantities to be calculated.
3 HisH 4 = 3 HisH2+ 4 # (10.0)
3 HisH2+ 4 (10.0) + 3 HisH2+ 4 = 0.100 M
3 HisH2+ 4 =
0.100
= 9.1 * 10 - 3 M
11.0
3 HisH 4 = 10 3 HisH2+ 4 = 9.1 * 10 - 2 M
3 His - 4 = (6.92 * 10 - 3)(9.1 * 10 - 2) = 6.3 * 10 - 4 M
3 HisH23 + 4 =
9.1 * 10 - 3
= 6.0 * 10 - 8 M
1.51 * 105
We have ignored activity coefficients, so the concentrations are at best accurate to
{10%. If the second approximation does not change values by more than 10%, we can
stop at the first approximation. The second approximation mass balance is
3 HisH 4 + HisH2+ 4 = 0.100 - 3 His - 4 - 3 HisH23 + 4
= 0.0994 M .
The first approximation is good enough.
Using the methods illustrated in example 4.15 we calculated the concentrations of all
histidine species as a function of pH as shown in figure 4.12. The pK’s correspond to the
pH values where two species cross—where their concentrations are equal. Note that the
0.1
2.0
His–
HisH
1.5
0.08
HisH2+
3
1.0
0.06
0.5
net charge
0.04
Net Charge
+
HisH2
Concentration (M)
FIGURE 4.12 The concentrations of histidine species as a
function of pH as calculated in
example 4.12. Note the sharp
change in concentration of
species near their pK’s (1.82,
6.00, 9.16), and that no more
than two histidine species are
present in significant concentrations at each pH. The net charge
decreases in steps from ⫹2 to
⫹1 (pH 4) to 0 (pH 8) to ⫺1. This
is analogous to what happens to
a protein molecule as a function
of pH. It will have a net positive
charge at low pH, a zero charge
at its isoelectric point, and a
negative charge at high pH.
0
0.02
-0.5
0
0
2
4
6
pH
8
10
12
Biochemical Applications of Thermodynamics | 135
concentration of each species reaches a maximum near 0.1 M (the total concentration of
all histidine species).
These methods can of course be used to calculate the concentrations of species needed
to produce a buffer of a given pH and salt concentration. For common buffers, the algebra
has already been done, and buffer tables can be found in such handbooks as the Handbook
of Biochemistry and Molecular Biology (CRC Press, Boca Raton, Florida), and online.*
Thermodynamics of Metabolism
Metabolism refers to the process by which cells use energy from their environment to
synthesize the building blocks of their macromolecules. Hundreds of reactions are
involved, and the path to a single product can involve many intermediates. An advantage
of thermodynamic analysis is that we can focus only on the beginning and the end states
of any process. However, we can also learn about the driving force for each of the steps.
Figure 4.13 presents a scheme for glycolysis, the metabolic process that converts
glucose into pyruvate and produces adenosine triphosphate, ATP. In aerobic organisms,
pyruvate then enters the mitochondria, where it is completely oxidized to CO2 and H2O in
rG°
Glucose
ATP
(kJ mol–1)
–16.7
ADP
Glucose-6-phosphate
+1.7
Fructose-6-phosphate
ATP
–14.2
ADP
Fructose-1,6-bisphosphate
+23.8
Dihydroxyacetone
phosphate
rG°
= +7.5
Glyceraldehyde 3-phosphate
NAD+ + Pi
2(+6.3)
NADH + H+
1,3-Bisphosphoglycerate
ADP
2(–18.8)
ATP
3-Phosphoglycerate
2(+4.6)
2-Phosphoglycerate
2(+1.7)
H2O
Phosphoenolpyruvate
ADP
2(–31.4)
ATP
Pyruvate
*http://microscopy.berkeley.edu/Resources/instruction/buffers.html, and http://www.sigmaaldrich.com/
life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html
Under less aerobic conditions,
in muscle during active exercise,
pyruvate is converted to lactate.
Under anaerobic conditions in
yeast, pyruvate is transformed
into ethanol.
FIGURE 4.13 The glycolytic
path for the conversion of
glucose to pyruvate in cells.
The ¢rG° for reactions from
glyceraldehyde-3-phosphate
to pyruvate are multiplied by 2
because 2 mol of these species
are formed per mole of glucose.
136 Chapter 4 |
Free Energy and Chemical Equilibria
the citric acid cycle. These reactions produce additional ATP. The overall oxidative reaction
(combustion) of glucose with O2(g) to form CO2(g) and H2O(l) is associated with a large
negative Gibbs free-energy change. Using ⌬ f G ⴰ values at 298 K from tables A.5–7 in the
appendix, we obtain for the reaction
ⴰ
C6H12O6(s) + 6O2(g) S 6CO2(g) + 6H2O(l) . (⌬ rG298
= -2878.5 kJ mol-1)
a-D-Glucose
Furthermore, most of this free-energy change arises from the enthalpy contribuⴰ
ⴰ
tion, ⌬ rH298
= -2801.6 kJ mol-1, and relatively little from the entropy, -T⌬ rS298
=
-1
-77.2 kJ mol . Thus, when glucose is burned directly in oxygen, nearly all the chemical
potential is released as heat. By contrast, in living cells a significant fraction of the
chemical potential is retained as chemical bond energy in the form of ATP and other
synthesized molecules.
TABLE 4.3 Standard free energies of reaction at 25 °C, pH 7 for steps in the
metabolism of glucose
¢rG°⬘(kJ molⴚ1)
Reaction
D-Glucose + ATP S D-glucose-6-phosphate + ADP
- 16.7
D-Glucose-6-phosphate S D-fructose-6-phosphate
+ 1.7
D-Fructose-6-phosphate + ATP S D-fructose-1,6-bisphosphate + ADP
- 14.2
Fructose-1,6-bisphosphate S dihydroxyacetone phosphate
+ glyceraldehyde-3-phosphate
Dihydroxyacetone phosphate S glyceraldehyde-3-phosphate
+ 23.8
S
Glyceraldehyde-3-phosphate + phosphate
1,3-bisphosphoglycerate NADH + H +
+ 6.3
+NAD+
1,3-Bisphosphoglycerate + ADP S 3-phosphoglycerate + ATP
+ 7.5
- 18.8
3-Phosphoglycerate S 2-phosphoglycerate
+ 4.6
2-Phosphoglycerate S phosphoenolypyruvate + H2O
+ 1.7
2-Phosphoenolpyruvate + ADP S pyruvate + ATP
+
Pyruvate + NADH + H S lactate + NAD
Pyruvate S acetaldehyde + CO2
+
Acetaldehyde + NADH + H S ethanol + NAD
- 31.4
- 25.1
+
- 19.8
+
- 23.7
*An
important reaction in many of these steps is the hydrolysis of ATP: ATP + H2O S ADP + phosphate.
⌬ rG⬚⬘ = - 31.0 kJ mol - 1. ⌬ rH⬚⬘ = 24.3 kJ mol - 1.
Source: Matthews, Christopher K., Biochemistry, 1st Ed., (c) 1990. Reprinted and electronically reproduced by permission of
Pearson Education, Inc., Upper Saddle River, NJ 07458.
The free energies at 25 ⬚C and pH 7 for the steps in the metabolic degradation
(catabolism) of glucose are given in table 4.3 and illustrated in figure 4.8. The biochemical
standard state is used in this table. This means that the pH is 7, the activity of water is
equal to 1, metabolites are solutes in aqueous solutions, and the activities of all solutes are
replaced by their total concentrations in mol L - 1. For example, ATP in the reaction refers
to the sum of all ionized species of ATP present at pH 7. Standard states of all solutes have
a total concentration equal to 1 M, but the solution is ideal.
The first step in the metabolism of glucose (table 4.3) is the formation of glucose-6phosphate, which has a relatively large positive standard free-energy change and will not
occur significantly under physiological conditions:
glucose+ phosphate S glucose- 6- phosphate + H2O
(⌬ rG⬚⬘ = + 14.3 kJ)
(1)
Biochemical Applications of Thermodynamics | 137
In the first step of glycolysis, however, this reaction is coupled to the hydrolysis of ATP to
form adenosine diphosphate, ADP, which is thermodynamically favorable,
ATP +H2O S ADP + phosphate .
( rG = - 31.0 kJ)
(2)
The sum of these two reactions has a negative standard free energy, and that process occurs
spontaneously. Adding (1) and (2), we obtain
glucose + ATP S glucose- 6- phosphate+ADP .
( rG = -16.7 kJ)
(3)
In this manner the strongly spontaneous hydrolysis of ATP is coupled to the otherwise
nonspontaneous glucose phosphorylation. This reaction is typical of the role played by
ATP in metabolism.
All reactions in the glycolytic path are enzyme catalyzed; the first step is catalyzed
by the enzyme hexokinase. The enzyme speeds the reaction and serves to control the
glycolytic path in response to the cell’s demand for metabolic energy, but the enzyme does
not affect the thermodynamics of the process. The free-energy changes are associated with
differences in chemical potential between reactants and products; they do not indicate how
fast the processes occur.
The second step in the metabolism of glucose is the isomerization of glucose-6phosphate to fructose-6-phosphate catalyzed by the enzyme phosphoglucose isomerase.
This reaction is also nonspontaneous in the standard state, but the positive free-energy
change is relatively small, rG = +1.7 kJ mol - 1. The equilibrium constant associated
with this free-energy change at 25C is 0.50, which means that significant amounts of
product are formed under glycolytic conditions. Furthermore, the glycolytic process is
drawn on by the third step in the process, which is the attachment of a second phosphate
group to fructose-6-phosphate to form fructose-1,6-diphosphate. This step is also driven
by phosphate transfer from ATP and has a strongly negative rG (-14.2 kJ mol - 1).
In the fourth step in the glycolytic path, the six-carbon sugar fructose-1,
6-bisphosphate is split into two three-carbon compounds, dihydroxyacetone phosphate
and glyceraldehyde-3-phosphate, catalyzed by the enzyme aldolase. This reaction has a
large positive free-energy change,
fructose@1,6@bisphosphate S dihydroxyacetone phosphate+ glyceraldehyde@3@phosphate , (4)
( rG298 = +23.8 kJ mol-1)
which would make it very nonspontaneous under standard conditions. But the concentrations of these metabolites in the cell are actually much less than 1 M, and this results in a
very small value for the reaction quotient, Q ⬇10 - 4, for this reaction. The effect of such
a small reaction quotient is to cause the actual free energy, rG, to be significantly less
positive. At 37 C each factor of 0.1 in the reaction quotient decreases the free energy by
about 6 kJ mol - 1 (RT ln 0.1 = -5.9 kJ mol - 1), so the effective value for the free-energy
change under actual metabolic conditions is rG ⬇0. This is a direct consequence of the
fact that in reaction (4) there are more product species than reactant species—all at low
concentrations. Le Châtelier’s principle tells us that a decrease in overall concentration for
such a reaction will favor products. Even though the system is distinctly not at equilibrium
under physiological conditions, that is the direction in which the spontaneous process is
driven under low-concentration conditions. An example of this behavior was presented in
example 4.2.
Dihydroxyacetone phosphate is in equilibrium with glyceraldehyde-3-phosphate. The
free energy change for the reaction
dihydroxyacetone phosphate m glyceraldehyde@3@phosphate ( rG298 = + 7.5 kJ mol-1)
138 Chapter 4 |
Free Energy and Chemical Equilibria
gives an equilibrium constant K = 0.05. Nevertheless, the next reactions in the glycolytic
path drain the pool of glyceraldehyde-3-phosphate sufficiently that essentially all dihydroxyacetone phosphate is converted to glyceraldehyde-3-phosphate.
The next step is the simultaneous oxidation and phosphorylation of glyceraldehyde-3phosphate to 1,3-diphosphoglycerate. The oxidation is carried out by the biological oxidant
nicotinamide adenine dinucleotide (NAD), and the phosphate is taken up directly from the
pool of “inorganic” phosphate, Pi:
glyceraldehyde@3@phosphate + NAD + + Pi S 1,3@bisphosphoglycerate + NADH + H +
( rG298 = 6.3 kJ mol-1)
Although this reaction has a positive standard free-energy change, the process continues
because the next step in glycolysis is very favorable, which is the transfer of one of the
phosphates of 1,3-diphosphoglycerate to ADP to generate ATP and 3-phosphoglycerate:
1,3@bishosphoglycerate + ADP S 3@phosphoglycerate + ATP ( rG298 = -18.8 kJ)
The remaining three steps on the path to pyruvate involve two isomerization reactions and a final step in which the remaining phosphate is transferred to ADP to form
ATP. No new principles are involved in these steps. The overall process by which one
molecule of glucose is metabolized all the way to pyruvate can be summarized by the
reaction
glucose + 2NAD + + 2ADP + 2Pi S 2pyruvate + 2NADH + 2ATP + 2H + + 2H2O
( rG298 = -80.6 kJ mol-1) .
In the overall glycolytic process to form pyruvate from glucose, there is a net formation
of 2 mol of ATP and 2 mol of the physiological reductant NADH per mole of glucose
consumed. In this way a significant fraction of the available chemical potential is
retained as chemical energy for use in other biosynthetic or energy-requiring reactions.
In yeast, pyruvate undergoes two-step fermentation to ethanol (table 4.3). For animals,
under strenuous exercise in muscle cells, NADH reduces pyruvate to lactate:
pyruvate + NADH + H + S lactate + NAD +
( rG298 = -25.1 kJ mol-1)
E X A M P L E 4 .16
The enzyme aldolase catalyzes the conversion of fructose-1,6-bishosphate (FDP) to
dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P). Under
physiological conditions in red blood cells (erythrocytes), the concentrations of these
species are [FDP] = 35 mM, 3 DHAP 4 = 130 mM, and 3 G3P 4 = 15 mM. Will the
conversion occur spontaneously under these conditions?
SOLUTION
For FDP S DHAP + G3P, we have rG298 = +23.8 kJ mol-1 and so
Q =
3 DHAP 4 3 G3P 4
(130 * 10 - 6)(15 * 10 - 6)
=
= 55.7 * 10 - 6 ,
3 FDP 4
(35 * 10 - 6)
rG = +23,800 J mol-1 + (8.314 J K-1 mol-1)(298 K) ln (55.7 * 10 - 6)
= -500 J mol-1 .
Under the actual conditions in the cell, the free energy of the reaction is negative, and
the reaction will occur spontaneously.
Isothermal Titration Calorimetry | 139
Isothermal Titration Calorimetry
Ligand Soln
A powerful method for studying equilibrium thermodynamics of biochemical reactions
focuses on detecting small heat changes upon noncovalent binding of very small quantities
of a ligand molecule, such as a drug or substrate, by a target host such as a protein or other
macromolecule:
+
M
L
.
M L
For an equilibrium of a ligand (L) interacting with a target macromolecule (M) to form a
noncovalent complex (M # L) it can be very difficult to make direct measurements of the
amounts of the reacting species. There may not be convenient fluorescent or absorptive
regions of the molecules to monitor by optical methods, for example. Importantly, in the case
of very strong binding, it may be very difficult to detect the very small amounts of unbound
reactants. Isothermal titration calorimetry (ITC, figure 4.14) presents a route to measure,
with good precision, the thermodynamic properties of very strong binding equilibria. We
will discuss a common case in which the macromolecule has one binding site. The fraction
of bound sites is written
fM =
3M # L 4
,
3M 4 + 3M # L 4
where we can substitute the equilibrium constant K to give
3L 4 K
.
1 + 3L 4 K
(4.55)
3 L 4 Tot = 3 L 4 + fM 3 M 4 Tot
(4.56)
fM =
The mass conservation relations are [L]Tot = [L] + [M # L], and [M]Tot = [M] + [M # L],
where free ligand ([L]) and free macromolecule ([M]) may be difficult to measure, but it is
straightforward by design to know the total amounts of each. The amount of bound ligand
can also be written as a function of fM:
Our goal is to express the fraction of bound sites with respect to quantities that are controlled by the experimentalist while avoiding quantities that would be experimentally inaccessible, such as the amount of free ligand [L] which could be below reasonable detection
limits for cases of very strong binding. Notice that eliminating [L] with Eqs. 4.55 and 4.56
results in a quadratic expression for fM which depends only on the total macromolecule
and total ligand concentration:
2
fM
- fM a
1
3 M 4 TotK
+
3 L 4 Tot
3 L 4 Tot
+ 1b +
= 0
3 M 4 Tot
3 M 4 Tot
(4.56A)
In ITC about 20–25 very small volumes of ligand solution are injected over time into
a small reaction vessel containing the macromolecule of interest with stirring to ensure
rapid mixing. The heat released with each injection of ligand is detected by comparing the
temperature of the reaction vessel to that of a reference cell that has been prepared with
buffer only and which receives no injections. As the amount of ligand increases, there will
be a steep drop in binding, and therefore heat evolution, when the macromolecule becomes
saturated with ligand. That is, the total heat produced for binding a fraction of the sites fM is
Q = fM( 3 M 4 TotV0) rH ,
(4.57)
Buffer
[M]Tot
FIGURE 4.14 Basic schematic
of an ITC instrument. Two small
reaction cells (ca. 1 ml each) are
adiabatically enclosed and the
binding reaction is conducted in
one using automated, carefully
calibrated injections from
a syringe containing ligand
solution. A feedback circuit
monitors the temperature
difference between the reference and reaction cells and
maintains both cells at the same
temperature throughout the
injections. Fins are attached to
the tip of the syringe and the
syringe tip is rotated to promote
mixing during the injections.
Corrections must be made
in simulations and in all data
analysis for dilution of the
macromolecule and ligand with
successive injections; these are
simple and are always implemented in commercial analysis
software.
A full ITC study includes running
several additional experiments
including injecting ligand into
buffer only (no macromolecule)
as a blank, whose data will be
subtracted from the binding
data.
Free Energy and Chemical Equilibria
FIGURE 4.15 Simulation of an
ITC experiment for a binding
constant of K 1 * 107, and
one site per macromolecule.
The data are computed based
on 5 μL injections of 0.3 mM
ligand (initial in syringe) to
a 10 μM (initial in reaction
cell) macromolecule solution.
Injections are conducted in
evenly spaced time increments,
but it is common to present
the data as a function of
the mole ratio L:M, which
can be calculated from the
experimental parameters.
0
0.5
(time)
Mole Ratio L : M
1.0
1.5
Joule/sec
140 Chapter 4 |
∆H
1:1 binding ratio
peak area = enthalpy per injection
where V0 is the initial volume of macromolecule solution. It can be seen that if the fraction of bound sites is 1, then the total heat is the product of the moles of binding sites
with the standard molar enthalpy, rH. ITC binding plots such as in figure 4.15 can be
fit to Eq. 4.57 (green line). Notably, ITC data yields the equilibrium constant K and thus
rG at constant temperature, and also the standard enthalpy change so that the standard
entropy change rS can then be found using rG = rH - T rS.
We should explain in more detail the use of Eq. 4.57, which is for one binding site,
to fit data such as in figure 4.15. Notice first that Eq. 4.57 is dependent upon fM, which
depends in turn on just the equilibrium constant K and the molar enthalpy of binding,
rH. Next, recognize that Eq. 4.57 is the total Q for all injections: if there have been j
injections then Eq. 4.57 is the sum of the areas (shaded for emphasis) of the first j peaks
in figure 4.15. Then we can use Eq. 4.57 to find the heat per injection so that the heat of
just the j’th injection is Q j-Q j–1 . The green curve is generated by finding the values of K
and the heat of binding which lead to the best fit to all of the individual heats of injection.
Double Strand Formation in Nucleic Acids
The formation of a double-stranded helix from two single strands of DNA is crucial in
identifying and locating specific sequences of nucleotides in a DNA. A probe oligonucleotide (a single strand of DNA typically containing 20 to 40 nucleotides) is added to
the single-stranded target DNA to form a double helix with its complementary sequence.
The double helix has Watson–Crick complementary base pairs in which adenine (A) pairs
with thymine (T) and guanine (G) pairs with cytosine (C). This technique is used to locate
genes in chromosomes, to diagnose infectious or genetic diseases, and to amplify DNA
by the polymerase chain reaction (PCR). In PCR amplification, the reaction mixture contains a minute amount of single- or double-stranded DNA to be amplified, a pair of oligonucleotides called the primers, a DNA polymerase stable at a temperature high enough
to denature the DNA in its double-stranded form, and deoxynucleoside triphosphates that
the polymerase uses in DNA synthesis. Samples for PCR are placed in an instrument programmed to cycle the incubation temperature through various settings. In each cycle, the
temperature is first raised to denature double-stranded DNA, and then lowered to allow
the binding of one primer to one DNA strand and the other primer to the complementary
Isothermal Titration Calorimetry | 141
strand. The DNA polymerase then extends the 3 ends of the bound primers to copy the
DNA strands. Following this extension, a second cycle is started. As each cycle doubles
the number of DNA molecules, rapid amplification can be obtained.
Knowledge of the equilibrium constant for formation of a DNA double strand of any
sequence at any temperature is of great interest. Systematic studies were done by mixing
complementary single strands in 1 M NaCl at pH 7 (Allawai and SantaLucia 1997). The
amount of double helix—and thus the equilibrium constant—was determined from the
decrease in ultraviolet light absorbance when a DNA double helix forms (see chapter 10).
The standard free-energy changes were calculated from the equilibrium constants; calorimetric measurements provided the enthalpy changes.
When a series of similar reactions is studied, one looks for simple patterns in the
data. The obvious question to ask for formation of DNA double strands is, does the thermodynamics just depend on the number of A•T base pairs and the number of G•C base
pairs formed? The answer is no; this simple assumption does not fit the data. The next
approximation is to assume that the sequence is important only through the base-pair nearest neighbors. This approximation is valid. We can rationalize it by noting that the main
interaction involved in forming a double strand is hydrogen bonding between base pairs,
plus London–van der Waals interaction between neighboring (stacked) base pairs. Thus,
rG, rH, rS—and also K as a function of temperature—can be calculated from ten
Watson–Crick nearest-neighbor parameters, plus parameters for the formation of the first
base pair of the double strand as shown in table 4.4. The general expression is
rG = rG(initiation) + rG(nearest neighbors)
(4.65)
and similar expressions for rH and rS. Consider the reaction
5-ATAGCA -3 + 5-TGCTAT-3 N
5-ATAGCA -3
|| | | | |
.
3-TATCGT-5
The double strand contains antiparallel complementary strands; the structures of the
molecules are shown in figures 3.6 and 3.7 and table A.8 in the appendix. To calculate the
thermodynamic values, we use Eq. 4.65. The rG (initiation) is positive—unfavorable. It
mainly represents the loss of entropy for bringing the two strands together; the enthalpy
of forming the first base pair is small. The rG (nearest neighbors) are all negative—
favorable; they characterize the hydrogen bonding and stacking interactions of adding
successive base pairs. It does not matter which base pair is assumed to form first. However, it is convenient to start at the left-hand end for counting nearest neighbors. For the
example given, the result is
A
rG = rG(initiation) + rG ° 兩
T
G
+ rG° 兩
C
T
T
兩 ¢ + rG ° 兩
A
A
C
C
兩 ¢ + rG ° 兩
G
G
A
A
兩 ¢ + rG ° 兩
T
T
G
兩¢
C
A
兩 ¢.
T
Using table 4.4
rG(kJ mol-1) = 8.1- 3.7- 2.4 - 5.4- 9.3- 6.0 = - 18.7 kJ mol-1 (K 310 = e - rG/RT = 1416).
For equilibrium constants at other temperatures, calculate rH and rS from table 4.4;
then, assuming rH and rS are independent of temperature, rGTⴰ = rH - T rS.
142 Chapter 4 |
Free Energy and Chemical Equilibria
TABLE 4.4 Thermodynamic parameters* for calculating double-strand stability
in DNA (pH 7, 1 M Nacl)
rGⴗ (kJ molⴚ1) at 37ⴗC
rHⴗ(kJ molⴚ1)
rSⴗ(JKⴚ1 molⴚ1)
A–
5 –A
·
3 –T
T–
–A–A
· ·
–T–T
-4.2
-33.1
-93.0
T–
5 –A
·
3 –T
A–
–A–T
· ·
–T–A
-3.7
-30.2
-85.4
–T–A
· ·
–A–T
-2.4
-30.2
-89.2
–A–C
· ·
–T–G
-6.0
-35.2
-93.8
–C–A
· ·
–G–T
-6.0
-35.6
-95.0
–A–G
· ·
–T–C
-5.4
-32.7
-87.9
–G–A
· ·
–C–T
-5.4
-34.3
-93.0
–C–G
· ·
–G–C
-9.1
-44.4
-113.9
–G–C
· ·
–C–G
-9.3
-41.0
-102.2
–C–C
· ·
–G–G
-7.7
-33.5
-83.3
A–
5 –T
·
3 –A
T–
C–
5 –A
·
3 –T
G–
A–
5 –C
·
3 –G
T–
G–
5 –A
·
3 –T
C–
A–
5’– G
·
3 –C
T–
G–
5 –C
·
3 –G
C–
C–
5’– G
·
3’– C
G–
C–
5 –C
·
3 –G
G–
The initial interaction of two strands contributes a loss of entropy and an unfavorable free energy that must be included in
calculations:
rG(initiation) = + 8.1 kJ mol - 1, rS(initiation) = - 23.4 J K - 1 mol - 1, rH(initiation) = + 0.8 kJ mol - 1.
*Based on data from H. T. Allawai and J. SantaLucia Jr., “Thermodynamics and NMR of Internal G-T Mismatches in DNA”,
Biochemistry 36, 10581–10594. Copyright © 1997 American Chemical Society.
Summary | 143
Ionic Effect on Protein–Nucleic Acid Interactions
Macromolecules often carry many charged groups. A double-stranded DNA, for example,
has two phosphate groups every 3.4 Å along the molecule, with each phosphate carrying
a negative charge. In aqueous solution, the negatively charged phosphate groups of the
polyelectrolyte interact with positive-charged small ions that are present. We call these
small ions counterions. For simplicity, we consider a DNA solution containing only Na +
as the counterions.
The Na + counterions interact with DNA in two ways. Because of the very large
negative-charge density of the DNA, there is a direct condensation of a fraction of the
positive counterions on the polyelectrolyte. The remaining unneutralized negative charges
are screened by a mobile ion atmosphere (Manning 1979). We let c be the effective fraction of each phosphate charge neutralized by Na + ions condensed on the DNA. When a
protein L binds to a particular site of the DNA, a number of Na + originally bound to the
DNA are displaced. If L interacts with n phosphate groups of the DNA, nc of Na + originally condensed on the DNA are released. We express the reaction as
K
P + L m PL + ncNa + .
The equilibrium constant K, which is independent of the Na + concentration, is
K =
⬵
(aPL)(aNa + )nc
(aP)(aL)
3 PL 4 3 Na + 4 nc
,
3P 4 3L 4
where 3 L 4 = concentration of protein, M
3 P 4 = concentration of DNA, mol phosphate L - 1
3 PL 4 = concentration of complex, M .
The concentration quotient
3 PL 4
K Kobs
3P 4 3L 4
is usually measured in media containing different amounts of Na + . We note that
Kobs = K 3 Na + 4 - nc.
Clearly, the observed apparent equilibrium constant Kobs will be strongly dependent on
3 Na + 4 if nc is large. For double-stranded DNA, c 苲 0.88. Then from a plot of log Kobs
versus log 3 Na + 4 , n can be calculated from the slope; this tells us how many phosphate
groups on the DNA the protein L interacts with (see problem 29 and Record et al. 1981).
Summary
Name
Symbol
Activity Coefficient
gA
Activity
aA
An effective concentration, always a product of
gA and a composition variable
pA,xA,cA,mA
Pressure, mole fraction, concentration, molality
Composition
Description
gA (a positive number less than or equal to 1),
measures nonideality of substance
144 Chapter 4 |
Free Energy and Chemical Equilibria
Chemical Potential
Gibbs Free Energy
Gibbs free energy depends on temperature, pressure, and composition variables
G = G(T, P, nA, nB, nC, . . .),
(4.1)
where nA, nB, nC, . . . = the number of moles of A, B, C, and so on.
The chemical potential (partial molar Gibbs free energy) of a substance A:
mA K a
0G
b
0nA T,P,ni ⬆ nA
Sum Rule for Chemical Potentials
At constant T and P, the Gibbs free energy of a system is equal to the sum of the products
of the number of moles of each component times its chemical potential:
G = nAmA + nBmB + g
(4.5)
Chemical Potentials of Reactions at Equilibrium
aA + bB S cC + dD
At equilibrium, 3 (amA + bmB) - (cmC + dmD) 4 = 0 , or amA + bmB = cmC + dmD.
At equilibrium, the sum of the chemical potentials of the reactants, each weighted according to the
stoichiometry of the reaction, is equal to the sum of the chemical potentials of the products similarly
weighted.
Chemical Potential and Activity
mA = mAⴰ + RT ln aA
(4.19)
ⴰ
aA = e(mA - m A)/RT
The activity of a substance A at temperature T is aA, and mA and mAⴰ are its chemical potential and
standard chemical potential, respectively, at T.
Standard States and Activities
Ideal Gases
Standard state is gas at 1 bar pressure:
aA =
pA
,
1 bar
(4.23)
where pA is the partial pressure of the ideal gas.
Real Gases
Standard state is a hypothetical state where the pressure is 1 bar, but the properties are those extrapolated from low pressures. For a real gas,
gApA
aA =
,
(4.24)
1 bar
where gA, the activity coefficient, depends on total pressure.
Summary | 145
Pure Solids and liquids
Standard state is the pure substance at 1 bar. The activity is equal to 1 for a pure solid or liquid at 1
bar, and it changes only slightly with pressure:
aA = 1
(4.25)
Solvent of a solution
Standard state is the pure solvent at 1 bar. The activity is
aA = gAxA .
(4.31)
Solutes in a Solution
Standard state of a solute is a hypothetical one where the concentration of the solute is equal to
1 molar (M) or 1 molal (m), but the properties are those extrapolated from very dilute solutions:
aB = gBcB (gB S 1 as cB S 0)
(4.35)
aB = gBmB (gB S 1 as mB S 0)
(4.41)
Ions
The Debye–Hückel Limiting Law for the activity coefficient of an ion in water at 25 C :
loggion = - 0.509 Z2i I1/2
(4.47)
Zi = charge on ion ( {1,{2,{3, etc.)
I = ionic strength =
1
ciZ2i
2a
i
(4.43)
Biochemical Standard State
At pH 7, aH + = 1; the activity of each of the other species is set equal to the total concentration of
all species of that molecule at pH 7.0:
aA =
a ci,A
(at pH 7.0)
(4.42)
species i
ΔG and the Equilibrium Constant for a Chemical Reaction
aA + bB S cC + dD
rG = rG + RT ln Q
(4.20)
rG = Gibbs free-energy change for the reaction at T and concentrations specified in Q.
rG = standard Gibbs free-energy change for the reaction at T with all reactants and products at
their standard states.
(aC)c(aD)d
Q
=
aA,aB,aC,aD
=
activities of reactants and products at the concentrations specified
rG
=
- RT ln K
K
=
equilibrium constant
=
b eq a
(aeq
B ) (aA )
c eq d
(aeq
C ) (aD )
aeq
A ,etc.
=
b eq a
(aeq
B ) (aA )
activities of reactants and products at equilibrium
2.3026RT
=
5708 J mol-1 at 298 K (K = e - rG>RT = 10 - rG/2.303RT)
(4.14)
(4.22)
146 Chapter 4 |
Free Energy and Chemical Equilibria
Temperature Dependence of Equilibrium Constant
c
ln
- ⌬ rH⬚
0(ln K)
d =
011>T2 p
R
K2
- ⌬ rH⬚ 1
1
=
b
a
K1
R
T2
T1
(van’t Hoff)
(4.52)
(⌬ rH⬚ = constant)
(4.53)
Mathematics Needed for Chapter 4
Review partial derivatives in chapter 3 and the mathematics appendix.
References
Standard biochemistry texts will describe metabolism in depth.
1. Alberty, R.A., A. Cornish-Bowden, R.N. Goldberg,
G.G. Hammes, K. Tipton, and H.V. Westerhoff. 2011
Recommendations for terminology and databases for
biochemical thermodynamics. Biophys. Chem. 155: 89–103.
2. Freyer, M.W., and E.A. Lewis. 2008 Isothermal titration
calorimetry: experimental design, data analysis, and
probing macromolecule/ligand binding and kinetic
interactions. Meth. Cell Biol. 84: 79–113.
Suggested Reading
Allawai, H. T., and J. SantaLucia Jr. 1997. Thermodynamics and
NMR of Internal G # T Mismatches in DNA. Biochemistry
36:10581–10594.
Manning, G. S. 1979. Counterion Binding in Polyelectrolyte
Theory. Acc. Chem. Res. 12:443–447.
Moser, C. C., J. M. Keske, K. Warnke, R. S. Farid, and P. L. Dutton.
1992. Nature of Biological Electron Transfer. Nature 355:
796–802.
Record, M. T. Jr., S. J. Mazur, P. Melançon, J.-H. Roe, S. L. Shaner,
and L. Unger. 1981. Double Helical DNA: Conformation,
Physical Properties and Interactions with Ligands. Annu.
Rev. Biochem. 50:997–1024.
SantaLucia, J. Jr., Allawi, H. T., and Seneviratne, P. A. 1996.
Improved Nearest-Neighbor Parameters for Predicting DNA
Duplex Stability. Biochemistry 35:3555–3562.
Turner, D. T., N. Sugimoto, and S. M. Freier. 1988. RNA Structure
Prediction. Annu. Rev. Biophys. Biophys. Chem. 17:167–193.
Problems
1. A key step in the biosynthesis of triglycerides (fats) is the
conversion of glycerol to glycerol-1-phosphate by ATP:
glycerol + ATP
Mg2+
h
a - glycerol kinase
glycerol@1@phosphate + ADP
At a steady state in the living cell, (ATP) = 10 - 3 M and
(ADP) = 10 - 4 M. The maximum (equilibrium) value of the
ratio (glycerol@1@P)/(glycerol) is observed to be 770 at 25⬚C
and pH 7.
a. Calculate K⬘ and ⌬ rG298⬚⬘ for the reaction.
b. Using the value of ⌬ rG298⬚⬘ for the reaction
ADP + Pi S ATP + H2O, together with the answer
to part (a), calculate ⌬ rG298⬚⬘ and K⬘ for the reaction
glycerol + phosphate S glycerol@1@phosphate + H2O .
2. In the frog muscle rectus abdominis, the concentrations of ATP, ADP, and phosphate are 1.25 * 10 - 3 M,
0.50 * 10 - 3 M, and 2.5 * 10 - 3 M, respectively.
a. Calculate the Gibbs free-energy change ⌬ rG⬘ for the
hydrolysis of ATP in muscle. Take the temperature and
pH of the muscle as 25⬚C and 7, respectively.
b. For the muscle described, what is the maximum amount of
mechanical work it can do per mole of ATP hydrolyzed?
c. In muscle, the enzyme creatine phosphokinase catalyzes
the following reaction:
phosphocreatine + ADP S creatine + ATP
The standard Gibbs free energy of hydrolysis of phosphocreatine at 25 ⬚C and pH 7 is - 43.1 kJ mol - 1 :
phosphocreatine + H2O S creatine + phosphate
Calculate the equilibrium constant of the creatine phosphokinase reaction.
3. An important step in the glycolytic path is the phosphorylation of glucose by ATP, catalyzed by the enzyme hexokinase
and Mg2 + :
2+
Mg
glucose + ATP h glucose@6@P + ADP
hexokinase
In the absence of ATP, glucose-6-P is unstable at pH 7, and
in presence of the enzyme glucose-6-phosphatase, it hydrolyzes to give glucose:
glucose@6@P + H2O
h
G - 6 - phosphatase
glucose + phosphate
Problems | 147
a. Using data from table 4.3, calculate ⌬ rG⬚⬘ at pH 7 for
the hydrolysis of glucose-6-P at 298 K.
b. If the phosphorylation of glucose is allowed to proceed
to equilibrium in the presence of equal concentrations of
ADP and ATP, what is the ratio (glucose@6@P)/(glucose)
at equilibrium? Assume a large excess of ATP and ADP:
(ATP) = (ADP) W [(glucose) + (glucose @6@P)].
c. In the absence of ATP (and ADP), calculate the ratio
(glucose@6@P)/(glucose) at pH 7 if phosphate = 10 - 2 M.
4. In the red blood cell, glucose is transported into the cell
against its concentration gradient. The energy for this transport is supplied by the hydrolysis of ATP:
ATP + H2O S ADP + Pi (⌬ rG⬚ = - 31.0 kJ mol - 1)
Assume that the overall transport reaction is 100% efficient
and given by
ATP + H2O + 2glucose(out) S 2glucose(in) + ADP + Pi .
a. At 25 ⬚C, under conditions where 3 ATP 4 ,[ADP 4 , and
3 Pi 4 are held constant at 1 * 10 - 2 M by cell metabolism, find the maximum value of
3 glucose(in) 4
.
3 glucose(out) 4
Assume all activity coefficients are equal to 1.
b. If the stoichiometry of transport were 1 mol of glucose
transported per mole of ATP hydrolyzed, what would be
the maximum concentration gradient of glucose under
the same conditions as part (a)?
c. In an actual cell, the glucose inside the cell may have
an activity coefficient much less than 1 due to nonideal
behavior. Would this increase or decrease the maximum
concentration gradient obtainable? (Assume that all
other activity coefficients are equal to 1.)
5. The following reactions can be coupled to give alanine and
oxaloacetate:
glutamate + pyruvate m Ketoglutarate + alanine
(⌬ rG303⬚⬘ = - 1004 J mol - 1)
glutamate + oxaloacetate m ketoglutarate + aspartate
(⌬ rG303⬚⬘ = - 4812 J mol - 1)
a. Write the form of the equilibrium constant for
pyruvate + aspartate m alanine + oxaloacetate and
calculate the numerical value of the equilibrium constant
at 30⬚C.
b. In the cytoplasm of a certain cell, the components are
at the following concentrations: pyruvate = 10 - 2 M,
aspartate = 10 - 2 M, alanine = 10 - 4 M, and oxaloacetate = 10 - 5 M. Calculate the Gibbs free-energy change
for the reaction of part (a) under these conditions. What
conclusion can you reach about the direction of this reaction under cytoplasmic conditions?
6. a. Use the following equilibrium constants to calculate
⌬ r H⬚ from the slope of ln K vs. (1/T) for the reaction
3-phosphoglycerate to 2-phosphoglycerate at pH 7.
T (ⴗC)
0
20
30
45
K
0.1535
0.1558
0.1569
0.1584
b. Calculate ⌬ rG⬚ at 25⬚C.
c. What is the concentration of each isomer of phosphoglycerate when 0.150 M phosphoglycerate reaches equilibrium at 25⬚C?
7. Oxides of sulfur are important in atmospheric pollution, arising particularly from burning coal. Use the thermodynamic
data at 25 ⬚C given in the appendix to answer the following
questions.
a. In air, the oxidation of SO2 can occur: 12O2(g) +
ⴰ
SO2(g) S SO3(g). Calculate ⌬ rG298
.
b. Find the equilibrium ratio of partial pressures of SO3(g)
to SO2(g) in air at 25 ⬚C; the partial pressure of O2(g)
is 0.21 bar.
c. SO3(g) can react with H2O(g) to form sulfuric acid,
H2SO4(g). Air that is in equilibrium with liquid water at
25 ⬚C has a partial pressure of H2O(g) of 0.031 bar. Find
the equilibrium ratio of partial pressures of H2SO4(g) to
SO3(g) in air at 25⬚C. (see Appendix A.5)
8. An important metabolic step is the conversion of fumarate
to malate. In aqueous solution, an enzyme (fumarase) allows
equilibrium to be attained:
fumarate + H2O m malate
at 25 ⬚C; the equilibrium constant K = (aM/aF) = 4.0.
The activity of malate is aM, and the activity of fumarate is aF, defined on the molarity concentration scale
(a = c in dilute solution).
a. What is the standard Gibbs free-energy change for the
reaction at 25 ⬚C?
b. What is the Gibbs free-energy change for the reaction
at equilibrium?
c. What is the Gibbs free-energy change when 1 mol of
0.100 M fumarate is converted to 1 mol of 0.100 M malate?
d. What is the Gibbs free-energy change when 2 mol of
0.100 M fumarate are converted to 2 mol of 0.100 M
malate?
e. If K = 8.0 at 35 ⬚C, calculate the standard enthalpy
change for the reaction; assume that the enthalpy is independent of temperature.
f. Calculate the standard entropy change for the reaction;
assume that ⌬ rS⬚ is independent of temperature.
9. Use the Gibbs free energy of hydrolysis of ATP under
standard conditions at 25 ⬚C, 1 bar, to answer the following:
a. Find ⌬ rG of the reaction when 3 ATP 4 = 10 - 2 M,
3 ADP 4 = 10 - 4 M, and [phosphate] = 10 - 1 M.
b. Calculate the maximum available work under the conditions of part (a) when 1 mol of ATP is hydrolyzed. This
work can be used, for example, to contract a muscle and
raise a weight.
c. Calculate ⌬ rG and the maximum available work
3 ATP 4 = 10 - 7 M, 3 ADP 4 = 10 - 1 M,
if
and
[phosphate] = 2.5 * 10 - 1 M.
148 Chapter 4 |
Free Energy and Chemical Equilibria
10. What is the pressure of CO(g) in equilibrium with the CO2(g)
and O2(g) in the atmosphere at 25 ⬚C? The partial pressure
of O2(g) is 0.2 bar and the partial pressure of CO2(g) is
3 * 10 - 4 bar. CO is extremely poisonous because it forms
a very strong complex with hemoglobin. Should you worry?
11. For a reaction, aA + bB S cC + dD,
⌬ rGTⴰ = cmCⴰ + dmDⴰ - amAⴰ - bmBⴰ
⌬ rGTⴰ = - RT ln K .
Combining these two expressions and dividing each term
by RT gives
dmDⴰ
amAⴰ
bmBⴰ
1 cmCⴰ
ln K = - a
+
b.
R T
T
T
T
a. Show that
ⴰ
Hm,i
d miⴰ
a b = - 2 ,
dT T
T
ⴰ
where Hm,i
is the standard partial molar enthalpy of species i
(at constant T, ⌬ rG⬚ = ⌬ rH⬚ - T⌬ rS⬚, m⬚i = H⬚i - TS⬚i )
b. From the result obtained in part (a) and the relation
between ln K and the chemical potentials, show that
d ln K>d(1>T) = - ⌬ rH⬚>R even if ⌬ rH⬚ is temperature
dependent.
12. a. From the ionization constant, calculate the standard
Gibbs free-energy change for the ionization of acetic
acid in water at 25 ⬚C.
b. What is the Gibbs free-energy change at equilibrium for
the ionization of acetic acid in water at 25 ⬚C?
c. What is the Gibbs free-energy change for the following
reaction in water at 25 ⬚C? (Assume that activity coefficients are all equal to 1.)
H + (10 - 4 M) + OAc - (10 - 2 M) S HOAc (1 M)
d. What is the Gibbs free-energy change for the following
reaction in water at 25⬚C? (Assume that activity coefficients are all equal to 1.)
H + (10 - 4 M) + OAc - (10 - 2 M) S HOAc (10 - 5 M)
e. What is the Gibbs free-energy change for transferring 1 mole of acetic acid from an aqueous solution of
1 M concentration to an aqueous solution of 10 - 5 M?
(Assume all activity coefficients are equal to 1.)
13. You want to make a pH 7.0 buffer using NaOH and phosphoric acid. The sum of the concentrations of all phosphoric acid species is 0.100 M. The equilibrium constants for
concentrations given in apparent units of mol/L are
H3PO4 m H + + H2PO4- (K1 = 7.1 * 10 - 3) ,
H2PO4- m H + + HPO24 - (K2 = 6.2 * 10 - 8) ,
HPO42 - m H + + PO43- (K3 = 4.5 * 10 - 13) .
a. Write the equation that specifies that the solution is
electrically neutral.
b. Compute the concentrations of all buffer species.
c. Use data in table 4.2 to calculate K2 at 37 ⬚C.
14. You made a pH 9.0 buffer solution at 25 ⬚C by mixing NaOH
and histidine (HisH) to give a solution that is 0.200 M in
total concentration of histidine.
a. Calculate the concentrations of all species at 25 ⬚C.
b. What is the pH of the buffer at 40 ⬚C? The two most
important ⌬ rH⬚ values for histidine are given in
table 4.2; assume that ⌬ rH⬚ = 0 for the first ionization. You can ignore the volume change of the solution.
c. Calculate the concentrations of all species at 40 ⬚C.
15. An important reaction in visual excitation is the activation
of an enzyme that catalyzes the hydrolysis of guanosine
triphosphate:
GTP + H2O S GDP + Pi (K⬘298 = 1.9 * 105)
a. Typical concentrations of GTP, GDP, and Pi in the retinal rod cell are 50 mM, 5 mM, and 15 mM, respectively.
What is ⌬ rG298 ⬘ for the above reaction in the rod cell?
b. If the GTP concentration were suddenly doubled, by
how much would ⌬ rG298 ⬘ change?
c. The solution described in part (a) is allowed to come to
equilibrium. What are the final concentrations of GTP,
GDP, and Pi?
16. In general, native proteins are in equilibrium with denatured forms:
protein (native) m protein (denatured)
For ribonuclease (a protein), the following concentration
data for the two forms were experimentally determined for
a total protein concentration of 1.0 * 10 - 3 mol L - 1:
Temperature
(ⴗC)
Native
(mol Lⴚ1)
Denatured
(mol Lⴚ1)
50
9.97 * 10 - 4
2.57 * 10 - 6
100
8.6 * 10 - 4
1.4 * 10 - 4
Determine ⌬ rH⬚ of the denaturation reaction, assuming it to
be independent of temperature.
17. A single-stranded oligonucleotide that has complementary
ends can form a base-paired hairpin loop.
C
C
AAAAAACCCCCCUUUUUU
K1
C C
C
C
A·U
A·U
A·U
A·U
A·U
A·U
a. At 25 ⬚C, the equilibrium constant K1 = 0.86. What are
the concentrations of loop and single strand at equilibrium if the initial concentration of single strands is
1.00 mM? Will increasing the initial concentration of
oligonucleotide increase or decrease the fraction of hairpin loop? Explain.
Problems | 149
b. At 37⬚C, the equilibrium constant K1 = 0.51. Calculate
⌬ rH⬚,⌬ rS⬚, and ⌬ rG⬚ at 37⬚C.
c. As the concentration of oligonucleotide increases,
another reaction becomes possible. A double-stranded
molecule with an internal loop can form:
C6
2 A6C6U6
K2
A6
·
U6
U6
·
A6
C6
18.
19.
20.
21.
At 25 ⬚C, K2 = 1.00 * 10 - 2 M - 1 (and K1 = 0.86).
Calculate the concentrations of all three species (single
strand, SS; double strand, DS; and hairpin loop, H)
at equilibrium in a solution of initial concentration of
single strands = 0.100 M.
Write out short answers describing how the chemical potentials of the species relate to the following processes.
a. A perfectly thermally insulated container has ice and
liquid water in equilibrium at 0 ⬚C. (i.e., ice water). Salt
(NaCl) is added to the container and allowed to dissolve
fully; some of the ice spontaneously melts and the temperature drops slightly. Did the chemical potential of
the liquid water increase or decrease upon adding salt?
Did the chemical potential of the solid ice increase or
decrease throughout this process?
b. A glass vial containing some liquid water is attached
to a vacuum pump that is turned on and allowed to
run for the entire experiment. The vial is not thermally
insulated but keep in mind that glass is a poor heat conductor. You make these observations. First, the liquid
water level decreases as you begin to watch. Second, the
liquid water begins to spontaneously freeze entirely to
ice. Finally, the water is completely removed from the
vial. The entire process requires a few hours of time.
Explain, using concepts of chemical potential, why
the liquid water is evaporated so rapidly. Explain why
the water is observed to spontaneously freeze. Finally,
describe the disappearance of the solid ice.
Find the dissociation constant, K, for pure water at pH 7
using the molarity standard state for water. You may take
the density of water to be 1.0 g/ml. Will K have the same
value in this standard state at 60 ⬚C? Explain.
An elastomer, such as a rubber band, maintains a constant
volume when it is stretched since the band thins, to a first
approximation. What thermodynamic quantity represents
the maximum non-PV work when a rubber band is stretched
at constant temperature?
The binding of oligomers (lys)n of the amino acid
lysine, where n indicates the number of lysines in each
oligomer, to a synthetic double-stranded RNA poly(rA # rU) has been studied by S. A. Latt and H. A. Sober
3 Biochemistry 6:3293- 3306 (1967) 4 . The dependence
of the apparent binding constant Kobs on 3 NaCl 4 in the
medium is tabulated as follows:
3 NaCl], M
0.06
0.10
0.25
0.39
log Kobs, n = 4
log Kobs, n = 5
4.62
–
–
4.71
2.39
2.79
1.60
1.88
Plot log Kobs against log 3 Na + 4 for each of the oligolysines
and interpret the data. The value of c for poly (rA # rU) is
close to that of duplex DNA.
22. Self-complementary oligonucleotides can form doublestranded helices in aqueous solution stabilized by Watson–
Crick base pairs. For example,
2(5⬘ - CGCGATATCGCG - 3⬘) m
5⬘ - CGCGATATCGCG - 3⬘
# # # # # # # # # # # #
3⬘ - GCGCTATAGCGC - 5⬘
Derive an expression for the equilibrium constant K as a
function of c, the initial concentration of single strands
and f, the fraction of single strands that are in the doublestranded helix at equilibrium. The equilibrium is
2S m D
K =
Note that
f =
3D 4
.
3S 4 2
2 3D 4
2 3D 4
=
.
c
3
4
3
4
2 D + S
23. a. Use nearest-neighbor values (table 4.4) for the thermodynamics of double-strand formation to calculate
⌬ rG⬚(37 ⬚C), ⌬ rH⬚, and ⌬ rS⬚ for
5⬘ - GGTTCC - 3⬘
and
# # # # # #
# # # # # #
3⬘ - CCCGGG - 5⬘
3⬘ - CCAAGG - 5⬘
b. The melting temperature, Tmelt, is defined as the temperature where f, the fraction of single strands in the helix at
equilibrium, is 1/2. Calculate the melting temperature for
each double helix above at 1.0 * 10 - 4 M initial concentrations of each single strand (the initial concentration of
strands, c, is 1.0 * 10 - 4 M for the helix on the left and
2.0 * 10 - 4 M for the helix on the right). Note that the
oligonucleotides in the double helix on the right are not
self-complementary; its equilibrium constant in terms of
f and c is slightly different from that in problem 22.
24. Genomic DNA sequences are detected by hybridization
with a radioactively labeled probe that is complementary to
the target sequence and forms a double strand.
a. For the equilibrium (probe + target S double strand)
express K, the equilibrium constant for the association
of the probe with the target, in terms of cprobe, the initial
probe concentration, ctarget, the initial concentration of
the target sequence, and f, the fraction of target hybridized to the probe to form a double strand. Assume that
the probe is present in large excess.
5⬘ - GGGCCC - 3⬘
Free Energy and Chemical Equilibria
b. The target sequence is
5⬘ - GGGGAATCA - 3⬘ .
Calculate ⌬ rG⬚ (use table 4.4) and K at 25 ⬚C for the
formation of a double strand with the above sequence
at pH 7.0, 1 M NaCl.
c. Find the melting temperature of the probe from the
target, defined as the temperature at which half of
the target is hybridized to the probe. Assume that
cprobe = 1.00 * 10 - 4 M and ctarget = 1.00 * 10 - 8 M.
d. In practice, annealing reactions are carried out in much
lower salt concentrations. A simple, common correction is
Tm( 3 Na + 4 ) = Tm(1 M) + 12 log 3 Na + 4 .
Find the Tm for [Na+] = 0.5 M, 0.05 M, and 0.005 M.
Can you rationalize the reason for these results?
e. Does the melting temperature as defined in part (c)
go up, down, or remain the same when the following
changes are made? Give reasons for your answers.
macromolecule, etc.) and differ only in the strength of binding (K). In which curve(s) did the initial injection essentially fully bind? In which curve(s) can you confidently see
the binding stoichiomietry by inspection without the need
for fitting? Rank the curves marked (a–c) according to the
strength of binding.
(a)
(b)
heat (J/sec)
150 Chapter 4 |
1. The probe concentration is doubled.
(c)
0.5
2. The target concentration is doubled.
1. ATAT TAAT TTAA TATA
2. GCCG CGCG GGGG CCCC
3. GCTA AGCG GTAT CAGT
4. GCTC AGCG GTAG CAGT
Now calculate the Tm’s more rigorously by the method in
question 24 (use 0.25 μM probe). Apply the salt correction in 24d assuming 50 mM salt. Compare the ‘back of
the envelope’ estimates to those you calculated and draw
conclusions.
26. Assume that the equilibrium constant K is so large in an
ITC binding experiment that for very early injections of
ligand L into the macromolecule solution, essentially all
ligand binds the macromolecule. Suppose the first injection adds an amount of ligand that is equal to 10% of the
macromolecule, and that the heat of the first injection was
measured to be -35 * 10-6 J. You prepared 2.0 ml of macromolecule at a concentration of 1.0 * 10-5 M. Find the molar
enthalpy of binding assuming a one site binding model.
27. The three titration curves measured below were obtained
under identical conditions (e.g., same number and
volume of injections, same concentrations of ligand and
2.0
2.5
28. When the macromolecule offers n independent binding sites
for the ligand L, the equations for ITC binding curves are
easily modified. Recognize that fM is still a number between
0 and 1, but now [ML] = nfM[M]Tot.
a. Determine the analogue to Eq. 4.65A for the case of
n independent binding sites. How should Eq. 4.57 be
modified for n independent binding sites?
b. Typical appearances of ITC binding curves for n
independent sites are illustrated below. Compared to
figure 4.15, which cases correspond to n = 1, 2, and 3
and how do you know?
(a)
heat (J/sec)
25. For each sequence give the Tm predicted by a popular “back
of the envelope” method of counting 4 degrees per GC pair
and two degrees per AT pair:
1.5
[Ltot]/[Mtot]
3. The probe contains a single base mismatch.
4. The salt concentration is decreased.
1.0
(b)
(c)
1.0
2.0
3.0
[Ltot]/[Mtot]
4.0
Chapter 5
The Statistical Foundations of
Biophysical Chemistry
“This is therefore a possible form for the final distribution of velocities; it is also the
only form”
— James Clerk Maxwell, 1868, quoted in Ludwig Boltzmann’s Lecture on Gas
Theory, 1895.
Concepts
We have already seen in chapter 2 that the internal energy of a monatomic ideal gas is
equal to its translational kinetic energy and depends only on the temperature. In this chapter, we go on to show that all thermodynamic properties, including energy, entropy, and
free energy, are directly related to the average energy of all molecules in a system and to
the distribution of molecules among the possible energies. The heat and work done when
there is a change in the system can be calculated from the changes in the energy levels of
the system and the changes in distribution of the molecules among the energy levels. These
concepts connect molecular interactions and molecular properties to the thermodynamic
laws we explored in chapter 3.
The same statistical averaging among energies that we can use to get the thermodynamic properties of simple systems can also be applied to the conformations of
macromolecules. All molecules have a range of conformations produced by vibrations of
all the bonds and torsional rotations around the single bonds. Macromolecules can have
a very wide range of conformations because they contain so many bonds. DNA from the
bacterial virus T4 is a double helix with a molecular weight of about 108; in solution this
threadlike molecule can range from a tight coil to an extended form. Some properties of
the molecule, such as how fast it can diffuse or sediment in solution, depend on the average
dimension of the molecule. Other properties depend on the range of possible conformations of the molecule. For example, the probability that the molecule will be broken if the
solution is stirred or poured depends on the fraction of molecules in an extended conformation as well as the average dimension. Statistical methods can provide the distribution of
molecular conformations and thus allow calculation of measurable properties. The concept
of a random walk is very useful in understanding the possible conformations of a flexible
macromolecule. The same logic can be applied to the statistical binding of small molecules
to sites on macromolecules.
Applications
If DNA in solution is heated, at a certain temperature the double helix begins to “melt.”
Base pairing is disrupted, regions of single-stranded loops form, and, if the temperature
is high enough, the two strands of the double helix dissociate completely. A schematic
151
1
152 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
illustration of the melting phenomenon is shown in figure 5.8. The transition from the
helix to the dissociated strands occurs in a temperature range of only a few degrees;
hence, the term melting is used to describe this transition. Statistical analysis provides an
explanation for the sharp transition. Furthermore, some of the thermodynamic parameters
for the transition can be evaluated by a statistical analysis. These parameters are helpful
in understanding the replication of the DNA, because the replication of DNA molecules
involves the dissociation of the two parent strands and the formation of new double strands
for each of the two daughter DNA molecules.
Either inside or outside the cell, the DNA interacts with a variety of small and not
so small molecules. Among the small molecules, the antibiotic actinomycin binds to
many sites of the DNA and can block transcription of the DNA into RNA. Many heterocyclic dye molecules and polycyclic aromatic hydrocarbons, such as 9-amino-acridine
and benzo[a]pyrene, can bind to the DNA and cause mutations. Such molecules are often
carcinogenic. The binding of small molecules to DNA can be understood by statistical
analyses of the binding experiments. Among the larger molecules, such as proteins, that
interact with the DNA, some bind to only a few specific sites; others bind with much less
discrimination.
Maxwell Boltzmann Statistics
The first and most fundamental question we might address with statistical thermodynamics is the following: how are the velocities (and thus the kinetic energies) of molecules
distributed around the average translational energy of (3/2)RT ? More generally, what law
governs the distribution of energies between molecules or modes, given a particular average energy? This question was addressed by Ludwig Boltzmann, considered the founder of
statistical mechanics and one of the most brilliant physicists of all time. Boltzmann found
a breathtakingly simple answer.
The Boltzmann Distribution
Consider two values of energy, E1 and E2. They can be the energies of quantum states
if we are discussing quantum mechanics, or they can be infinitesimally narrow bands
of energy, E1 S 1E1 + dE2 and E2 S 1E2 + dE2, in the classical mechanics that prevailed during Boltzmann’s lifetime. For any molecule in a large collection of molecules,
the probability of it being in energy or energy band 1 is p1 and of it being in energy or
energy band 2 is p2.
Without any math at all, we can first conclude that the relative probability p2 >p1 of
being in energy state E2 relative to energy state E1 cannot depend on the absolute values of
the energies, since as we discussed in chapter 2, there is no practical zero of energy. (There
was no zero of any kind in Boltzmann’s time, which predated relativity and E = mc2.) The
p2 >p1 can therefore only depend on the difference between E2 and E1. Therefore, we can
write the relative probability as a function of (E2 - E1):
p2 >p1 = f1E2 - E1 2
(5.1)
We don’t yet know what function f is. But we do know that the probability of a particle having infinite energy is zero, because if one particle in a system has infinite energy,
the whole system has infinite energy, which is impossible. Therefore, p2 S 0 as E2 S ⬁,
and so:
f1⬁2 = 0
(5.2)
Maxwell Boltzmann Statistics | 153
Finally, let’s introduce a third energy state E3, as shown in figure 5.1. Because energy
is conserved:
E3 - E1 = 1E3 - E2 2 + 1E2 - E1 2
p3
E3
p2
E2
And by simple algebra:
p3
p2
p3
=
*
p1
p2
p1
p1
Therefore, when two arguments of f are added, the result must be a product of the functions
of the two separate arguments:
f1E3 - E1 2 = f1E3 - E2 2f1E2 - E1 2
(5.3)
E1
FIGURE 5.1 The probability of
a molecule being in a state with
a given energy declines as that
energy increases.
The only function which has all three properties described by Eqs. 5.1–5.3 is the
exponential of a multiple of -E. (Since we shouldn’t be taking exponentials of dimensioned
quantities, the multiplication factor must have the units of reciprocal energy.) Let’s call
the multiplication factor b. Then, in general, we can write for any energy state or band of
energy denoted by i:
pi = N exp1-bEi2
(5.4)
We have to include the normalization factor N because the total probabilities of all
states i, or the integral over all energy bands, must add up to 1. We will examine how to
evaluate N in the next section.
The Maxwell-Boltzmann Distribution
Unfortunately, energy alone does not completely specify the motion of a molecule. A
particular kinetic energy, E = 12mv2, is associated with an infinity of different velocity
vectors, v: v # v = v2. These vectors can be depicted as arrows, all of which terminate on
the surface of a sphere of radius v. We can call the magnitude of velocity v speed v, which
is a scalar (figure 5.2).
Even though the number of vectors corresponding to any speed vi is infinite, we can
still compare the number of vectors between speed v1 and v1 + dv, with the number of
vectors between speed v2 and v2 + dv, because the number of vectors in each case is
proportional to the surface area of the sphere on which the vectors terminate. The concept
that two infinite numbers, divided by each other, can have a result that is finite and not
unity is perhaps unfamiliar, but it is well-founded in the branch of mathematics called set
theory. We call the number of microscopic states associated with a single value of energy
the degeneracy and give it the symbol g. If the number of vectors between speed v1 and
v1 + dv is g1 and the number with speed between v2 and v2 + dv is g2, then the value of
g2 >g1 = 4pv2 2dv>4pv1 2 dv = v2 2 >v1 2 = E2 >E1. For example, if v2 = 2v1, then g2 = ⬁
and g1 = ⬁, yet g2 >g1 = 4. Infinity is truly a strange concept!
For practical purposes, this means that when we compare the number of molecules
of an ideal gas with two different energies, we have to consider both the exponential
dependence of probability on energy discussed in the previous section, and the relative
degeneracy associated with each energy. Therefore, we can write for velocity vi, the number
of velocity vectors between vi and vi + dv is
gi ⬀ 4pvi 2 dv ,
dv
v
dv
(5.5)
and the probability of a particle having velocity between vi and vi + dv is
pi dv ⬀ 4pvi 2dv * exp1- bEi 2 = 4pNvi 2dv exp1- 12 bmvi 2 2 ,
v
(5.6)
FIGURE 5.2 The number of velocity
vectors for a given narrow range of
speeds from v to v + dv is proportional
to v2.
154 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
where N is our normalization factor, and b is still unknown. We can evaluate N by integrating the probabilities over all possible energies; the integral must equal 1.
⬁
⬁
1
2
2
3 p(v)dv = 1 = 3 4pNv exp a- 2 bmv b dv
0
(5.7)
0
A little math, a good table of integrals, or a symbolic algebra program will give us the
result:
N = a
bm 3>2
b
2p
(5.8)
All that remains is to evaluate b. We can do this by noticing that if we have any quantity
Y(v) dependent on v, the average value of Y per molecule over the whole collection of
molecules is just
⬁
8 Y 9 = 3 Y (v)p(v)dv.
(5.9)
0
Of course, the kinetic energy of the molecules is such a quantity Y.
⬁
bm
8 E 9 = 3 E(v)p(v)dv = a
b
2p
0
3>2
⬁
1
1
2
2
2
3 c 2 mv * 4pv exp a- 2 bmv b d dv
(5.10)
0
This looks fearsome, but after cancellation and simplification, it has a remarkably simple
result.
8E9 =
3
2b
(5.11)
But from the kinetic theory of gases, which we explored in chapter 2, we already know
the average value of the kinetic energy of a molecule in a monatomic ideal gas; it’s just
Um >NA, the molar internal energy divided by Avogadro’s number! Since, from Eq. 2.25,
3
Um
2 RT
=
= 32k BT,
NA
NA
(5.12)
we can conclude that b = 1>kBT. It would seem we have proven this only for translational
motion in a monatomic ideal gas, but recall that the equipartition theorem requires that
energy is distributed equally between all thermally accessible modes. This means that the
value of b we obtained above is not exclusive to translational motion or to monoatomic or
even ideal gases; it is true in general.
Finally, we should go back and substitute for b in our various equations. From Eq. 5.8,
we get
N = a
3>2
3>2
m
M
b
b ,
= a
2pkBT
2pRT
(5.13)
since M = NAm and R = NAkB. Incorporating this and b = 1>kBT into Eq. 5.7, we get
p(v)dv =
2 M 3>2 2
- Mv2
a b v exp a
b dv .
A p RT
2RT
(5.14)
This is the Maxwell-Boltzmann distribution. A graph of p(v) vs. v is shown in
figure 5.3 for two diatomic gases, N2 and H2. The average translational kinetic energy for
Maxwell Boltzmann Statistics | 155
FIGURE 5.3 Distribution of
molecular speeds for two gases,
nitrogen and hydrogen, at
298 K.
0.002
N2
H2
p(v)
0.0015
0.001
0.0005
500
1000
v (m
1500
2000
s–1)
both gases is the same, but because H2 is much lighter, its average speed is higher [by a
factor of 128>22 1>2]. At low speeds, the probability of molecules at that speed increases
proportional to v2, because the exponential term is still close to 1. As the speed increases,
the decaying exponential begins to dominate the plot, and at high speeds, the number of
molecules at that speed drops exponentially, as can be seen very clearly for N2.
The Maxwell-Boltzmann Distribution and the Speed
In the same way that we computed the average energy using Eq. 5.10, we can compute any
other quantity that depends on the speed of the molecule. In chapter 2, when we discussed
the kinetic theory of gases, we found that we could not calculate the average speed of a
molecule, but only the root mean square speed. Now, however, we have the tools we need
to compute the average speed, and any other quantity dependent on speed. From Eqs. 5.10
and 5.14, we have:
⬁
2
M
8 v 9 = 3 vp(v)dv =
a b
A p RT
0
3>2
⬁
- Mv2
2
c
v
*
v
exp
a
b d dv
3
2RT
0
(5.15)
8RT
A pM
(5.16)
Integrating and simplifying, this becomes:
8v9 =
In the same way, the average squared speed is
8 v2 9 =
⬁
2 M 3>2
- Mv2
3RT
a b 3 c v2 * v2 exp a
b d dv =
p
A
RT
2RT
M
0
(5.17)
This agrees, as it should, with Eq. 2.25. Finally, we can ask what is the most probable speed. This corresponds to the top of the Maxwell-Boltzmann distribution, where
156 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
the slope is zero, and we can therefore obtain it by differentiating Eq. 5.14 and setting the
result to zero:
dp(v)
2 M 3>2
- Mv2
Mv
- Mv2
=
a b c 2v exp a
b - v2 a
b exp a
bd = 0
dv
A p RT
2RT
RT
2RT
(5.18)
This gives
2v exp a
- Mv2
Mv
- Mv2
b = v2 a
b exp a
b,
2RT
RT
2RT
which, on dividing by the common factor, becomes
2 = v2 a
M
2RT
b 1v =
.
RT
A M
(5.19)
So vmp, the most probable speed, is less than < v >, the average speed, which is less
than < v2 >1>2, the root-mean-square speed, in the ratios 22: 2(8/p): 23.
Statistical Thermodynamics
As we have seen, an analysis of the distribution of the velocities of molecules based on
purely classical mechanics can be reconciled with the simple result for the internal energy
of gases. The mathematics of such systems is, however, far more complicated than it
appears, particularly since, in Boltzmann’s day, the very existence of atoms was controversial. In working through this intensely mathematical field, Boltzmann had the insight
that if, instead of the continuous distributions of velocities that give rise to the infinities
that we had to grapple with in the previous section, the energies of molecules were confined to discrete values, the calculations would become far more tractable. For this reason,
Boltzmann is sometimes given credit for being one of the progenitors of quantum mechanics, 25 years before Max Planck and his theory of black body radiation. We will delve
into the details of quantum theory in chapter 11; but for the moment all we need to do, as
Boltzmann did, is to assume that the atoms or molecules in a system, instead of having a
continuous spectrum of energies and velocities, have discrete energy states, with energies
Ei and occupation numbers Ni. The integrals we have in the Maxwell-Boltzmann distribution will be replaced by summations; the infinities by extremely large but finite numbers.
To illustrate the basic concepts in simple terms, our quantitative discussions are limited to a system of noninteracting particles (that is, an ideal gas). The purpose of these
discussions is to provide a molecular basis for the thermodynamic laws. We show how the
macroscopic thermodynamic properties—internal energy U, reversible heat qrev, reversible
work wrev, and entropy S —are related to the energy levels of the molecules. In the section “Statistical Mechanical Entropy,” some qualitative examples are provided for more
complex systems.
Statistical Mechanical Internal Energy
In chapter 2 we discussed the internal energy for a macroscopic system and the relation
between the internal energy U, the heat absorbed by the system q, and the work done on
the system w. A macroscopic system, such as 1 L of air, 1 drop of water, or 1 carat of
diamond, contains many microscopic particles. One liter of air at room temperature and
atmospheric pressure contains approximately 5 * 1021 molecules of O2, approximately
four times as many molecules of N2, and many other molecules. A tiny drop of water
1 μm 110 - 6 m2 in diameter contains about 17.5 billion molecules, and 1 carat (200 mg)
Statistical Thermodynamics | 157
of diamond contains about 1022 carbon atoms. Thus, the numbers we encounter, when
thinking about the statistics of molecules, are immense.
In chapter 11 we shall discuss the energy levels for a single molecule. For very simple
systems (such as ideal gases), the energy levels can be calculated from quantum mechanics.
For more complex systems, the energy levels can be obtained in principle, although the
actual computation can be very difficult and, in many cases, may not yet be possible. But
it is possible, as Boltzmann did, to discuss the statistics of systems in which molecules are
confined to discrete energy levels, without knowing very much about the values of those
energies, and we shall do that here.
For an ideal gas, the energy levels of a molecule can be assigned with good approximation to translational, rotational, vibrational, and electronic motion. Translational energy
levels are particularly closely spaced, and in fact can only be studied in macroscopic systems at tiny fractions of a Kelvin.
Each monatomic gas molecule in a container will have an energy Ei determined by its
detailed quantum mechanical properties. We can describe the energy distribution of the
monatomic gas by specifying the number of molecules N 1, N 2, c N i in each possible
energy level E1, E2, c Ei. Because the molecules of an ideal gas do not interact with one
another, the total energy E of the system is simply the sum of the energies of the molecules:
E = N 1E1 + N 2E2 + c N i Ei
= a Ni Ei .
(5.20)
i
The numbers N 1, N 2, c N i are called occupation numbers. Equation 5.20 tells us
that the macroscopic energy of an ideal gas can be calculated from the energy levels of
each molecule and the number of molecules in each energy level.
The energy of a system with interacting molecules (such as a real gas, a liquid, or a
solid) can still be written in the same form as Eq. 5.20. However, now the energies Ei are
the energy levels of the entire system, and Ni is replaced by the probability of finding the
system in energy level Ei.
There is a very useful distinction that prevails whether the molecules are isolated or
interacting. The energy levels Ei depend on the contents of the system and the volume of
the system, but not on the temperature. The temperature controls the occupation numbers
Ni, or, for a system with interacting molecules, the distribution of the whole system among
its energy levels. Changing the temperature of a system changes the distribution of the
system among its energy levels; it does not change the energy levels themselves. Changing
the volume of a system changes its energy levels.
Work
Consider a gas in a container with a movable wall (a piston). Let the external force in the
direction of movement of the piston be Fex. If we move the piston by a distance dx in the
direction of the external force, the work done on the gas by the surroundings is
dw = Fex dx .
(2.1)
If we assume reversibility, then the external force is balanced by an internal force f.
This force, as discussed in chapter 2, is made up of tiny impulses exerted by the collision
of molecules of gas with the piston wall. If we group the molecules according to energy
states Ei, then the force exerted by molecules along the direction x parallel to the travel of
the piston, in any individual energy state, is Ni Fix, and the total force against the piston
exerted by all molecules is
F = -Fex = a Ni Fix .
i
(5.21)
158 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
The work done by a reversible process is
dw = Fexdx = - a Ni Fix dx .
(5.22)
i
Force is the negative of the derivative of energy with respect to distance.
Fix = -
dEi
dx
(5.23)
Substituting Eq. 5.23 into Eq. 5.22, we obtain
dw rev = - a N i a
i
dEi
b dx = a N i dEi .
dx
i
(5.24)
Equation 5.24 states that the work done on the system in a reversible process is proportional to the change in the energy levels due to the change in the size of the container.
This is true for any system; it is not limited to an ideal gas. For an ideal gas, changing the
dimensions of the container affects only the translational energy. For real gases, liquids,
and solids, changing the size of the system affects the molecular interactions as well.
Heat
Starting with Eq. 5.20, we can write the differential expression for the change of E with
the change of Ei and Ni
dE = a N i dEi + a Ei dN i .
i
(5.25)
i
Combining Eqs. 5.24 and 5.25 we obtain
dE = dw rev + a Ei dN i .
(5.26)
i
If we identify the total energy of the system with the thermodynamic internal energy U —
that is, we set the zero of energy to the ground state 0 K energy of the system—we obtain
from the First Law of Thermodynamics, Eq. 2.26:
dE - dwrev = dqrev
Therefore,
dqrev = a Ei dN i .
(5.27)
i
Equation 5.27 states that the heat absorbed by a system undergoing a reversible change
is related to changes in the number of molecules in the various energy levels. This immediately suggests that dqrev is the part of the total energy that is related to the distribution
of the molecules among the various energy levels. As heat is absorbed, the number of
molecules in the higher-energy levels increases relative to the number in the lower-energy
levels. At 0 K, all the molecules are in the lowest-energy level. As the temperature is raised,
the population of the higher-energy levels increases as heat is absorbed.
Most Probable (Boltzmann) Distribution
We have shown how the thermodynamic energy, the reversible heat, and the reversible work
are related to the energy levels, Ei, and the number of molecules, Ni, in each energy level. In
principle, we can calculate the energy levels from quantum mechanics, although this is very
difficult except for the simplest systems. However, to obtain thermodynamic properties we
must also be able to calculate the values of the Ni and how the Ni change with temperature.
Equation 5.20 was written for a system of ideal gas molecules, where it is clear that
because the molecules do not interact, the energy of the system is the sum of the energies
Statistical Thermodynamics | 159
of the individual molecules. The same kind of equation applies to a system containing
interacting molecules, but we have to be more careful how we define Ei. It is useful to
think of a huge number of identical systems in thermal equilibrium; all have the same T.
This is called an ensemble of systems. The average energy of the system is equal to the
total energy of the ensemble divided by the total number of systems. However, each system
may have a different individual energy Ei. In effect, to deal with interacting systems, we
replace the idea of a collection of independent particles, each particle with an energy Ei,
with a collection of independent systems, each with a system energy Ei. The energy of the
ensemble is given by
E = a N iEi ,
(5.28)
i
which looks deceptively like Eq. 5.20; but now Ei is the energy of each system in the
ensemble and Ni is the number of systems in energy level Ei. The average energy < E >
(which is the thermodynamic energy) is
8E9 =
a N iEi
i
N
N = a Ni .
(5.29)
i
There are many combinations of Ni that give the same energy < E >, but we will find
that the most probable distribution of Ni is so much more probable than all the others that
we need to consider only this most probable distribution. For example, in an ensemble of
ten systems, it is possible that nine systems have zero energy (the lowest-energy level) and
one system has ten times the average energy. This corresponds to N i = 1 for Ei = 10 <E>
and N i = 9 for Ei = 0. This is very improbable. It is much more probable that a wide range
of Ei values be represented among the systems of the ensemble. As the number of systems in the ensemble increases, all improbable distributions become negligible. We chose
the ensemble as a thought experiment to allow us to consider the energies of interacting
molecules in a system. Therefore, in our thought experiment, we can have the number of
systems become infinite. This means that the most probable distribution of energies among
the systems is the only one we need to consider.
Consider a simple ensemble where it is easy to calculate all the possible distributions
of Ni. We need to find all the distributions that give an arbitrary value of average energy
E0. To simplify the example, we choose a system in which the energies are evenly spaced
and the spacing equals E0. Figure 5.4 shows the possible distributions of systems among
energy levels in an ensemble containing one system 1N = 12, two systems 1N = 22, and
three systems 1N = 32. For these small ensembles, we can count all possible distributions
consistent with our choice of average energy < E > = E0. Notice that some distributions
are more probable than others. For two systems in the ensemble, the distribution with
one system in the ground level 1E = 02 and one system in the level E = 2E0 is twice as
probable as the distribution with both systems in the level E = E0. There are two ways
of assigning two systems to two different levels; there is only one way of assigning two
systems to one level. Similarly, for three systems in the ensemble, the distribution with
one system in each of the three lowest energy levels is the most probable; there are six
ways of forming it. As the number of systems in the ensemble increases, the most probable
distribution (labeled W* in figure 5.4) has increasingly more ways of being formed than
any other distribution. The most probable distribution is the only one we need consider.
For any distribution, the number of ways of assigning systems to energy levels is
W =
N!
N!
=
,
N 1!N 2!N 3! . . .
⌸i N i!
(5.30)
160 Chapter 5 |
N=1
＜E＞ = ε0
N=2
＜E＞ = ε0
The Statistical Foundations of Biophysical Chemistry
E4 = 3ε0
E3 = 2ε0
E2 = ε0
E1 = 0
E4 = 3ε0
E3 = 2ε0
E2 = ε0
E1 = 0
＜E＞ = ε0
E4 = 3ε0
E3 = 2ε0
E2 = ε0
E1 = 0
｝
W* = 1
N4 = 0
N3 = 0
N2 = 2
N1 = 0
N4 = 0
N3 = 1
N2 = 0
N1 = 1
E4 = 3ε0
E3 = 2ε0
E2 = ε0
E1 = 0
N=3
N4 = 0
N3 = 0
N2 = 1
N1 = 0
｝
W=1
｝
W* = 2
N4 = 0
N3 = 0
N2 = 3
N1 = 0
E4 = 3ε0
E3 = 2ε0
E2 = ε0
E1 = 0
｝
W=1
N4 = 1
N3 = 0
N2 = 0
N1 = 2
E4 = 3ε0
E3 = 2ε0
E2 = ε0
E1 = 0
｝
W= 3
N4 = 0
N3 = 1
N2 = 1
N1 = 1
｝
W* = 6
FIGURE 5.4 Possible distributions of systems in an ensemble among energy levels. For simplicity, the energy spacing for each system is made
constant and equal to ε0. All possible distributions are shown for three different numbers (N = 1, 2, 3) of systems in the ensemble. The average
energy of all the systems is chosen to be ε0. For each distribution, the number of ways (W) of forming the distribution is shown. The most
probable distribution is labeled W*. The green circles depict which systems are in which energy level (0, ε0, 2ε0, 3ε0, and so on). Note that for each
distribution, N = gi Ni (the total number of systems in the ensemble) and <E> = ε0.
where ∏ means the product over all i. This is the number of ways that N distinguishable
objects can be arranged into boxes with N1 in one box, N2 in another, and so on. As seen
in figure 5.4, the possible values of Ni depend on the value of the average energy chosen,
and clearly the sum of Ni must equal N.
E X A M P L E 5 .1
Find all the possible distributions of Ni for four systems in an ensemble (N = 4) for
the energy-level pattern in figure 5.4 and < E > = E0. There are five distributions.
Use Eq. 5.30 to calculate the number of ways of forming each possible distribution to make sure that it agrees with your direct counting. There are two equally
most probable distributions: N 1 = 1, N 2 = 2, N 3 = 1, N 4 = 0, N 5 = 0, and
N 1 = 2, N 2 = 1, N 3 = 0, N 4 = 1, N 5 = 0.
The most probable distribution is the one with the largest number of ways of forming
it (designated W* in figure 5.4). As N gets large, it becomes tedious to find all possible
Statistical Thermodynamics | 161
distributions and then to calculate all the W ’s to find the maximum, but there is a simple
method using Eq. 5.30. Let’s first take the natural log of both sides:
ln W = ln a
N!
b = ln N! - ln N1! - ln N2! - ln N3! N= ln N! - a ln N!
N1!N2!N3! . . .
i
(5.31)
This will have a maximum value of ln W = ln N! if all the ln Ni terms are zero; or,
in other words, if N i = 1 for all values of i. However, this is not the distribution we need
because it does not fit the necessary constraints that the average energy 6 E7 and the total
number of systems N must be kept constant, see Eq. 5.29. To find the maximum in W when
these constraints are added requires consideration of the relationship between energy conservation and probabilities, similar to what we used in deriving the Maxwell-Boltzmann
distribution. The result, in fact, is very similar; the most probable distribution is
e - bEi
Ni
=
N
.
- bEi
ae
i
We have already shown for the classical Maxwell-Boltzmann problem that agreement with
the average energy predicted by the kinetic theory of gases (as well as the measured heat
capacities of real gases) requires that b = 1k BT 2 - 1. The Boltzmann distribution (or most
probable distribution) is therefore written
Ni
e - Ei /kBT
=
,
- Ei /kBT
N
a ie
(5.32)
where kB is the Boltzmann constant. This equation is essentially a discrete version of
Eq. 5.4 and is among the most important in statistical thermodynamics; it tells us the probability Ni /N of finding a system with a particular value of energy Ei. Like all of Boltzmann’s
work, it was obtained before quantum mechanics, but it applies to quantum mechanical
systems with minor modification. We can use Eq. 5.32 as it stands if we let the sum over
i refer to quantum states instead of quantum energy levels. An equivalent way of writing
it is by explicitly including the degeneracy of each energy level. The degeneracy gi is the
number of quantum states having the same energy. The Boltzmann distribution is now
Ni
gi e - Ei /kBT
=
,
- Ei /kBT
N
a i gi e
(5.33)
where the sum is over all energy levels. This equation is also written as
Ni
gi e - Ei /kBT
=
,
N
Q
(5.34)
Q = a gi e - Ei /k BT
(5.35)
where Q is the partition function:
i
In earlier work, particularly in the biophysical literature, Q is often replaced by the
symbol Z. The partition function is, in qualitative terms, the weighted average number of
states occupied at a given temperature. Equations 5.32 and 5.34 can be used in a very wide
range of applications. From the energy-level distribution (obtained quantum mechanically)
and the temperature, we can calculate the average energy (the thermodynamic energy)
from Eq. 5.29:
- Ei /k BT
a i gi Ei e
a i N i Ei
8E9 =
=
- Ei /k BT
N
a i gi e
(5.36)
162 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
This, of course, is the exact analog of Eq. 5.10 for the classical Maxwell-Boltzmann distribution, but with the integrals over a continuous probability function replaced with summations over discrete states. The ratio of the number of molecules in two energy levels is
Nj
Ni
=
gj e - Ej>k BT
gi e
- Ei >k BT
=
gj
gi
e - (Ej - Ei)>k BT .
(5.37)
We see that at absolute zero temperature (0 K) all molecules (or all systems) are in the
lowest-energy level, because the exponential factor in Eq. 5.37 is zero. As the temperature
is raised, higher-energy levels are populated and, as the temperature approaches infinity, the distribution becomes independent of the energies—the exponential factors are all
equal to 1.
Equation 5.37 can be used to measure temperature. The ratio of the number of particles
in two electronic energy levels can be measured from the intensities of emitted light at two
different wavelengths or frequencies. The frequencies of the emitted light give the energies of the two energy levels. Thus, the temperature is the only unknown in Eq. 5.37. The
temperatures of stars have been determined with this method.
The significance of the magnitude of k BT is apparent from Eq. 5.32 through 5.37. As
we discussed in a qualitative way in chapter 2, if the energy spacing ⌬E = Ej - Ei is
small compared to k BT, many energy levels will be populated. This applies to translational
energy and rotational energy levels at room temperature. If the energy-level spacing is large
compared to k BT, only the lowest-energy level is significantly populated. This usually
applies to electronic energy levels. Vibrational energy levels constitute an intermediate
case, where several of the lowest-energy levels are significantly populated at room temperature. At room temperature 1T = 300 K2, k BT is 4.14 * 10 - 21 J. It is more useful to
remember the value per mole: RT = N Ak BT ⬵ 2.5 kJ>mol at 300 K.
The statistical mechanics of Boltzmann distributions had to be modified slightly with
the advent of quantum mechanics; briefly, Boltzmann assumed particles were distinguishable, whereas in quantum mechanics they are indistinguishable and either fermions (which
are subject to the Pauli exclusion principle) or bosons (which are not). However, this modification has few consequences except at very low temperatures.
Statistical Mechanical Entropy
We have used the qualitative concept that entropy is a measure of the disorder of a system
(chapter 3). Boltzmann showed that the entropy is related to W, the number of ways of
forming the most probable distribution. We discussed earlier that as long as the number of
particles in the system is more than a few hundred, the most probable distribution is the
only one we need to consider. All other distributions are negligible. Boltzmann deduced
that entropy is proportional to the logarithm of the number of ways of forming the most
probable distribution W.
S = k B ln W
(5.38)
Equation 5.38 provides a molecular interpretation of entropy. Let’s see how this definition
is consistent with the laws of thermodynamics.
As we have discussed, at ordinary temperatures W is very large. When the absolute
temperature T approaches zero, however, the energy approaches a minimum, and all particles will be in the lowest-energy levels available to them. Thus, W is either unity or a very
small number compared with the values of W at ordinary temperatures. In other words,
as T approaches zero, ln W is either rigorously zero or vanishingly small compared with
ln W at ordinary temperatures. From the Third Law of Thermodynamics, S is also zero
as T approaches zero for a perfectly ordered crystalline substance. From the point of view
Statistical Thermodynamics | 163
of statistical mechanics, the Third Law is a natural consequence of the occupation of the
lowest available energy levels by the particles as T approaches zero.
One statement of the Second Law of Thermodynamics is that for an isolated system,
the equilibrium state is the one for which the entropy is a maximum. From the statistical
mechanical point of view, the equilibrium state of an isolated system is one that represents
the most probable distribution and has the maximum randomness.
Examples of Entropy and Probability
The relation S = kB ln W contains useful qualitative insights for many problems. In this
section we consider a number of quantitative examples.
EXAMPLE 5.2
Consider the two-chamber apparatus depicted in Example 3.1. We showed in that
example that when the stopcock between two equal-sized chambers, one filled with
gas, one containing a vacuum, is opened, gas flows adiabatically between the two
chambers, causing an entropy change of ⌬S = nR ln 2. Let’s now consider the process
in molecular terms. According to Eq. 5.38:
⌬S = S2 - S1 = kB ln W2 - kB ln W1 = kB ln
W2
,
W1
where the subscripts 1 and 2 refer to the initial and final states, respectively.
If there is only a single molecule, before we open the stopcock, it can only be in the
left chamber, so we’ll call that distribution L. There is only one possibility, so W 1 = 1.
After we open the stopcock, it can either be on the left, or the right, (L or R); there are
two distributions, so W 2 = 2. If there are two molecules, before we open the stopcock,
there is still only one possibility — LL — so W 1 = 1. After we open it, there are now
four possible distributions — LL, LR, RL, and RR, so W 2 = 22 = 4. If we have nNA
molecules, before the stopcock is open, W1 is still 1 (LLLLL…); after it is open:
W2 = 2nNA
Applying Eq. 5.38:
⌬S = kB ln
W2
2nNA
= kB ln
= nNAkB ln 2 = nR ln 2 ,
W1
1
which is the same result obtained by classical thermodynamics.
EXAMPLE 5.3
Suppose we have nD mol of an ideal gas D and nE mol of an ideal gas E, at the same
temperature and pressure and separated initially by a closed stopcock, as shown in
figure 3.5. If the stopcock is opened, mixing of the two gases occurs spontaneously.
Let the volumes occupied initially by D and E be VD and VE, respectively. Because the
gases are at the same temperature and pressure,
VD
nD
=
.
nE
VE
To calculate ⌬ S for this process, we can divide VD into yD and VE into yE boxes of
equal size. The number of ways of placing nDNA molecules of D in yD boxes is ynDD NA
164 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
and the number of ways of placing nENA molecules of E in yE boxes is y nEEN A. Initially,
with the barrier present, the total number of ways W1 is therefore
W 1 = y nDD N A y nEE N A .
If the stopcock is opened, there are y D + y E boxes for the gases, and
W 2 = 1y D + y E 2 nDN A 1y D + y E 2 nEN A .
Thus,
- nDN A
- n EN A
yD
yE
W2
= a
b
a
b
.
W1
yD + yE
yD + yE
However,
yD
VD
nD
=
=
,
yE
nE
VE
so
yD
nD
=
= xD
yD + yE
nD + nE
and
yE
nE
=
= xE ,
yD + yE
nD + nE
where xD and xE are the mole fractions of D and E, respectively, in the mixture.
Thus,
W2
= xD- nD NAxE- nE NA
W1
and
⌬S = k B ln
W2
= - (nDk BN A ln x D + nEk BN A ln x E) = - (nDR ln x D + nER ln x E ) .
W1
This is the same result given in Eq. 3.22.This equation can be generalized to give the
ideal entropy of mixing for any number of components:
⌬S = - R a ni ln x i
(5.39)
i
Partition Function: Applications
In our treatment leading to Eq. 5.34, we have chosen an isolated system. The partition
function
Q = a gi e - Ei >k BT
(5.35)
i
sums over all the energy states that a molecule can occupy and is therefore called the
molecular partition function.
For simple molecules, the various energy levels can be obtained by quantum mechanical calculations or by spectroscopic measurements, and the partition functions can be
obtained. If the partition function of a system is known, all the thermodynamic properties
of the system can be calculated. For a system of N noninteracting particles, the pressure p,
internal energy U, and entropy S are related to the partition function Q by the relations
p = Nk BT a
0 ln Q
b
0V T
(5.40)
Statistical Thermodynamics | 165
0 ln Q
b
0T V
U
S = kB N ln Q +
.
T
U = Nk BT 2 a
(5.41)
(5.42)
EXAMPLE 5.4
We will derive Eq. 5.41. The average energy of one molecule is given by
a i gi Ei e
8E9 =
Q
- Ei /kBT
Q = a gi e - Ei /k BT .
(5.36)
(5.35)
i
The derivative of the partition function Q with respect to T, keeping the energy levels
constant, is specified by keeping V constant, since with V constant there is no work:
a
0Q
1
b =
gi Ei e - Ei >k BT .
0T V
k BT 2 a
i
We see that <E> can be written in terms of this derivative:
8E9 =
But
k BT 2 0Q
a
b
Q
0T V
dQ
= d ln Q, so
Q
8 E 9 = k BT 2 a
0 ln Q
b .
0T V
This is the average energy of one molecule; for N molecules, we multiply by N to
obtain the internal energy U = N<E>. For derivations of Eqs. 5.40 and 5.42, a
standard physical chemistry text should be consulted.
Another important property of the partition function can be seen better if we elect to
sum over all quantum states rather than energy levels; that is, the gi degenerate states are
summed up individually:
Q = a e - E/k BT
(5.43)
all states
To a high degree of approximation, the energy of a molecule in a particular state is the
simple sum of various types of energy, such as translational energy Etr, rotational energy
Erot, vibrational energy Evib, electronic energy Eel, and so on. If
E = Etr + Erot + Evib + Eel ,
it follows immediately that
Q = a a e - Etr >k BT b a a e - Erot >k BT b a a e - Evib >k BT b a a e - Eel >k BT b = QtrQrotQvibQel (5.44)
In other words, if the energy can be expressed as a sum of terms, the partition function can
be expressed in corresponding terms; the product of which gives the total partition function.
Systems other than an isolated system can also be treated by statistical mechanics.
Because systems of chemical or biological interest are seldom isolated systems, it is more
useful to obtain the partition functions for closed or open systems. However, it is beyond
the scope of this book to treat such problems.
166 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
For a system consisting of complex molecules, it is not yet possible to obtain the
partition function rigorously. Consider a polypeptide. It has translational, rotational, vibrational, and electronic degrees of freedom, although usually the molecule is entirely in the
electronic ground state, so Qel = 1. Since the atoms interact, the energy levels of one
are strongly affected by many others. It is obviously not practical to obtain the complete
molecular partition function. On the other hand, if we are interested only in, say, the transition between an a-helix and a random coil, there is no need to know all the details of
the energy levels. All we really need are the relative contributions to the partition function
of an amino-acid residue in the helix and the coil states of the molecule. In the next section, we will discuss how to use these more limited partition functions in the context of
conformational transitions of macromolecules.
The Random Walk
We now discuss an important problem in statistics: the random-walk problem. The simplest
example of such a problem involves someone walking along a sidewalk but unable to decide
at each step whether to go forward or backward. The direction of each step is random.
This is a one-dimensional random walk. The two-dimensional random walk is analogous
to a dazed football player who may take steps randomly in any direction on the field. The
problem is also easily generalized to three (or more) dimensions.
Although the random-walk problem may seem abstract and have little to do with
molecules and physical chemistry, this is not so. A number of problems of interest, such
as molecular diffusion and the average size of a long polymer molecule, can be treated as
random-walk problems in three dimensions.
To introduce the random-walk problem, we start with the one-dimensional version. If
we use the symbol 1 for a step forward and the symbol 0 for a step back, the record of a
walk of N steps can be represented by an N-digit number, such as 11000101011 . . . 0011.
Let p be the probability of a step forward and q be the probability of a step back. When
we said that the direction of each step was “random,” we implied that p = q. In general,
though, p does not have to be the same as q.
Before going further, let’s elaborate on what we mean by “the probability of a step
forward being p.” Imagine that N similar fleas (N is a very large number) are hopping
randomly forward or backward. These N fleas form an ensemble. After each has taken
one hop, m are found to have moved forward, and the rest, N - m, to have moved backward. The fraction of fleas that moved forward, m >N, is the probability p. Similarly,
1N - m 2 >N = q. Obviously, p + q = 1, a consequence of the fact that we allow only
two possibilities, either a hop forward or a hop backward. This imaginary experiment
can be performed in a different way. Instead of having N fleas take one hop each, we can
allow one flea to take N hops. We expect that, if there are no time-dependent changes, the
fraction of steps forward is p. It is a basic postulate of statistical mechanics that the time
average of a property of a system is the same as the ensemble average at any instant. This
means that watching one molecule over a long period of time and averaging its properties
will give the same result as an average at one time over many molecules. Until recently
it was not practical to observe a single molecule to test the basic postulate of statistical
mechanics. Now we can follow the diffusion of a single molecule, as will be discussed in
chapter 8, and we can measure the kinetics of a single enzyme (see chapter 10). In most
cases, the basic postulate is correct—the time average of a single molecule equals the
ensemble average of many molecules—but there do seem to be exceptions; for example,
when individual enzymes have different kinetic properties.
Now we come back to the one-dimensional random-walk problem. We need to be
able to calculate the probability of taking any number of steps forward and any number
The Random Walk | 167
backward. Then we will be able to find how far we can expect to move from the starting
point after N random steps. The answer is that after N random steps we will just as likely
have moved forward as backward. However, this does not mean that we will always, or
even often, end up at the starting point after N random steps.
For a given sequence of N steps represented by 11000101011 … 0011, the probability
for this sequence to occur is ppqqqpqpqpp … qqpp = pmqN-m, where m is the number of
1’s. There are many ways that we can take N steps with m of them forward. For example,
if N = 4 and m = 2, the possible ways are 0011, 0110, 1100, 1010, 0101, and 1001. Several
additional examples are presented in table 5.1.
In table 5.1, we see that the sum of the probabilities of all possible outcomes of an
N-step walk can be represented in the form 1q + p2 N . Because q + p = 1, this sum is
equal to 1, but we want to know how the components of the sum depend on q, p, and N.
The sum can be expanded by the binomial theorem:
(q + p)N = q N + Nq N - 1p +
N(N - 1) N - 2 2
N!
q
p + g+
qN - m pm + g + pN (5.45)
2!
m!(N - m)!
Remember that
N! = N(N - 1)(N - 2) g (3)(2)(1) .
TABLE 5.1 Some Random-Walk Examples
Total number
of steps
1
2
3
N
Record of random
walk
Probability
Sum of probabilities
of all possibilities
q + p
0
q
1
p
00
q2
q2 + 2qp + p2
01, 10
2qp
= (q + p)2
11
p2
000
q3
q3 + 3q2p + 3qp2 + p3
001, 010, 100
3q 2p
= (q + p)3
011, 110, 101
3qp2
111
p3
0000 …
qN
…
…
…
…
1111 …
pN
(q + p)N
Each term in the expansion gives the probability of the corresponding outcome. The
term q N, for example, is the probability that all N steps are backward. The term
W(m) =
N!
q N - mpm
m!(N - m)!
(5.46)
gives the probability that m steps are forward and N - m steps are backward. The
coefficient
N!
m!(N - m)!
(5.47)
168 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
is the number of ways one can take N steps with m of them forward. Because q + p = 1,
the numerical value of 1q + p2 N is of course always 1.
Calculation of Some Mean Values for the Random-Walk Problem
The important equation is Eq. 5.46. Knowing the probability of a random walk of N steps
with m steps forward, we can now calculate averages of interest, such as the average number
of steps forward,
N
8 m 9 = a mW (m) ,
m =0
(5.48a)
and the mean square number of steps forward,
N
8 m 2 9 = a m 2W (m) ,
m =0
(5.48b)
Mean Displacement
First, let’s calculate the average of the displacement. If the length of each step is l, the net
displacement for m steps forward is [m - 1N - m 2]l, or 12m - N 2l. Since N and l are
constant, if we determine 8 m 9, the mean displacement forward is 12 8 m 9 - N 2l.
Substituting Eq. 5.46 into Eq. 5.48a, we obtain
8 m 9 = 0 * q N + 1 * Nq N - 1p + 2
N 1N - 12
2!
q N - 2p 2 + g + m
N!
q N - m pm + g + Np N .
m! 1N - m 2!
(5.49)
We can use this equation to calculate the average number of steps forward, but we note that
its terms look like those in Eq. 5.46, except that each term is multiplied by the exponent
of p in the term. This suggests that the expression for 8 m 9 is related to the derivative of
Eq. 5.46 and can be greatly simplified. We denote Eq. 5.45 by Q because it is a sum of the
number of ways each path may be constructed (equivalent to g) multiplied by the probability of each way, and so is, in effect, a partition function.
Q = (q + p)N = q N + Nq N - 1p +
N(N - 1) N - 2 2
N!
q
p + g+
q N - m pm + g + pN
2!
m!(N - m)!
0Q
N(N - 1) N - 2
N!
= Nq N - 1 + 2
q
p + g+ m
q N - m pm - 1 + g + Np N - 1
0p
2!
m!(N - m)!
pa
0Q
N(N - 1) N - 2 2
N!
b = Nq N - 1p + 2
q
p + g+ m
q N - m pm + g + Np N
0p
2!
m!(N - m)!
(5.45)
(5.50)
Comparing Eqs. 5.50 and 5.49, we see that
8 m 9 = pa
0Q
b
0p
(5.51)
but
Q = (q + p)N .
Therefore,
a
and
8 m 9 = pa
0Q
b = N(q + p)N - 1
0p
0Q
b = Np(q + p)N - 1 = Np .
0p
(5.52)
The Random Walk | 169
The last equality comes from the fact that 1q + p2 = 1; this leads to the simple and intuitively reasonable result. The average number of steps forward is the probability of a forward step times the total number of steps. For the random (unbiased) walk, p = 1>2, and
8m 9 =
N
,
2
and the net displacement forward is
12 8 m 9 - N 2l = 0 .
(5.53)
(5.54)
This seems completely reasonable. If the probability of going forward is the same
as that of going backward, on the average half of the total number of steps taken will be
forward. After taking N steps, sometimes a person might be a few paces in front of the
origin, sometimes a few paces behind the origin. The average, or mean, displacement
comes out to be zero.
Mean-Square Displacement
After taking a larger and larger number of steps N at random, it becomes increasingly
unlikely that we will end up exactly where we started. The mean-square displacement is
a measure of how far we can expect to be from the origin, on average. The importance of
the difference between the average distance from the origin and the average net displacement forward is that in calculating the average distance we do not distinguish whether we
finally end up in front of or behind the origin. For example, if after N steps person A ends
up three steps forward of the origin and person B ends up three steps behind the origin,
the average of their net displacements is zero; but the average of their distances from the
origin is clearly not zero; it is three steps.
One way to characterize the average distance from the origin is to calculate the meansquare displacement. In a single journey of N steps, if m steps are forward, the net displacement forward is
d = 12m - N2l
and the square of the displacement is
d2 = 12m - N 2 2l2 .
(5.55)
(5.56)
Whether m 7 N >2 (a net displacement forward) or m 6 N >2 (a net displacement backward), d2 is positive.
The mean value of 12m - N 2 2 can be calculated as follows, where 812m - N 2 2 9
designates the average of that quantity:
812m - N 2 2 9 = 4 8 m 2 9 - 4 8 m 9 N + N 2
(5.57)
But 6 m 7 = N >2 for a random walk; thus,
812m - N 2 2 9 = 4 8 m 2 9 - 2N 2 + N 2 = 4 8 m 2 9 - N 2.
To calculate 8 m 2 9 , we start with its definition given in Eq. 5.48b,
(5.58)
N
8 m2 9 = a m2W(m)
m=0
= 0 * q N + 1 * Nq N - 1p + 22
N(N - 1) N - 2 2
N!
q
p + g+ m 2
q N - m pm + g+ N 2p N .
2!
m!(N - m)!
(5.59)
If we differentiate Eq. 5.50 with respect to p, we obtain
0Q
N(N - 1) N - 2
0
N!
c pa
b d = Nq N - 1 + 22
q N - m pm - 1 + g+ N 2pN - 1 .
q
p + g+ m 2
0p
0p
2!
m!(N - m)!
(5.60)
170 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
Comparing Eqs. 5.59 and 5.60, we note that multiplying the right side of Eq. 5.60 by p
gives the right side of Eq. 5.59. Therefore,
8 m 2 9 = pe
0Q
0
c pa
bdf .
0p
0p
(5.61)
From Eq. 5.52,
pa
0Q
b = Np(q + p)N - 1 .
0p
Therefore,
0(q + p)N - 1
0Q
0
c pa
b d = Ncp
+ (q + p)N - 1 d
0p
0p
0p
= N 3 p(N - 1)(q + p)N - 2 + (q + p)N - 1 4
= N 3 p(N - 1) + 1 4
and
p
0Q
0
c pa
b d = Np 3 p(N - 1) + 1 4 .
0p
0p
(5.62)
0Q
N(N + 1)
0
N N - 1
c pa b d = a
+ 1b =
0p
0p
2
2
4
(5.63)
For p = 1>2,
8 m2 9 = p
but
812m - N 2 2 9 = 4 8 m 2 9 - N 2
(5.58)
and, after all these complicated manipulations, we obtain the strikingly simple result that
812m - N 2 2 9 = N 1N + 12 - N 2 = N .
Therefore, the mean-square displacement 8 d2 9 = 812m - N 2 2 9 l2 is
8 d2 9 = Nl2 .
The root-mean-square displacement 8 d2 9 1>2 is
8 d2 9 1>2 = N 1>2l .
(5.64)
(5.65)
(5.66)
Equation 5.66 says that after N random steps, on the average, one is 2N paces from
where one started. Although we have derived this equation for a one-dimensional random
walk, it can be shown that it holds for a random walk in two- or three-dimensional space as
well. We presented the derivations for the random-walk displacements in detail to illustrate
how average properties are obtained from partition functions.
Diffusion
Molecules, either in the gas phase or the liquid phase, are always undergoing collisions.
Let’s assume that the average distance traveled by a molecule between two successive
collisions (the mean free path) is l. Whenever a molecule has a collision, its direction
changes, depending on the direction of the molecule hitting it. The total number N of collisions for a given molecule is proportional to the time t. There is a close relation between
the diffusional process and the random walk. The fact that the individual “steps” in the
diffusion process are not all of the same length turns out to be unimportant as long as the
measurements are made over distances that are large compared with the mean free path.
For a given spatial distribution of molecules at time zero, the spatial distribution of the
The Random Walk | 171
molecules at time t can be obtained by considering each molecule as a random walker.
The average value of the square of the displacement for the diffusional process, based on
Eq. 5.65, is expected to be proportional to N and therefore to the elapsed time. The proportionality constant is a measure of how fast the molecule diffuses and is directly related
to the diffusion coefficient D, as we shall see in chapter 7.
Average Dimension of a Linear Polymer
A flexible polymer can assume many conformations which differ little in energy. Consider
a polypeptide,
O
H
C
C
H
R
R
C
N
C
H
O
H
O
N
C
C
H
N
R
H
for example. Along the backbone of the polymer, rotation around the amide bonds N—C
has a comparatively high-energy barrier because of the partial double-bond character of
these bonds. Rotation around the other bonds is relatively free, however, resulting in many
polymer conformations of similar stability. Some of these conformations may be highly
extended, whereas others may be much more compact. Thus, in discussing the dimensions
of such a polymer, it is useful to specify average dimensions. There is an ensemble of polymers with a range of conformations at any time; each polymer will change its conformation
over a period of time to sample the range of conformations in the ensemble.
A quantity frequently used to express the average dimension of a flexible polymer is
the root-mean-square end-to-end distance, 8 h2 9 1>2. h is the straight-line distance between
the ends of the molecule; the root-mean-square end-to-end distance is the square root of the
average of its square. If a polymer molecule is highly extended, its end-to-end distance is
much greater than for the same molecule in a compactly coiled form. A simple and useful
model for the evaluation of the average end-to-end distance is the random-coil, or Gaussian,
model. In this model, the linear polymer is considered to be made of N segments, each
of length l, linked by 1N - 12 universal joints (unrestricted bending) so that the angle
between any pair of adjacent segments can take any value with equal probability.
The reader may recognize immediately that the random-coil model is exactly the same
as the random-walk problem we have discussed. From Eq. 5.65, the mean-square end-toend distance is 8 h2 9 = 8 d2 9 = Nl2. For a random-coil polymer, 8 h2 9 is proportional to
N and therefore proportional to the molecular weight of the polymer. We can also write
8 h2 9 = Nl2 = Nl1l2 = Ll ,
(5.67)
where L K Nl is the contour length of the molecule. The contour length of a polymer
molecule is the length measured along the links of the polymer.
Real polymer molecules are, of course, made of monomers linked by chemical bonds
rather than segments linked by universal joints. The relation between bond lengths and
lengths of statistical segments is not simple. In general, the length l of a statistical segment
must be determined experimentally.
Let’s consider polyethylene,
H
H
C
C
H
C
H H
H
172 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
For simplicity, we consider only the carbon backbone chain,
θ
C
C
C
with a bond angle u between two adjacent bonds and bond length b for each C—C bond.
If u could assume any value, you could imagine a universal joint at each carbon atom and
the mean-square end-to-end distance 8 h2 9 = Nl2 = Nb2, where N is the number of carbon
atoms per polymer chain. We know, however, that the bond angle u cannot assume any
value. Rather, the most stable conformation for
C
θ
C
C
bonds has an angle u close to the tetrahedral angle of 109°. The restriction in u makes the
molecule a little less flexible. If the random-coil model is used for polyethylene, we expect
that l 7 b; the effective distance l between carbon atoms will be greater than the bond
length b. In fact, it can be shown that l = 22b for polyethylene.
For a polymer such as a double-stranded DNA, the effective segment length l of an
equivalent random coil cannot be related directly to the bond lengths. Hydrodynamic measurements give l ~100 nm for DNA. Thus, for a flexible molecule such as polyethylene, the
segment length is of the order of the bond length l = 22b ~ 0.22 nm; for a stiff molecule
such as a double-stranded DNA, the effective length is several orders of magnitude larger
than a bond length.
A quantity closely related to mean-square end-to-end distance is the mean-square
radius, which is defined by
2
a im i ri
2
8R 9 =
(5.68)
a im i
for any collection of mass elements, where mi is the mass of the ith element and ri is the
distance of this element from the center of mass of the collection.
It can be shown that for an open-ended random coil,
Nl2
,
(5.69)
6
and for a circular random coil (formed by linking the two ends of the open-ended chain),
8 R2 9 =
8 R2 9 =
Nl2
.
12
(5.70)
Debye showed that 8 R2 9 can be measured experimentally by light scattering.
Note that the random-coil model predicts that 8 h2 91>2 or 8 R 2 91>2 is proportional
to 2N or the square root of the molecular weight. This relation can be tested experimentally. If it is found to be true, the polymer is said to behave as a random coil, or a
Gaussian chain.
For real polymers, in general, interactions occur between adjacent segments,
and they have preferred orientations instead of equally probable orientations, as
required by the random-coil model. At a given temperature, the random-coil model
is expected to be applicable only in solvents (“u solvents”) in which the interactions
between polymer segments are balanced by solvent−polymer interactions. Otherwise, the interactions between the polymer segments will cause deviations from the
random-coil model.
Helix–Coil Transitions | 173
Helix–Coil Transitions
Helix–Coil Transition in a Polypeptide
In a polypeptide chain, the partial double-bond character of the N—C bonds for each
amide residue in the chain requires the group of six atoms
H
N
Cα
C
Cα
O
to lie in a plane under most circumstances. It was first suggested by Pauling and his
coworkers in 1951 that the chain of planar groups can be rotated around the a-carbons
(C atoms between the CO group of one amino acid and the NH group of the next) into a
helix such that a hydrogen bond is formed between each C“O group and the fourth following NH group. The locations of the hydrogen and oxygen atoms involved in hydrogen
bonding are illustrated in figure 5.5 (the resulting helix, called an a-helix, is shown in
figure 3.10). The polypeptide backbone of the a-helix follows closely along the contour
described by a helical spring made from a single strand of wire. The locations and amounts
of a-helices in a protein are deduced by X-ray diffraction and NMR studies.
For a number of synthetic polypeptides, the entire chain of the molecule can assume
the a-helix structure, depending on the solvent, temperature, pH, and so on. In the nonhydrogen-bonded form, the polypeptide chain is flexible and can assume many conformations, as discussed previously. Such a chain is frequently said to be in the “coiled” form.
The transition from the coiled form to the helical form (or vice versa) is referred to as a
coil–helix transition. For polypeptides of high molecular weight, if one of the parameters
(such as temperature) is changed, the transition from the coiled form to the a-helix form
(or vice versa) generally occurs within a narrow range.
We can understand the basic features of the helix−coil transition using an approach
that is similar to our discussion of the binding of small molecules by a polymer. The
conformation of the chain can be specified by the states of the carbonyl oxygen atoms.
We use the symbol h (for helix) to represent a carbonyl oxygen if it is hydrogen-bonded
and the symbol c (for coil) if it is not. Whenever an oxygen is in a bonded state, it should
be understood that it is hydrogen-bonded to the NH group in the fourth residue ahead
in the chain (see figure 5.5). We number starting at the amino end of the polypeptide;
thus, the CO in the residue 1 can bond to the NH in residue 5, the CO in residue i can bond
to the NH in residue i + 4, and so forth. The last four residues in a chain by definition
cannot be designated as h; their carbonyl groups cannot bond to an appropriate NH. It is
further assumed that the sense of the helix is unique (either left-handed or right-handed,
but not both). In this way, the conformation of a chain of N residues is represented by
N letters, with the last four letters always c. For example, the conformation shown in
figure 5.5 is designated hhcccc.
Now consider the transition from a state such as cccccccc . . . to a state cchccccc . . . ,
which represents the transition of the third residue from the nonbonded state to the a-helix
conformation. In the nonbonded state, there are two covalent bonds in each residue around
which rotation may occur. They are the bonds on each side of the a-carbon labeled with
arrows in figure 5.5. In a helical form, however, the chain assumes a much more rigid conformation. Because the oxygen atom of the third residue is hydrogen-bonded to the NH
group of the seventh residue, the transition of the third residue from the coil state to the
helix state means that the fourth, fifth, and sixth residues must be ordered into the helical
174 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
FIGURE 5.5 A polypeptide
chain showing the hydrogen
bonding between the C = O and
the NH groups in an α-helix.
Residues are numbered starting
from the amino terminal; the
residues are separated by
dotted lines. Arrows show the
two bonds (on either side of
each α-carbon) around which
rotation can occur in the
coil form.
H
R
C
N
C
H
O
H
O
N
C
C
H
R
H
R
C
N
C
H
O
H
O
N
C
C
H
R
H
R
C
N
C
H
O
H
O
N
C
C
H
R
conformation (even though their oxygen atoms are not hydrogen bonded and they are designated c). In other words, the formation of the first helical element involves the ordering of
three residues; the orientation around six bonds is fixed. The transition of the state cchccccc.
. . to the state cchhccccc. . ., however, requires the ordering of only one more residue, the
seventh, because the fourth, fifth, and sixth are already ordered. The orientation around only
two more bonds is fixed. Intuitively, then, we expect that the formation of the first helical
element is more difficult than the formation of the next adjacent helical element.
Based on this discussion, we assign statistical weights according to the following rules:
1. For each c, the statistical weight is taken as unity.
2. For each h after an h, the statistical weight is s.
3. For each h after a c, the statistical weight is ss.
Note that s is the equilibrium constant for the reaction
cchccccc c S cchhcccc c
for the addition of a helical element to the end of a helix; ss is the equilibrium constant
for the reaction
cccccccc c S cchcccc c
for the initiation of a helical element. We expect s to be less than 1 because initiation of
an a-helix is more difficult than propagation of the helix.
However, the very nature of the a-helix structure dictates more than nearest-neighbor
influence. For example, if the third residue goes into a helical state, the fourth, fifth, and
sixth residues are brought into the helix conformation. The consequence of this is that the
conformations of three adjacent residues are closely related, and conformations with two
h’s separated by no more than two c’s are expected to be rare. To see the last point more
clearly, let’s consider a reaction
chhhhccccc c S chcchccccc c
c123456789 c S c123456789 c
as an example. For c hhhhccccc c the second to seventh residues are held in
helical geometry by the hydrogen bonds from residues 1 to 5, 2 to 6, 3 to 7, and 4 to 8.
In c hcchccccc c, the second to seventh residues are still held in helical geometry by
hydrogen bonds from residues 1 to 5 and 4 to 8.
Thus, for this reaction, although two hydrogen bonds are broken, there is no gain in
freedom for any of the residues. Therefore, the reaction is not favored. This leads to the
fourth rule for the assignment of statistical weights:
4. For two h’s separated by no more than two c’s, the statistical weight is zero.
To assign a statistical weight of zero is of course equivalent to saying that such conformations are not permitted. The correct statistical weight might be 10 -7, not 0, but the
rule is a good approximation for our purpose.
Helix–Coil Transitions | 175
With these four rules, we can assign the statistical weight for any conformation. The
statistical weight for the conformation chhhccccchccchhcccc is s3s6, for example.
With these rules for the statistical weights, which correspond to the exponential in
Eq. 5.35, the partition function Q can be formulated and the expression 10lnQ>0lns2 s gives
the average number of monomer units in the helical conformation per polymer molecule.
We do not give the mathematical details here. Certain features of the coil−helix transition
are discussed in the following examples.
EXAMPLE 5.5
Results of the classical treatment of Zimm et al. (1959) for the coil−helix transition of
poly-g-benzyl-L-glutamate are shown in figure 5.6. The curves shown were obtained
by the statistical analysis that we outlined. The value of s used to calculate the curves
is 2 * 10 - 4. As we have discussed, the parameter s represents the equilibrium constant
for the change of a coil element at the end of a helical segment into a helical segment:
s
chhhhhc cm chhhhhh c
If the polypeptide molecule is very long, the free-energy difference between a monomer unit in the coil form and in the helix form is zero at the temperature corresponding
to the midpoint of the transitions (Tm, the melting temperature) shown in figure 5.6.
Thus, s is equal to 1 at Tm. At any other temperature, s can be calculated from
ln
s2
⌬H ⬚ 1
1
= a
b,
s1
R T2
T1
where ⌬H ⬚ is the enthalpy change for the reaction shown above.
When N is large, the transition is very sharp. The polypeptide changes from 80% coil
to 80% helix over a temperature range of ~7⬚C. In this temperature range, the change in
s is rather small, from 0.97 to 1.03. The cooperative transition is the result of a small s.
If we consider the formation of a helix from a coil, a small s means that to start a helical
region in the middle of a coil region is difficult, but once a helical region is initiated, it is
1.0
n = 1500
0.8
f
0.6
n = 46
0.4
n = 26
0.2
0.0
10
20
30
40
T (°C)
50
60
70
FIGURE 5.6 Temperature
dependence of the fraction of
monomer units in the helix form
for poly-γ -benzyl-L-glutamate
(in a 7:3 mixture of dichloracetic
acid and 1,2-dichloroethane
as the solvent) with polymer
lengths of 26, 46, and 1500
monomer units. In this solvent
the helix is favored at high
temperatures.
(Based on data from B. H. Zimm
and J. K. Bragg, “Theory of the
Phase Transition between Helix and
Random Coil in Polypeptide Chains,”
Journal of Chemical Physics, 31,
p. 526. Copyright (c) 1959 American
Institute of Physics; B. H. Zimm,
P. Doty, and K. Iso, “Determination of
the Parameters for Helix Formation
n Poly-y-Benzyl-L-Glutamate,”
Proceedings of the National
Academy of Sciences, 45(11),
pp. 1601–1616. Copyright (c) 1959
National Academy of Sciences.)
176 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
much easier to extend it. This can be seen from the statistical weights of the species for
the following transition:
cccccccc c m ccccchcc c m ccccchhc c
Statistical
weight:
1
ss
ss2
For the first reaction, the equilibrium constant is ss>1, or ss. For the second reaction, the
equilibrium constant is ss2 >ss = s, which is much greater than ss.
Conversely, if we consider the formation of a coil from a helix, to start a coil region
in the middle of a helix is difficult, as follows:
cccchhhhhhh cm cccchhccchh cm cccchhcccch c
Statistical
weight:
ss7
s 2 s4
s 2 s3
The equilibrium constant for the first reaction is s2s4 >ss7, or s>s3. The equilibrium
constant for the second reaction is s2s3 >s2s4 = 1>s. Since, in the region of interest, s ⬇ 1
and s = 1, to initiate a coil region in the middle of a helix is more difficult. A consequence
of this is that, for short polypeptides, coil regions are expected to be at the ends of the
molecules. For long polypeptides, there are so many sites in the middle of the molecules
that, in the transition zone, coil regions are expected to be present in the middle of the
molecules as well.
The sharpness of the transitions shown in figure 5.6 differs for different N, although
all curves can be fitted by the same parameters. For the shorter polypeptides, the transition is broader.
EXAMPLE 5.6
As shown in figure 5.7, heating ribonuclease A (in 0.04 M glycine buffer, pH 3.15)
causes the unfolding of the protein. The unfolding is monitored by the change in
light absorption. The transition from the folded form to the unfolded form occurs in
a narrow temperature range. The unfolding of a protein is more complex than the helix−coil transition of a simple polypeptide, however. The folded protein has structural
features other than the a-helix. In addition, factors such as disulfide bridges, ionic
bonds, and interactions between the nonbonded groups can contribute to the stability
of the folded form. However, because most of these factors contribute significantly
only if the protein is in a correctly folded form, the unfolded to folded or folded to
unfolded transition is highly cooperative. In fact, the all-or-none model has been used
in such studies until very recently, when kinetic studies indicated that there are more
than two states involved.
Helix–Coil Transition in a Double-Stranded Nucleic Acid
In the preceding discussion on the helix−coil transition in an a-helix, we have seen that the
difficulty in initiating the formation of a helical segment 1s = 12 is the essential reason
for the cooperative transition. In a double-stranded nucleic acid, a somewhat similar situation exists. Double-stranded DNA is held together by the interactions between complementary bases in two single strands of DNA, as illustrated in figure 3.12. Let’s represent a base
pair (in the hydrogen-bonded form) by the symbol 1 and a pair of bases in the nonbonded
Helix–Coil Transitions | 177
FIGURE 5.7 Temperature
dependence of ε - εN of
ribonuclease A. ε is the molar
absorptivity of the protein
at 287-nm wavelength and
temperature T; εN is the same
quantity when the protein is in
its native (folded) form.
–1.5
ε – εN
–1.2
–0.9
(Based on data from J. Brandts and
L. Hunt, “Thermodynamics of protein
denaturation. III. Denaturation
of ribonuclease in water and
in aqueous urea and aqueous
ethanol mixtures,” Journal of the
American Chemical Society, 89(19),
pp. 4826–4838. Copyright (c) 1967
by American Chemical Society.)
–0.6
–0.3
–0.0
20
30
40
50
60
70
Tm (°C)
form by the symbol 0. Let s be the equilibrium constant for the formation of an additional
bonded pair at the end of a helical segment:
s
11110000 cm 11111000 c.
If we compare the reaction above with the reaction
ss
11110000 cm 11110100 c,
we note that there is an important difference. In the first reaction, the newly formed base
pair is directly “stacked” on the base pair at the end of the original helix. For polynucleotides, either DNA or RNA, the stacking of base pairs is thermodynamically favored. In
the second reaction, there is little stacking interaction between the newly formed pair
and the last pair in the preceding helical segment. If the equilibrium constant is ss for
the second reaction, we expect that s V 1 if stacking interaction is thermodynamically
favored. Thus, the helix−coil transition in a double-stranded nucleic acid is similar to that
in a polypeptide.
There are several new features in the nucleic acid case, however. First, let’s consider
the two reactions
11110000 cm 11110100 c
and
11110000…0000
j zeroes
11110000…0010 .
j zeroes
If j is very large, it means that the pair of bases that are to form the additional base pair
are, on the average, far apart in the nonbonded form. From Eq. 5.66, the root-mean-square
average distance between this pair of nonbonded bases is proportional to 2j. It is therefore
expected that the larger j is, the more difficult it is to bring the two together to form a base
pair. Thus, unlike the case for a polypeptide, s for a nucleic acid is not a constant but is a
function of j. The equilibrium constant for the first reaction is sis and that for the second
is sjs. This complicates the situation quite a bit. To compare this case with the binding
of small molecules by a polymer, the dependence of s on j is equivalent to the situation
where we have to consider not only the nearest-neighbor interactions 1j = 02 but also the
next-nearest-neighbor interaction 1j = 12, the next, …, and so on.
178 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
Second, the formation of the first base pair between two complementary chains means
that the two chains, which are free to move about in solution independently of each other,
must be brought together. After the formation of the first base pair, one of the two chains
can still be free to be anywhere, but the second chain is constrained by the pairing to move
with the first chain. Thus, for the reaction
000c000 h 0000100c000
all zero
all zeros except one
the equilibrium constant is taken as ks, with k expected to be much less than 1. The parameter k is dependent on total concentration. The lower the concentration, the harder it is to
bring two chains together, and the smaller the value of k.
Third, there are two major types of base pairs in a nucleic acid. For a DNA, these are
A · T pairs and G · C pairs; for an RNA, they are A · U pairs and G · C pairs. Because the
stabilities of the two kinds of base pairs are different, in general, we use the parameters
sA and sG for these types rather than a single s. Strictly speaking, the stacking interaction
between two adjacent pairs is dependent on what kinds of pairs they are, and therefore sA
and sG are further dependent on at least the nearest-neighbor base sequence. In chapter 4,
we discussed a different approach to this issue, and showed that using a set of empirical
parameters, derived from melting data on short DNA double-stranded oligomers, we could
predict the thermodynamics of short strand melting from the base sequence.
To include all these considerations in a theoretical analysis of the helix−coil transition
is beyond the scope of this text. Nevertheless, from our analyses of the problem, a number
of features of the helix−coil transition in a nucleic acid can be understood:
1. When N is large, the transition from the completely helical form to the completely coiled
form in a given solvent occurs within a narrow range of temperature. This cooperative
process is frequently referred to as the melting of a nucleic acid. The temperature
corresponding to the midpoint of the transition is designated as Tm, the melting
temperature. The cooperativity of the transition is primarily a result of the stacking
interactions between the adjacent base pairs, which make the initiation of a helical
segment, as well as the disruption of a base pair inside a helical segment, difficult.
2. The enthalpy change for the reaction
s
11110000 cm 11111000 c
is negative; heat is evolved. In most solvents, sG 7 sA and so Tm for a nucleic acid rich
in G + C is higher than that for a nucleic acid rich in A + T1or A + U2. If the base
composition of a nucleic acid is intramolecularly heterogeneous—that is, some segments
of each molecule are richer in A + T1or A + U2 than the other portions—the melting
profile is broadened, with the melting of the regions richer in A + T1or A + U2
preceding the melting of the 1G + C2 -rich regions. This is depicted in figure 5.8. Because
of the cooperative nature of the transition, segments heterogeneous in base composition
must be sufficiently long to show independent melting profiles.
3. For a large molecule, the parameter k contributes negligibly to ln Q. Because k is
the only concentration-dependent term in Q, Tm for a large molecule is expected to
be independent of the concentration. For short helices, this is not true. Tm is lower at
lower concentration. Similarly, at the same total concentration of nucleotides, Tm for
a high-molecular-weight nucleic acid is higher than that of a low-molecular-weight
nucleic acid of the same base composition. This dependence is evident only in the
molecular-weight range where k contributes significantly to ln Q.
4. As discussed for the helix−coil transition in a polypeptide, if N is small, the coiled
regions should be at the ends of the molecules. This is frequently referred to as “melting
from the ends.”
Helix–Coil Transitions | 179
(a)
FIGURE 5.8 Schematic illustration of the thermal “melting”
of a double-stranded DNA (a)
to its complementary single
strands (c). At an intermediate
temperature, single-stranded
loops and/or ends can form, as
illustrated in (b).
(b)
(c)
+
EXAMPLE 5.7
The transition temperatures for the helix−coil transitions of DNAs correlate well with
the base composition of the DNA. Plots of a large number of such measurements for two
sets of experimental conditions are shown in figure 5.9. Analogously, for a particular
DNA molecule, the range of temperatures (transition width) over which melting occurs
is related to the uniformity of base composition along major segments of the chain.
Synthetic DNAs with regular repeating sequences will have sharp melting transitions.
EXAMPLE 5.8
The effect of sequence length on DNA melting temperature is shown in figure 5.10. For
palindromic DNA consisting of back-to-back poly(dA) tracts, denoted as AnTn, increasing the value of n increases Tm and makes the transition more cooperative, thus making
the melting curve steeper. Ultimately, poly1dA2 * poly1dT2, where n becomes
effectively infinite, melts at 76.5⬚C 1Tm = 349.6 K2.
110
Tm (°C)
100
0.15 M sodium chloride
+ 0.015 M sodium citrate
90
80
70
60
30
0.01 M sodium phosphate buffer
+ 0.001 M EDTA
40
50
60
% (G + C)
70
80
FIGURE 5.9 Melting temperatures of double-stranded DNAs
are plotted against the mole
fraction of G · C pairs in the
DNAs. In a given ionic medium,
Tm increases with increasing
G · C content, indicating that
sG > sA. Note also that for a given
base composition, Tm is lower
in a medium with a lower Na+
concentration. This is because
the repulsion between the
negatively charged phosphate
groups on the two complementary strands is decreased by the
positively charged counterions
(Na+ in this case). Increasing the
Na+ concentration increases the
stability of the double helix and
therefore increases its Tm
(Based on data from J. Marmur, P.
Doty, “Determination of the base
composition of deoxyribonucleic
acid from its thermal denaturation
temperature,” Journal of Molecular
Biology, 5(1), pp. 109–118, ISSN
0022–2836, 10.1016/S0022–
2836(62)80066–7. Copyright (c)
1962 Elsevier, Inc. (http://www.
sciencedirect.com/science/article/
pii/S0022283662800667))
The Statistical Foundations of Biophysical Chemistry
FIGURE 5.10 Length dependence of the melting curves
for short double-stranded
DNA helices. As the sequence
length increase, the melting
temperature and the cooperativity also increase, although
the increase gets smaller with
each step. Eventually, the curves
will converge toward that for
poly (dA) * poly (dT) ss-DNA
concentration 0.001 M; NaCl
concentration 1 M.
100
80
% single stranded
180 Chapter 5 |
A4T4
A5T5
60
A6T6
A7T7
40
20
0
280
290
300
T (K)
310
320
330
EXAMPLE 5.9
The difference in Tm for short helices of identical base composition but different
sequence shows that s is primarily dependent on whether a base pair is A · T or G · C
but also depends on the sequence of the base pairs (figure. 5.11). The nearest-neighbor
approximation to the sequence dependence of formation of DNA double strands was
discussed in chapter 4. For large double-stranded DNAs and RNAs, the parameter sA is
therefore the average s of an A · T pair. Similarly, sG is the average s of a G · C pair.
$7$7
VLQJOHVWUDQGHG
FIGURE 5.11 Sequence
dependence of the melting curves
for short double-stranded DNA
helices. Replacing an A and a T in
the middle of the sequence with
a C and a G causes a ~10° increase
in the melting temperature;
replacing a CG with GC or TA with
AT increases it by an order of
magnitude less. sA is substantially
smaller than sG; within sequences of
the same overall base composition,
the ordering of the bases makes a
smaller but nonzero contribution.
ss-DNA concentration is 0.001 M;
NaCl concentration is 1 M.
$$77
$*&7
$&*7
7 .
Binding of Small Molecules by a Polymer
When two simple species, A and B, interact, we found in chapter 4 that the system at equilibrium can be described by an equilibrium constant K and that the standard free-energy
Binding of Small Molecules by a Polymer | 181
change ⌬G⬚ is related to the equilibrium constant K by the equation ⌬G⬚ = -RT ln K.
As an example, for the combination of A and B to form AB,
A + B L AB ,
the equilibrium constant and the standard free-energy change are simply
K =
3 AB 4
3A 4 3B 4
(5.71)
and
⌬G⬚ = -RT ln K .
(5.72)
In many cases of biological interest, one of the two interacting species is a complex
macromolecule, and the other is a relatively small molecule. Usually, the macromolecule
has more than one site where it can bind the small molecule. As a result, many different molecular complexes can form, depending on which sites of the macromolecule are
occupied by the small molecules. For example, a hemoglobin molecule has four sites for
oxygen. A high-molecular-weight DNA has numerous sites for the binding of actinomycin.
In such cases, an expression such as Eq. 5.71 is not very descriptive in comparison with
the richness of detail of the different species at the molecular level. We could write an
equilibrium constant for each possible species, but the numbers often are very large and
such cases can be treated more easily by the use of statistical methods. In the following
sections, we explore a number of examples.
Identical-and-Independent-Sites Model
Suppose that a polymer molecule P has a number of identical and independent sites for
binding a smaller molecule A. If the sites are identical (each has the same affinity for A)
and independent (the occupation of one site has no effect on the binding to another site), a
very simple description of this system can be obtained. Examples for which the identicaland-independent-sites model is applicable are the binding of DNA polymerase molecules
to a single-stranded DNA, the binding of the drug ethidium at low concentrations to a
DNA double helix, and the binding of substrates to certain enzymes or proteins containing
several identical subunits per molecule.
As a simple example, let’s consider a polymer P with four identical and independent
sites. This polymer molecule might be a tetrameric protein, with each of the four subunits
having one site for the binding of a substrate A. Suppose that the four sites are distinguishable and can be labeled 1, 2, 3, and 4. Since the sites are identical, the equilibrium
constants of the following reactions are identical:
P + A m PA(1)
K =
3 PA(1) 4
3P 4 3A 4
P + A m PA(2)
K =
3 PA(2) 4
3P 4 3A 4
P + A m PA(3)
K =
3 PA(3) 4
3P 4 3A 4
P + A m PA(4)
K =
3 PA(4) 4
3P 4 3A 4
182 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
The superscript (1, 2, 3, or 4) in each reaction specifies which site of the polymer is
involved. The total concentration of species with one A molecule per P molecule, [PA], is
3 PA 4 = 3 PA(1) 4 + 3 PA(2) 4 + 3 PA(3) 4 + 3 PA(4) 4
= 4K[P][A].
In other words, for the reaction
K1
P + A m PA (any site)
K1 =
3 PA 4
= 4K .
3P 4 3A 4
(5.73)
The factor 4 in Eq. 5.73 indicates there are four ways for the formation of a PA (by binding of an A to any one of the four sites of P), but only one way for any PA to dissociate
into P and A.
For the reaction
K2
PA (any site) + A m PA2 (any two sites)
K2 =
3 PA2 4
4 - 1
3
= a
bK = K ,
2
2
3 PA 4 3 A 4
(5.74)
there are (4 - 1), or 3, ways of picking one of the three remaining sites in PA for the
formation of a PA2, and there are two ways for the dissociation of a PA2 into PA and A.
Similarly, the equilibrium constants K3 and K4 are:
K3 = a
4 - 2
2
bK = K
3
3
(5.75)
K4 = a
4 - 3
1
bK = K .
4
4
(5.76)
and
The concentrations of the various species are
3 PA 4 = 4K 3 P 4 3 A 4 = 4 3 P 4 (K 3 A 4 )
3 PA2 4 = K 3 PA 4 3 A 4 = a
3
2
4#3 2
b K 3 P 4 3 A 4 2 = 6 3 P 4 (K 3 A 4 )2
2
3 PA3 4 = K 3 PA2 4 3 A 4 = a
2
3
4#3#2 3
b K 3 P 4 3 A 4 3 = 4 3 P 4 (K 3 A 4 )3
3#2
3 PA4 4 = K 3 PA3 4 3 A 4 = K 4 3 P 4 3 A 4 4 = 3 P 4 (K 3 A 4 )4.
1
4
(5.77)
Let K[A] = S. The total concentration of the polymer is
3 P 4 T = 3 P 4 + 3 PA 4 + 3 PA2 4 + 3 PA3 4 + 3 PA4 4
= 3 P 4 (1 + 4S + 6S2 + 4S3 + S4)
= [P] 11 + S 2 4 .
(5.78)
The binomial coefficients are exactly those seen in the random-walk problem
(Eq. 5.45). In fact, n sites on a macromolecule which can be either occupied or unoccupied
looks mathematically almost identical to a random walk of n steps that can be taken either
Binding of Small Molecules by a Polymer | 183
forward or backward. The total concentration of A molecules that are bound to the polymer
molecules is
3 A 4 bound = 3 PA 4 + 2 3 PA2 4 + 3 3 PA3 4 + 4 3 PA4 4
= 3 P 4 3 4S + 12S2 + 12S3 + 4S4)
= 4 3 P 4 S(1 + 3S + 3S 2 + S 3)
= 4[P]S11 + S2 3.
(5.79)
The number of bound A molecules per polymer molecule is n:
n =
3 A 4 bound
4 3 P 4 S(1 + S)3
4S
=
=
1 + S
3 P 4 total
3 P 4 (1 + S)4
(5.80)
In general, if there are N identical and independent sites:
n =
=
NS
1 + S
NK 3 A 4
1 + K 3A 4
(5.81)
From Eq. 5.81, it follows that
n(1 + K 3 A 4 ) = NK 3 A 4
n = (N - n)K[A]
and
n
3A 4
= K(N - n)
(5.82a)
= KN - Kn.
(5.82b)
or
n
3A 4
Ƭ>$@
This is the Scatchard equation. When N is greater than 2, the parameter n (the number
of bound A molecules per polymer molecule) is usually used to express the extent of binding. If n/[A] is plotted versus n, according to Eqs. 5.82a or 5.82b, a straight line results. The
slope of this line is -K, where K is the equilibrium constant for binding to one site. The
intercept of the line with the n-axis is N, where N is the total number of sites per polymer
molecule. Such a plot is called a Scatchard plot (figure 5.12).
Langmuir Adsorption Isotherm
A binding curve (amount of material bound versus concentration) at a constant temperature
is called an isotherm. In the independent-sites model, because the occupation of one site
has no effect on binding to other sites, the spatial distribution of the sites has no effect on
the final binding Eq. 5.82.
Equation 5.82 can also be changed into the following form:
n>N
1 - n>N
= K 3A 4
VORSH ².
Ƭ
Ƭ 1
FIGURE 5.12 A Scatchard
plot for a simple system with
independent binding sites.
The slope is the negative of
the equilibrium constant; the x
intercept is the number of sites
per macromolecule.
184 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
Because n is the number of sites occupied per polymer molecule and N is the total number
of sites per polymer molecule, n/N is simply the fraction of sites occupied, which is designated as f. We obtain, then,
f
= K[A] .
1 - f
(5.83)
Equation 5.83 was first derived by Irving Langmuir in 1916 for the adsorption of a gas on
a solid surface.
Nearest-Neighbor Interactions and Statistical Weights
In the previous sections, the several binding sites on a macromolecule are assumed to be
independent or noninteracting. Frequently, however, the binding of a molecule to one site
affects the binding of molecules to other sites. The interactions between the sites greatly
increase the complexity of the problem, and analytical solutions of the binding isotherms
can be obtained only for special cases. However, a general method for simplifying the
problem involves the use of statistical weights. The statistical weight vi of species i is
defined as its concentration relative to a reference concentration:
vi =
ci
cref
Statistical weights can simplify the algebra in problems that contain many species. In
equations involving ratios of concentrations, statistical weights can replace concentrations.
For example, n can be written in terms of statistical weights instead of concentrations.
Because the statistical weight of each species is the relative concentration of that species,
the mole fraction xi of any species is the statistical weight of the species i divided by the
sum of the statistical weights of all species:
vi
xi =
(5.84)
a ivi
The denominator, the sum over all statistical weights, is equivalent to the partition function
in statistical mechanics; here we call it the sum over states:
Q = a vi
i
(5.85)
Now, we consider a special case with N identical but interacting sites arranged in a
linear array. In other words, the identical sites form a one-dimensional lattice. First used
by Ising, this model is called the Ising model. To simplify matters, we first consider only
nearest-neighbor interactions. Longer-range interactions will be discussed in a later section.
The N identical sites in a linear array can be expressed by an N-digit number of 0’s
and 1’s such as 00111000 . . . 1011100011, in which the symbol 0 represents a free site and
the symbol 1 represents an occupied site. The number written represents a linear polymer
with N identical sites, such that the first two sites are unoccupied, the next three sites are
occupied, and so on. Let K be the equilibrium constant for the binding of an A molecule
to a site with no occupied nearest neighbors and let tK be the equilibrium constant for
the binding of an A molecule to a site with one adjacent site occupied. The parameter t
accounts for the interaction between the two adjacent occupied sites. If t is less than 1,
the binding is anticooperative (it is more difficult to bind next to an occupied site); if t is
greater than 1, the binding is cooperative (it is easier to bind next to an occupied site); if
t is equal to 1, there is no interaction between sites.
Binding of Small Molecules by a Polymer | 185
With these definitions, the equilibrium constant for any reaction can be written
quickly. A few examples follow for a simple case with N = 3:
Reaction
000
001
001
101
101
+
+
+
+
+
Equilibrium constant
A L 001
A L 101
A L 011
A L 110
A L 111
K
K
tK
t
t2K
The ratio of concentrations of [001] to [000], the statistical weight of [001], is
vi =
3 001 4
= K 3A 4 .
3 000 4
Similarly, the statistical weights of [101], [011], and [111] are (K[A])2, t(K[A])2, and
t 2(K[A])3, respecively. The free species [000] has a statistical weight of one by convention. Defining the product K[A] as S, the statistical weight of each state for the case N = 3
is given in table 5.2.
TABLE 5.2 Statistical Weights of Species Resulting from Binding to a Trimer
Number of sites occupied
Species
Statistical weight
0
1
000
001
010
100
101
011
1
S
S
S
S2
tS2
110
tS2
111
t2S3
2
3
Note that the statistical weight of each state can be written as tiS j, where j is the
number of 1’s (occupied sites) in the state and i is the number of 1’s following 1 (number
of nearest-neighbor interactions).
For our present case, summing the statistical weights listed in table 5.2 gives
Q = 1 + 3S + S 2 + 2tS 2 + t2S 3 .
(5.86)
The average number of bound A per P is
n =
1 3 PA 4 + 2 3 PA2 4 + 3 3 PA3 4
3 P 4 + 3 PA 4 + 3 PA2 4 + 3 PA3 4
=
(3S) + 2(S2 + 2tS2) + 3(t2S3)
1 + 3S + S2 + 2tS2 + t2S3
=
3S + 2(S2 + 2tS2) + 3t2S3 .
Q
Equation 5.87 can be written in a simple form. Because
Q = 1 + 3S + S2 + 2tS2 + t2S3
a
0Q
b = 3 + 2S + 4tS + 3t2S 2
0S t
(5.87)
186 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
and
Sa
0Q
b = 3S + 2(S 2 + 2tS 2) + 3t2S 3 .
0S t
(5.88)
The right side of Eq. 5.88 is identical to the numerator in Eq. 5.87. Thus,
n = S
= c
= a
(0Q>0S)t
Q
0Q>Q
0S >S
d
t
0 ln Q
b .
0 ln S t
(5.89)
f
The form of Eq. 5.89 is identical to the expression we gave for the number of helical
units in a random coil polymer, but of course this must be so; the helix–coil equilibrium is
essentially the same problem as the binding of small molecules to a macromolecule. Binding to one site affects adjacent sites just as presence of one residue in a helical conformation
affects the conformation of adjacent residues! Equation 5.89 tells us that we can calculate
the number of sites bound, n, from the sum of states Q. The derivative of Q with respect
to S(= K[A]) for any amount of cooperativity t leads to an equation for n as a function
of S. This equation is the binding curve; it characterizes the number of sites bound versus
concentration of free ligand [A]. Although Eq. 5.89 is derived for a special case, it applies
to any system in which there is binding to N identical sites with interaction only between
adjacent sites.
1.0
0.8
0.6
0.4
0.2
0
E X A M P L E 5 .1 0
Show that for the case of four identical and independent sites discussed previously,
Eq. 5.89 gives the same result as Eq. 5.80.
0
1
2
3
4
5
f
S
1.0
0.8
0.6
0.4
0.2
0
0
0.1 0.2 0.3 0.4 0.5
S
FIGURE 5.13 Fraction of sites
occupied f as a function of the
substrate concentration times
the binding constant for a linear
array with three binding sites
(N = 3). The curves shown are
for the cases of no interaction
between occupied sites (t = 1;
top) and for positive interaction
between adjacent sites with a
value of t = 20; bottom.
Cooperative Binding, Anticooperative Binding,
and Excluded-Site Binding
It is informative to examine the shape of a plot of the fraction of binding sites occupied,
f = n>N, as a function of S = K[A]. With N = 3, f is plotted against S for the cases t = 1
and t = 20 in figure 5.12. t = 1 of course means independent sites, where f increases
gradually with S and approaches 1 asymptotically (figure 5.13 top). The case t = 20, or
in general, for t 7 1, is termed cooperative binding because the binding of A to one site
makes it easier to bind another A to an adjacent site. Note that the shape of the f versus S
plot is sigmoidal (S-shaped; figure 5.13 bottom). As S increases, f increases slowly at first,
showing an upward curvature, and then increases more rapidly. At still higher values of f,
a decrease occurs in the slope of the f versus S plot with increasing S showing a downward
curvature. Such a sigmoidal plot of f versus S is indicative of cooperative binding.
For the independent-sites model, the shape of the f versus S curve is not affected by the
number of sites N per polymer molecule because N is absent in the binding Eq. 5.83. For
cooperative binding, the shape of the f versus S curve is dependent on N. As N increases,
the number of terms in the sum of states Q increases, and the binding curve obtained from
Eq. 5.89 changes. For the same value of the interaction parameter t, the difference in S for
a given change in f is less for larger N. Thus, the abruptness of the cooperative binding is
even more extreme. If N is very large, it can be shown that f changes from essentially zero to
Binding of Small Molecules by a Polymer | 187
essentially 1 in the range 11 - 2> 2t2 6 tS 6 11 + 2> 2t2. Thus, a sharp transition
occurs at tS = 1 for large values of t. At low concentrations of [A], tS = tK[A] is less
than 1, and very little binding occurs; however, as the concentration of [A] is increased
so that tS becomes slightly greater than 1, the cooperative binding is suddenly triggered,
and nearly all sites are filled.
The relative amounts of the various species are another important difference between
independent-sites binding (noncooperative binding) and cooperative binding. For N = 3,
let’s examine the situation when v = 3>2 (when half of the sites are occupied). The mol
fractions of the species with 0, 1, 2, and 3 sites occupied are from table 5.2: 1>Q, 3S>Q,
1S2 + 2tS2 2 >Q, and t2S3 >Q, respectively. These quantities are calculated for t = 1
1v = 3>2 at S = 12 and t = 20 1v = 3>2 at S = 0.1312 and are plotted in figure 5.14.
The difference between the two distributions is evident. In the case t = 1, the most
abundant species center around n = 3>2. In the case t = 20, however, the predominant
species are the one with all sites free (33 mol %) and the one with all sites occupied
(30 mol %). Such an uneven distribution is a consequence of the cooperative nature of binding. Because the occupation of one site makes the binding to adjacent sites more favorable,
the bound A molecules tend to cluster. When t is even larger, only the species with all sites
occupied and the species with all sites free are of importance. This is the all-or-none limit.
In the all-or-none limit, for our case of N = 3, we can simply consider the reaction as
0.4
0.3
xν 0.2
0.1
0
0.4
0.3
xν 0.2
0.1
0
0
1
0
1
ν
ν
2
3
2
3
FIGURE 5.14 Distribution of
species at half saturation
for N = 3. Independent sites
(t = 1; top); cooperative binding
(t = 20; bottom). Note that
the cooperative binding tends
to lead to an all-or-none-type
binding.
P + 3A S PA3
because neither of the other species, PA and PA2, is present in significant amounts. The
equilibrium constant for the reaction is
K⬘ =
and
n =
3 PA3 4
,
3P 4 3A 4 3
3 3 PA3 4
3 P 4 + 3 PA3 4
=
3K ⬘ 3 A 4 3
1 + K ⬘ 3A 4 3
.
(5.90)
We use K⬘ in these equations to remind us that this is not the equilibrium constant
for binding at a single site.
Because n/3 = f the fraction of sites occupied, simple algebra gives
f
= K⬘ 3 A 4 3
1 - f
(5.91)
d log[ f/(1 - f )]
= 3.
d log[A]
In general, if there are N sites per polymer molecule, the all-or-none model (maximum
cooperativity) yields
d log[ f/(1 - f )]
= N.
d log[A]
log [f/(1– f)]
and
slope = n
(5.92)
For no cooperativity 1t = 12, the binding curve is independent of N and the left-hand side
of Eq. 5.92 is equal to 1. An experimental measure of cooperativity in binding is obtained
by a plot of log [ f >11 - f 2] versus log[A]. This plot is a Hill plot (see figure 5.15), and
the slope of the plot is n, the Hill constant. If the slope n is 1, the sites are independent. If
log[A]
FIGURE 5.15 Hill plot. If
n = N, the binding is all-ornone. If n = 1, the binding is
noncooperative.
188 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
FIGURE 5.16 Binding of
adenosine A to polyuridylic acid.
The fraction of sites occupied f
is plotted against the concentration of free adenosine [A].
0.8
0.6
f
(Based on data from Wai Mun
Huang, Paul O.P. Ts’o, “Physicochemical basis of the recognition
process in nucleic acid interactions: I.
Interactions of polyuridylic acid and
nucleosides,” Journal of Molecular
Biology, 16(2), pp. 523–543.
IN29, ISSN 0022–2836, 10.1016/
S0022–2836(66)80189–4. Copyright
(c) 1966 Elsevier, Inc. (http://www.
sciencedirect.com/science/article/
pii/S0022283666801894))
1.0
0.4
0.2
0.0
–3.5
–3.0
–2.5
–2.0
–1.5
log10[A]
the slope is greater than 1, there is some cooperativity. If the slope is equal to N, the binding can be approximated by the all-or-none model.
Also of interest is the case t 6 1, when binding to one site reduces the affinity of
adjacent sites (anticooperative or interfering binding). In the limit t = 0, the occupation of
one site excludes binding to adjacent sites, and the case is referred to as the excluded-site
model.
N Identical Sites in a Linear Array with Nearest-Neighbor Interactions
In the preceding section, we discussed quantitatively the binding of small molecules to a
polymer molecule with three identical sites in a linear array. A number of important concepts came out of the discussion. These include cooperative and anticooperative binding,
the all-or-none limit, and the excluded-site model. The transition from a polymer with
N = 3 to one with a large number of monomer units requires no new physical concepts,
but the mathematics is more involved. This is described in detail in the references given
at the end of the chapter. Instead of repeating the lengthy derivations here, we examine an
illustrative example.
E X A M P L E 5 .11
Adenosine molecules can form hydrogen-bonded complexes with the bases of polyuridylic acid. In figure 5.16, the fraction of sites occupied in polyuridylic acid is shown
as a function of the concentration of free adenosine. The shape of the curve clearly
indicates cooperative binding. The interaction parameter t is calculated to be ~104. The
cooperativity is interpreted as a result of the favorable “stacking” interactions between
adjacent adenosine molecules.
Identical Sites in Nonlinear Arrays with Nearest-Neighbor Interactions
We have described binding to linear arrays such as nucleic acids, but more complex arrays
can be treated by similar methods. Here we give only one example of the differences that
occur with different arrays.
Binding of Small Molecules by a Polymer | 189
E X A M P L E 5 .12
Glyceraldehyde-3-phosphate dehydrogenase is an enzyme consisting of four
identical subunits. It first binds NAD+ and then catalyzes the oxidation and
phosphorylation of glyceraldehyde-3-phosphate to 1,3-diphosphoglycerate.
The binding of NAD+ to the enzyme isolated from muscle cells was
investigated by A. Conway and D. E. Koshland, Jr. (1968, Biochemistry 7:411)
and was found to occur in four stages, with equilibrium binding constants
K1 = 1011 M - 1, K2 = 109 M - 1, K3 = 3.3 * 106 M - 1, and K4 = 3.9 * 104 M - 1.
Because the binding of successive NAD+ molecules to a single tetrameric enzyme
becomes progressively weaker, this is an example of anticooperativity. (In contrast, the
enzyme derived from yeast exhibits positive cooperativity in binding NAD+.)
To model this binding pattern, it is reasonable to assume that the binding of the first
ligand (NAD+) results in a structural change that weakens the binding to the neighboring subunits. The predicted pattern of binding will be different, depending on whether
the subunit interactions occur as (1) a linear array, where each site in the center can
have two interactions and each site at each end can have only one; (2) a square array,
where each site can have two interactions; or (3) a tetrahedral array, where each site can
have three interactions:
(1)
(2)
(3)
Negative binding interactions are assumed to occur whenever a neighboring subunit
has a binding site that is already occupied by a ligand. Because the four experimental
binding constants decrease progressively by factors of the order of 10-2, this is consistent with the pattern expected for a tetrahedral array. Let’s test this hypothesis.
In the following table representing the site occupation of the tetrahedral model, we
use 0’s to indicate free sites and 1’s to indicate bound sites. We assume that when two
ligands are bound, each of the six pairwise interactions is identical and characterized by
the parameter t.
Reaction
+ A
Equilibrium constant
K
4 possible species
+ A
tK
6 possible species
+ A
t2K
4 possible species
+ A
t3K
1 possible species
The table makes it evident that the second ligand to bind involves one interaction,
the third ligand involves two interactions, and the fourth ligand involves three
190 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
interactions with subunits having ligands already bound. The fact that each experimentally measured equilibrium constant Ki differs by a factor t from the previous one is the
property that we are seeking to match qualitatively to our theoretical calculations. To
test this quantitatively, we need to consider the number of binding arrangements that
is possible for a given number of ligands bound and to write the experimental Ki ⬘s in
terms of the parameters t and K given in the table.
a. Use reasoning, like that which led to Eq. 5.86, to show that for the tetrahedral model
of glyceraldehyde-3-phosphate dehydrogenase the sum of states is
Q = 1 + 4S + 6tS2 + 4t2S3 + t6S4 ,
where S = K[A].
b. Now calculate expressions for K1, K2, K3, and K4 in terms of t and S. For example,
K1 =
K2 =
3 EA 4
4S
1
=
*
= 4K
1
3E 4 3A 4
3A 4
3 EA2 4
6tS2
1
3tS
3
=
*
=
= tK.
4S
2
3 EA 4 3 A 4
3A 4
2 3A 4
Next we need to establish the numerical values for t and K that can be used to best
simulate the experimental data. Because the values of K3 and K4 were determined with
greater precision, we begin with them.
c. Show that your results for K3 and K4 in part (b) are consistent with the relation
K4
3
3.9 * 104 M - 1 .
= t =
K3
8
3.3 * 106 M - 1
Therefore, t = 3.2 * 10 - 2 and K = 5.0 * 109 M - 1.
Calculate values for K1 and K2 and compare them with the experimental values to
evaluate the suitability of the tetrahedral, pairwise-interaction model. Remember that
only the order of magnitude of these binding constants was determined experimentally.
(Answer: K2 = 2.4 * 108 M - 1 and K1 = 2.0 * 1010 M - 1)
Finally, construct a table of interactions for the four-site linear model and the square array
model. Based on the behavior of your expressions for the ligand binding equilibrium
constants, what is your conclusion about the suitability of the alternative to the tetrahedral
model? (For the linear array, Q = 1 + 4S + 3 1t + 12S 2 + 21t2 + t2S 3 + t3S 4; for
the square array, Q = 1 + 4S + 212t + 12S 2 + 4t2S 3 + t4S 4.2
Summary
Statistical Thermodynamics
The Most Probable Distribution:
The properties of a system at equilibrium are represented, to a high degree of approximation, by the properties of the most probable
distribution:
Ni
gie - Ei/kBT
=
N
Q
(5.34)
Q = a gie - Ei/kBT = partition function
(5.35)
i
Summary | 191
gi
Ni
= e - (Ei - Ej)/k BT
gj
Nj
(5.37)
Ni = number of molecules in energy level Ei
Nj = number of molecules in energy level Ej
N = total number of molecules
gi, gj = degeneracy, the number of states with energy level Ei, Ej
kB = Boltzmann’s constant = R>NA (R = Gas constant and NA = Avogadro’s constant)
Q = molecular partition function, the weighted average number of thermally populated states in the system
Entropy:
S = k ln W
(5.38)
Random Walk and Related Topics
8 h2 9 = Nl2 = Ll
(5.67)
8 h2 9 = mean end-to-end distance of a polymer molecule
N = number of segments in a polymer molecule
l = statistical length of a segment; l is a measure of a stiffness of the polymer
L = contour length of the polymer
Nl2
8R2 9 =
1open@ended random coil2 ,
6
(5.69)
where 8 R 2 9 is the mean square radius,
8 R2 9 =
Nl2
1circular random coil2.
12
(5.70)
Helix−Coil Transitions
Simple Polypeptides:
The a-helix is characterized by hydrogen bonds between CO groups and NH groups in the fourth residue ahead in a polypeptide chain.
The initiation of a helical element in a coiled region is more difficult than the addition of a helical element to the end of a helix. For
polypeptides of high molecular weight, helix−coil transitions can be very sharp. Statistical methods have been used successfully to
treat the helix−coil transition.
Proteins:
Native proteins are in a highly ordered or folded structure. Unfolding of a protein involves the disruption of hydrogen bonds, disulfide
bridges, ionic bonds, and nonbonded interactions. The transition from a folded to an unfolded form (and vice versa) is usually highly
cooperative and approaches the all-or-none limit.
Double-Stranded Nucleic Acids:
Owing to favorable interactions (“stacking interactions”) between neighboring base pairs in a double helix, the formation of an
additional base pair at the end of a helical region is favored compared with the formation of a base pair in the middle of a coiled
region. This leads to a cooperative transition between the helical and coiled forms. The basic features of the transition—such as the
sharpness of the transition, the temperature dependence, the effect of the molecular weight of the polynucleotide, and the effect of
base composition—can be understood from statistical mechanical considerations of the problem.
Binding of Small Molecules by a Polymer
Polymer Molecule with N Identical and Independent Sites for the Binding of A, a Small Molecule:
n
= K(N - n)
3A 4
f
= K 3A 4
1 - f
v = number of bound A molecules per polymer molecule
(5.82)
(5.83)
192 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
f = fraction of sites occupied
N = number of sites per polymer molecule
K = intrinsic binding (equilibrium) constant
Polymer with N Identical Sites in a Linear Array with Nearest-Neighbor Interactions:
n = a
0 ln Q
b
0 ln S t
(5.89)
v = number of bound A molecules per polymer molecule
Q = sum of statistical weights of species (partition function)
S = K[A]
t = cooperativity parameter
Cooperative Binding:
If t 7 1, the binding of an A molecule to one site makes it easier to bind another A to an adjacent site. An f versus A plot is sigmoidal in
shape. If t is very large, the binding approaches the all-or-none limit, and the predominant species have f = 0 and f = 1. In this limit,
d log[f/(1 - f)]
= N.
d log[A]
(5.90)
Anticooperative Binding:
If t 6 1, the binding of an A molecule to one site makes it more difficult to bind another A to an adjacent site. In the limit t = 0,
binding to one site excludes binding to adjacent sites.
Mathematics Needed for Chapter 5
We use permutations and combinations in calculating distributions of systems among energy levels (Eq. 5.30) and in random walks
(Eq. 5.47). The number of permutations of n objects is
Pn = n! = n(n - 1)(n - 2) c # 3 # 2 # 1 .
A simple example is the 3! = 6 permutations of the three letters, a, b, c.
A combination of objects is a group of objects without respect to order. The three letters (a, b, c) form one combination. The
number of combinations of n objects taken r at a time is
n
n!
.
nCr = a b =
r
r!(n - r)!
A simple example is the number of ways of arranging three pennies into groups of two heads and one tail:
3C2
=
3!
= 3
2!1!
Stirling’s approximation for factorials is useful for large N:
ln N! = N ln N - N
If a group of particles can be arranged in W1 ways and independently, they can also be arranged in W2 ways; the total number
of arrangements is W 1 * W 2. Similarly, if the probability of an event occurring is p1 and if the probability of an independent event
occurring is p2, then the joint probability of the two independent events occurring is p1 * p2. For n independent arrangements of n
independent probabilities, the expressions are
W = q Wi = W1 # W2 # W3 c Wn
n
i=1
p = q pi = p1 # p2 # p3 c pn.
n
i=1
Problems | 193
References
Textbooks on Statistical Mechanics
1. Chandler, D. 1987. Introduction to Modern Statistical
Mechanics. New York: Oxford University Press.
2. Davidson, N. 1962. Statistical Mechanics. New York:
McGraw-Hill.
3. Hill, T. L. 1960. An Introduction to Statistical
Thermodynamics. Reading, MA: Addison-Wesley.
4. McQuarrie, D. A. 1976. Statistical Mechanics. New York:
Harper & Row.
5. Nash, L. K. 1970. Introduction to Statistical
Thermodynamics. Reading, MA: Addison-Wesley.
For useful books that treat some of the material of this chapter in
more detail, see
6. Cantor, C. R., and P. R. Schimmel. 1980. Biophysical
Chemistry. Part III: The Behavior of Biological
Macromolecules. San Francisco: Freeman.
The relation between entropy and probability is discussed entertainingly in
7. Atkins, P. W. 1984. The Second Law. New York: Scientific
American Books.
8. Bent, H. A. 1965. The Second Law. New York: Oxford
University Press.
Suggested Reading
Ackers, G. K., M. L. Doyle, D. Myers, and M. A. Daugherty. 1992.
Molecular Code for Cooperativity in Hemoglobin. Science
255:54–63.
Creighton, T. E. 1983. An Empirical Approach to Protein
Conformation Stability and Flexibility. Biopolymers 22:49–58.
Crothers, D. M., J. Krak, J. D. Kahn, and S. D. Levene. 1992.
DNA Bending, Flexibility, and Helical Repeat by Cyclization
Kinetics. Methods Enzymol. 212B:3–29.
Levene, S. D., and D. M. Crothers. 1986. Ring Closure
Probabilities for DNA Fragments by Monte Carlo Simulation.
J. Mol. Biol. 189:61–72; Topological Distributions and the
Torsional Rigidity of DNA: A Monte Carlo Study of DNA
Circles. J. Mol. Biol. 189:73–83.
McGhee, J. D., and P. H. von Hippel. 1974. Theoretical
Aspects of DNA–Protein Interactions: Cooperative and
Non-Cooperative Binding of Large Ligands to a OneDimensional Homogenous Lattice. J. Mol. Biol. 86:469–480.
Petsko, G. A., and D. Ringe. 1983. Fluctuations in Protein
Structure from X-Ray Diffraction. Annu. Rev. Biophys.
Bioeng. 13:331–371.
Wang, J. C. 1982. DNA Topoisomerases. Sci. Am. 247 (July):
94–109.
Problems
1. The atmosphere of Venus is almost pure carbon dioxide,
M = 44.00 g>mol, and at the surface has a temperature of
740 K and a pressure of 93 bars. Compute: (a) the mass of
1 L of Venusian atmosphere, (b) the average speed of a gas
molecule on the surface of Venus, (c) the most probable
speed of a gas molecule on the surface of Venus.
2. The first excited vibrational energy level of diatomic chlorine (Cl2) is 558 cm-1 above the ground state. Wavenumbers, the units in which vibrational frequencies are
usually recorded, are effectively units of energy, with
1 cm - 1 = 1.986445 * 10 - 23 J. If every vibrational energy
level is equally spaced, and has a degeneracy of 1, sum over
the lowest 4 vibrational levels to obtain a vibrational partition
function for chlorine. Determine the populations of each of the
four levels at 298 K and the average molar vibrational energy
8 Em.vib 9 for chlorine at 298 K.
3. Mass spectrometers often use ‘drift tubes’ in which molecular ions can come into equilibrium with a gas at a particular
temperature. Calculate the most probable speed, the average speed, and the root mean square velocity at 298 K for
lysozyme (m.w. 14,300 Da) in a drift tube. What is the most
probable translational kinetic energy of lysozyme?
4. A certain macromolecule P has four identical sites in a linear
array for the binding of a small molecule A. Prepare a table
listing all species with half of the sites occupied (species
with two occupied sites and two unoccupied sites) and their
statistical weights for the following cases:
a. The sites are independent.
b. Nearest-neighbor interactions are present.
c. Binding to one site excludes the binding to a site immediately adjacent to it.
d. Give also, for part (b), an expression for the concentration ratio [PA4]>[PA2].
5. A DNA has short single-stranded ends that can join either
intramolecularly to form a ring or intermolecularly to form
aggregates. Discuss briefly and concisely under what condition you expect ring formation will predominate and under
what condition you expect intermolecular aggregation will
predominate.
194 Chapter 5 |
The Statistical Foundations of Biophysical Chemistry
6. For a dicarboxylic acid where the two groups are far
apart, using statistical methods show that the ratio of
the acid dissociation constants is expected to be ~4. (The
acid dissociation constants K1 and K2 are the equilibrium
constants for the reactions
K1
HO2C苲R苲CO2 m
and
-
O2C苲R苲CO2H + H +
K2
O2C苲R苲CO2H m - O2C苲R苲CO2- + H + ,
respectively.)
7. At a height of 200 km above the earth’s surface, in the thermosphere, the atmosphere is almost entirely nitrogen (N2).
The pressure is approximately 70 nPa (7 * 10-8 Pa) and the
temperature approximately 900 K. The earth’s escape velocity is 11.2 km s-1.
a. What fraction of the nitrogen molecules in the thermosphere have velocities greater than the earth’s escape
velocity?
b. Molecules which have velocities in excess of the earth’s
escape velocity in the radial (outward) direction can
escape the earth’s gravity into space (if they do not collide). What fraction of molecules have radial components of velocity in excess of the earth’s escape velocity?
c. How many nitrogen molecules will there be in a 1 km
thick shell of the thermosphere, if the inner distance of
the shell from the earth’s surface is 200 km? (The earth’s
radius is 6353 km.) How many of these molecules will
have enough radial velocity to escape?
8. In proteins, valine has three possible rotamers, corresponding to different configurations about the Ca–Cb bond. The
figure below shows the three configurations in Newman projection, and computed energies associated with them.
-
H3C
+H
3N
H
H
COO–
CH3
A (0 kJ mol–1)
H
+H
3N
H
CH3
COO–
CH3
H 3C
+H
3N
H
H
CH3
COO–
B (3.655 kJ mol–1) C (6.127 kJ mol–1)
a. Calculate the value of the molar partition function for an
equilibrium mixture of these three conformers, at 298 K.
b. Using Eq. 5.41 and 5.42, calculate the molar internal
energy and the molar entropy associated with an equilibrium mixture of the conformers, at 298 K.
9. Calculate the average distance, and the average square
distance travelled for a random walk of 10 steps, if the
probability of going forward on each step is twice the probability of going backward.
10. Find the mean-square radius of a polyethylene chain of
molecular weight 1400 g mol-1, if the C-C bond length is
0.154 nm.
11. A non-self-complementary DNA tetramer has four base
pairs and thus three distinct sites where an intercalating agent
such as proflavin can slide in between adjacent base pairs at
low temperature. However, two intercalating agents cannot
occupy adjacent sites. If the equilibrium constant for binding to a single site is K, calculate the statistical weight of
each bound species, the partition function Q, and the average
number of bound proflavins per tetramer, as a function of
proflavin concentration and K.
12. Calcium ions bind to the SERCA Ca2+-ATPase, which has two
identical calcium sites, in two stages with apparent equilibrium
constants K1 = 7 * 105 M-1 and K2 = 2 * 106 M-1. Calculate
K and t, and plot the fraction of sites occupied over the range
of calcium concentrations 10-5 M 7 [Ca2+] 7 10-7 M-1. Is the
binding positively or negatively cooperative?
13. An oligopeptide has seven amide linkages (or eight amino acid
residues, including the terminal carboxyl group). With the rules
for the statistical weights that we have discussed for the formation of an a-helix, obtain the function Q, which is the sum of
the statistical weights of all species. Also obtain an expression
for v, the average number of helical residues per molecule, as
a function of s and S.
14. The H+ titration curve of 1,2,3,4-tetracarboxyl cyclobutane
–OOC
COO–
HC
CH
HC
CH
–OOC
COO–
should be interesting. We will try to estimate it by using a
simple, nearest-neighbor interaction model with the binding
constant for a single COO-, K = 10-5 and t = 10-2.
a. There are 16 possible species involving 0, 1, 2, 3, and
4 H+ bound. Give the statistical weight for each in
terms of t and S = K[H+]
b. Write an expression for Q, the sum over states, and use
Q to obtain an expression for v, the number of H+ bound
per molecule.
c. Calculate f = v>4 for pH 4, pH 5, and pH 6 and compare
it with the titration curve (f vs. pH) of a single COO-.
15. Consider the coil–helix transition for a polypeptide containing 50 amides. At a certain temperature, the equilibrium
constant for adding an amide to a helical region is s = 1;
the helix initiation parameter is s = 10-4. Statistical weights
are relative to the species that is 100% coil.
a. Write an expression in terms of s and s for the statistical
weight of the species that has all the possible amides in
the a-helix conformation.
b. Write an expression in terms of s and s for the statistical
weight of the species that has 20 amides in the a-helix
conformation in three separate helical regions.
c. Which of all the possible species is in the highest concentration? That is, which of all the possible species has
the highest statistical weight?
Problems | 195
d. If s were 10, which of all possible species would have
the highest concentration?
e. Roughly sketch a plot of f = fraction of amides in the
a-helix conformation vs. temperature. The ΔH° for the
reaction helix to coil is negative. Label which temperature region corresponds to the helix and which to the coil.
f. Write an equation for f in terms of s, s, and Q.
16. An RNA oligonucleotide has the sequence A6C7U6. It can
form a hairpin loop held together by a maximum of six
A ⴢ U base pairs:
C
C
C CC
A
A
A
A
A
A
C
C
U
U
U
U
U
U
Assume that the helical region can melt only from either end.
Use the notation s = equilibrium constant for adding a base
pair to a helical region and sj ws = equilibrium constant for
initiating the first base pair forming a loop. The subscript
j = 7, 9, 11, 13, 15, 17 characterizes the number of unpaired
bases in the loop.
a. Calculate the statistical weight of each species that can
be present.
b. Calculate the mole fraction of each species that can be
present.
c. Assume that the molar absorptivity of each species depends
on only the number of base pairs formed: e0 = absorptivity
per mole of mononucleotide for species with no base pairs,
e1 = absorptivity of all species with one base pair, etc.
Write an expression for the absorbance in a 1-cm cell of a
solution of A6C7U6 in terms of s, sj, ej and c = concentration of A6C7U6 in moles of nucleotides per liter.
17. The statistical effective segment length l of a DNA
molecule is 100 nm. Calculate the mean-square end-toend distance <h2> and the contour length L for a bacterial
DNA with 107 base pairs. The distance between base pairs
is 0.34 nm.
18. a. How many 3-letter “words” can be made from 26 letters?
A “word” is any sequence of 3 letters from AAA to ZZZ.
b. How many different basketball teams of 5 players can be
chosen from a group of 100 people?
c. How many different proteins containing 100 amino acids
can be made from the 20 commonly-occurring amino
acids?
19. What is the change of entropy for the following reactions?
a. 100 pennies are changed from all heads to all tails.
b. 100 pennies are changed from all heads to 50 heads plus
50 tails.
c. 1 mol of heads is mixed with 1 mol of tails to give 2 mol
of half heads and half tails.
20. A system of a particle in a magnetic field has a simple
energy-level diagram that looks like this:
e2 = 0
g2 = 1
e1 = -D
g1 = 2
a. Write an expression for the partition function Q for the
system. Your answer should be given in terms of D, T,
and universal constants.
b. Write an expression for the average energy of the system. Your answer should be given in terms of D, T, and
universal constants.
c. Calculate the average energy in the high-temperature
limit.
d. Calculate the average energy in the low-temperature
limit.
e. Write an expression for the ratio of the probabilities of
finding the particle in the two states. Your answer should
be given in terms of D, T, and universal constants.
f. Calculate the entropy of the particle in the hightemperature limit.
Chapter 6
Physical Equilibria
Concepts
Biological organisms are highly inhomogeneous; different regions of each cell have different concentrations of molecules and different biological functions. The storage and expression of genes occur in the nucleus; protein synthesis occurs in the cytoplasm. The ATP
to drive many of these processes comes from oxidative phosphorylation in mitochondria.
Transport of O2 and essential nutrients into cells and the removal of CO2 and waste products are required for all these processes. In this chapter, we focus on the thermodynamics
underlying the different concentrations of molecules among different regions of a system.
Examples of these regions include cell compartments separated by membranes, the inside
and outside of a membrane, a liquid or solid in contact with a gas, and two immiscible
liquids. In such systems the chemical potential (partial molar free energy) is the thermodynamic property that tells us whether two or more compartments are in equilibrium with
one another (chemical potential is the same in all compartments at equilibrium) or whether
they are not in equilibrium (unequal chemical potentials).
If a species has a different chemical potential in two phases (or compartments) in
contact with one another, it will move to the phase with the lower chemical potential. This
will continue until the species reaches equilibrium and its chemical potential becomes
the same in all phases. This simple idea allows us to distinguish and characterize active
and passive transport in cells, the equilibrium concentrations of molecules separated by
semipermeable membranes, and the equilibria of molecules between solids, liquids, and
gases that determine freezing points, solubilities, boiling points, and osmotic pressure.
Applications
Membranes and Transport
Living cells are separated from their surroundings by membranes and contain a variety of
subcellular particles— organelles—also enclosed by membranes. Most biological membranes consist of a lipid bilayer that contains proteins and other molecules that serve as
recognition sites, signal transmitters, or ports of entrance and exit. Membranes are so thin,
having thicknesses of only one or two molecules, that they are often considered to be twodimensional phases. The thermodynamic properties of membranes are then described in
terms of surface properties, such as the surface chemical potential and the surface tension
or pressure. Membranes not only separate the contents of a compartment from its surroundings but also permit the controlled transport of molecules and signals between the inside
and outside. Differences between the inside and outside of a cell influence the exchange
of metabolites and electrical signals, the flow of heat, and changes in shape. Temperature
differences cause heat flow, pressure differences cause changes in shape, and electrochemical potential differences cause molecular transport and electrical signals.
196
Phase Equilibria | 197
Ligand Binding
Noncovalent interactions that bind ligands like O2 to hemoglobin, substrates to enzymes, and
complementary strands of DNA or RNA to one another underlie essential dynamic processes
in living cells. We have explored the theory of ligand binding in chapter 5. Here we will
examine its experimental measurement. Equilibrium dialysis provides a method of exploring
the binding between macromolecules and small ligand molecules. In equilibrium dialysis a
semipermeable membrane allows a ligand to reach equilibrium between two phases, one of
which contains a macromolecule. The difference in concentrations of the ligand on opposite
sides of the membrane depends on the interaction of ligand and macromolecule. This provides a very useful and easy method for studying equilibrium binding constants.
Colligative Properties
The chemical potential must be the same for each component present in two or more phases
at equilibrium with one another. The component can be the solvent for each of the solutes
in a solution. The phases are solids, liquids, gases, or solution compartments separated by
a semipermeable membrane. Any change in a property such as temperature, pressure, or
activity in one phase that results in a change in chemical potential must be accompanied
by an equal change in chemical potential in the other phases, for the system to remain in
equilibrium. This fundamental fact allows us to explain the colligative properties, such as
the freezing-point lowering, the boiling-point elevation, the vapor-pressure lowering, and
the increase of osmotic pressure when a solute is dissolved in a solvent. Colligative properties play essential roles in biological cells. Osmotic pressure that results from the activities of components present in the cytoplasm needs to be balanced by a suitable external
pressure lest the cell rupture and burst. Colligative properties are used to determine the
concentrations and molecular weights of solutes in solution; they can be used to measure
association and dissociation equilibrium constants of biopolymers.
Phase Equilibria
The transfer of a chemical from one phase to another is illustrated by the evaporation of
liquid water into the vapor phase; the heat removed from our bodies by evaporation of
sweat is essential to survival in hot climates. Another example is the transport of ions from
inside a cell to outside, which is vital to nerve conduction. We can consider the inside and
the outside of the cell as two different phases.
Equilibrium is a state where nothing seems to happen macroscopically—nothing apparently changes with time. For living systems to exist, they must be out of equilibrium; dynamic
processes need to occur to maintain the living state. When an organism dies, it approaches
closer to equilibrium. The usefulness of considering equilibrium in connection with living
organisms is that it helps define the direction of dynamic processes. One of the consequences
of the Second Law of Thermodynamics (chapter 3) is that spontaneous processes result in the
system moving toward a state of equilibrium. For an open system, spontaneous processes are
accompanied by a decrease in chemical potential, ⌬mT, p 6 0. At equilibrium, ⌬mT, p = 0.
One-Component Systems
The Exact Clapeyron Equation
The master equation that allows us to understand phase changes is the differential of free
energy as a function of the differentials of temperature and pressure:
dG = Vdp - SdT
(3.35)
Since V and S are always positive, this means G always slopes downward with increasing
temperature and upward with increasing pressure. If we have a good understanding of the
dependence of volume and entropy on temperature and pressure, we can integrate this
198 Chapter 6 |
Physical Equilibria
equation, setting an arbitrary zero of free energy at some reference temperature and pressure
T0 and p0, and plot the free energy on a three-dimensional graph as a function of temperature
and pressure as independent variables. These are depicted as the green surfaces for water
liquid in figure 6.1(a) and for water ice in figure 6.1(b). In both cases, the free energy is a surface that slopes upward with increasing pressure and downward with increasing temperature.
In figure 6.2 we plot the difference between the two free energy surfaces displayed
in figure 6.1:
(6.1)
fGm = Gm, water - Gm, ice
fGm, the molar change in free energy in the phase transition, looks superficially
similar to the two surfaces shown in figure 6.1, but it decreases with increasing pressure,
because the difference in molar volumes,
fVm = Vm, water -Vm, ice ,
(6.2)
is negative. We also plot in gray the horizontal plane for which G = 0. The sloping fGm
surface and the horizontal G = 0 surface intersect at a line. Because the fGm surface
slopes downward with increasing pressure, the line of intersection tilts backward; as pressure increases, the temperature decreases. At every point on this line, the free energy for
the phase change between ice and water, fGm = 0, is zero; therefore the line describes
the pressure versus temperature dependence of the spontaneous melting point — in other
words, the values of pressure and temperature for which the transition between water and
ice is reversible. We can derive the equation for the p versus T line as follows. Subtracting
dGice = Vice dp - Sice dT
FIGURE 6.1 Molar free-energy
surface of (a) water (b) ice as a
function of temperature and
pressure relative to Gm for H2O
at 273 K and 1 bar.
(a)
Gm
(kJ mol–1)
2.0
1.0
0.0
–1.0
FIGURE 6.2 (Light green) The
molar free energy of fusion of
water, ΔfusGm, as a function of
T and p. (Gray) The G = 0 surface
on which reversible transitions
must lie. (Green line) The
intersection of the two surfaces,
showing the dependence of
the temperature of fusion, Tfus,
on pressure. Note that because
the ΔfusGm tilts downward with
increasing pressure, the value
of Tfus decreases with increasing
pressure.
2.0
1.0
0.0
–1.0
200
270
280
T (K)
(b)
Gm
(kJ mol–1)
300
260
(6.3)
300
200
260
100
p (bar)
270
280
T (K)
290 0
100
p (bar)
290 0
fusGm
(kJ mol–1)
0.5
300
0.0
–0.5
200
p (bar)
260
100
270
T (K)
280
290
0
Phase Equilibria | 199
from
dGwater = Vwater dp - Swater dT
(6.4)
fGm = fV m dp - f S m dT ,
(6.5)
we obtain
where fVm and fSm are defined in the same way as fGm in Eq. 6.1. Now, since on
the line of intersection, fGm = 0, we have anywhere on that line
fV m dp = f S m dT .
(6.6)
But at a reversible phase transition
fS m = f Hm /T f ,
(6.7)
where Tf is the equilibrium phase transition temperature. This gives, on insertion into
Eq. 6.6 and rearrangement,
T f fV m
0T
,
a b =
(6.8)
0p f
fH m
where the left hand side denotes the dependence of T on p along the trajectory of the reversible phase transitions. We can use either intensive or extensive values for the volume and
enthalpy on the right-hand side, as long as we are consistent, since if we use extensive
values the moles will cancel above and below.
This expression for the derivative of the phase transition temperature with respect to
pressure contains no approximations or limiting assumptions, and it is therefore known
as the exact Clapeyron equation. As long as we know the phase transition temperature at
one pressure and either know the volume and enthalpy change as a function of pressure
and temperature, or can assume they are constant, we can integrate the exact Clapeyron
equation.
E X A M P L E 6 .1
The enthalpy of melting of ice at 1 bar is 6.007 kJ>mol; the density of water at 0C is
999.9 kg m - 3, while that of ice is 915.0 kg m - 3. Assuming fusVm and fusHm (the
molar volume and enthalpy change on fusion) are constant, determine the freezing
point of water at 100 bar.
SOLUTION
From the densities and the molar mass we can obtain the molar volumes
Vm = M>r; Vm, ice = 0.018015 kg mol - 1 >(915.0 kg m - 3) = 1.969 * 10 - 5 m3 mol - 1
Vm, water = 0.018015 kg mol - 1 >(999.9 kg m - 3) = 1.802 * 10 - 5m3 mol - 1
fus Vm = Vm, water - Vm, ice = -1.67 * 10 - 6 m3 mol - 1
a
Tfus fusVm
0T
273.15 K * -1.67 * 10 - 6m3mol - 1
b =
=
= -7.60 * 10 - 8K Pa - 1 .
0p f
fusHm
6007 J mol - 1
T
0T
We assume that a
b ⬵ a b ; since p = 99 bar = 9.9 * 106 Pa, we
p
0p fus
fus
obtain
T = 9.9 * 106 Pa * -7.60 * 10 - 8 K Pa - 1 = -0.75 K .
So the 100 bar freezing point is -0.75 K or 272.40 K.
Note that the freezing point decreases with pressure; this is a consequence of ice being
less dense than water (ice contracts when it melts). Most substances expand when they
melt, and so T>p > 0. The enthalpy of melting is always positive. (Why?)
Physical Equilibria
FIGURE 6.3 Phase diagram
of water. Tt is the triple point
(T = 273.16 K, p = 0.006 bar)
while Tc is the critical point
(T = 647 K, p = 221 bar), where
water vapor and water liquid
cease to be distinct phases.
Note the y axis is logarithmic.
100 bar
Ice Ih
Tc
Water
10 bar
p
200 Chapter 6 |
1 bar
0.1 bar
Water vapor
1000 Pa
Tt
100 Pa
10 Pa
1 Pa
200
300
400
T (K)
500
600
The Gibbs Phase Rule
All pure substances have an equation of state that, given a particular temperature and pressure, defines a molar volume. They thus have three variables of state, and one constraint,
leading to two degrees of freedom. As we have seen, however, if two phases are present at
equilibrium, there is an additional constraint: ⌬ fGm = 0. This leaves only a single degree
of freedom. J. Willard Gibbs found a general equation linking the number of degrees of
freedom, the number of phases, and the number of chemical components present in the
system. This is called the Gibbs phase rule,
F=C-P+2,
where F is the number of degrees of freedom, C the number of components, and P the
number of phases. If we have two degrees of freedom, then two variables of state are free
to vary over their whole range, and any three variables of state, plotted against each other,
can be described by a two-dimensional surface. With a single degree of freedom, the two
variables of state are related by a line; while if there are no degrees of freedom, the system
is described by a point—a single set of fixed values for all variables of state.
Most substances have three possible phases—solid, liquid, and gas—and therefore we
can have three lines, describing the dependence of temperature on pressure for the solidliquid, liquid-gas and solid-gas transitions. These lines need not be straight, since volume
and enthalpy changes will always depend at least slightly on pressure and temperature.
At the point where all three lines meet—that is, where solid, liquid, and gas are all at
equilibrium with each other and coexist, we have the triple point, a value of temperature
and pressure characteristic for the substance. The phase diagram for water is shown in
figure 6.3; the triple point values for water are T = 273.15 K, p = 0.006 bar.
Vapor-Liquid Equilibria and the Clausius-Clapeyron Equation
The pressure dependence of the vapor-liquid equilibrium temperature has some special
properties. At pressures close to ambient, we can make three simplifying assumptions,
which allow the exact Clapeyron equation to be integrated analytically. These assumptions are
• The molar volume of the vapor (gas) is much larger than that of the liquid:
⌬ vapV = Vm, gas - Vm, liquid 苲 Vm, gas
(6.9)
Phase Equilibria | 201
• The gas is ideal:
15
(6.10)
• The enthalpy of vaporization is independent of pressure and temperature. This is the
weakest of the three assumptions, but as we shall see, it is also the least necessary.
10
ln pv
V m, gas = RT >p
5
0
–5
Inserting Eq. 6.9 and Eq. 6.10 into the exact Clapeyron equation (Eq. 6.8), we obtain
a
T 1RT >p2
0T
RT 2
b
=
=
.
0 p vap
⌬ vapHm
p⌬ vapHm
(6.11)
We can now separate variables:
dpvap
pvap
=
⌬ vapHm dT
R
T2
(6.12)
Let’s assume for the moment that the enthalpy of vaporization is independent of temperature. We can perform either a definite or an indefinite integration. Doing a indefinite
integration first, we obtain
⌬ vapHm
ln pvap = + C.
(6.13)
RT
pvap is, as usual, strictly speaking, divided by the standard pressure p⬚ = 1 bar. We can
leave it as p as long as we remember that p must always be expressed in bars.
At pvap = 1 bar pressure, ln pvap = 0, and so, since the vaporization is under reversible conditions:
⌬ vapHm
⌬ vapSm
C =
=
(6.14)
RT
R
Inserting this back in Eq. 6.13, we obtain
ln pvap = -
⌬ vapHm
RT
⌬ vapSm
+
R
.
(6.15)
A plot of the vapor pressure against 1>T should therefore yield a slope of - ⌬ vap Hm >R
and an intercept of ⌬ vapS m >R . This is shown in figure 6.4. Alternatively, we can perform
a definite integral between two temperatures T1 and T2, giving
ln a
pvap,2
pvap,1
b = -
⌬ vapHm
R
a
1
1
b.
T2
T1
(6.16)
This second and probably more common form of the equation is usually called the
Clausius-Clapeyron equation, or sometimes the inexact Clapeyron equation. It is extremely
useful, since it allows us to determine the boiling point of a liquid as a function of pressure,
or, alternatively, the vapor pressure of the liquid as a function of temperature. Its similarity
to the van’t Hoff equation (Eq. 4.53) is not coincidental; the vapor pressure is actually the
equilibrium constant for the process
Liquid L Gas ,
since the activity of a pure liquid is defined as 1.
We can make a minor modification to the Clausius-Clapeyron equation that substantially increases its accuracy. We relax the assumption that the enthalpy of vaporization is
temperature-independent, and instead make it a linear function of temperature by using
the difference in heat capacities between the liquid and vapor (Eq. 2.66):
⌬ vapH = ⌬ vapH⬚ + ⌬ vapCp⬚1T - T ⬚2 ,
(6.17)
0.001 0.002 0.003
1/T (K–1)
FIGURE 6.4 Plot of ln pvap
versus 1>T (Clausius-Clapeyron
plot) for the water>water vapor
equilibrium. The fit gives an
intercept ≤vap Sm >R = 13.27 and
a slope -≤vap Hm >R = 4968 K.
Values at the 1 bar boiling
point are 13.11 and 4891 K,
respectively.
Physical Equilibria
FIGURE 6.5 Vapor pressure
of water as a function of
temperature. The light green line
is a fit assuming the enthalpy
of vaporization is constant
(Eq. 6.15); the darker green line
is a slightly better fit obtained
assuming it has a linear
dependence on temperature
(Eq. 6.19). The 1 bar boiling
point of 372.86 K is also shown.
5.0
Tb
4.0
pvap (bar)
202 Chapter 6 |
3.0
2.0
1.0
300
350
400
450
T (K)
where ⌬ vapCp⬚ = Cp⬚(gas) - Cp⬚ (liquid) and T⬚ is some standard temperature where
⌬ vapCp⬚ and ⌬ vapH ⬚ are known; usually the 1 bar boiling point. Inserting this in Eq. 6.12,
we obtain
dpvap
1⌬ vapH⬚ - ⌬ vapCp⬚T ⬚2dT
⌬ vapCp⬚dT
=
+
,
(6.18)
2
pvap
RT
RT
which integrates to
ln a
pvap
pvap⬚
b = -
1⌬ vapH ⬚ - ⌬ vapCp⬚T ⬚2 1
⌬ vapCp⬚
1
T
a b +
ln a b .
R
T
T⬚
R
T⬚
(6.19)
This is accurate over a wider range of temperatures, as shown in figure 6.5.
Solutions of Two or More Components
It is useful, when we think about systems with more than one chemical substance, to express
equilibrium in terms of chemical potentials. Let’s consider a two-component system consisting of gasoline and water. The two components are not fully miscible—hence the two
phases—but at the same time, a small quantity of water (A) will dissolve in the gasoline (B),
and a small amount of gasoline in the water. We can write the total free energy of either the
water phase (phase 1) or the gasoline phase (phase 2) in terms of the chemical potential of
each component, at constant temperature and pressure. From Eq. 4.3, for two components:
dG = mA,1dnA,1 + mB,1dnB,1 + mA,2dnA,2 + mB,2dnB,2
(6.20)
Let’s assume we are at equilibrium. If we keep the number of moles of B constant in both
phases, then
dG = mA,1dnA,1 + mA,2dnA,2 .
(6.21)
If we allow A to change only by moving from phase 1 to phase 2, then this becomes:
dG = mA,1dnA,1 - mA,2dnA,1 = 1mA,1 - mA,2 2dnA,1
(6.22)
By choosing the sign of dnA,1 (which amounts to deciding the phase from which to transfer A) we can always make this negative, and thereby lower the free energy. But if we are
at equilibrium, the free energy change must be zero! The only way the right-hand side of
Phase Equilibria | 203
Vapor Pressure
Raoult’s Law governs the vapor-liquid equilibrium of solutions. Raoult’s Law is a limiting
law; while real solutions may not exactly obey it, in the limit of high dilution, all solutions
obey it. It is best described by the equation
pvap
lim a
b = xA ,
(6.23)
x AS1 pvap⬚
where pvap is the vapor pressure of the solvent in the solution, pvap⬚ is the vapor pressure
of the pure solvent, and xA the mole fraction of solvent in the solution. Figure 6.7 shows a
Raoult’s Law plot for a solution of glucose solution in water. Even though the vapor pressure drops below the ideal Raoult’s Law diagonal at higher mole fractions of glucose (or
lower mole fractions of water), nonetheless at high mole fractions, the curve linking the
experimental points approaches the slope of 1 required by Raoult’s Law.
Raoult’s Law is one of two requirements for an ideal solution; the solvent in an ideal
solution describes Raoult’s Law. Since the system is at equilibrium, the chemical potential
of the solvent in the solution is equal to that in the gas phase. The chemical potential of
the vapor, which we shall assume is an ideal gas, is given by
m = m⬚ + RT ln pvap ,
(6.24a)
where m⬚ is the chemical potential of the real gas standard state of 1 bar. Because the
standard state is one bar, to be valid, the vapor pressure must be expressed in bars. If we
wish to be correct, we can write the equation instead as
pvap
m = m⬚ + RT ln a
b,
(6.24b)
p⬚
with the standard pressure p⬚ = 1 bar. For the solution, we can substitute Eq. 6.23 into
Eq. 6.24a, giving
mA = m⬚ + RT ln x A + RT ln pvap⬚ .
(6.25)
Here m⬚ is the chemical potential of the gas (1 bar). However, if we want to redefine the
equation in terms of a standard state of a pure liquid, we can substitute mA⬚ (the chemical
potential of the pure liquid) using mA⬚ = m⬚ + RT ln pvap⬚ (Eq. 6.24a), giving
mA = mA⬚ + RT ln x A .
(6.26)
To be clear about a frequently confusing point: Equation 6.26 gives the chemical potential
of the solvent relative to the chemical potential of a pure liquid standard state; Eq. 6.25
gives it relative to the chemical potential of the vapor as an ideal gas at 1 bar. Any nonideality in the gas at pressures near 1 bar can be ignored, because it will be small compared
to nonidealities in solution.
What if the solution does not obey Raoult’s Law? In that case, instead of the mole
fraction, xA, we substitute the rational activity axA in Eq. 6.26:
mA = mA⬚ + RT ln axA
(6.27)
And, of course, axA = gAxA, where gA is the activity coefficient. This is equivalent to
replacing Eq. 6.23 by
pv
= axA .
pv⬚
(6.28)
1.0
pvA/p°vA
Eq. 6.22 can be zero is if mA,1 - mA,2 = 0, or in other words, if the two chemical potentials of component A are identical. The same argument can be applied to component B,
and therefore, at equilibrium, not only does the total free energy of each phase have to be
identical — the chemical potential of each component must also be identical.
0.9
0.8
0.8
xA
0.9
FIGURE 6.6 Vapor pressure
pvap of water in equilibrium with
glucose solutions, divided by
that of pure water pvap°, plotted
against the mole fraction of
water xA.
(Based on data from J. B. Taylor and
J. S. Rowlinson, “The thermodynamic
properties of aqueous solutions
of glucose,” Transactions of the
Faraday Society, 51, p. 1183–1192.
DOI: 10.1039/TF9555101183.
Copyright (c) 1955 Royal Society
of Chemistry.)
1.0
204 Chapter 6 |
Physical Equilibria
We can now relate vapor pressures to the activity of the solvent. Combining Eqs. 6.27
and 6.28,
pvap
mA = mA⬚ + RT ln a
b.
(6.29)
pvap⬚
Vapor-pressure lowering is a practical method for determining the amount of a solute
in solutions. The vapor pressure of normal saline solution (0.9% by wt. NaCl), for example,
is 36 Pa lower than pure water; this is readily measured using a sensitive manometer.
Another example of the use of this technique is in determining the molecular mass of a
protein in solution. If a weighed sample of a pure protein is dissolved in a known volume
of water, then measurement of the vapor-pressure lowering of the solution allows the mole
fraction of the protein to be calculated. Knowing both the weight concentration and the
mole fraction allows us to calculate the molecular mass of the protein in the solution.
Henry’s Law
Oxygen (O2) is necessary to support life in multicellular organisms (fish, animals, humans)
and many microorganisms. Oxygen is supplied to the atmosphere by photosynthesis and is
consumed by a variety of biological and nonbiological processes. At present, the concentration
of O2 in the atmosphere is about 20%, and lakes and oceans also store a significant amount
as dissolved O2. At equilibrium with the atmosphere, at 25°C, the mole fraction of dissolved
O2 in water, xO2 is 4.7 * 10 - 6. Although this seems like a very small number, the amount of
O2 dissolved in 1 L of water is about 3% of that contained in 1 L of air at 1 bar and 25°C.
The equilibrium described by the expression
O2(g) L O2(aq)
is governed by Henry’s Law. Henry found experimentally that the solute (O2) vapor pressure is proportional to the solute mole fraction in solution. By convention, just as the
solvent is designated A, the solute is designated B. Henry’s Law is
pvap,B = k Bx B ,
(6.30)
where pvap,B is the vapor pressure of the solute, xB is mol fraction of B in solution, and kB
is the (temperature dependent) Henry’s Law coefficient for solute B in solvent A.
Thus, Henry’s Law shows that gas solubility is directly proportional to the gas pressure.
Henry’s Law coefficients for some common gases in water are given in table 6.1. Among these
gases, He is the least soluble, and C2H2 and CO2 are the most soluble. Gas solubility is also a
function of temperature and depends on the solvent. Oxygen is about 20% less soluble at 37°C
(physiological temperature) than at 25°C. Note that lower numbers mean higher solubility.
TABLE 6.1 Henry’s Law coefficients k B ⴝ pv,B/x B, in bar, for aqueous solutions
Gas
0°C
25°C
37°C
139 *
103
138 * 103
84 *
103
98 * 103
He
131 *
103
N2
50 *
103
CO
35 * 103
57 * 103
67 * 103
O2
26 * 103
42 * 103
50 * 103
CH4
23 *
38 *
103
46 * 103
Ar
24 * 103
38 * 103
45 * 103
103
CO2
0.71 *
C2H2
0.71 * 103
103
103
1.59 *
2.13 * 103
1.32 * 103
1.69 * 103
Source: Based on data from Harvey, A. H., “Semiempirical correlation for Henry’s constants over large temperature ranges,”
AIChE Journal, 42, pp. 1491–1494. DOI: 10.1002/aic.690420531. Copyright (c) 1966 John Wiley and Sons, Inc.
Phase Equilibria | 205
Note that Raoult’s Law and Henry’s Law express the same proportionality between
vapor pressure and mole fraction, but the proportionality constant is different. For dilute
solutions, where these laws hold best, xA is typically near 1.0. In this limit the solvent vapor
pressure must approach that of pure solvent pvap⬚, which is the proportionality constant for
Raoult’s Law. By contrast, solute B in the same solution is surrounded almost entirely by A
molecules, which is typically a very different environment from that in pure liquid B. As
a consequence, Henry’s Law coefficients, kB, are typically different in different solvents
and different from pvap⬚. All of this becomes complicated, however, if both components
have stable liquid phases at the particular temperature, and the mole fractions approach
0.5. Then the distinction between solute and solvent is arbitrary. However, solutions with
mole fractions near 0.5 are usually far from ideality, and so Raoult’s and Henry’s Laws
do not apply.
Proving this is beyond the scope of this text, but it can be shown that if the solvent in
a solution obeys Raoult’s Law, the solute (if volatile) obeys Henry’s Law, and the solution
is thus ideal.
EXAMPLE 6.2
Use Raoult’s Law to calculate the vapor pressure of water saturated with air at 1 bar
pressure and 25°C. For this purpose, we can consider dry air to be a mixture of 78.08%
N2 (by volume), 20.95% O2, and 0.93% Ar. Using Henry’s Law and the data in table 6.1,
we calculate the total mole fraction of dissolved air:
xair = xN2 + xO2 + xAr =
xair =
pN2
kN2
+
pO2
kO2
+
pAr
kAr
0.7808
0.2095
0.0093
+
+
= 1.45 * 10 - 5
84 * 103
42 * 103
38 * 103
The vapor pressure of pure water at 25°C obtained from standard tables is 3167 Pa. The
vapor-pressure lowering for air-saturated water at 25°C and 1 bar is
⌬pH2O = poH2O - pH2O = poH2O(1 - xH2O) = poH2Oxair
⌬pH2O = (3167 bar)(1.45 * 10 - 5) = 0.046 bar .
Thus, the effect of the dissolved air is to decrease the water vapor pressure by only
0.00145%. This is a completely negligible change from the vapor pressure of pure
water. The presence of saturated water vapor in the air (the definition of 100% humidity) does, however, change the percent composition of the humid air, which contains
0.03167 bar or 3.167 mol % of water vapor. The percentages of the other gases relative
to dry air need to be decreased accordingly.
For a system like air-saturated water to be in equilibrium, each component must have
the same chemical potential in each phase. If the concentration of O2 in the liquid-water
phase is so low that the chemical potential μ of the dissolved O2 is less than that of the
O2 in the gas phase, then O2 will be transferred from the gas phase until equilibrium is
reached. At that point, Δμ for the transfer is zero, and no further change occurs. Therefore,
at equilibrium,
mO2(g) = mO2(aq) .
This is also true for water in the system and for nitrogen and argon.
206 Chapter 6 |
Physical Equilibria
EXAMPLE 6.3
The Henry’s Law coefficient kN2 for nitrogen gas in water at 25°C is 8.4 * 104 bar.
Calculate the number of moles of N2 dissolved in 1 mole of water at N2 partial
pressures of 0.8 bar and 4 bar pressure, and from this, calculate what volume of N2 at
0.8 bar pressure will come bubbling out of 1 L of water that has previously been equilibrated with N2 at 4 bar pressure.
SOLUTION
At pN2 = 0.8 bar, x N2 = 0.8 bar>(8.4 * 104 bar) = 9.5 * 10 - 6.
At pN2 = 4.0 bar, x N2 = 4.0 bar>(8.4 * 104 bar) = 4.76 * 10 - 5.
Since in both cases, xN2 V 1, we can to a good approximation assume
x N2 = nN2 >(nN2 + nH2O) @ nN2 >nH2O, and so, per mol of water at 0.8 bar,
nN2 = 95 mmol, while at 4 bar, nN2 = 476 mmol.
1 L of water at 25 C contains (1 L × 997.1 g>L)>(18.015 g>mol) = 55.35 mol.
So at 4 bar, there will be 4.76 * 10 - 5 * 55.35 mol = 0.00264 mol N2. 80% of this
bubbles out.
V = RTnN2 >p = 8.31447 J mol - 1K - 1 * 298.15 K * 0.8 * 0.00264 mol>105
Pa = 5.2 * 10 - 5 m3 = 52.7 mL N2 gas per L of water.
This effect is the reason for the phenomenon known as the diver’s bends. If a diver
breathes air at 5 bar pressure (the pressure at a depth of about 40 m or 130 feet below
the sea surface), corresponding to a nitrogen partial pressure of about 4 bar, and then
surfaces too quickly, the nitrogen will simply bubble out of his or her body fluids, like
carbon dioxide out of soda. Tiny nitrogen bubbles inside the muscles and nerves cause
excruciating pain, and can even cause tissue damage leading to death. The solution is
to put the diver in a chamber pressurized with pure oxygen. The extra pressure on the
nitrogen bubbles forces them back into solution, and then they can be flushed from the
body through the lungs, by exchanging with a gas with very low partial pressure of
nitrogen.
Question to ponder: why does dissolved O2 not present the same problem?
Partition Equilibria
Partitioning is the distribution of a solute between two immiscible phases, often two
liquid phases, but sometimes (as in chromatography) a liquid phase and the surface of
a solid. Let’s say the two phases are two solvents 1 and 2, and B is a solute. Then at
equilibrium
mB1 = mB2 .
(6.31)
Since mB1 = m⬚B1 + RT ln aB1 and mB2 = m⬚B2 + RT ln aB2, we have
m⬚B1 + RT ln aB1 = m⬚B2 + RT ln aB2 .
Rearranging:
RT (ln aB2 - ln aB1) = -(m⬚B2 - m⬚B1)
aB2 >aB1 = exp[ -(m⬚B2 - m⬚B1)>RT] = exp[ -(⌬m⬚B)>RT]
(6.32)
This ratio of activities is called the partition coefficient, and it can be used to determine
the free energy of transfer of a solute from phase 1 to phase 2.
Phase Equilibria | 207
EXAMPLE 6.4
The hexane>water partition coefficient of phenol at 298.15 K is 0.20; calculate the free
energy of transfer of phenol from water to hexane.
SOLUTION
aB2 >aB1 = 0.20 1 ⌬m⬚B = -RT ln (aB2 >aB1) = -8.31447 J mol-1 K-1
* 298.15 K * ln (.2) = 4.0 kJ mol -1
There is an interesting and sometimes useful relation between the partition coefficient
and the solubility. Since in a saturated solution, the activities of dissolved and solid solute
are equal and the solute and solution are at equilibrium, two phases saturated in the same
solute must be at equilibrium.
EXAMPLE 6.5
The solubility of phenol in water is 83 g>L. What is its solubility in hexane?
SOLUTION
Assuming the partition coefficient is based on molarity, then the solubility of phenol in
hexane must be 0.20 * the solubility in water, or in other words, 16.6 g>L.
EXAMPLE 6.6
1-octanol and water are immiscible solvents, and octanol is often thought to have
solvent properties similar to biological lipids. The partition equilibria between water
and octanol have therefore been extensively tabulated to indicate the tendency of
chemical compounds to concentrate in lipids. Benzoic acid, a common preservative, M = 122.12 g mol-1, has a partition coefficient between octanol and water
ao >aw = 76, and the solubility of benzoic acid in water is 3.5 g>kg water. What is the
solubility in octanol?
SOLUTION
The solubility of benzoic acid in water is 0.029 m. This is the molality of the solution
that is in equilibrium with solid benzoic acid. If we assume that molalities equal activities, then the molality of a solution of benzoic acid in octanol in equilibrium with this
benzoic acid solution in water is 0.029 m * 76 = 2.2 m. Since this is the molality in
equilibrium with a saturated solution in water, it is in equilibrium with solid octanol,
and therefore it is the molality of a saturated solution in octanol.
Relatively hydrophobic solutes, like the fatty acid compounds CnH2n+1COOH with n 7 4,
are more soluble in a hydrocarbon like liquid n-heptane than in a dilute aqueous buffer.
For palmitic acid (hexadecanoic acid, C15H31COOH) solubility measurements indicate that
m⬚heptane - m⬚water = -38 kJ mol-1. Solubility data for fatty acids with numbers of carbon
atoms nC, where nC = 8 (octanoic), nC = 10 (decanoic), nC = 12 (dodecanoic), and so on
to nC = 22 (doeicosanoic) can be used to generate the relation:
m⬚heptane - m⬚water = (17.82 - 3.45 nc) kJ mol-1
(6.33)
This equation gives the standard free-energy change in transferring 1 mol of fatty acid from
liquid water to liquid heptane. The positive value for the first term on the right side of
Eq. 6.33 reflects the preference of the polar COOH group for water as solvent. The addition
208 Chapter 6 |
Physical Equilibria
TABLE 6.2 Hydropathy Index for the sidechains of amino acids in proteins
Side chain
Hydropathy index
Side chain
Hydropathy index
Isoleucine
4.5
Tryptophan
-0.9
Valine
4.2
Tyrosine
-1.3
Leucine
3.8
Proline
-1.6
Phenylalanine
2.8
Histidine
-3.2
Cysteine/cystine
2.5
Glutamic acid
-3.5
Methionine
1.9
Glutamine
-3.5
Alanine
1.8
Aspartic acid
-3.5
Glycine
-0.4
Asparagine
-3.5
Threonine
-0.7
Lysine
-3.9
Serine
-0.8
Arginine
-4.5
The hydropathy index is the relative hydrophobicity-hydrophilicity of the side-chain groups in proteins at neutral pH.
Positive values characterize hydrophobic groups which tend to be buried in water-soluble proteins; negative values
characterize hydrophilic groups which appear on the surface of the proteins. The relative values were deduced from free
energies of transfer of side-chain molecules from water to organic solvents or from water to the gas phase.
Source: Reprinted from Journal of Molecular Biology, Vol. 157(1), Jack Kyte, Russell F. Doolittle, A simple method for
displaying the hydropathic character of a protein, Pages No. 105–132, Copyright (1982), with permission from Elsevier.
(http://www.sciencedirect.com/science/article/pii/0022283682905150)
of CH2 groups to increase the length of the hydrocarbon part of the molecule then results
in a decrease of the strength of the interaction with water relative to that with heptane by
3.45 kJ mol-1 per CH2 group. For nC 7 5 the fatty acids are more soluble in hydrocarbons,
whereas the shorter chain carboxylic acids are more soluble in water.
Another useful application of this approach is to determine the relative hydrophobicities
of the side chains of the amino acids that are the constituents of proteins. Many structures
of water-soluble proteins are now known in detail, based on X-ray crystallography or highresolution nuclear magnetic resonance. Almost uniformly, charged or polar amino acid side
chains are exposed to the aqueous medium at the surface of the protein molecules, whereas
nonpolar hydrocarbon side chains are buried in the interior, lowering the lower free energy.
On the other hand, in proteins that are embedded in lipid membranes, the hydrocarbon side
chains are on the outside of the protein, and the polar groups are hidden inside.
Amino acid side chains can be characterized as hydrophilic (polar) or as hydrophobic
(hydrocarbon-like). A quantitative evaluation is provided by a hydropathy index based
on measurements of the solubility of the different amino acid side chains in polar versus
nonpolar solvents. One such set of indices is presented in table 6.2.
ligand
+macromolecule
in solution
ligand
in solution
FIGURE 6.7 Equilibrium dialysis.
Equilibrium Dialysis
If we had to rely only on normal saline solution flowing in our veins and arteries, the
amount of O2 transferred from our lungs to muscle cells would be far too little to support
life as we know it. The dissolved oxygen concentration is too low, even if we breathe pure
O2. Red blood cells (erythrocytes) solve this problem for us by packaging hemoglobin (Hb)
molecules, which are proteins containing a heme (Fe protoporphyrin IX, see appendix
table A.9) prosthetic group that binds O2 strongly. Muscle cells contain a related protein,
myoglobin (Mb), that binds O2 and stores it until needed. Red blood cells are packed full
of Hb and contained by a plasma membrane boundary. The plasma membrane is freely permeable to small, neutral molecules like O2 but is impermeable to large protein molecules
like Hb and to many charged ions. Of the dozen or so proteins associated with lipids in the
membrane, one is an anion channel that enables bicarbonate to be exchanged for chloride.
A second function of the red blood cells is to carry the product of oxidative metabolism,
Phase Equilibria | 209
bicarbonate, from the muscle cells back to the lungs, where it is converted to CO2 gas and
exhaled. Despite all of this complexity, red blood cells are small enough to squeeze through
the tiniest capillaries in the circulatory system.
The operation of the red blood cell is illustrative of a technique called equilibrium
dialysis that is a laboratory method of studying the binding of a ligand (small molecule
like O2) by a macromolecule (protein, nucleic acid, etc.) quantitatively. We have examined
the theory of such binding events in chapter 5. Materials made from reformulated cellulose
are typically used instead of a biological membrane. These sheet polymers contain small
pores that allow the free passage of small molecules, including ions, but prevent the large
macromolecules from passing. The small molecule is sometimes called the ligand. The
tubes typically have a nominal molecular mass cut-off (e.g., 1000 Daltons); molecules
smaller than this pass through; molecules larger than this are prevented from passing
through. Tubes of cellulose can be filled with 100 mL or so of a macromolecular solution
and tied off. The closed specimen solution in the dialysis bag can then be immersed in a
beaker of solution containing the ligand. Immediately, the ligand starts to diffuse across the
dialysis membrane, and this continues until equilibrium is reached. Stirring the contents
speeds the approach to equilibrium.
Once equilibrium is reached, all concentrations remain constant. Portions of the
contents of the dialysis bag and of the outside medium may then be removed for analysis.
In a “control” where no macromolecule is present in the dialysis bag, the concentrations of all species will be the same inside and outside the bag. In a different “control”
where a protein like albumin that does not bind is in the bag, the O2 concentration at
equilibrium will still be the same inside and outside. If a protein like myoglobin is in
the bag, then the total concentration of O2 inside will be significantly larger than the
concentration of O2 outside, where Mb is absent. The presence of the Mb inside serves to
concentrate the O2 by binding it. By a series of quantitative measurements, it is possible
to determine both the binding equilibrium constant and the number of ligand binding
sites per macromolecule.
A convenient way of treating the data to extract these values is through the Scatchard
equation, which was discussed in chapter 5. There are three experimentally measurable
quantities; the concentration of the small molecule (we’ll call it A) outside; the concentration of A inside, which is the sum of both free and bound A inside; and the concentration
of the macromolecule inside, which we know because we put it there!
The Scatchard equation is
n
= K(N - n) .
(5.82a)
3A 4
Since A is free to move across the membrane, [A] = [A]outside is the concentration of free
A both inside and outside. In chapter 5 we defined n as the number of bound A molecules
per polymer molecule (Eq. 5.80). Since we know the total concentration of A inside is the
sum of bound A and free A, and the concentration of free A is the same inside and outside,
the concentration of bound A inside is [A]inside - [A]outside, and n is given by
3 A 4 inside - 3 A 4 outside
.
(6.34)
[M]
n can therefore be calculated from three known or measurable quantities. The left-hand
side of the equation is known, and on the right we know the value of n, so we can, using
experimental data alone, create a Scatchard plot of n>[A] versus n, extracting N from the
x-intercept and K from the negative of the slope.
Data illustrating binding of O2 to Mb are shown in figure 6.8. Taking advantage of the
different visible absorption spectra of myoglobin with and without O2 bound, the fraction
of Mb having O2 bound can be determined experimentally for a range of O2 pressures.
n =
Physical Equilibria
FIGURE 6.8 Binding of O2 to
human myoglobin. (top) Data
obtained at 30°C are plotted
as the fraction of binding
sites occupied, ν, against the
pressure of O2 (Pa). (bottom) Scatchard plots of the
binding of O2 to Mb at several
temperatures. [Data recalculated
from A. Rossi-Fanelli and E.
Antonini, Arch. Biochem. Biophys.
77 (1958): 478–492.]
1.0
0.8
0.6
ν
0.4
0.2
0.0
200
400
pO2 (Pa)
600
800
0.010
10°C
0.008
20°C
ν/pO2 (Pa–1)
210 Chapter 6 |
0.006
0.004
30°C
35°C
0.002
40°C
0.2
0.4
ν
0.6
0.8
1.0
These values are plotted directly in figure 6.8(a). The fraction of binding sites occupied
increases toward a maximum with increasing pressure of O2.
The same data are plotted as Scatchard plots in figure 6.8(b). All of these lines were
calculated from a single pair of parameters; the enthalpy and entropy change of the binding reaction
Mb + O2(g) K MbO2
decreases with increasing temperature which allows us to calculate the equilibrium constant
as a function of temperature using the van’t Hoff equation.
Phase Equilibria | 211
FIGURE 6.9 Binding of fluoride
ion (F-) to ferrimyoglobin.
A least squares fit to the
data gives N = 0.95 ~ 1 and
K = 27.2 M-1
25
ν/[A], M–1
20
(Based on data from A. Rossi-Fanelli,
E. Antonini, “Studies on the
oxygen and carbon monoxide
equilibria of human myoglobin,”
Archives of Biochemistry and
Biophysics, 77(2), pp. 478–492,
ISSN 0003–9861, 10.1016/0003–
9861(58)90094–8. Copyright (c)
1958 Elsevier, Inc. (http://www.
sciencedirect.com/science/article/
pii/0003986158900948)).
15
10
5
0.2
0.4
0.6
0.8
1.0
ν
FIGURE 6.10 Binding of
Mg–ATP by tetrahydrofolate
synthetase from
C. cylindrosporum.
Measurements were carried out
at 23°C and a pH of 8.0.
50
ν/[Mg-ATP] (mM–1)
40
(Based on data from N.P.
Curthoys and J.C. Rabinowitz,
“Formyltetrahydrofolate Synthetase:
BINDING OF ADENOSINE
TRIPHOSPHATE AND RELATED
LIGANDS DETERMINED BY PARTITION
EQUILIBRIUM,” Journal of Biological
Chemistry, 246, pp. 6942–6952.
Copyright (c) 1971 American Society
for Biochemistry and Molecular
Biology.)
30
20
10
0
0
1
2
ν
3
4
A similar use of the Scatchard plot to show binding of the fluoride ion to the oxidized
form of myoglobin is shown in figure 6.9. Binding of the fluoride ion to the heme iron
causes a substantial change in the absorption spectrum.
EXAMPLE 6.7
The formyltetrahydrofolate synthetases can utilize the energy of hydrolysis of ATP
(to ADP and phosphate) for the formation of a carbon–nitrogen bond between
(l)-tetrahydrofolate and formate. Binding of the Mg complex of ATP to such an
enzyme from C. cylindrosporum has been measured; the data are plotted in Scatchard
form in figure 6.10.
5
212 Chapter 6 |
Physical Equilibria
The data fit the identical-and-independent-sites model well. The slope of the plot
gives an equilibrium constant K = 1.37 * 104 M-1. The intercept gives N = 4.2. Since
the enzyme is known to have four identical subunits (each has a molecular weight of
60,000), the binding results are consistent with one Mg–ATP binding site per subunit.
In the discussion leading to Eq. 5.16, we defined the quantities n and N as follows:
n = average number of molecules of A bound to the macromolecule
N = number of binding sites on the macromolecule
An oligomeric protein such as formyltetrahydrofolate synthetase discussed in example 6.7
is made up of a small number of monomeric units. A DNA molecule is a polynucleotide consisting of nucleotides as its monomer building blocks. Sometimes it is more convenient to
define the binding parameters on a per monomer unit basis as follows:
r = average number of molecules of A bound per monomer unit of the macromolecule
n = number of binding sites per monomer unit of the macromolecule
The Scatchard equation is unchanged by these new definitions:
r
= K(n - r)
(6.35)
3A 4
Thus, the Scatchard plot can also be presented with r and n as the parameters.
In figure 6.11, data are plotted for the binding of the trypanocide drug ethidium
(ethidium is sometimes used to treat animals infected by parasitic trypanosomes) to a
double-stranded DNA. Note that the data are represented reasonably well by the identicaland-independent-sites model. The r-intercept gives n = 0.23 or approximately one binding
site per four nucleotides (two base pairs). This is consistent with the fact that ethidium
binds to DNA by intercalation. Planar, aromatic molecules like ethidium slide between
two adjacent base pairs in DNA so that the hydrophobic parts are stacked on the base
pairs; this is intercalation. Note also that at the lower temperature the line is steeper, indicating a greater binding constant. This temperature dependence of the binding constant
yields a negative ⌬H⬚ or about -25 kJ mol-1 for the binding of the drug. The fairly strong
interaction (K = 2.66 * 104 mol-1 and 5.60 * 104 mol-1 at 20⬚C and 0⬚C, respectively)
between the drug and DNA is evidence that the target of the drug in vivo is the DNA of
the trypanosome.
9
8
r/[ethidium] (mM–1)
FIGURE 6.11 Scatchard plot of
the binding of ethidium to DNA
in a medium containing 3 M
CsCl and 0.01 M Na3EDTA.
7
0°C
6
20°C
5
4
3
2
1
0
0
0.1
r
0.2
Phase Equilibria | 213
FIGURE 6.12 Fraction of
hemes of Mb or Hb occupied by
oxygen, f, as a function of the
pressure of oxygen. Myoglobin
has only one heme group
and therefore cannot show
cooperativity. Hemoglobin
has four hemes; the binding is
cooperative.
1
0.9
0.8
Myoglobin
0.7
0.6
(Based on data for Mb from
figure 5.6; data for Hb from
P. Astrup, K. Engel, J. W.
Severinghaus, and E. Munson,
“The Influence of Temperature and
Ph on the Dissociation Curve of
Oxyhemoglobin of Human Blood,”
Scandinavian Journal of Clinical
& Laboratory Investigation, 17(6),
pp. 515–523. Copyright (c) 1965
Informa Healthcare.)
f 0.5
p50 = 200 Pa
0.4
p50 = 2200 Pa
0.3
Hemoglobin
0.2
0.1
0
0
2000
4000
6000
Cooperative Binding and Anticooperative Binding
As we discussed at length in chapter 5, binding to multiple sites on a macromolecule can
be independent, or it can be cooperative or anticooperative. In cooperative binding, the
first ligand bound makes it easier for the next one to be bound. This may be caused by a
conformational change in a multisubunit protein that makes it easier for successive ligands
to be bound. The limiting case of cooperative binding is all-or-none binding. In all-ornone binding the first ligand increases the binding of the other ligands so much that all
N ligands are bound at once. As the concentration of ligand is increased, the number of
ligands bound increases sharply to the maximum possible. The macromolecule has either
no ligands bound or N ligands bound. In anticooperative binding, each succeeding ligand
is bound less strongly than the previous one.
A useful way to plot binding data and to learn about the cooperativity (or anticooperativity) of binding is to plot fraction of sites bound, f, versus concentration of free
ligand. A vital example of cooperative binding is the binding of oxygen by hemoglobin.
Hemoglobin has four heme-binding groups that bind four oxygen molecules cooperatively.
The binding is cooperative so that hemoglobin releases most of its bound oxygen at the
low oxygen pressure in the tissues but binds the maximum amount of oxygen in the lungs.
Figure 6.12 compares the binding of oxygen to myoglobin with only one heme-binding
site and to hemoglobin with four interacting binding sites. The cooperative-binding curve
for hemoglobin shows a characteristic sigmoidal shape; sigmoidal means “shaped like the
letter S.” Myoglobin, which has only one binding site so it cannot bind cooperatively, has
a binding curve characteristic of a molecule with any number of identical and noninteracting sites. That is, the data for myoglobin are represented by the Scatchard equation; a
linear Scatchard plot is obtained. On the other hand, if we plot the hemoglobin data on a
Scatchard plot, we obtain a very different curve (figure 6.13). While the correct x-intercept
of N = 4 is obtained, the Scatchard plot is not linear, but convex. Convexity in a Scatchard
plot is indicative of cooperative binding, even without a Hill plot (figure 5.15). Conversely,
as you might guess, if the Scatchard plot is concave, this suggests anticooperative or
exclusionary binding.
ν/pO2, kPa–1
pO2, Pa
1.0
0.8
0.6
0.4
0.2
1
2
ν
3
4
FIGURE 6.13 Scatchard plot of
the hemoglobin data presented
in figure 6.12.
214 Chapter 6 |
Physical Equilibria
Membranes
When we think of phases, we think of the usual solid, liquid, or gaseous states. Whenever
there are two phases in contact, there is also a surface, or interface, between them. This
surface has properties different from those in the two bulk phases and therefore will have
different behavior. The surface has thermodynamic properties specified by its free energy,
enthalpy, and so on, just as bulk phases have. However, there are differences. A surface is
two-dimensional instead of three-dimensional. This means that concentration units for a
surface are mol m-2 instead of mol m-3.
Biological cells are small volumes surrounded by membranes. (Membranes are
present inside the cells, also.) The membranes typically contain lipids, proteins, glycolipids, and other amphiphilic molecules. Membranes constitute a different phase from
the rest of the cell, and for some purposes they can be thought of as a surface phase.
The membranes of course have a finite thickness and definite volumes, but it may
sometimes be more useful to think of them as two-dimensional surfaces rather than
three-dimensional.
Lipid Molecules
Oil droplets or particles of greasy dirt are hydrophobic. When they are suspended in
water, the surface or interface is attractive to amphiphilic molecules (see chapter 3).
Amphiphilic molecules consist of two parts or regions, one of which interacts favorably
with the hydrophobic phase and the other with the aqueous phase. Amphiphilic molecules
locate preferentially at the interface. This happens to molecules that have a hydrocarbon
tail and a polar head; they are called surface-active molecules, or surfactants. Sodium
dodecylsulfate (see figure 6.14), a principal component of many commercial detergents,
orients at the surface of dirt particles and makes them soluble in water. The hydrocarbon
tail is attracted to the oily dirt “phase,” and the polar sulfate head faces the water (see
figure 6.15). The hydrocarbon part is hydrophobic; the polar group is hydrophilic, which
is typical of amphiphilic molecules.
The structures of several surfactants are shown in figure 6.14. There is a large variety
of hydrophilic head groups: some are polar because of exposed -OH or -NH2 groups;
others are negatively charged (phosphatidylglycerol) or zwitterionic (phosphatidylcholine
or phosphatidylethanolamine), bearing both positive and negative charges at pH 7. There
is also variety among the hydrocarbon tails: chain lengths range between 12 and 18 or
more carbon atoms and double bonds or large multiple-ring structures as in cholesterol
may be present. The number of hydrocarbon tails attached to a head group may also vary,
but usually there are one or two. Such variations in composition are important for defining
the properties of the surfaces, especially in membranes and vesicles. They determine the
thermal stability, fluidity, and curvature of the membranes, as well as the interactions with
proteins and other molecules associated with or embedded in the membrane. An illustration of the variety of forms and interactions exhibited by lipids and other surfactants is
shown in figure 6.15.
Lipid Bilayers
A typical cell membrane consists of a lipid bilayer into which are incorporated other
amphiphilic molecules, such as cholesterol, sphingolipids (found in brain tissue), proteins,
and glycolipids (molecules containing sugar or saccharide residues). One can think of the
lipid bilayer as a kind of two-dimensional solvent; however, its thickness cannot be ignored.
As illustrated in figure 6.15, the bilayer is arranged as a sandwich with the hydrophobic
portions of the lipids turned inward and the hydrophilic portions on the outer surfaces in
contact with the aqueous phases. Thus, just as a soap bubble is a bilayer that separates two
Membranes | 215
FIGURE 6.14 Some common
surface-active molecules.
O–
Sodium palmitate
Na+
O
Sodium oleate
O
O–
Na+
O–
NH3+
O
Dipalmitoylphosphatidylserine
O
(DPPS)
O
O
O
O
O
P
O–
Dipalmitoylphosphatidylcholine
(DPPC)
O
O
O
O
O
O
O
P
Dipalmitoylphosphatidylethanolamine
O
(DPPE)
N+
O–
O
O
O
O
O
O
NH3+
O–
P
O
O
S
O
O–
Na+
Sodium dodecylsulfate (SDS)
HO
Cholesterol
regions of air, the membrane bilayer serves as a boundary between two aqueous phases
that, in biological cells, typically differ from one another in solute concentrations, ionic
strength, pH, and so on.
Membrane proteins, glycolipids, and other components associated with the lipid
bilayer serve important functions, such as gating the transport of molecules across the
bilayer from one aqueous phase to the other, receiving and transmitting signals, defining
the electrical properties of the cell, decorating the surface so other cells can identify it,
and many others. Most membrane proteins span the bilayer; that is, they have a dumbbell
shape with a hydrophobic waist and polar ends that extend into the aqueous environment
on either side. The polar ends are often not identical in structure, which helps define distinctions between inside and outside the cell or organelle. Although the lipid bilayer serves
effectively to block the passive transport of ions or of large molecules and even smallmolecule metabolites, nevertheless water moves readily across the hydrophobic interior of
the bilayer, thanks to a family of specialized water-transport proteins called aquaporins.
Physical Equilibria
O–
Na+
O
O
–
O
O
–
–
O
O
O
Na+
Na+
Na+
O
Na+
O
Na+
(b)
O–
air
(a)
–
O
O
–
Na+
Na+
O
–
O
O
O–
Na+
Na+
grease
O
Na+
O
O–
O–
O
Na+
H
H
–
O
O
O
O–
O
O
water
+
Na+ O
O
–
Na
water
(d)
Na+
– + – + – + – + – + – + – + – +
Na+
O
–
O–
O
O
O
Na+
–
Na+
O
Na
O
–
O
Na+
+
+
Na
–
O
water
Na
+
O
+
Na
(c)
O
OO
OO
OO
O
H
+
Na
O
H
O–
–
O
–
O
–
O
Na+
O
O–
O
–
O
Na+
O
O–
Na+
O–
O
O
–
O
Na+
O
Na+
Na+
–
O
O–
O
O
–
O
Na+
– + – + – + – + – + – + – + – +
water
water
– +
– +
(e)
–
+
+
water
+
–
–
+
–
– +
+
–
– +
+
+ –
+ – +– + –
water
– +
+ –
+ – + – +
–
–
+
–
+
– +
+
–
+
–
–
FIGURE 6.15 Diagrammatic
representation of some
common lipid phases. (a) Lipid
monolayer at an air–water
interface. (b) Detergent
solubilizing a grease particle.
(c) Lipid or detergent micelle
in water. (d) Lipid bilayer
membrane separating two
aqueous phases. (e) Vesicle with
bilayer separating an internal
aqueous phase from the
external aqueous medium.
O
216 Chapter 6 |
Phase Transitions in Lipids, Bilayers, and Membranes
Lipid bilayers are complex structures that undergo a number of phase transitions. At very
low temperatures, the lipid molecules are locked in place, and the hydrocarbon tails are
immobile. As the temperature increases, the hydrocarbon tails begin to ‘melt’, allowing
interconversion between gâuche and trans conformations about the CH2 i CH2 single
bonds, but the lipid molecules as a whole are packed into a reasonably well-ordered twodimensional lattice. At a higher temperature still, the important gel-liquid crystal transition
occurs. In the lower temperature gel phase, the lipid molecule as a whole is frozen into a
single location in the membrane. In the liquid-crystalline phase, the entire lipid molecule is
free to move both rotationally, about the normal to the membrane plane, and translationally
within the plane. The liquid-crystalline phase is the only biologically relevant one; organisms usually cannot tolerate their membranes freezing into the gel phase.
Membranes | 217
FIGURE 6.16 Differential
scanning calorimetry of a
suspension of dipalmitoylphosphatidylcholine (0.508 mg g-1)
in water.
dq/dT (J K–1glipid–1)
500
(Based on data from N. Albon and
J. M. Sturtevant, “Nature of the gel to
liquid crystal transition of synthetic
phosphatidylcholines,” Proceedings of
the National Academy of Science, 75,
pp. 2258–2260. Copyright (c) 1978
National Academy of Sciences.)
400
300
200
100
0
41.0
41.2
41.4
41.6
41.8
42.0
T (°C)
Lipid-phase transitions can be studied using a technique called differential thermal
analysis (DTA), using a differential scanning calorimeter. In this method, heat is added to
the sample electrically at a constant rate, and the temperature is continuously monitored
using a small probe. As long as the lipid remains a single solidlike phase, the temperature
rises smoothly and with a rate determined by the heat capacity of the lipid. When the
temperature reaches the “melting point” of the lipid, extra heat is required to induce the
transition from solidlike to liquidlike behavior. Once the transition is completed, the temperature once again rises smoothly in proportion to the heat capacity. Figure 6.16 shows
a plot of the excess specific heat associated with the main transition in the lipid dipalmitoylphosphatidylcholine (DPPC). The peak of the curve at 41.55°C indicates the gel-liquid
crystal transition temperature for this lipid. The width of the peak, which becomes broader
as the heating rate is increased, is a measure of the sharpness of the transition. Very sharp
peaks are associated with first-order phase transitions in pure lipids, where the transition
from one phase to the other is highly cooperative.
Of course, 42°C is above most mammalian body temperatures, and so if our cell membranes were composed of pure DPPC, we would not live long! Organisms lower their gelliquid crystal transition temperatures by incorporating single or multiple cis double-bonds
in the hydrocarbon chains, which kink the chains. These kinked chains are intrinsically
more disordered and so have lower transition temperature. In addition, membranes are
composed of mixtures of lipids. Just as in ordinary solutions, mixing in ‘solutes’ such as
cholesterol lowers the melting temperature and also broadens the transition. Interestingly,
trans-double bonds tend to keep the hydrocarbon chains straight, making the chains more
rigid. This may be one reason why so-called ‘trans fats’ are unhealthy.
Some biological organisms, including single-celled bacteria and flowering plants
like oleander, are able biochemically to alter the composition of their membrane lipids
in response to large changes in growth temperature. In this fashion, membranes of cells
maintain their fluidity when grown at temperatures so low that the membranes of the hightemperature grown organism would solidify. This is an example of thermal adaptation.
218 Chapter 6 |
Physical Equilibria
The mobility of molecules such as DPPC in bilayers has also been investigated using
techniques such as deuterium nuclear magnetic resonance and electron paramagnetic spin
labeling. It is evident from these studies that mobility increases significantly from outside
to inside the bilayer. The headgroups are held relatively rigidly in their network, whereas
the hydrocarbon tails exhibit more mobility. Even this mobility is graded from the regions
close to the headgroup (mostly rotation about an axis normal to the bilayer surface in this
region) to the much greater, liquidlike flexibility of the terminal lipid region in the interior
of the bilayer sandwich.
Surface Tension
Because of intermolecular attractions, the molecules at the surface (exposed to air) of a
bulk liquid are attracted inward. This creates a force in the surface that tends to minimize
the surface area. If the surface is stretched, the free energy of the system is increased. The
free energy per unit surface area, or the force per unit length on the surface, is called the
surface tension. The units for energy per unit area and force per unit length are identical.
The SI units are millinewtons per meter (mN m-1) or mJ m-2. Table 6.3 gives the surface
tensions of a diverse range of liquids.
What happens when a substance is dissolved in the liquid or added to the surface? The
surface tension typically either decreases or does not change very much. It never increases
greatly. There is a thermodynamic reason for this, but before stating it let’s consider what
would happen if we could greatly increase the surface tension of water. Water drops from
your faucet could become the size of basketballs because the size of the drop is directly
proportional to the surface tension. You might have to cut water with a knife and chew it
well before swallowing. Water would not wet anything, so processes depending on capillary action would not work. We are saved from these possible catastrophes. Any substance
that tends to raise the surface tension of a liquid raises the free energy of the surface. The
substance therefore concentrates less at the surface.
Just as in three dimensions the pressure is (0G>0V)T , so analogously in two dimensions the surface tension, or surface free energy, is just (0G>0A)T,p where A is the surface
area. Substances that lower the surface tension also lower the free energy of the surface;
they preferentially migrate to the surface. Thus, substances that lower the surface tension
concentrate at the surface and give large decreases in surface tension, but substances that
raise the surface tension avoid surfaces and give only small increases in surface tension.
The quantitative expression for this is called the Gibbs adsorption isotherm:
⌫ = -
1 dg ,
1 dg
⬵ RT d ln a
RT d ln c
(6.36)
TABLE 6.3 Surface tensions of pure liquids in air
Substance
Platinum
Surface tension (mN m−1)
Temperature (°C)
1819
2000
Mercury
487
15
Water
71.97
25
58.85
100
Benzene
28.9
20
Acetone
23.7
20
n-Hexane
18.4
20
5.2
-247
Neon
Membranes | 219
FIGURE 6.17 Effect of solute
concentration on surface
tension in aqueous solution
at 20°C.
100
LiCl
γ (mN m–1)
80
60
40
ethanol
20
0
0
0.2
0.4
0.6
0.8
1.0
xB
where ⌫ is the surface excess concentration in mol m-2; g is the surface tension in N m-1; R
is the gas constant; a is the activity of the solute in bulk solution, and c is the concentration
of the solute in bulk solution, relative to some standard state concentration c°.
The sign of the excess surface concentration ⌫ is opposite to the sign of the change
of the surface tension with concentration (or activity) of solute in the solution. Figure 6.17
shows the change in surface tension of water when LiCl or ethanol is added. Ionizing salts
are almost the only solutes that raise the surface tension of water.
There are many naturally occurring surfactants in plants and animals. DPPC (see
figure 6.14) is the major component of pulmonary surfactant, a layer that lowers the surface
tension at the surface of the alveoli — the smallest air-filled compartments — in the lung
and allows one to breathe. A large surface area is necessary for the efficient exchange
of gases in the lung. DPPC lowers the surface tension of water to nearly zero and allows
large aqueous surface areas to exist in the lung. Premature babies lack this vital surfactant;
treatment with an artificial surfactant via a breathing tube has greatly increased their rate
of survival.
The surface-tension decrease caused by surface-active molecules can be measured
with a Langmuir film balance, illustrated in figure 6.18. Some surfactant, usually in a
volatile or aqueous solvent, is added to the surface of water between a movable barrier and
a float sensor. As the barrier is slowly moved to compress the film, an increasing force
must be applied to the float from the opposite direction to prevent the float from being
displaced from its null position. The surface pressure necessary to maintain the film in the
compressed state is just the difference between the surface tension g0 of the water beyond
the float and the surface tension g of the coated water:
ps = g0 - g
The surface concentration in mol m-2 depends on the surface pressure. The experiment
is the two-dimensional analog of a pressure versus volume experiment in three dimensions. It is traditional to plot surface pressure versus area per molecule in nm2>molecule.
220 Chapter 6 |
Physical Equilibria
FIGURE 6.18 A Langmuir
film balance measures surface
tension. The force per unit
length F/l required to compress
a surface containing a surfactant
layer is known as the surface
pressure πs. The float is
connected to a torsion wire (not
shown) to measure the force.
0
Movable
barrier
Balance
indicator
Float
Scale
Pure
water
Water plus
surfactant layer
Force
Pure
water
Surface pressure π = γ0 – γ
60
π (mN m–1)
FIGURE 6.19 Surface
pressure versus area curves
for two amphiphilic molecules
(surfactants) at an air–water
interface. The molecular areas
correspond to average cross
sections of the molecules
projected on the surface.
The steeply rising portions
of the curves occur when the
molecules in the surface layer
have been compressed into a
compact arrangement.
40
20
0
0.2
0.4
A
(nm2
0.6
0.8
molecule–1)
The area occupied by the film at various stages of compression is determined using a
calibrated distance scale associated with the movable barrier. By knowing the amount of
surfactant added, one can calculate the area per molecule, which is just the reciprocal of
the surface concentration. Some representative results are shown in figure 6.19. The experiment typically begins with an expanded film, corresponding to large values of the area per
molecule. Under these conditions, the surfactant molecules are far from one another, on
average, giving a film with low surface pressure, corresponding to a surface tension equal
to that of pure water.
As the film is compressed, the surface pressure rises. In some cases, discontinuities
appear in the plots owing to phase transitions that occur in some surfactants. The high
surface pressures at low area per molecule (high surface concentration) result from the low
surface tensions produced by the surface-active molecules. The maximum possible surface
Membranes | 221
pressure, equal to the surface tension of pure water, occurs when the surface tension of
the film falls to zero. This is implicit in the preceding equation. The experiment typically
ends with the collapse of the film, whereupon the surface pressure drops abruptly to zero.
Surface Free Energy
Amphiphilic molecules containing hydrocarbon chains prefer the surface of an aqueous
solution, which lowers the surface tension. The longer the chain, the higher is the surface
concentration of the amphiphile. To understand the thermodynamic consequences of the
interactions of lipids in an aqueous environment, we need to look more closely at the free
energy of transfer of molecules between bulk phases.
Hydrophobic Effect
Hydrocarbons are almost completely insoluble in water, and long hydrocarbon-chain
amphiphiles typically have low solubility. The thermodynamic consequence is that for
most nonpolar molecules the free energy is positive for transfer from an organic solvent to
water. Contrary to what one might expect, the enthalpy of transfer for small hydrocarbons
is negative. A large drop in entropy occurs when these small hydrocarbons enter water,
accounting for the positive free-energy change.
To explore this situation thermodynamically, consider a liquid hydrocarbon phase
(phase 1) in equilibrium with a water phase (phase 2). The mole fraction of hydrocarbon
in the hydrocarbon phase, xH1, is close to 1. Therefore,
mH1 = m⬚H1 + RT ln aH1 = m⬚H1 + RT (ln x H1 + ln gH1) ⬵ m⬚H1 + RT ln x H1 , (6.37)
since gH1 S 1 as xH1 S 1; the pure hydrocarbon is the standard state for phase 1.
For hydrocarbon in the water phase, the mole fraction xH2 is close to zero:
mH2 = m⬚H2 + RT ln aH2 = m⬚H2 + RT (ln x H2 + ln gH2) ⬵ m⬚H2 + RT ln x H2 , (6.38)
since gH2 S 1 as xH2 S 0; the hydrocarbon is the solute, and the solute standard state is
obtained at infinite dilution.
The two terms in the chemical potential reflect two different origins, termed unitary
and cratic. The cratic contribution, RT ln xH, is statistical in nature and arises from the
entropy of mixing between solute and solvent. The unitary contributions to the chemical
potential are m⬚H1 and m⬚H2. ⌬m⬚ = m⬚H2 - m⬚H1 contains only the changes of the internal
energy and the entropy of the hydrocarbon and its interactions with solvent as the hydrocarbon environment changes from hydrocarbon to water.
But at equilibrium, mH1 = mH2, and so we can equate Eqs. 6.37 and 6.38. This gives
m⬚H1 + RT ln x H1 = m⬚H2 + RT ln x H2
or
⌬m⬚ = m⬚H2 - m⬚H1 = RT ln x H1 - RT ln x H2 ⬵ -RT ln x H2 .
(6.39)
xH2 is the solubility of the hydrocarbon in water, which is typically very small, and so
⌬m⬚ values are significantly positive. If we plot ⌬m⬚ against temperature, we can separate enthalpic and entropic terms using ⌬m⬚ = ⌬H ⬚ - T⌬S⬚. The unitary enthalpy ⌬H ⬚
can also be obtained experimentally from calorimetric measurements. Data for several
hydrocarbons are shown in table 6.4. The values for ⌬H⬚ are significantly negative for
small hydrocarbons, but they rise as the size of the hydrocarbon molecule increases and
are positive for aromatic molecules. The corresponding entropies are negative for both
aliphatic and aromatic hydrocarbons and do not show a marked dependence on the size of
the molecule. In parallel studies, it has been shown that aliphatic alcohols behave similarly
to the aliphatic hydrocarbons. Furthermore, investigations of the partial molar volume
222 Chapter 6 |
Physical Equilibria
TABLE 6.4 Thermodynamic data for the transfer of hydrocarbons to water at 25°C
Hydrocarbon
⌬m⬚ (kJ mol-1)
⌬Hⴗ (kJ mol-1)
⌬Sⴗ (J mol-1 K-1)
C2H6
16.3
-10.5
-88
C3H8
20.5
-7.1
-92
C4H10
24.7
-3.3
-96
C5H12
28.6
-2.0
-103
C6H14
32.6
0
-109
C6H6
19.3
+2.1
-58
C6H5CH3
22.8
+1.7
-71
C6H5C2H5
26.2
+2.0
-81
C6H5C3H7
28.8
+2.3
-89
Source: Based on data from C. Tanford, The Hydrophobic Effect: Formation of Micelles and Biological Membranes,
2nd Edition. Copyright (c) 1980 John Wiley and Sons, Inc.
show that ⌬V ⬚ 6 0 for these hydrophobic or amphiphilic molecules. This means that the
molecules occupy less volume in the aqueous environment than in the pure organic phase.
Clearly, this behavior is somehow tied to the counterintuitive decrease in entropy (increase
in ordering) that occurs when amphiphiles are added to water.
Explanations for the behavior just described are usually made in terms of the effect
on the structure of liquid or “solvent” water. As we have seen, there is evidence from thermodynamic and physical measurements on pure water that the hydrogen bonding in water
gives rise to an organized but fluctuating network of water molecules in the liquid state.
The decrease in entropy associated with introducing amphiphiles must then result in a further ordering of the water of solvation, at least in the immediate vicinity of the introduced
solute molecule. These have been described as “microscopic icebergs”; however, that may
be an exaggeration of the degree of increased structure. In any event, the solvation water
has decreased entropy and apparently a decreased volume in comparison with bulk water.
At extremely low concentrations, translational entropy (cratic) ensures dispersion
of the lipid as a homogeneously distributed solute in water. Above a small threshold of
concentration called the critical micellar concentration, phospholipids form aggregates,
including bilayers, to lower the total free energy. In this fashion, the hydrophobic effect
serves to keep membranes intact in an aqueous environment. This may be demonstrated in
a dramatic way by taking a red blood cell suspension and adding an organic solvent such
as acetone to the medium. The acetone destabilizes the cell membranes and spills out the
hemoglobin, presumably by disrupting the hydrogen-bonding networks that are normally
present in the aqueous phase. The hydrophobic effect’s contribution to the stability of a
lipid bilayer in water is equivalent to a lateral pressure squeezing the phospholipids together
and thus preventing the hydrocarbon chains from coming in contact with water.
Vapor Pressure and Surface Tension
If small droplets and a much larger drop of water are placed in a closed container saturated
with water vapor, the small droplets eventually disappear, and the large one becomes larger.
This phenomenon is a result of surface tension; the small droplets have a higher vapor pressure than the large one. When water or any other substance is finely divided, the surface
effect becomes significant, and we need to include a term gdA in the expression for the
change in free energy associated with increase in surface area A. We replaced
dG = Vdp - SdT + mdn
(6.40)
Membranes | 223
for bulk water by
dG = Vdp - SdT + mdn + gdA
(6.41)
for the small droplets. The free energy per mole of bulk water (water in the interior of the
liquid) is μ. Equation 6.41 shows that the free energy of a liquid is increased by increasing its surface area; thus, a liquid will minimize its free energy by decreasing its surface
area. This means a liquid will form a sphere if there are no other forces acting on it, such
as gravity or other surfaces. Combining two spheres into one lowers the total surface area
and therefore the total free energy. Thus, small spheres will tend to combine to form larger
spheres. Mercury with its higher surface tension illustrates this better than water. We usually ignore the effect of surface area and the contribution of surface free energy to the total
free energy; the effects are too small. However, significant effects can occur as shown by
the higher vapor pressure of small—micron-sized—droplets. We can combine the terms
mdn + gdA of Eq. 6.41 to obtain the total chemical potential of the liquid water; this takes
into account the effect of the surface on the water molecules. Water in small droplets (with
a high surface-to-volume ratio) has a higher total chemical potential and higher vapor pressure than bulk water and will therefore transfer spontaneously into a larger drop.
Biological Membranes
The composition and orientation of the molecules in biological membranes determine
their function. Along with the diversity in composition comes a richness of properties that
correlates with the variety of membrane functions, such as the rate of transport of water,
ions, or other solutes across the surface. Usually, membranes contain an assortment of lipid
molecules with diverse chemical structures, together with proteins and sometimes polysaccharides. The lipids are typically fatty acid esters that differ in the length of the fatty acid
chain, the degree of unsaturation, the charge or polarity of the esterifying group, and the
number of fatty acids esterified per molecule. The proteins may be intrinsic or integral, in
the sense that they are incorporated directly into the membrane structure, or extrinsic, if
they are attached to the membrane surface or interact strongly with it. Other components,
such as cholesterol, may also be present. The Fluid Mosaic Model of membrane structure
was developed by Singer and Nicholson (1972). Not only is the biological membrane characterized by a high degree of fluidity, but it is also constituted of a heterogenous set of
membrane-associated proteins. Application of this model to the chloroplast membrane is
shown in figure 6.20.
The variety of membrane composition results in a range of physical properties. For
example, the mobility of the molecules in two dimensions in a membrane may be that of a
typical liquid or that of a solid. As we have discussed, it is relatively common to encounter
phase transitions in membranes or artificial surface films, and these transitions are closely
analogous to their three-dimensional counterparts. The transition temperature or melting
point, for example, is sensitive to the lipid composition; the thermodynamic analysis of
two-dimensional solutions can be carried out just as for three-dimensional systems.
The relation between the intrinsic proteins and the lipids is more complex and is not
yet well understood. Many proteins are amphiphilic and interact with both the hydrophobic
lipids and the polar aqueous interface. These hydrophobic and hydrophilic sites are located
in different regions of the protein molecule, resulting in a strong orientation with respect
to the membrane surface. Apart from this orientation, the protein may behave much like
a solute in the lipid solvent; essentially a two-dimensional solution is formed. There is
even a two-dimensional analog of precipitation, in which the solution becomes saturated
with respect to a protein constituent and the protein separates as a distinct phase. Electron
microscopy (see chapter 15) of membrane surfaces provides a useful method for detecting
and characterizing the phase changes.
224 Chapter 6 |
Physical Equilibria
Lumen
Stroma
ATPase
PS2
b6/f
PS1
FIGURE 6.20 The Fluid Mosaic Model of membrane structure applied to chloroplast thylakoids. Immersed
in the lipid bilayer of the internal membranes in the grana stacks (right side) are photosystem 1 (PS1) and
cytochrome b6/f/Rieske (b6/f) protein complexes. The external membranes in the grana stacks and the stroma
lamellae (left side) also contain photosystem 2 (PS2) and the phosphorylation coupling factor (ATPase). Each
complex has a particular orientation between the lumenal and the stromal faces of the thylakoid membrane.
EXAMPLE 6.8
A spherical cell 1 μm in diameter divides into two spherical cells. Calculate the work necessary to increase the surface area if its surface tension is 12.3 * 10-3 N m-1. Assume that
the volume of the daughter cells taken together is the same as that of the parent.
SOLUTION
The surface area of the parent cell is A 1 = pd2 = 3.14 * 10-12 m2. The volume of the parent cell is V 1 = pd3 >6 = 0.524 * 10-18 m3 and that of each
daughter cell is V2 = 0.262 * 10-18 m3. Rearranging the equations for the area
and surface of a sphere, we obtain A 2 = (36pV 22)1>3 = 1.980 * 10-12 m2 and
2A 2 = 3.960 * 10-12 m2 for the two daughter cells. Thus, the increase in surface
area associated with the cell division is ⌬A = 2A 2 - A 1 = 0.82 * 10-12 m2. The
work necessary to carry out this process is non-pV work, and is therefore given by
w = ⌬G = g⌬A = 12.3 * 10-3 N m-1 * 0.82 * 10-12 m2 = 1.0 * 10 - 14 J.
To put this in perspective, we may compare it with the free energy involved in the
hydrolysis of ATP under physiological conditions, for which ⌬G ⬵ 49 kJ>mol =
8.1 * 10-20 J>molecule. Thus, the energy from about 125,000 ATP molecules is
required to provide the work involved in dividing the new membrane associated with a
dividing cell.
Colligative Properties | 225
Colligative Properties
The normal boiling point of a liquid is the temperature at which the pure liquid is in
equilibrium with its vapor at 1 bar. The normal freezing point is the temperature at which
the liquid and solid are in equilibrium at 1 bar. The boiling point and freezing point of a
solution depend on the concentration of any solutes dissolved in the liquid. The solubility
of gases and solids in a solvent depends on the temperature and pressure. These properties
involve phase equilibria between gas and liquid or between solid and liquid. They can be
understood by straightforward thermodynamic analysis.
The osmotic pressure of a solution—the pressure that must be applied to a solution
to keep it in equilibrium with pure solvent—depends on the concentrations of the solutes
dissolved. If a solution is separated from pure solvent by a semipermeable membrane
(permeable to solvent but not to solutes), solvent will flow through the membrane into the
solution as long as equilibrium is not attained. This effect is vital in transporting water
from the roots up to the leaves of trees. It also causes rupture of red blood cells in contact
with pure water or with a solution that is too dilute (hypotonic).
The properties described above are called colligative because they are tied together by
a common origin. In dilute solution, they depend only on the number of solute molecules,
not on the kinds of solute molecules. This means that colligative properties can be used
to count the number of molecules present in a solution. We can easily prepare a solution
containing a known weight of solute molecules; thus, we can use a colligative property to
determine their molecular weight.
For sufficiently dilute solutions, the vapor pressure, boiling point, freezing point, and
osmotic pressure of a solution are linearly dependent on the concentration of solutes present. Furthermore, they are all closely related to one another. If one is measured, the others
can be calculated. For example, if the osmotic pressure of a solution is measured, the vapor
pressure of the solvent is immediately known. For dilute or ideal solutions, from the measured boiling-point rise, the freezing-point lowering can be calculated. We find that in ideal
solutions the colligative properties depend only on the number of solute molecules — the
mole fraction. We have already seen how the mole fraction of solute determines the vapor
pressure of an ideal solution, via Raoult’s Law. So if we measure the molar concentration of
the solute in solution, we can calculate any colligative property. The effective link between
the measured colligative properties is the activity of the solvent. For a real solution, we can
obtain the activity of the solvent in the solution from one measured colligative property.
Then any other colligative property at the same temperature can be calculated.
Boiling-Point Elevation and Freezing-Point Depression
When a small amount of solute is added to a solvent, the boiling point of the solution is
raised, and the freezing point of the solution is lowered, relative to those of the pure solvent.
The changes in the boiling point and the freezing point are found to be directly proportional
to the concentration of the solute in dilute solution:
Tf⬚ - Tf = Kf m
(6.42)
T b - T b⬚ = K bm ,
(6.43)
where Tf⬚ and Tb⬚ are the (pressure dependent) freezing and boiling points of the pure solvent, Tf and Tb are the freezing and boiling points of the solution, Kf and Kb are freezingpoint depression and boiling-point elevation constants that depend only on the solvent,
and m is the molality of the solution. Eqs. 6.42 and 6.43 can be used to measure total
concentration of solutes in complicated mixtures of solutes in solution. For example, the
freezing point of seawater immediately gives a measure of the total number of moles of
ions and other solutes of all kinds present in the seawater. In using Eqs. 6.42 and 6.43,
226 Chapter 6 |
Physical Equilibria
we must remember that the solutions should be dilute and that only the pure solvent must
freeze or evaporate. The solutes should not be volatile at the boiling point or soluble in the
solid solvent at the freezing point for the simple equations to apply.
We have used three different concentration scales, so it is well to know the relations
among them. From the definition of mole fraction x, molality m, and molarity c, we can
convert from one to the other. Here we give the relations that apply to dilute solutions:
c = r⬚m ,
(6.44)
xB = MAm>(1 kg)
(6.45a)
where r⬚ is the density of the solvent,
(if MA, the solvent molecular mass, is expressed in kg>mol), and
xB = MAm>(1000 g)
(6.45b)
(if MA, the solvent molecular mass, is expressed in g>mol).
We note that for dilute aqueous solution
xB = m> (55.6 mol>kg) and c = m .
(6.46)
The activity of the solvent is the link that connects the seemingly unrelated colligative
properties. We use the solvent standard state for the activity of the solvent, so the chemical
potential of the standard state m⬚ is chosen to be that of the pure solvent.
At the normal freezing point, the chemical potential of the pure liquid and solid
(frozen) solvent are identical: m⬚1 = m⬚s or ⌬ fus m⬚ = m⬚1 - m⬚s = 0. ⌬ fus m⬚ depends on
temperature:
⌬ fus m⬚ = ⌬ fus H ⬚ - T ⌬ fus S ⬚
(6.47)
If we assume ⌬ fus H ⬚ and ⌬ fus S ⬚ are independent of temperature over small intervals,
then, because ⌬ fusS⬚ = ⌬ fus H ⬚/Tf⬚, this becomes
⌬ fus m⬚ = ⌬ fus H ⬚ - T⌬ fus H ⬚/Tf ⬚ = ⌬ fus H ⬚(Tf ⬚ - T)>Tf ⬚ .
(6.48)
When we add solute to the solution, the chemical potential is lowered:
m = m⬚1 + RT ln xA
(6.26)
At the freezing point of the solution, this must be equal to the chemical potential of
the solid at the lower temperature, which is given by
m⬚s = m⬚1 - ⌬ fus m⬚ = m⬚1 - ⌬ fusH⬚(Tf ⬚ - Tf)>Tf ⬚ .
(6.49)
Equating 6.26 and 6.48, and substituting Tf , the freezing point of the solution, for T,
we obtain
RTf ln xA = - ⌬ fus H⬚(Tf⬚ - Tf)>Tf ⬚ ,
(6.50)
which we can immediately rearrange to give
T f⬚ - T f = -
RT fT f ⬚ ln x A
.
⌬ fus H o
(6.51)
We can make a few more useful approximations. First, because we usually do the experiments under conditions where the lowering of freezing point is small, we can assume
Tf * Tf⬚ ⬵ (Tf⬚)2. Second, as xA S 1, ln xA S (xA -1) = -xB, and according to Eq. 6.45a,
x B = M Am/(1 kg). Inserting these approximations into Eq. 6.50, we obtain
T f⬚ - T f =
R(T f⬚)2M Am .
1 kg * ⌬ fusH o
(6.52a)
Colligative Properties | 227
The 1 kg is often omitted, but it is useful as a reminder to express molecular mass of solvent
in kg mol - 1. Alternatively, we can write
Tfo - Tf =
R(T f o)2M Am
,
1000 g * ⌬ fus H o
(6.52b)
where, of course, MA must be expressed in g mol - 1.
The derivation of the boiling point elevation expression follows very similar lines,
and yields
T b - T bo =
R(T bo)2M Am
1000 g * ⌬ vap H o
or T b - T bo =
R(T bo)2M Am ,
1 kg * ⌬ vap H o
(6.53)
depending, again, on what units are used for molecular mass.
These methods for measuring molecular weights are limited to temperatures near the
boiling point or freezing point of the solvent. As a consequence, they have limited applicability to purely biological systems, but in special cases they may be quite useful. A study of
the molal concentration of total solute particles in different cells of the kidney was made
by putting a kidney slice on the cold stage of a microscope. The freezing point determined
for each separate cell immediately gave the total molality of its contents.
Osmotic Pressure
(a)
Hospital patients have (on rare occasions) died because pure water was accidentally injected
into their veins. The osmotic pressure of the solution inside the blood cells is high enough
to cause them to break under these circumstances. Osmotic pressure is defined as the pressure that must be applied to a solution to keep its solvent in equilibrium with pure solvent
at the same temperature. If a solution is separated from a solvent by a semipermeable
membrane (such as the membrane of a red blood cell), solvent will flow into the solution
side until equilibrium is reached. At constant temperature, equilibrium can be attained by
increasing the pressure on the solution or by diluting the solution until it is essentially pure
solvent. A red blood cell placed in pure water or in a dilute salt solution will burst before
equilibrium is reached.
Osmotic pressure is a colligative property closely related to the vapor pressure, freezing point, and boiling point. The activity of the solvent is the parameter that is most easily
related to the osmotic pressure.
Adding a solute to a solution lowers the activity of the solvent; by raising the pressure
on the solution, the activity of the solvent is increased. Van’t Hoff found empirically that
for dilute solutions the osmotic pressure ⌸ is directly proportional to the concentration of
the solute. Because of the importance of osmotic pressure in biology, we will derive the
relevant equations.
Figure 6.21 shows the condition at equilibrium for a solution in contact with pure solvent through a semipermeable membrane that allows only solvent to pass. The equilibrium
occurs in an osmometer in which the extra pressure ⌸ necessary for equilibrium is produced by the hydrostatic pressure of a water column on the solution side. At equilibrium,
there will be no net flow of solvent across the membrane, so the chemical potential of the
pure solvent (phase 1) at pressure p must be equal to the chemical potential of the solvent
in solution (phase 2) at pressure p + ⌸, where ⌸ = rgh:
mA1(p) = mA2 (p + ⌸)
The temperature is constant throughout. If we let p = 1 bar, then mA1 = m⬚A. Because of
the presence of the solute in solution, the chemical potential of A due to osmotic pressure is
Solution
Solvent
Membrane
(b)
h
FIGURE 6.21 Schematic
diagram of an osmotic pressure
apparatus. h is the equilibrium
difference in height between
the solution and solvent
chambers.
228 Chapter 6 |
Physical Equilibria
decreased by an amount that can be calculated using Eq. 6.26, assuming an ideal solution:
mA2(p) = m⬚A + RT ln x A
⌬mosm = mA2(p) - m⬚A = RT ln x A
(6.54)
To compensate for this change and restore equilibrium with the solvent, we need to increase
the chemical potential of A by an equal amount by applying external pressure ⌸ to the
solution. Remembering that, at constant temperature,
p2
⌬m = m2 - m1 = 3 V m dp = V m(p2 - p1) ,
(6.55)
p1
we obtain, at ambient pressure p,
p
⌬mosm = RT ln x A = 3
V m,A dp = - V m,A⌸ .
(6.56)
p+⌸
Obviously this assumes that the pressure is low enough that the molar volume of solvent
is not affected. Equating 6.54 and 6.56:
- Vm,A⌸
(6.57)
ln xA =
RT
For dilute solutions, we can use the same approximations we did above for freezing point
depression:
nB
nB
ln xA ⬵ xA - 1 = - xB = ⬵(6.58)
nA
nA + nB
Then,
RT
⌸ =
n .
(6.59)
nAV m,A B
In dilute solutions, nAVm,A is essentially equal to V, the volume of solution (the volume of
solute is negligible). Therefore, for dilute solutions,
RT
⌸ =
n = cRT .
(6.60)
V B
Equation 6.60 is the relation that was first found empirically by van’t Hoff. These equations
state that the osmotic pressure of a solution is directly proportional to the concentration
of solute in dilute solution. Equation 6.60 is written to look like the ideal gas equation,
to make it easy to remember. However, ⌸ is not the pressure of the solute, but rather the
pressure that must be applied to the solution to keep solvent from flowing in through a
semipermeable membrane.
To gain an appreciation for the magnitude of osmotic pressure, note that, according to
Eq. 6.60, a solution with c = 0.5 mol L - 1 = 500 mol m-3 at 310 K will have an osmotic pressure of 500 mol m - 3 * 8.31447 J mol - 1 K - 1 * 310 K = 1.26 * 106 Pa or 12.9 bar. This
is typical of the osmotic concentration of the cytoplasm inside animal cells. When such cells
come into contact with pure water, the osmotic pressure may rupture the cell membranes. This
happens in the hemolysis of red blood cells when the blood serum is diluted with pure water.
Osmotic-pressure measurement is the most useful colligative method for determining
molecular weights. It is accurate for molecular weights approaching 1 million, and small molecule impurities, buffers, or salts do not interfere if suitable membranes are used. The molecular weight of only the solutes that do not pass the semipermeable membrane is obtained.
Equation 6.60 can be rewritten as
w
⌸ =
RT ,
(6.61)
M
-3
where w is the concentration of solute in kg m and M its molecular mass in kg mol-1.
Colligative Properties | 229
∏/c (bar m3 mol–1)
Positive deviations
from ideality
(usual situation)
RT
Ideal solution
Negative deviations
from ideality
(aggregation)
c (m3 mol–1)
Usually, a series of concentrations is measured, and extrapolation is made to zero concentration. A plot of ⌸ versus w gives RT>M as the intercept; the slope characterizes deviations
from ideality (see figure 6.22). A positive slope is usually found for these plots; a negative
slope indicates an increase in molecular weight with increasing concentration. This occurs
if the solute undergoes aggregation. For real ionic solutions, the relation ⌸ = cRT begins
to fail for concentrations greater than 0.01 M. Osmotic pressure, as well as the other colligative methods, is used to study aggregation in solution at higher concentration.
Both vapor pressure and osmotic pressure can be measured at any temperature. They
are therefore more useful methods than boiling-point or freezing-point measurements when
working with biological molecules.
Molecular-Weight Determination
The molecular weight is an important property of any molecule, but is particularly important for proteins, nucleic acids, and other macromolecules. The most accurate molecular
weights come from precise knowledge of the amino acid sequence of a protein or the base
sequence of a nucleic acid, or from mass spectrometry. Nevertheless, sequence information
is not informative about posttranslational processing or aggregation of proteins, and mass
spectrometry does not detect weak noncovalent interactions such as exist in oligomeric
proteins. These are detectable by measurements based on colligative properties.
Osmotic pressure is the most useful colligative property to use for macromolecules. In
addition to its high sensitivity (see example 6.9), it is relatively immune to low-molecular
mass contaminants, which pass through the membrane, equilibrate, and therefore do not
contribute to the osmotic pressure. With other colligative methods, 1 mmol = 0.182 g of a
glucose contaminant has the same colligative effect as 20 g = 1 mmol of a 20 kD protein!
One caveat is that osmotic pressure measurements must be carried out at the pH at which
the protein is neutral (the pI). If the protein is either positively or negatively charged, charge
neutrality of the solution will require that it be accompanied by some number of lowmolecular mass counterions such as chloride or sodium, each of which has, per molecule,
the same effect on the osmotic pressure as the protein itself!
EXAMPLE 6.9
At 298 K, a solution of 0.5 g of bovine pancreatic ribonuclease A in 100 mL of buffer
solution of density 1.020 g mL-1 gives difference of height h = 9.23 cm in an osmotic
pressure apparatus similar to that shown in figure 6.21. What is the molecular weight of
ribonuclease A?
FIGURE 6.22 Dependence of
osmotic pressure on weight
concentration.
230 Chapter 6 |
Physical Equilibria
SOLUTION
The hydrostatic pressure is given by
⌸ = rgh = 1020 kg m - 3 * 9.802 m s - 2 * 0.0923 m = 922 Pa. With M = cRT>⌸
and w = 5 kg m - 3, we obtain M = 13.4 kg mol - 1, or 13.4 kD, or 13,400 g mol - 1.
Note that there is a substantial rise in height in the apparatus for a very modest
concentration of a moderately-sized protein.
Summary
Exact Clapeyron Equation
a
Tw ⌬ wVm
0T
b =
0p w
⌬ wH m
(6.8)
Gibbs Phase Rule
F = C - P + 2,
where F is the number of degrees of freedom, C the number of components, and P the number of phases.
Three Forms of the Clausius-Clapeyron Equation
ln pvap = ln a
ln a
pvap
pvap⬚
b = -
pvap,2
pvap,1
⌬ vapHm
b = -
RT
⌬ vapS m
+
R
⌬ v Hm 1
1
a
b
R
T2
T1
1⌬ vapH ⬚ - ⌬ vapCp⬚T ⬚2 1
⌬ vapCp⬚
1
T
a b +
ln a b ,
R
T
T⬚
R
T⬚
(6.15)
(6.16)
(6.19)
where ⌬ vapHm and ⌬ vap S m are the enthalpy and entropy of vaporization, and pvap is the vapor pressure. In Eq. 6.19 ⌬ vapCp⬚ = Cp⬚
= Cp⬚ (gas) - Cp⬚ (liquid), the difference between the gas and liquid heat capacities, at a standard temperature T⬚, ⌬ vapH⬚ is the
enthalpy of vaporization at that temperature, and pvap⬚ is the vapor pressure at that temperature.
Raoult’s Law
pvap
Lim a
b = xA ,
S
x A 1 pvap⬚
(6.23)
where pvap is the vapor pressure of the solvent in the solution, pvap° is the vapor pressure of the pure solvent, and xA the mole fraction
of solvent in the solution.
Solvent Chemical Potential in a Solution Obeying Raoult’s Law
mA = mA⬚ + RT ln x A ,
(6.26)
whereμA° is the chemical potential of the pure liquid and xA its mole fraction.
Solvent Chemical Potential in a Solution not Obeying Raoult’s Law
mA = mA⬚ + RT ln axA ,
(6.27)
where axA = gAxA, where gA is the activity coefficient.
Henry’s Law
pvap,B = k Bx B ,
where pvap,B is the vapor pressure of the solute, xB is its mole fraction, and kB is the Henry’s Law coefficient.
(6.30)
References | 231
Partition Coefficient
aB2 >aB1 = exp[ - (⌬m⬚B)>RT]
(6.32)
aB2 >aB1 = exp[ - (m⬚B2 - m⬚B1)>RT] = exp[ - (⌬m⬚B)>RT]
(6.32)
The ratio of activities aB2>aB1 is called the partition coefficient.
Gibbs Adsorption Isotherm
⌫ = -
1 dg
1 dg
⬵ RT d ln a
RT d ln c
(6.36)
⌫ is the surface excess concentration in mol m - 2; g is the surface tension in N m - 1; R is the gas constant; a is the activity of the
solute in bulk solution; and c is the concentration of the solute in bulk solution, relative to some standard state concentration c⬚.
Standard Chemical Potential Change on Solution of a Weakly Soluble Species
⌬ sm⬚ = - RT ln x B ,
(6.39)
where xB is the mole fraction of solute.
Depression of Freezing Point
R(T fo)2M Am
,
(6.52a)
1 kg * ⌬ fusH o
where MA, the molecular mass of the solvent, is expressed in kg mol - 1, m is the solute molality, Tf is the freezing point of the solution, Tf⬚ is the freezing point of the pure solvent, and ⌬ fusH⬚ its enthalpy of fusion at Tf°:
T fo - T f =
R(T fo)2M Am
,
1000 g * ⌬ fusH o
where MA, the molecular mass of the solvent, is expressed in g mol-1.
T fo - T f =
(6.52b)
Elevation of Boiling Point
T b - T bo =
R(T bo)2M Am
R(T bo)2M Am
o
,
or
T
T
=
b
b
1000 g * ⌬ vapH o
1 kg * ⌬ vapH o
(6.53)
where MA, the molecular mass of the solvent, is expressed in kg mol-1 (first equation) or g mol-1 (second equation), m is the solute
molality, Tb is the boiling point of the solution, Tf° is the boiling point of the pure solvent, and ⌬ vap H⬚ its enthalpy of vaporization at Tb⬚.
Osmotic Pressure
⌸ = cRT ,
and ⌸ is the osmotic pressure.
w
⌸ =
RT ,
M
where w is the concentration of solute in kg m-3 and M its molecular mass in kg mol-1.
where c is the concentration in mol
(6.60)
m-3
(6.61)
Mathematics Needed for Chapter 6
Review the power series expansion of ln x as x S 1!
References
1. Branton, D., and D. Deamer. 1972. Membrane Structure.
New York: Springer-Verlag.
3. Hille, B. 1992. Ionic Channels of Excitable Membranes,
2d ed. Sunderland, MA: Sinauer.
2. Gennis, R. D. 1989. Biomembranes: Molecular Structure
and Function. New York: Springer-Verlag.
4. Koshland, D. L. Jr., G. Nemethy, and D. Filmer, 1966.
Comparison of Experimental Binding Data and Theoretical
Models in Proteins Containing Subunits. Biochemistry
5:365–385.
232 Chapter 6 |
Physical Equilibria
5. Nicholls, D. G. 1982. Bioenergetics, an Introduction to
Chemiosmotic Theory. New York: Academic Press.
7. Tanford, C. 1980. The Hydrophobic Effect: Formation of
Micelles and Biological Membranes, 2d ed. New York: Wiley.
6. Silver, B. L. 1985. The Physical Chemistry of Membranes,
an Introduction to the Structure and Dynamics of
Membranes. Boston: Allen & Unwin.
8. Wyman, J., and S. J. Gill, 1990. Binding and Linkage,
Functional Chemistry of Biological Macromolecules. Mill
Valley, CA: University Science Books.
Suggested Reading
Dickerson, R. E., and I. Geis. 1983. Hemoglobin: Structure,
Function, Evolution, and Pathology. New York: Benjamin>
Cummings.
Linder, M. E., and A. G. Gilman. 1992. G Proteins. Sci. Am. 267:56–65. (G proteins are membrane-bound proteins that cause
cellular responses to chemical messengers.)
Phillips, J. L. 1998. The Bends: Compressed Air in the History
of Science, Diving and Engineering. New Haven, CT: Yale
University Press.
Singer, S. J., and G. L. Nicholson. 1972. The Fluid Mosaic Model
of the Structure of Cell Membranes. Science 175:720–731.
Ward, P., L. Greenwald, and O. E. Greenwald. 1980. The Buoyancy
of the Chambered Nautilus. Sci. Am. 243:190–203.
Wolfenden, R. 1983. Waterlogged Molecules. Science 222:
1087–1093.
Internet
White, S. 1999. Membrane Proteins and Bilayers. [Biophysical
Society, Online Biophysics Textbook.]
Problems
1. a. How many kilograms of water can be evaporated from
a 1 m2 surface of water in a lake on a hot, clear summer
day, if it is assumed that the limiting factor is the Sun’s
heat of 1800 J s-1 m-2 for an 8-hour day? Assume that
the air and water temperatures remain constant at 40⬚C.
b. What volume of air is needed to hold this much water
vapor at 40⬚C?
c. When the Sun sets, the air temperature drops to 20⬚C .
Assuming that the excess water vapor is precipitated as
rain, calculate the weight of the rain and the amount of
heat released when the water condenses. When it starts
raining, do you expect the temperature of the air to rise
or fall? Explain.
d. Calculate the vapor pressure of liquid water at 200⬚C.
Assume that the heat of vaporization does not change
from its value at 100⬚C. (Table 2.2 is useful for this
problem.)
2. The vapor pressure of pure, solid pyrene at 25⬚C is 91.4 μPa.
The solubility at 25⬚C of pyrene in water is 1.00 * 10-4 M.
The solubility of pyrene in ethanol at 25⬚C is greater than
10-4 M. Neither water nor ethanol dissolves significantly in
pyrene.
a. What is the vapor pressure of pyrene above a saturated
solution of pyrene in water at 25⬚C?
b. If ethanol is added to the saturated aqueous solution,
more pyrene dissolves. When equilibrium with the solid
pyrene is reached again, will the vapor pressure of the
pyrene above the new solution be greater than, less than,
or the same as in part (a)?
c. The solubility of pyrene at 10-4 M is 1.10 * 10-3 M in an
aqueous solution of 0.10 M cytosine. This increase in solubility is due to the formation of a complex between pyrene
and cytosine. Calculate the values of the equilibrium constants for the following reactions at 25⬚C, where P = pyrene
and C = cytosine. Assume activities equal concentrations.
P(solution) + C(solution) m P.C(solution)
P(solid) + C(solution) m P.C(solution)
3. a. Which of the following has the lowest standard chemical
potential at 25⬚C: CH4(s), CH4(l), CH4(g), or CH4(aq)?
Explain your answer.
b. Calculate the solubility in mol L-1 for CH4 in water at
25⬚C when the CH4 gas pressure is 1 bar.
c. Will the solubility be greater or less at 0°C? Explain
your answer.
d. Henry’s Law constants for nearly all of the gases listed
in table 6.1 increase with increasing temperature. What
can you conclude from this about an important thermodynamic property (for example, enthalpy, heat capacity,
entropy, or free energy) associated with the dissolution
of gases in water?
4. The standard free energy for the reaction of oxygen binding
to myoglobin
Mb + O2(g) m MbO2
is ⌬G⬚ = - 30.0 KJ mol - 1 at 298 K and pH 7. The standard
state of O2 is the dilute solution, molarity scale; therefore
the concentration of O2 must be in M.
Problems | 233
What is the ratio MbO2/Mb in an aqueous solution at equilibrium with a partial pressure of oxygen pO2 = 400 Pa? Assume
ideal behavior of O2 gas and for the protein in solution.
5. The binding of a ligand A to a macromolecule was studied by
equilibrium dialysis. In each measurement a 1.00 * 10 - 6 M
solution of the macromolecule was dialyzed against an
excess amount of a solution containing A. After equilibrium
was reached, the total concentration of A on each side of the
dialysis membrane was measured. The following data were
obtained:
Total concentration of A (μM)
The side without
macromolecules
The side with
macromolecules
5.1
6.7
10.2
12.8
20.1
23.4
52.2
56.2
105.0
109.1
200.0
204.7
Using these data, calculate n for each value of [A]. Make
a Scatchard plot and evaluate the intrinsic equilibrium constant K and the total number of sites per macromolecule,
if the independent-and-identical-sites model appears to be
applicable.
6. Plot v/[A] versus n for the following cases:
a. There is a total of ten identical and independent sites
per polymer and the intrinsic binding constant is
K = 5 * 105 M - 1.
b. There is a total of ten independent sites per polymer.
Nine of the ten sites are identical, with 5.0 * 105 M - 1.
The binding constant for the tenth site is 5.0 * 106 M - 1.
For more than one type of site, the Scatchard equation
can be generalized to
n
3A 4
rium dialysis at the same temperature. In one run, after the
binding equilibrium was established, the concentration of
free ATP outside the bag was found to be 1.00 * 10 - 4 M,
and the total ATP concentration inside the bag was found
to be 3.00 * 10 - 4 M.
a. On the basis of the information above, calculate K,
the equilibrium constant for the binding of ATP to the
enzyme at 20⬚C.
b. The equilibrium dialysis experiment was repeated for
a series of outside ATP concentrations at two different
temperatures (20⬚C and 37⬚C). The following Scatchard
plots were obtained:
NiKi
= a
.
i 1 + Ki 3 A 4
[Note that the deviation from linearity is small for part
(b). Because experimental error in such measurements
is usually appreciable, be cautious in concluding from a
linear Scatchard plot that the sites are identical.]
c. Repeat the calculation and plot of part (b) for five identical sites of each type.
7. The enzyme tetrahydrofolate synthetase has four identical
and independent binding sites for its substrate ATP. For an
equilibrium dialysis experiment, a solution of the enzyme
was prepared. First the osmotic pressure of this solution was
measured to be 2.4 * 10 - 3 bar at 20⬚C using an osmometer.
Then the enzyme solution was placed in a dialysis bag, and
the binding of ATP to the enzyme was measured by equilib-
= T(20°C)
= T(37°C)
ν
[ATP]
ν
Using this information, calculate ⌬H ⬚ and ⌬S⬚ for the binding of ATP to its site on the enzyme. (Assume that ⌬H ⬚ and
⌬S⬚are independent of temperature.)
8. a. Using Henry’s Law constants in table 6.1, calculate
the solubility in M of each gas in water at 25⬚C if
PO2 = 0.2 bar, PN2 = 0.75 bar, and PCO2 = 0.05 bar.
b. What will be the vapor pressure at 25⬚C of the water in
this solution if Raoult’s Law holds? The vapor pressure
of pure water at 25°C is 3167 Pa.
9. a. Assume that Raoult’s Law holds and calculate the boiling
point of a 2 M solution of urea in water.
b. Urea actually forms complexes in solution in which two
or more molecules of hydrogen bond to form dimers and
polymers. Will this effect tend to raise or lower the boiling point?
10. When cells of the skeletal vacuole of a frog were placed in
a series of aqueous NaCl solutions of different concentrations at 25⬚C, it was observed microscopically that the cells
remained unchanged in 0.7 wt% NaCl solution, shrank in
more concentrated solutions, and swelled in more dilute
solutions. Water freezes from the 0.7% salt solution at
-0.406°C.
a. What is the osmotic pressure of the cell cytoplasm relative to that of pure water at 25⬚C?
b. Suppose that sucrose (M = 342 g mol - 1) was used
instead of NaCl (M = 58.5 g mol - 1) to make the isoosmotic solution. Estimate the concentration (wt%) of
sucrose that would be sufficient to balance the osmotic
pressure of the cytoplasm of the cell. Assume that
sucrose solutions behave ideally.
234 Chapter 6 |
Physical Equilibria
11. On the planet Taurus II, ammonia plays a role similar to that
of water on Earth. Ammonia has the following properties:
Normal (1 bar) boiling point is -33.4°C where its enthalpy
of vaporization is 23.34 kJ mol-1. Normal freezing point is
-77.7°C.
a. Estimate the temperature at which the vapor pressure of
NH3 (l) is 8 kPa.
b. Calculate the entropy of vaporization of NH3 at its normal
boiling point. Compare this value with that predicted by
Trouton’s rule (ΔvapS° = 88 J mol-1 K-1) for unassociated liquids. Give a likely reason for any discrepancy.
c. Ammonia undergoes self-dissociation according to the
reaction
15.
2 NH3 K NH4+ (am) + NH2-(am) ,
where (am) refers to the “ammoniated” ions in solution.
At -50°C, the equilibrium constant for this reaction is
K = [NH4+][NH2-] = 10-30 .
Calculate Δμ° for the self-dissociation.
d. When 1 mol of ammonium chloride, NH4Cl, is dissolved
in 1 kg of liquid ammonia, the boiling point at 1 bar
is observed to occur at -32.7°C. What conclusions can
you reach about the nature of this solution? State the
evidence for your answer.
12. One of the important factors responsible for geologic evolution (erosion in particular) is the pressure developed by the
freezing of water trapped in enclosed spaces in rock formations. Estimate the maximum pressure that can be developed
by water freezing to ice on a cold night (T = -10°C). The
densities of ice and water at -10°C may be taken as 0.92
and 1.00 g mL-1, respectively.
13. Proteins can be hydrolyzed to amino acids by heating in
very dilute aqueous NaOH solution. The protein itself is also
very dilute. The reaction is carried out at 120°C by heating
the solution in a sealed tube that was evacuated prior to sealing. What internal pressure must the tube withstand at this
temperature? The normal boiling point and ΔvapH° of water
may be taken as 100°C and 40.67 kJ mol-1, respectively.
14. Indicate whether each of the following statements is true or
false as it is written. Alter each false statement to make it
correct.
a. If two aqueous solutions containing different nonvolatile
solutes exhibit exactly the same vapor pressure at the
same temperature, the activities of water are identical in
the two solutions.
b. The most important reason that foods cook more rapidly
in a pressure cooker is that equilibrium is shifted in the
direction of products at the higher pressure.
c. If solutions of a single solute are prepared at equal molalities but in different solvents, the freezing-point lowering
will be the same for the different solutions, assuming that
each behaves ideally.
16.
17.
18.
d. If solutions of a single solute are prepared at equal
mole fractions but in different solvents, the fractional
vapor-pressure lowering will be the same for the different solutions, assuming that each behaves ideally.
e. If two liquids, such as benzene and water, are not completely miscible in one another, the mixture can never
be at equilibrium.
A solution of the sugar mannitol (mol wt 182.2) is prepared
by adding 54.66 g of mannitol to 1000 g of water. The vapor
pressure of pure liquid water is 2338 Pa at 20°C. Mannitol
is nonvolatile and does not ionize in aqueous solutions.
a. Assuming that aqueous mannitol solutions behave
ideally, calculate the vapor-pressure lowering (the difference between the vapor pressure of pure water and
that of the solution) for the above solution at 20°C.
b. The observed vapor-pressure lowering of the mannitol
solution above is 12.40 Pa. Calculate the activity coefficient (based on mole fraction) of water in this solution.
c. Calculate the osmotic pressure of the mannitol solution
of part (b) when it is measured against pure water and
compare it with the osmotic pressure of the ideal solution.
a. Assuming that glucose and water form an ideal solution,
what is the equilibrium vapor pressure at 20°C of a solution of 1.00 g of glucose (mol wt 180 g mol-1) in 100 g
of water? The vapor pressure of pure water is 2338 Pa at
20°C.
b. What is the osmotic pressure, in Pa, of the solution in
part (a) versus pure water?
c. What is the activity of water as a solvent in such a
solution?
d. What would be the osmotic pressure versus pure water of
a solution containing both 1.00 g of glucose and 1.00 g of
sucrose (mol wt 342) in 100 g of water at 20°C?
Workers in underwater diving suits necessarily breathe
air at greater than normal pressure. If they are returned to
the surface too rapidly, N2, dissolved in their blood at the
previously higher pressure, comes out of solution and may
cause emboli (gas bubbles in the bloodstream), bends, and
decompression sickness. Blood at 37°C and 1 bar pressure
dissolves 1.3 mL of N2 gas (measured for pure N2 at 37°C
and 1 bar) in 100 mL of blood. Calculate the volume of N2
likely to be liberated from the blood of a diver returned to
1 bar pressure after prolonged exposure to air pressure at
300 m of water (below the surface). The total blood volume
of the average adult is 4.7 L; air contains 78 vol% N2.
Hexachlorobenzene, C6Cl6, is a solid compound that is
somewhat volatile but not very soluble in water; water is not
soluble in C6Cl6. Solid C6Cl6 and liquid water are allowed to
reach equilibrium at 20°C to form a solid phase, an aqueous
phase, and a gaseous phase. The vapor pressure of pure solid
C6Cl6 is 1.45 mPa at 20°C and 8.85 mPa at 40°C.
a. A sample of the equilibrium aqueous phase was removed
and the equilibrium concentration of C6Cl6 was found
Problems | 235
to be mole fraction xB = 3 * 10-10. What are the vapor
pressures of H2O and C6Cl6 for this phase at 20°C? The
vapor pressure of pure water is 2338 Pa at 20°C.
b. C6Cl6 binds to proteins. A protein is added to the
aqueous solution, and equilibrium with solid C6Cl6
is established. State whether each of the following
increases, decreases, or remains unchanged compared
to part (a):
1. The vapor pressure of C6Cl6 for the aqueous solution.
2. The vapor pressure of water for the aqueous solution.
3. The concentration of dissolved (free) C6Cl6 in the
aqueous solution.
c. What is the mole fraction of C6Cl6 in the gas phase at
equilibrium at 20°C?
d. Calculate the heat of vaporization of solid C6Cl6.
19. The solubilities of two amino acids in two solvents at 25°C
are given below; they are the concentrations present in saturated solutions.
In water
In ethanol
Glycine
3.09 M
4.04 * 10-4 M
Valine
0.6 M
1.32 * 10-3 M
a. Calculate the standard free energy of transfer Δμ° of
1 mol of glycine from the solid to the aqueous solution at 25°C The standard state for the solid is the pure
solid; the standard state in the solution corresponds to
a = 1 extrapolated from a dilute solution; the molarity
concentration scale is used. Consider the solutions to
be ideal.
b. Calculate the standard free energy of transfer Δμ° of
1 mol of glycine from ethanol to the aqueous solution
at 25°C.
c. Assume that the effects of the backbone and side chain
are simply additive (glycine essentially has no side
chain) and calculate the standard free energy of transfer
of 1 mol of the (CH3)2CH- side chain of valine from
water to ethanol at 25°C.
d. Ethanol can be considered to mimic the interior of a
protein. Will the mutation of a glycine to a valine in
the interior of a protein favor the folding of the protein
if interaction with the solvent is the dominant effect?
Explain.
e. Assume that the protein-folding equilibrium is a twostate transition: folded or unfolded. Calculate the change
in equilibrium constant due to the mutation from glycine
to valine for folding at 25°C, if interaction with the solvent is the dominant effect.
20. The human red-blood-cell membrane is freely permeable
to water but not at all to sucrose. The membrane forms a
completely closed bag, and it is found by trial and error that
adding the cells to a solution of a particular concentration of
sucrose results in neither swelling nor shrinking of the cells.
21.
22.
23.
24.
In a separate experiment, the freezing point of that particular
sucrose solution is found to be -0.56°C.
a. If the exterior of the cell contains only KCl solution, estimate the KCl concentration. Assume that the membrane
is also impermeable to KCl.
b. If these cells are suspended in distilled water at 0°C,
what would be the internal hydrostatic pressure at
equilibrium, assuming that the cell did not change
volume?
a. What is the osmotic pressure in bar of a 0.1 M solution
of urea in ethanol compared with that of pure ethanol
at 27°C? State what assumptions you make in the
calculation.
b. What is the osmotic pressure in bar of a 0.1 M solution
of NaCl in water compared with that of pure water
at 27°C? State what assumptions you make in the
calculation.
c. An aqueous solution of osmotic pressure 5 bar is
separated by a semipermeable membrane from an aqueous solution of osmotic pressure 2 bar. Which way will
the solvent flow? How large a pressure must be applied
to prevent this solvent flow?
d. What is the free-energy change for transferring 1 mol
of water from a solution of osmotic pressure 5 bar to a
solution of osmotic pressure 2 bar?
A protein solution containing 0.6 g of protein per 100 mL
of solution has an osmotic pressure of 22 mm of water
at 25°C and at a pH where the protein has no net charge
(the isoelectric point). What is the molecular weight of the
protein?
A climber has carefully measured the boiling point of water
at the top of a mountain and found it to be 95°C. Calculate
the atmospheric pressure at the mountaintop. The boiling
point of water at 1 bar pressure is 99.82°C and the heat
absorbed in vaporizing 1 mol of water at constant pressure
is 40.67 kJ mol-1.
The water in freshwater lakes is supplied by transfer from
the oceans. Consider the oceans to be a 0.5 M NaCl solution
and freshwater lakes to be a 0.005 M MgCl2 solution. For
simplicity, consider the salts to be completely dissociated
and the solutions sufficiently dilute so that all the equations
for dilute solutions apply.
a. Calculate the osmotic pressure of the ocean water and of
the lake water at 25°C relative to pure water.
b. Give a simple relation between the osmotic pressure and
the freezing-point depression of a solution by which you
could calculate ΔfT without knowing the freezing-point
depression constant for the solvent.
c. How much free energy is required to transfer 1 mol of
pure water from the ocean to the lake at 25°C. (In the
actual process, the sun supplies this energy.)
d. Which solution, the ocean or the lake, has the higher
vapor pressure of water?
236 Chapter 6 |
Physical Equilibria
e. The observed water vapor pressure at 99.64°C for 0.5 M
NaCl is 0.9853 bar. What is the activity of water at
this temperature? The vapor pressure of pure water at
99.64°C is 1 bar.
25. For each of the following processes, state whether ΔG,
ΔH, and ΔS of the system increase, decrease, or do not
change. If there is not enough information given, state what
experimental measurements need to be made to allow you
to deduce the direction of the change.
a. A system contains a 1 M sucrose solution and pure
water separated by a membrane permeable to water
only. A small amount of water is transferred through the
membrane from the pure water to the sucrose solution at
constant temperature and pressure.
b. Liquid water at 99.64°C, 1 bar pressure is evaporated
to water vapor at 99.64°C, 1 bar pressure. The system
is the water.
c. A system contains pure water ice at 0°C suspended in a
0.01 M KCl solution at the same temperature. The pressure on the system is 1 bar. The system is the solid water
plus the KCl solution.
d. Water vapor at 1 bar pressure and 25°C is converted to
liquid water at the same temperature and pressure along
a reversible path. The system is the water.
26. An aqueous solution containing 0.1 g L-1 of pyruvic acid is
separated from pure H2O by a membrane semipermeable to
water, all at 25°C.
a. Calculate the osmotic pressure that you expect for this
solution. State any important assumptions that you
make.
b. The osmotic pressure observed is 4920 Pa. Use this value
to estimate the pK for the acid dissociation of pyruvic
acid at 25°C.
c. What fractional vapor-pressure lowering relative to pure
water do you expect for the solution of pyruvic acid at
80°C?
27. Which of the following has the largest (most positive) value
for the
a. molar entropy increase?
H2O (l, 0°C) S H2O (l, 100°C)
H2O (l, 0°C) S H2O (l, 0°C)
H2O (l, 100°C) S H2O (g, 1 bar, 100°C)
H2O (g, 1 bar, 100°C) S H2O (g, 0.1 bar, 100°C)
b. free energy of transfer of 1 mol of the substance from
inside to outside a cell, all at 37°C?
NaCl (a = 0.010 inside) S NaCl (a = 0.014 inside)
KCl (a = 0.100 inside) S KCl (a = 0.005 inside)
glucose (a = 0.10 inside) S glucose (a = 0.20 inside)
NaCl (a = 0.10 inside) S NaCl (a = 0.20 inside)
28. A 1.0 M sucrose solution and a 0.5 M sucrose solution
(both in water) in separate open vessels are placed next
to each other in a sealed container at constant temperature. What, if anything, will occur? Explain your answer
briefly.
29. The measured osmotic pressure of seawater is 24.8 bar at
273 K.
a. What is the activity of water in seawater at 273 K? Take
the partial molar volume of water in seawater to be
18.0 mL mol-1.
b. The vapor pressure of pure water at 273 K is 4.6 torr.
What is the vapor pressure of water in seawater at this
temperature?
c. What temperature would you expect to find for seawater
in equilibrium with the polar ice caps? The melting point
of pure water at 1 bar pressure is 0°C and Δf H° may be
taken as 6.0 kJ mol-1 independent of temperature.
30. The following device is a candidate for a perpetual motion
machine:
turbine
blade
pA
pA°
solution
solvent
1. All parts of the machine are at the same temperature T equal to that of the heat bath surrounding the
machine.
2. The compartment at the left contains a solution, separated from the compartment at the right (containing
pure solvent) by a semipermeable membrane, permitting solvent transfer only. A pressure difference Δp =
rgh balances the osmotic pressure difference ⌸ across
the membrane.
3. The vapor pressure pA° above pure solvent is higher
than the pressure pA above the solution (Raoult’s Law).
There is therefore a pressure difference Δp = pA° -pA in
the tube, and the flow of vapor in the upper connecting
tube, from right to left, drives a turbine blade.
4. Vapor moving through the connecting tube condenses at the surface of the solution. It tends to dilute
the solution, therefore reducing its osmotic pressure.
This means that solvent will flow back through the
semipermeable membrane, restoring the original concentration of the solution.
Problems | 237
5. The solvent is therefore set into perpetual motion,
flowing as vapor from right to left in the upper
part and from left to right in the lower part,
yielding energy.
Analyze the operation of this machine carefully. First, decide
whether this machine works or fails. Then defend your
answer as quantitatively and clearly as possible.
31. Most phospholipids have a gel-liquid crystal phase transition;
the integrity of biological membranes depends on the lipid
being in the liquid crystal phase. It has been suggested
that one function of cholesterol (its other function being
to scare middle-aged men) is to lower the temperature of
the gel-liquid crystalline temperature, so that the membrane
remains liquid crystalline.
Dipalmitoyl phosphatidylcholine has a transition
temperature of 41.55°C and a transition enthalpy of 36.2
kJ>mole. Its molecular weight is 734 g>mol. How much will a
5% w/w concentration of cholesterol (m.w. 386.7 g>mol)
depress its phase transition temperature, assuming the
solution to be ideal?
Chapter 7
Electrochemistry
“There is a powerful agent, obedient, rapid, easy, which conforms to every use, and reigns
supreme on board my vessel. Everything is done by means of it. It lights, warms, and is
the soul of my mechanical apparatus. This agent is electricity.”
— Jules Verne, 1870, in Twenty Thousand Leagues under the Sea, translated by
Gerard Harbison
Concepts
The first clear human records of bioelectricity are inscribed on 5th dynasty tombs in
Egypt, dating from about 2400 BCE. These carvings show depictions of the electric
catfish, Malaptererus electricus, native to the Nile. Our first encounter with bioelectricity, therefore, was likely painful; these fish can deliver jolts of up to 350 V. Somewhat
later, mysterious earthen jars found in middle-eastern archaeological sites may have been
the first chemical batteries, possibly used to do electroplating. However, systematic study
of the interaction of electric currents with biological organisms had to wait until the
18th century and the famous experiments of Luigi Galvani, who showed that electric shocks
applied to frog legs caused the muscles to contract, and inspired Mary Shelley’s novel
Frankenstein, and later an entire genre of horror movies.
Galvani’s compatriot, Alessandro Volta, was one of the first to appreciate that, while
biological organisms generate and are influenced by electricity, electric phenomena are not
exclusive to life, and can be generated by an apparatus as simple as two disks of different
metal in contact with each other and with an ionic solution. These simple electrochemical
cells possess a characteristic potential. Under standard conditions, the electrochemical
potential is an unvarying property of a specific pair of metals or other chemical substances,
one of which is oxidized and one reduced, and it is proportional to the standard free energy
of the redox (reduction > oxidation) reaction. There is in fact an identity between the electrical work done by an electrochemical cell and the free energy.
The Nernst equation is the second link between thermodynamics and electricity, and
relates the concentrations of chemical compounds at equilibrium and their electrochemical
potential under nonstandard conditions. This link underpins the generation of bioelectrical
potentials; cells pump ions across cell membranes, leading to electrical potentials, which
are in turn sustained by equilibria with ion concentration gradients across these membranes.
Biological organisms do not merely do chemistry; they also do electrochemistry.
Applications
238
The large voltages electric eels and catfish produce are generated by stacking literally
thousands of cell membranes, each of which individually creates a potential of the order
of 100 mV. The spectacular (one might even say shocking) voltages these electric fishes
create are a bizarre evolutionary adaptation of a much more general phenomenon: the
Basic Electricity | 239
membrane potential. All cells carry membrane potentials. These potentials are the foundation of nerve transmission, and also drive the transport of most chemical compounds
across biological membranes.
Frogs, seaweed, and other organisms that live in contact with water have semipermeable skins. Water and some ions and small molecules pass through the skins; macromolecules generally do not. The frog or the seaweed can selectively concentrate certain
molecules inside and selectively exclude or excrete other molecules. How do they do it?
If a molecule can easily pass through the skin, how can the inside concentration be maintained at a value that is different from the outside concentration, and still be consistent
with thermodynamics? The answer is to ensure that the free-energy change for transporting the molecule inside is negative. For example, the presence of a protein inside seaweed
that strongly binds the iodide ion ensures that the iodide concentration in the seaweed is
always higher than in the seawater. If the concentration of free iodide is the same inside
and outside, the bound iodide would account for the concentrating effect of the seaweed.
This effect, known as passive transport, does not depend on whether the seaweed is alive
or dead. Similarly, metabolism is not involved in the transport of ligands like O2 through
the walls of the alveoli to bind to macromolecules like hemoglobin in the blood.
Cells, however, also perform active transport. A number of different kinds of ion
pumps, of which the most important is the sodium > potassium ATPase, actively transport
ions across cell membranes, using the chemical energy provided by ATP hydrolysis to overcome unfavorable electrical and chemical potential gradients. The negative potential across
the cell membrane allows positive ions such as Ca2+ and Mg2+ to accumulate by simple
equilibration; it can also drive the transport of other, sometimes uncharged molecules using
specialized membrane proteins such as symporters and antiporters.
The electrical potential itself is exquisitely sensitive to the movement of small numbers
of ions, and can under the right circumstance change rapidly, allowing the phenomenon of
nerve conduction. It can also be brought into and out of equilibrium with much more robust
ion concentration gradients, allowing long-term maintenance of stable cell potentials.
Even more fundamentally, the processes of respiration and photosynthesis drive hydrogen ions across the cell membranes of bacteria and the inner membranes of mitochondria
and chloroplasts. This gradient of potential and of hydrogen ion concentration is, in turn,
used to drive the energetically unfavorable synthesis of ATP from ADP and an inorganic
phosphate. Thus the transduction of energy in all but a few organisms uses electrical
transmembrane potentials as intermediates.
Finally, humans have learned to use the Nernst equation to build sensors to convert
chemical concentrations of oxidizable or reducible molecules, such as glucose, into electrical
potentials, and thus build sensitive chemical sensors that allow rapid and reliable measurements of concentration of single chemical compounds in chemically complex mixtures such
as blood; this is the basis of the glucose meters used in management of diabetes. Development
of electrochemical sensors is an important contemporary area of applied chemical research.
Basic Electricity
The base electrical SI unit is the ampere or amp (A), which is technically defined as the
electrical current which, passing through two infinitely long parallel wires a meter apart,
gives a force between the wires of 2 * 10-7 N per meter of length. The amp is really
not a fundamental unit, however; rather, it corresponds to a current or flow of electrical
charge Q, which can be expressed as a differential:
I =
dQ
.
dt
The SI unit of electrical charge is the coulomb (C): 1 A = 1 C s-1.
(7.1)
240 Chapter 7 |
Electrochemistry
The smallest known quantity of charge on a free particle is the charge on the electron,
which is often given the special abbreviation e. This charge has been measured to high
accuracy; the best current estimate is 1.60217656 * 10-19 C. Even though the electron is
a negatively charged particle, e is defined as a positive quantity; the charge on an electron,
strictly speaking, is therefore -e. Ions have charges Z which are integer multiples of e; the
charge on a Mg2+ ion, for example, is Ze = 2e.
What about a mole of ions, instead of 1 ion? If a single ion has a charge of e (in other
words, Z = 1), the charge of a mole of ions, Qm = NAe = 96485.3 C mol-1. This quantity
of charge is given a special designation, the Faraday constant, or F.
Just as moving a mass in the Earth’s gravity requires work, so does moving charge in
an electrical potential. The SI unit of electrical potential is the volt (V). 1 joule of work
is required to bring a charge of one coulomb from an infinitely distant point (which is
assumed to have a potential of zero) to a point with an electrical potential of 1 V. In other
words, wE = -QV. Electrical potential is a state variable, as is charge, and therefore
electrical work wE, unlike mechanical work, is a state variable.
We can also define electric power P as electrical work done per second:
dw E
.
(7.2)
dt
Combining this with wE = QV and Eq. 7.1, we obtain P = VI. The SI unit of power is
the watt (W).
P=
Capacitance and Electrical Neutrality
Just as a mass carries with it its own gravitational field, so a charge creates an electrical
potential around it. For example, walking across a carpet in your socks rubs some electrons
onto you from the carpet, or onto the carpet from you, so that you become charged up to a
potential of perhaps thousands of volts, enough to cause sparks when you touch an object
at a different potential and to give you a shock. It is truly ‘shocking’ how few electrons it
takes to give you a potential of thousands of volts.
The quantity that determines the potential on an object per unit of charge it carries is the
capacitance (C ): C = Q>V. The SI unit of capacitance is the farad (F); as you might expect,
1 F = 1 C>1 V. Calculating capacitances of real objects is beyond the scope of this book,
but a simple formula gives the capacitance (actually, the self-capacitance) of a conducting
sphere: C = 4pe0r, where r is the radius of the sphere, and e0 is the electric constant:
e0 = 8.854187817620 * 10-12 F m-1. A little math indicates that the capacitance of a liter
of a solution of an electrolyte in water, in a round flask that occupies a sphere of radius
0.062 m, is 6.9 * 10-12 F. Therefore, unless they are specially designed to carry large
quantities of charge, the capacitances of real objects are tiny!
To charge that solution to 1 V requires 6.9 * 10-12 C of electrons or ions.
But 6.9 * 10-12 C corresponds to (6.9 * 10-12 C)>F mol of electrons or ions, or
7.2 * 10-17 mol, somewhat over 43,000,000 particles. A few tens of millions of electrons
or ions, in other words, is enough to bring a 1 L solution to a potential of 1 V.
So, if we have a 1 L salt solution that contains 1 mol of chloride ions and it contains
1 + (7.2 * 10-17) mol of sodium ions, that solution will have a potential of approximately
+1 V. This is a paradox which often confuses even quite advanced students. In electrochemistry, we always assume there are equal quantities of positive and negative ions (or
more correctly, equal positive and negative ionic charges) in a solution. But if the quantities
are exactly equal, the solution should be neutral! The resolution of the paradox is that the
excess of one species of ion, positive or negative, needed to create a potential of the order
of a few volts, is unmeasurably small, and can always be neglected. In all electrochemical calculations, the total Q of the positive ions is equal and opposite to the total Q of the
negative ions.
The Electrochemical Cell | 241
Ground and the Reference Potential
While, in theory, the zero of potential is defined as the potential at an infinite distance, in practice, this zero of potential cannot even be approximated. As we move away
from a charge, we encounter other charges. If we leave the Earth for what we ordinarily consider the vacuum of space, we are immersed in a wind of positively charged
protons streaming from the Sun. The galaxy itself has a large electromagnetic field,
meaning that the zero of potential, if it exists even approximately, is millions of lightyears away.
For this reason, all practical potentials must be measured not with respect to an absolute
zero of potential but to a reference potential. The reference potential in electronics is often
the so-called ground potential; basically, we run a wire to a metal post sunk in the ground.
In electrochemistry, as we will see, we often instead use a reference electrode to generate a
reference potential. But regardless, there is no practical absolute zero of potential. Potential
is in this respect like most other thermodynamic properties, such as energy, enthalpy, and
so on, which can only practically be defined relative to some standard value. It is unlike
entropy, for which a true zero value may be defined.
The Electrochemical Cell
Each of us is surrounded by dozens of electrochemical cells; we call them batteries, a term
coined by Benjamin Franklin. They power our cellphones, smoke detectors, thermostats,
our laptop or tablet computers, and our garage door openers. It is hard to believe the first
documented electrochemical cell was invented only slightly more than two centuries ago,
by the Italian scientist Alessandro Volta. Volta’s pile was made up of pairs of silver and
zinc disks, spaced by pieces of brine- or lye-soaked cloth. At the surface of the zinc disks,
zinc metal dissolved in the brine, leaving two electrons:
Zn S Zn2 + + 2eThe electrons flowed to the silver surface, where they reacted with water to give hydrogen gas:
2 H2O + 2e- S 2OH- + H2
Stacks of these disks gave estimated voltages between one end of the pile and the other
of up to 50 V.
Volta, in truth, didn’t really understand how his battery worked. He made some perceptive and accurate observations, such as that zinc gave a larger voltage than tin. But he
thought the electrical potential was being generated by the contact between the different
metals, rather than the solution. He thought the device could operate indefinitely, and did
not appreciate that the corrosion of the zinc, which he saw as a problem to be remedied,
was actually what was generating the electricity.
Reversibility in the Electrochemical Cell
It should be clear that Volta’s cell was inherently irreversible. Zinc dissolved in the aqueous electrolyte, but (at least at the beginning) the reverse process could not occur since
there were no zinc ions in the solution. Likewise, there was no supply of hydrogen gas at
the cathode, and so hydrogen could be generated but it could not react backwards to give
hydrogen ions.
Figure 7.2 shows a first attempt at a reversible electrochemical cell. We immerse a
zinc electrode — the anode — in a solution of zinc ions (zinc sulfate will do nicely). We
can similarly immerse a copper electrode — the cathode — in a solution of copper ions,
and connect the two electrodes with a wire, so electrons can flow between them. Finally,
FIGURE 7.1 Alessandro
Volta’s original drawing of his
electrochemical pile — the first
documented electrical battery.
The disks labeled Z and A are
zinc and silver; the dark disks
between them are brinesoaked cloth.
242 Chapter 7 |
Electrochemistry
FIGURE 7.2 A Zn>Cu electrochemical cell. Zinc dissolves at
the cathode, leaving two electrons, which flow to the anode,
where they react with Cu2+ to
plate out as copper metal. A
salt bridge allows ions to flow
between the two halves of the
apparatus and maintains electrical neutrality. If the resistance is
sufficient, the cell will develop
a potential of approximately
1.1 V, negative on the zinc side.
The exact value depends on the
concentrations.
1.1
V
Zn metal
salt bridge
Zn2+
Cu metal
Cu2+
we need a mechanism for ions to flow between the two solutions; as we calculated in the
previous section, a very small number of electrons flowing from one metal to the other
will create a substantial potential difference, which will shut the battery down. This is
usually done with a ‘salt bridge’, a piece of porous material or a U-tube allowing ions to
move between the two solutions, restoring electrical neutrality. If we make the zinc and
copper solutions equal in concentration and place a voltmeter between the zinc and copper
electrodes, the experiment shows us we will develop a potential of &1.1 V between the two
electrodes, with the zinc electrode negative.
This electrochemical cell is still not reversible; zinc will dissolve from the anode and
copper will ‘plate out’ on the cathode. However, if we apply an external positive potential to
the cathode, relative to the anode, we can slow down, stop, or even reverse the electrochemical process. At an external potential Vex of approximately 1.1 V on the cathode, relative
to the anode, zinc stops dissolving and copper stops plating out. At a potential of greater
than 1.1 V, zinc will, in fact, plate out on the anode and copper will dissolve at the anode.
At the point the reaction stops, we have achieved reversibility. The external potential
needed to achieve this condition depends on the metals used to build the cell, and to a
lesser extent on the concentration of the ions in the solution. We call this value of Vex the
electrochemical potential, and give it the symbol E(cursive E). If the experiment is done
in the standard state (1 bar pressure, pure zinc and copper metals, solutions of Zn2 + and
Cu2 + activity equal to exactly 1 m), at reversibility, we measure a standard electrochemical
potential E⬚.
Electrical Work, Electrochemical Potential, and Free Energy
The work done by the cell by moving a charge Q against the external potential Vex is the
electrical work, wE = QVex. If both Q and Vex have the same sign, the work is positive,
because it’s being done by the surroundings on the system.
Previously, when we discussed the First Law, we wrote ⌬U = q + w, with
dw = -pex dV. This was the situation when there was only pressure-volume work. Now,
however, when we also have electrical work, we have to add it to the pV contribution so
that the First Law includes all contributions to the work:
⌬U = q + w pV + w E .
(7.3)
The Electrochemical Cell | 243
We can now use ⌬H = ⌬U + ⌬(pV) and ⌬G = ⌬H - T⌬S, to obtain
⌬G = q + w pV + w E + ⌬(pV ) - T ⌬S .
(7.4)
At constant pressure, ⌬(pV) = p⌬V , and under reversible conditions, wpV = -p⌬V . At
constant temperature, under reversible conditions, ⌬S = q>T, or q = T⌬S. Making these
substitutions:
⌬G = T ⌬S - p⌬V + w E + p⌬V - T ⌬S = w E .
(7.5)
In other words, the electrical work at constant pressure and temperature, under reversible
conditions, is the free energy change of the reaction. Some books refer to this as the useful
work, as if pV work isn’t useful! It is probably more accurate to say that in the most general
case, what we call wE above is the ‘non-pV work’.
We can now insert our definition of electrical work to obtain
⌬G = QV ex .
(7.6)
Since we’re generally interested in chemical processes, the charge per mole is just 1 Faraday,
multiplied by the number of electrons, ne, involved in oxidizing or reducing one mole of
reactant, and by -1, since the electronic charge is negative. ne is always a positive integer.
If we’re in an electrochemical cell under reversible conditions, Vex = E, and so
⌬ rGm = -v eFE .
(7.7)
It’s usual to drop the subscripts denoting constant pressure and constant temperature,
because we almost always do electrochemistry under these conditions. If ⌬G is negative, E
is positive, and we can thence deduce that a positive electrochemical potential corresponds
to a reaction that goes spontaneously in the forward direction.
Finally, under standard conditions:
⌬ rG⬚ = -v eFE⬚ .
(7.8)
We can obtain expressions for ⌬S and ⌬H in a straightforward fashion:
a
0⌬ rGm
b = - ⌬rSm
0T
p
0E
b
0T p
(7.9)
0E
b d.
0T p
(7.10)
⌬ rS m = v eF a
or
⌬ rHm = ⌬ rGm + T ⌬ r S m = v eF c E+T a
By measuring the electrochemical potential as a function of temperature, therefore, we can
obtain the entropy, enthalpy, and free energy of reaction.
Standard Electrochemical Potentials
Because the electrochemical potential depends only on the free energy of reaction (Eq. 7.8),
it is a state function, and for that reason it is additive. So, for example, if we build a standard
cell with zinc as the anode and copper as the cathode — which is written in conventional
electrochemical notation as:
Zn 兩 Zn2 + (1m) ‘ Cu2 + (1m) 兩 Cu
E = 1.1037 V
The single vertical line denotes a phase boundary (in this case, the electrode surface); a
double line denotes a salt bridge. We can also build one with a copper anode and a silver
cathode:
Cu 兩 Cu2 + (1m) ‘ Ag + (1m) 兩 Ag
E = 0.4577 V
244 Chapter 7 |
Electrochemistry
H2(1 bar)
The potential of a zinc > silver cell can be obtained by adding these two together:
Zn 兩 Zn2 + (1m) ‘ Ag + (1m) 兩 Ag
Pt metal
E = 1.5614 V
Note that it is impossible to measure the potential for just a single reduction or oxidation.
A cell must always include a species being oxidized and another being reduced. To discuss
the electrochemical potentials of single species — a half-cell — therefore, we must define
a standard half-cell. Chemists have chosen the reduction of H+ to H2 gas under standard
conditions to have a standard electrode potential of zero:
H + (aq) + e - S ½ H2 (g)
H+(a = 1)
FIGURE 7.3 The standard
hydrogen electrode (SHE).
E⬚ = 0 V
To be a standard cell, the hydrogen ions and the hydrogen gas must each have an activity of 1; this means the molality of H+ is somewhere close to 1 m and the gas pressure is
approximately 1 bar.
It is obvious we cannot build an electrode out of hydrogen gas, so instead we use a
platinum electrode coated with finely divided platinum (platinum black). The oxidation
of platinum metal has an electrochemical potential far more negative (unfavorable) than
that of hydrogen gas; in addition, the platinum surface catalyzes the electrochemical
reaction between H + and H 2. The standard hydrogen electrode (SHE) is depicted in
figure 7.3.
In practice, because the activity coefficients for hydrogen ions at molal concentrations
are quite different from 1, it is preferable to measure cell potentials versus a SHE with a
much lower H+ concentration (ideal-dilute solution), and correct for concentration using
the Nernst equation (see below). The SHE is therefore something of a theoretical construct,
although recent research indicates an approximately 0.85 m solution of HCl in water has
a 苲 1 for H + .
In addition to measuring other half-cells coupled to a SHE, all standard potentials are tabulated as reductions, regardless of whether the potential is favorable or
unfavorable. So, for example, if we measure the half cell potential for Zn 兩 Zn2 + , it is
tabulated as
Zn2 + (aq, a = 1) + 2 e - S Zn (s)
E⬚ = -0.7618 V
even though, coupled to a SHE, the zinc half-cell reaction would spontaneously proceed
in the other direction, as shown by the sign of the voltage.
In order to obtain the electrochemical potential for a redox reaction from two half-cell
potentials, we do the following:
1. Reverse reactants and products for the oxidation half-reaction and multiply the
electrochemical potential of the oxidation half-reaction by -1.
2. Multiply the two reactions by integer(s) so as to make the number of elections equal
for oxidation and reduction steps.
3. Add the reactions and the electrochemical potentials.
When balancing the electrons, we do not multiply the electrochemical potentials by
the same integers. The number of electrons per mole is explicitly included in Eq. 7.8.
E X A M P L E 7.1
Obtain a balanced chemical equation, and calculate the standard electrochemical
potential for the reduction of Fe3 + to Fe (metallic iron) by Zn metal.
The Electrochemical Cell | 245
SOLUTION
The two half-reactions are:
Zn2 + + 2e - S Zn
E⬚ = -0.7618
Fe3 + + 3e - S Fe
E⬚ = -0.037
(1) Reverse the oxidation step and multiply the electrochemical potential by -1:
Zn S Zn2 + + 2 e -
E⬚ = 0.7618
(2) Multiply the two reactions by integers:
3 Zn S 3 Zn2 + + 6e -
E⬚ = 0.7618
2 Fe3 + + 6e - S 2 Fe
E⬚ = -0.037
(3) Add the reactions and the electrochemical potentials:
3 Zn + 2 Fe3 + S 2 Fe + 3 Zn2 +
E⬚ = 0.725
Whether we write the equation this way, or as Zn + 2/3 Fe3 + S 2/3 Fe + Zn2 + , the
electrochemical potential is the same.
Concentration Dependence of E
The standard potentials in table 7.1 can be converted to standard free energies using Eq. 7.8.
What if we are interested in the potentials at non-standard concentrations? We start with
the relation between standard Gibbs free energy and molar Gibbs free energy:
⌬ rGm = ⌬ rG⬚ + RT ln Q .
(7.9)
Substituting Eq. 7.7 for ⌬ rGm and 7.8 for ⌬ rG⬚:
-v eFE = -v eFE⬚ + RT ln Q .
(7.10)
Simplifying, we obtain the Nernst equation:
E = E⬚ -
RT
ln Q .
veF
(7.11)
The reaction quotient is as defined previously (Eq. 4.16). At 25°C, we can make use of
the identity that ln Q = ln 10 log10Q = 2.303 log10Q, and substitute 298 K for T, to get
the useful rule of thumb:
E = E⬚ -
0.0591 V
log Q .
ve
(7.12)
The Nernst equation can be used to calculate the potential for any concentrations of reactants and products in a cell. At equilibrium, ⌬ rG = 0 and therefore E = 0, so the left hand
side of Eq. 7.11 is zero and we can substitute K for Q. Thus, the standard potential gives
the equilibrium constant for the reaction in the cell:
E⬚ =
RT
ln K .
v eF
(7.13)
246 Chapter 7 |
Electrochemistry
TABLE 7.1 Standard Reduction Electrode Potentials at 25°C
Oxidant/Reductant
+
K >K
2+
Ca
>Ca
+
Na >Na
2+
Mg
2+
Mn
2+
Zn
>Mg
>Mn
>Zn
Acetate> acetaldehyde
2+
Fe
>Fe
+
H >H2 >Pt
CO2 >formate
Eⴗ(V)
Electrode reaction
K
+
-
+ e SK
2+
Na
–2.931
-
+ 2e S Ca
Ca
+
+ e S Na
2+
–2.372
-
–1.185
+ 2e S Mn
Mn
2+
-
+ 2e S Zn
Zn
CH3COO
-
-
+ 2e
2+
+ 2 e S Fe
+
-
2H
–2.71
-
+ 2e S Mg
Mg
Fe
–2.868
-
2+
–0.7618
+
+ 3H S CH3CHO + H2O
–0.105
-
+
–0.586
–0.447
+ 2e S H2
CO2 (aq) + H
Eⴗ⬘(V) (pH 7)
0
-
+ 2e S HCOO
3+
-
–0.181
2+
–0.414
–0.409
Ferredoxin
Fd[Fe ] S Fd[Fe ]
–0.395
FMN(Mitochondrial
complex I)
FMN + 2H + + 2e - S FMNH2
–0.380
Gluconolactone> glucose
C6H10O6 + 2H + + 2e - S C6H12O6
–0.345
L-cystine>L-cysteine
3+
Fe
>Fe
L@cystine + 2H
3+
+
-
+ 2e S 2 L@cysteine
+ 2e S Fe
Fe
-
–0.037
+ 2H
+
-
+ 2e S CH3CHOHCH2COO
Acetoacetate>
b-hydroxybutyrate
CH3COCH2COO
NADP + >NADPH
NADP + + H + + 2e - S NADPH
+
NAD >NADH
–0.355
-
NAD
+
+ H
+
-
–0.346
–0.339
-
+ 2e S NADH
+
–0.324
-
Glutathione(S-S)>
GlutathioneSH
GSSG + 2H
+ 2e S 2 GSH
–0.240
FMN>FMNH2
FMN + 2H + + 2e - S FMNH2
–0.213
FAD>FADH2
Acetaldehyde> ethanol
FAD + 2H
+
-
+ 2e S FADH2
CH3CHO + 2e
Pyruvate> lactate
CH3COCOO
Oxaloacetate> malate
-
-
-
–0.212
+0.221
+
+ 2H S CH3CH2OH
+ 2H
+
-
+ 2e S CH3CHOHCOO
-
+
-
-
-
OOCCOCH2COO + 2H + 2e S OOCCHOHCH2COO
FAD>FADH2(Complex II)
FAD
Fumarate> succinate
-
+
+ 2H
+
–0.184
-
–0.158
-
+ 2e S FADH2
OOCCH “ CHCOO
-
+ 2H
+
–0.193
–0.05
-
-
+ 2e S OOCCH2CH2COO
+0.045
-
Myoglobin
Mb[Fe ] + e S Mb[Fe ]
+0.046
Ubiquinone
UQ + 2H + + 2e - S UQH2
+0.052
Dehydroascorbate>
ascorbate
DHA + H + + 2e - S Asc -
+0.080
AgCl>Ag>Cl -
AgCl + e - S Ag + Cl -
3+
-
3+
2+
+0.222
-
Cytochrome c
Cyt@c[Fe ] + e S Cyt@c[Fe ]
Calomel
1
O2 >H2O2
-
>2 Hg2Cl2 + e S Hg + Cl
O2 + 2H
+
-
+ 2e S H2O2
+0.254
2+
-
+0.268
+0.695
+0.281
(continues)
Transmembrane Equilibria | 247
TABLE 7.1 Standard Reduction Electrode Potentials at 25°C (cont.)
Oxidant/Reductant
Electrode reaction
Eⴗ(V)
Cu2 + >Cu
Cu2 + + 2e - S Cu
+0.342
I2 >I
I2 + 2e S 2 I
+0.535
-
-
Ag >Ag
Ag
O2 >H2O
O2 + 4H
NO3- >NO2 -
NO3-
Br2(aq) >Br
+
+
-
+ e S Ag
-
+
+ 3H
+0.800
+ 4e S 2H2O
-
+
+1.229
+ 2e S HNO2 + H2O
-
+0.934
-
Br2 + 2e S 2Br -
+1.087
Cl2(g) >Cl -
Cl2 + 2e - S 2Cl -
+1.358
Mn3 + >Mn2 +
Mn3 + + e - S Mn2 +
+1.541
F2(g) >F
F2 + 2e S 2F
+2.866
-
-
-
-
E X A M P L E 7. 2
The enzyme glutathione reductase replenishes the cell’s supply of glutathione (GSH),
regenerating two molecules of GSH from a single molecule of oxidized glutathione
(GSSG), using NADPH as a source of two reducing equivalents. Using the data in
table 7.1 and a typical cellular NADP+>NADPH ratio of 0.005, calculate the equilibrium
cellular concentration of GSSG at pH 7 and 25°C, if the GSH concentration is 4 mM.
SOLUTION
The two half-reactions are:
GSSG + 2H + + 2e - S 2 GSH
E⬚⬘ = -0.240 V
NADP + + H + + 2e - S NADPH
E⬚⬘ = -0.339 V
(1) Reverse the second reaction and multiply the electrochemical potential by -1:
NADPH S NADP + + H + + 2e -
E⬚⬘ = +0.339 V
(2) The number of electrons is balanced.
(3) Add the reactions and the electrochemical potentials:
GSSG + H + + NADPH S 2 GSH + NADP +
K =
Eⴗ⬘(V) (pH 7)
[GSH]2eq[NADP + ]eq
[GSSG]eq[NADPH]eq
=
E⬚⬘ = +0.099 V
0.005 ( 4 * 10 - 3 ) 2
[GSSG]eq
We do not include the H concentration, since the activity of H + is defined as 1 at
pH 7 in the biochemists’ standard state.
Solving Eq. 7.13 gives us K = 2223, which in turn gives [GSSG] = 3.6 * 10 - 11
M = 36 pM. Healthy cells keep GSSG concentrations very low!
+
Transmembrane Equilibria
The membranes of cells and organelles, such as chloroplasts and mitochondria, are good
electrical insulators and are impervious to ions. They can therefore sustain transmembrane
potentials of hundreds of millivolts and large concentration gradients. Cells and organelles
in fact dedicate a considerable part of their genomes to proteins that pump ions across cell
+0.815
248 Chapter 7 |
Electrochemistry
membranes, that act as channels (often gated channels) across cell membranes, and that
can use these gradients of concentration and potential to transport other materials.
As before, the work done in transporting a charge Q across a cell membrane with
potential V is wE = QV. As before, we can equate wE with ΔG, and adapt Eq. 7.7:
⌬ rGm = ZFV .
(7.14)
Since we are now discussing the movement of ions, rather than electrons, we have replaced
-ve with Z, the charge on the ion; and E with V, the transmembrane potential. Moving a
positive charge to a more positive potential requires work and a positive free energy change.
This molar free energy difference can be written as the electrical part of an electrochemical
potential difference, ⌬m:
⌬mE = ZFV .
(7.15)
If there is also a concentration gradient across the membrane, the electrochemical potential
also has a chemical component (equivalent to the regular chemical potential discussed in
chapter 4):
⌬mC = ⌬m⬚ + RT ln Q .
(7.16)
Since both the inside and outside of cells or organelles are aqueous media, the standard
chemical potentials are the same. We can therefore write the total electrochemical potential
as a sum of electrical and chemical terms:
⌬m = ⌬mE + ⌬mC = ZFV + RT ln Q .
(7.17)
In this case, Q = ain/aout, the ratio of ion activities inside and outside the cell. Again, since
these environments are often very similar, the activity coefficients are often nearly identical, and we can replace activities by concentrations. Finally, at equilibrium, the difference
in chemical potentials is zero, and we have:
ZFV = -RT ln K or V = -(RT/ZF) ln K
(7.18)
Compare this with Eq. 7.13.
E X A M P L E 7. 3
The plasma membrane of mammalian cells typically has a potential V = Vin - Vout 苲
-70 mV. It also contains pores that allow potassium to selectively equilibrate across the
membrane. If extracellular potassium is typically 5 mM, calculate the concentration of
potassium inside the cells at 37°C, assuming activities can be replaced by concentrations.
SOLUTION
For K+, Z = +1, and therefore RT>ZF = 8.31447 * 310K>(1 * 96485.3) = 0.0257 V.
Since V = -0.07 V, by Eq. 7.18, ln Q = 2.62, or Q = [K + in]>[K +out] = 13.7. This
means [K + in] = 13.7 * 0.005 = 0.069 M or 69 mM.
Donnan Effect and Donnan Potential
In chapter 6, we described equilibrium dialysis as an easy way to measure binding of a
small molecule ligand by a macromolecule. The method becomes more complicated if the
macromolecule and ligand are charged. The requirement that the solutions on each side
Transmembrane Equilibria | 249
of the dialysis membrane must be electrically neutral means that there can be an apparent
increase in binding of a ligand with the opposite charge to that of the macromolecule and
a decrease in binding of a ligand with the same charge. These effects depend on the net
charge on the macromolecule and are not caused by binding at specific sites. The effect of
the net charge of the macromolecule on the apparent binding of a ligand can be minimized
by using high concentrations of a salt not involved in binding. Thus, the charged macromolecule and ligands are not the main contributors to the total concentration of ions in
the solutions.
When equilibrium (except for the macromolecule) is reached for charged species, a
voltage is developed across the membrane. The asymmetric distribution of ions caused by
the charged macromolecule is called the Donnan effect, and the transmembrane potential is
called the Donnan potential. It is observed experimentally when a charged macromolecule
is dialyzed in the presence of an electrolyte; it also contributes slightly to the membrane
potential of most cells. The experimental apparatus is shown in figure 7.4. A dialysis membrane (a semipermeable membrane with a pore size that allows small molecules, typically
under 3000 Da, to pass through, but impermeable to large molecules) encloses a solution
of a macromolecule in an electrolyte solution; outside the membrane is the same solution
without the macromolecule.
For simplicity, we consider the dialysis of a macromolecule with a net charge of ZM
and concentration cM against NaCl solution. Outside the membrane, there must be charge
balance, and so the electrolyte (say, NaCl) must have equal numbers of positive (Na + ) and
negative (Cl - ) ions. Electrical neutrality requires that
cNa, out = cCl, out = c ,
(7.19)
where c is the concentration of NaCl outside the membrane at equilibrium.
Inside the membrane, we also have electrical neutrality, but the macromolecule charge
contributes to it. Summing up over all charges inside:
cNa, in - cCl, in + Z McM = 0 .
(7.20)
At equilibrium, because the membrane is permeable to Na+ and Cl - ,
mNa, in = mNa, out
(7.21a)
mCl, in = mCl, out .
(7.21b)
From Eq. 7.17, we get:
⌬mNa = mNa, in - mNa, out = FV + RT ln (aNa, in >aNa, out) = 0
(7.22a)
⌬mCl = mCl, in - mCl, out = -FV + RT ln (aCl, in >aCl, out) = 0 .
(7.22b)
If we make the reasonable assumption that the activity coefficients are approximately equal
inside and outside the membrane, then:
FV = -RT ln (cNa, in >cNa, out) = RT ln (cCl, in >cCl, out) .
(7.23)
-ln (cNa, in >cNa, out) = ln (cNa, out >cNa, in) = ln (cCl, in >cCl, out) .
(7.24)
and therefore
mV
electrolyte
+macromolecule
electrolyte
FIGURE 7.4 Equilibrium
dialysis apparatus for measuring
the Donnan effect.
250 Chapter 7 |
Electrochemistry
Exponentiating and cross multiplying, and substituting Eq. 7.19:
cNa, out cCl, out = cNa, in cCl, in = c2 .
From Eq. 7.20 we have cCl, in = cNa, in + ZMcM, so
c2 - cNa, in(cNa, in + Z McM) = 0 .
(7.25)
This is a quadratic equation, whose positive root is
cNa, in =
-Z McM + 2(ZMcM)2 + 4c2
.
2
The ratio r of cNa, in to cNa, out is
r=
cNa, in
Z McM 2
-Z McM
=
+ a
b +1 .
c
2c
B 2c
(7.26a)
From Eq. 5.31, we can also calculate the ratio of Cl - inside to outside:
cCl - , in
c
1
=
= .
cCl - , out
cNa, in
r
(7.26b)
The equations above show that for a positively charged macromolecule the concentration of
positive ions inside will be less than that outside (r 6 1) and the concentration of negative
ions inside will be greater than outside.
For a negatively charged macromolecule, r 7 1, and the concentration of positive
ions is greater inside than outside. For example, for a macromolecule with a net positive
charge ZM = 10 at a concentration of 1 mM (cM = 10 - 3), dialyzed against 0.100 M NaCl
(c = 0.100), Eq. 7.26a gives r = 0.95. This means that the ratio of Na + inside to outside is
0.95, and the ratio of Cl - inside to outside is 1/r = 1.053. Increasing the positive charge
or concentration of a macromolecule will make r decrease; increasing the concentration
of NaCl will make r approach 1.
The asymmetry in Na + or Cl - concentration on the two sides of the membrane is surprising because the membrane is permeable to Na + and Cl - and we might think that the relations
mNa, in = mNa, out and mCl, in = mCl, out would predict equal concentrations of Na + and Cl on the two sides of the membrane. There is, however, a potential difference V, the Donnan
potential, across the membrane, as illustrated in figure 5.13. From Eq. 7.22a we obtain
V = -
RT cNa, in
RT
RT aNa, in
⬇ ⬇ ln
ln
ln r .
F aNa, out
F cNa, out
F
(7.27)
From the example above with r = 0.95, for Na + at 298 K, we can calculate a transmembrane
voltage of +1.3 mV. The plus sign means that the electrical potential is higher inside than
outside the membrane because the positively charged macromolecule is inside. We of
course calculate the same value for the voltage if we use the ratio of Cl - concentration
inside and outside.
Plasma Membrane Potentials and the Naⴙ-Kⴙ ATPase
In the example above, potassium concentrates passively inside the cell, responding to a
gradient of electrical potential. This raises the question, however, of how the membrane
potential itself is set up. The answer is that we have identified systems that undertake active
transport of ions, a process that is dependent on active cellular metabolism. Active transport
Transmembrane Equilibria | 251
is defined as the transport of a substance from a lower to a higher chemical potential.
Because the total free-energy change of the process must be negative, active transport is
tied to a chemical reaction that has a negative free-energy change. In a biological system,
this means that metabolism is occurring and driving the pump. Therefore, an experimental
test of whether active transport is involved is to poison the metabolic activity to see whether
the transport also stops. A classic example is the membrane Na + -K+ ATPase, sometimes
called the sodium–potassium pump, which uses the free energy of hydrolysis of ATP to
pump Na + ions out of the cell and K + ions into the cell. The net reaction for the active
transport is thought to be:
3Na + in + 2K + out + ATP S 2 K + in + 3Na + out + ADP + Pi
Representative concentration differences across a typical animal cell membrane are illustrated in figure 7.5. There is also a voltage difference of about -0.07 V across the cell
membrane; the inside is negative relative to the outside as shown. This membrane potential
arises from the action of the Na + -K + ATPase pump, and also from the permeability of
the membrane to potassium and the impermeability to sodium. Thus, the magnitude of the
potential is determined largely by the ratio [K + in]>[K +out].
The Na + -K+ ATPase is inhibited by cardiotonic steroids, such as digoxin, obtained
from the foxglove plant, or ouabain, a dart poison obtained from African trees. Cardiotonic
steroids at low levels make the heartbeat more powerful; at high levels, they shut down the
ion pump completely, paralyzing the heart.
The crystal structures of the Na + -K+ ATPase in various stages of its reaction cycle,
and in its ouabain-inhibited form, have recently been determined. A mechanism for the
reaction is given in figure 7.6. Binding of ATP causes a conformational change in the
protein that narrows the ion channel, thus favoring the exchange of smaller Na + ions for
larger K + ions inside the cell. The Na + ions catalyze covalent attachment of the ATP,
whose hydrolysis, followed by detachment of the ADP, causes the channel to widen and
open on the outside, favoring exchange of the bound Na+ for extracellular K + . Detachment
(a)
(b)
(c)
(d)
(e)
(f)
Inside
[Na+] = 10 mM
[K+] = 70 mM
V = –70 mV
Outside
[Na+] = 140 mM
[K+] = 5 mM
FIGURE 7.5 Typical ion
concentrations inside and
outside the plasma membrane.
(g)
Na+
Extracellular fluid
Na+
K+
Na+
Na+
Na+
Na+
Na+
Na+
ATP
Na+
Na+
K+
Pi
ADP
Pi
Na+
Pi
K+
Pi
Cytosol
Na+
K+
K+
ADP
K+
FIGURE 7.6 The Na+-K+ ATPase pump cycle. In state (a), the pore contains three sodium ions and is covalently linked to an ATP. In step (b), the ATP is hydrolyzed to bound ADP and Pi. Dissociation of the bound
ADP in step (c) leads to a conformational change where the pore widens and opens up on the extracellular
face of the membrane, allowing Na+ ions to escape and be replaced by K+ ions (d). These induce dephosphorylation (e) and cause the pore to close. An ATP can then bind (f) causing the pore to open on the cytosolic side, and narrow, forcing out the K+ ions and allowing Na+ to bind (g). These then catalyze a covalent
link between ATP and the protein, which in turn causes the pore to close again (back to a).
ATP
Cell membrane
ATP
252 Chapter 7 |
Electrochemistry
of the covalently bound phosphate then closes the channel on the outside and readies it for
binding of another ATP.
It is important to make sure that the detailed mechanism is consistent with the laws of
thermodynamics. If it is not, we know that we have left something out.
The free energy for the net reaction must be negative. This means that the positive
free energy of actively transporting ions against concentration and voltage gradients must
be more than balanced by the negative free energy of ATP hydrolysis.
Let’s divide the process into three parts: the transport of Na + ions out, the transport
of K + ions in, and the hydrolysis of ATP. The free energy per mole difference between
Na + inside and outside is given by Eq. 7.19:
⌬m = ZFV + RT ln
ain
.
aout
(7.19)
We replace ratios of activities by ratios of concentrations and use a temperature of 37°C:
10
⌬m = 1 * 96485.3 * -0.07 + 8.31447 * 310.15 * ln
= -13559 J>mol
140
But the pump works in the opposite direction (it propels the ion from inside to outside)
and so ⌬m = 13.6 kJ>mol.
The free energy per mole for the transport of K + ions is
100
⌬m = 1 * 96485.3 * -0.07 + 8.31447 * 310.15 * ln
= 971 J>mol .
5
The electrical potential difference of -70 mV favors transport of positive ions in; thus, the
favorable effect of the electric field, attracting positive ions toward the inside of the cell,
nearly cancels the effect of the concentration gradient for potassium ions, which tends to
cause them to diffuse out of the cell. In contrast, moving positively charged sodium ions
out of the negatively charged cell interior is unfavorable.
The free energy for the hydrolysis of ATP to ADP and inorganic phosphate (Pi) is
⌬m = ⌬m⬚ + RT ln
[ADP][Pi]
.
[ATP]
⌬m⬚ is -31.3 kJ mol-1 for ATP hydrolysis at 310 K. Typical concentrations in healthy
muscle cells are [ATP] = 1 mM, [ADP] = 40 μM, and [Pi] = 25 mM. This gives Q = 0.001:
⌬m = -31300 + 8.31447 * 310.15 * ln(0.001) = -49133 J>mol
We can now account for all three free-energy changes:
⌬m
⌬ rGm
(kJ mol-1) (kJ mol-1)
3
13.56
40.7
n
Na + in S Na +out
K + out S K + in
ATP + H2O S ADP + Pi
ATP + H2O + 3Na
+
+
in + 2K out
2
0.97
1.9
-49.1
S ADP + Pi + 3Na
+
+
out + 2K in
-6.5
Clearly, the free energy of ATP hydrolysis is sufficient to account for the active transport of the ions. However, the free-energy calculation does not prove the mechanism. It
only shows that the mechanism is a possible one that does not violate thermodynamic
principles. This is a very important test, however, because if the calculation did not
give a net negative free energy, the mechanism would have been immediately disproved.
In fact, any sign this reaction is departing from the canonical conditions of low ADP,
Transmembrane Equilibria | 253
high intracellular potassium, and low intracellular sodium is usually a sign of imminent
cell death!
The transmembrane electrical and chemical potential differences are linked but complementary in function. The electrical potential acts on all ions, and can therefore facilitate
the accumulation of positively charged ions such as Ca2 + and Mg2 + in the cytoplasm. On
the other hand, it is exquisitely fragile; movement of a few thousand ions can collapse or
even reverse the membrane potential (see problem 7.1)! The chemical potential difference
is far more robust, since it involves a macroscopic concentration difference and therefore
trillions or quadrillions of ions per cell. In some cases, it is directly coupled to the electrical potential difference by, for example, potassium channels that allow equilibration of K +
across the membrane. This allows the robust chemical potential difference to buffer the
electrical potential. In other cases, the chemical potential for a particular ion is decoupled
from the electrical potential, as is the case most of the time for Na + ions.
Excitable membranes, such as those in the sensory nervous system of animals, take
advantage of the possibilities available by alternately coupling and decoupling the electrical potential with the chemical potential of an ion. These membranes contain separate
Na + and K + channels, both of which are voltage-gated. In response to the membrane
potential reaching a certain value, the Na + channels open, allowing inflow of the farfrom-equilibrium Na + ions and causing the membrane potential to collapse and even
reverse. This causes the gate to close, and potassium channel gates to open, restoring
the normal cell potential. The resulting wave of (relatively) positive electrical potential
can travel down the nerve cell axon, carrying a neuronal signal. These channels can also
respond to activators such as Ca2 + , and can be blocked by inhibitors such as tetrodotoxin
(the puffer fish poison).
Using the techniques of modern electrophysiology, the currents associated with single
channels can be monitored, as illustrated in figure 7.7. By recording the current flowing
across a small region of membrane-containing neuronal channels, the pattern of opening
and closing of individual channels is seen as pulses of current that occur intermittently and
with a range of duration. When one channel is open, an ionic current (upward deflection)
flows; when two channels are open simultaneously, twice as much current flows. When
the channel gates close, the current falls to zero. The effect of applying a depolarizing
transmembrane potential on the magnitude of current flowing in an open channel is also
seen in figure 7.7.
+20 mV
5 pA
+50 mV
+100 mV
0 ms
2.5 ms
FIGURE 7.7 Simulation of single
channel currents in a small area
of cell membrane that includes
two potassium channels. A
single channel opening is seen
as a current step upward from
zero (the baseline value). As the
applied (depolarizing) potential
increases, the channels are more
likely to open, and also the
flow of ions increases, because
the potassium electrochemical
potential is further from
equilibrium. Occasionally, two
channels open at the same time,
particularly on the bottom trace.
254 Chapter 7 |
Electrochemistry
Photosynthetic energy conversion involves H+ gradients (ΔpH) across the thylakoid
membrane (see figure 6.21). Light activation of PS1 and PS2 pumps H+ ions from the
stromal to the lumenal phase. A transmembrane electric field V (positive inside) also
results from the electron transport induced by PS1 and PS2. Both the ΔpH and the V
provide the source of electrochemical potential to power the ATPase, a tiny molecular
machine that covalently links ADP with phosphate to produce high chemical potential
ATP. The ATP in turn provides a source of chemical potential for a variety of biochemical reactions, including those involved in converting CO2 to sucrose, cellulose, and other
plant materials.
Biological Redox Reactions and Membranes
Oxidation–reduction reactions are essential for energy storage and conversion in biological organisms. For example, the pyruvate produced as a product of glycolysis (figure 4.13)
undergoes oxidative decarboxylation to form acetyl coenzyme A, which then enters the
citric acid cycle. In the citric acid cycle, the acetate is coupled to oxaloacetate to form
the six-carbon molecule, citrate. Citrate then undergoes a series of at least eight reactions that involve progressive oxidation of two of the carbon atoms to CO2 and return
of the remaining four-carbon portion as oxaloacetate for reentry into the cycle. Coupled
to the oxidation of one mol of acetate to 2 mol CO2 is the reduction of 3 mol of NAD +
to NADH and the production of 1 mol of reduced flavin adenine dinucleotide (FADH2).
This reducing power generated in the citric acid cycle is subsequently used for biosynthetic reactions and for the formation of ATP by oxidative phosphorylation (respiration)
in mitochondria.
The step that completes the citric acid cycle is the oxidation of malate to oxaloacetate coupled with the reduction of NAD + to NADH by the enzyme malate
dehydrogenase:
COO–
HO
COO–
malate
dehydrogenase
C H
C O
+ NAD+
H
+ NADH + H+
C H
H
COO–
C H
COO–
malate
oxaloacetate
To calculate the ⌬ rG⬚⬘ associated with this reaction, we can use E⬚⬘ values for the
corresponding redox half-reactions listed in table 7.1:
oxaloacetate + 2H + + 2e - S malate
E⬚⬘ = -0.158 V
NAD
E⬚⬘ = -0.324 V
+
+ H
+
+ 2e S NADH
-
E⬚⬘ = -0.324 - (-0.158) = -0.166 V
⌬ rG⬚⬘ = -2 * (96485.3 C mol - 1) * -0.166 V = 32.0 kJ>mol
This step in the cycle is not spontaneous under standard conditions. However, it will
become spontaneous under conditions where the demand for NADH is high; that is, when
[NADH]/[NAD + ] is low. Under these conditions, the malate will undergo conversion to
oxaloacetate, which serves to turn on the citric acid cycle. This is an example of metabolic
regulation whereby supply and demand are kept in reasonable balance.
Biological Redox Reactions and Membranes | 255
complex I
complex II
complex III
complex IV
Mitochondrial intermembrane space
4 H+
2 H+
UQ
UQH2
UQ
UQH2
FMNH2
FMN
FADH2
FAD
NAD+
NADH
+ H+
2 cyt-c
(Fe3+)
2 cyt-c
(Fe2+)
2 cyt-c
(Fe2+)
2 cyt-c
(Fe3+)
UQH2
UQ
½ O2
+ 2H+
H2O
fumarate succinate
Mitochondrial matrix
FIGURE 7.8 The mitochondrial respiratory chain, which contains two pathways that merge at complex III.
The first pathway begins at NADH, proceeds through reduced ubiquinone (UQH2) and reduced cytochrome-c
(cyt-c) and ends in the reduction of oxygen to water. The second pathway begins at succinate, but also ends
in the reduction of water using UQH2 and cyt-c as intermediates.
Oxidative Phosphorylation
A principal source of cellular ATP is the process of oxidative phosphorylation (respiratory electron transport) carried out in mitochondria.* This is a series of electron transfer
reactions catalyzed by four membrane-associated enzyme complexes whereby NADH is
oxidized, ultimately by O2. These complexes are depicted in fig 7.8.
The complexes I, III, and IV form a super-complex in the mitochondrial inner membrane called a respirasome. However, it is useful to analyze the thermodynamics of the
respirasome in terms of each of the four enzyme complexes individually.
NADH-Q Reductase (Complex I)
NADH, which is a soluble small molecule reductant produced in the citric acid cycle
and elsewhere, is imported into the mitochondria. It undergoes oxidation at the enzyme
complex I, formally called NADH-Q reductase. Complex I is a large (⬃100 kDa), complex
protein, with up to 46 subunits, that contains a flavin mononucleotide, (FMN) cofactor,
and nine iron-sulfur centers. Initially, NADH reduces the FMN, which then feeds the pair
of electrons singly down a chain of iron-sulfur (FeS) centers. Finally, the electrons are
passed on to reduce ubiquinone (UQ). The reduced ubiquinone is then released to travel to
complex III; being hydrophobic, it probably diffuses predominantly within the membrane
itself. All of the redox activity takes place in the ‘peripheral’ part of the complex, which
protrudes into the mitochondrial matrix (the center of the mitochondrion); it is however
coupled to a membrane-bound region that translocates four protons from the matrix to the
outside of the mitochondrial inner membrane (the intermembrane space). The reduction
half-reactions taking place in the complex are:
*Mitochondria
2 H+
NAD + + H + in + 2e - S NADH
E⬚⬘ = -0.324 V
UQ + 2H +in + 2e - S UQH2
E⬚⬘ = +0.052 V
are found in eukaryotic cells. They are membrane-surrounded structures in which oxidative
metabolism occurs.
256 Chapter 7 |
Electrochemistry
Reversing the first half-reaction, and adding, we see the overall reaction is strongly
favorable:
UQ + H +in + NADH S NAD + + UQH2
E⬚⬘ = +0.376 V
The standard free-energy difference ⌬ rG⬚⬘ = -v eFE⬚⬘ = -2 * 96485.3 C mol - 1 *
0.376 V = -72.6 kJ mol - 1. Q for this reaction is surprisingly hard to measure in viable
mitochondria, but it is probably close to 1.
This highly favorable reaction is now thought to be coupled to the translocation of
no fewer than four protons from the mitochondrial matrix to the intermembrane space
(which for small molecules is in equilibrium with the cytoplasm). Using typical values
of -160 mV for the membrane potential (Vin - Vout ) and a ⌬pH of 0.5 (Matrix pH 7.8;
cytoplasmic pH 7.3), we can obtain the free-energy change for the translocation of a single
proton from the mitochondrial membrane interior to the exterior from Eq. 7.19, assuming
a temperature of 25°C. The chemical potential difference is
ain
= ZFV - 2.303 RT⌬pH
⌬m = ZFV + RT ln
aout
= 1 * 96485.3 * -0.16 + 8.31447 * 298.15 * 2.303 * -0.5 = -18292 J mol - 1.
Translocating a mol of protons from inside to out therefore takes -⌬m, or 18.3 kJ>mol.
Adding the two coupled processes:
⌬m
⌬ rG m
4
+18.3
-73.2
1
-72.6
-72.6
n
H +in S H +out
UQ + H +in + NADH S NAD + + UQH2
UQ +
5H +in
+ NADH S NAD
+
+ UQH2 +
4H +out
(kJ mol-1) (kJ mol-1)
-0.6
The ⌬ rGm value is near zero, and so the reaction is essentially at equilibrium. This can
be confirmed experimentally; it has been shown that artificially increasing the membrane
potential in mitochondria can cause the complex I reaction to run in reverse, reducing
NAD + using UQH2. That reaction would ordinarily be overwhelmingly unfavorable.
The gradient of proton electrochemical potential is used to drive ATP synthesis
(see below).
Succinate Dehydrogenase (Complex II)
The electrochemical potential for the fumarate reduction half-reaction to succinate
-
OOCCH = CHCOO - + 2H + + 2e - S - OOCCH2CH2COO - E⬚⬘= +0.040V
is positive, and if reversed, could not feasibly drive the reduction of NAD + . The mitochondrion therefore has a separate membrane-bound succinate dehydrogenase complex that uses
reducing equivalents from succinate to reduce ubiquinone, via a flavin-adenine dinucleotide (FAD) intermediate (and a chain of FeS centers similar to those seen in complex I).
The full reaction is
UQ + -OOCCH2CH2COO- S OOCCH = CHCOO- + UQH2 E⬚⬘ = +0.012 V .
The reaction is obviously very close to equilibrium at Q ~ 1, and therefore does not
drive any proton translocation. In fact, the membrane-bound part of complex II seems to
be little more than a stalk or anchor for the enzyme, all of whose reactive elements lie in
the mitochondrial matrix.
Biological Redox Reactions and Membranes | 257
Coenzyme Q – Cytochrome c Oxidoreductase (Complex III)
Both complex I and complex II produce UQH2, a hydrophobic quinol that probably partitions into the mitochondrial inner membrane, and thus diffuses to the next element in
the respiratory chain, complex III. This enzyme, via a complex mechanism involving
half-reduced ubisemiquinone intermediates, uses one molecule of UQH2 to reduce two
molecules of cytochrome-c, a relatively small, soluble, basic protein that resides in the
mitochondrial intermembrane space. Cytochrome-c contains a heme iron that has an Fe3 +
(oxidized) and an Fe2 + (reduced) state. We can once again determine the standard biochemists’ electrochemical potential for the full reaction from the two-half reactions:
UQ + 2H +in + 2e - S UQH2
E⬚⬘ = +0.052 V
cyt@c(Fe3 + ) + e - S cyt@c(Fe2 + )
E⬚⬘ = +0.254 V
Reversing the first half-reaction, multiplying the second by two, and adding:
2 cyt@c(Fe3 + ) + UQH2 S cyt@c(Fe2 + ) + UQ + 2H +out
E⬘ = +0.202 V
Notice the two protons are released on the intermembrane side. Notice also that while
we reverse the sign of the electrochemical potential, we do not multiply its value by 2!
The standard free-energy difference ⌬ rG⬚⬘ = -v eFE⬚⬘ = -2 * 96485.3 C mol - 1 *
0.202 V = -39.0 kJ mol - 1. Q is once again probably close to 1.
This reaction is coupled with the transport of two protons across the cell membrane
per UQH2 oxidized. Again assuming each mol of protons translocated requires 18.3 kJ,
the overall process is favored by -39.0 + 2 * 18.3 = 2.4 kJ mol - 1 under standard
conditions. The reaction, like that of complex I, runs close to equilibrium.
Cytochrome c Oxidase (Complex IV)
The fourth mitochondrial membrane complex, cytochrome c oxidase, receives electrons
one at a time from reduced cytochrome-c [cyt@c(Fe2 + )], which diffuses approximately
10 nm from the active site of complex III, and transmits them to molecular oxygen, which
requires four electrons overall to become reduced to two water molecules. The cytochrome
c oxidase complex accomplishes this one-electron to four-electron transfer through the
mediation of two heme centers. Each heme is associated with a copper ion near the heme
iron. The net reaction (per two electrons transferred) is
2 cyt @c(Fe2 + ) + 1>2 O2 + 2H +in S 2 cyt@c(Fe3 + ) + H2O .
The protons come from the matrix side of the membrane via a proton channel. From the
two half-reactions
cyt@c(Fe3 + ) + e - S cyt@c(Fe2 + )
E⬚⬘ = +0.254 V
>2 O2 + 2H + + 2e - S H2O
E⬚⬘ = +0.815 V
1
we obtain a value for E⬚⬘ = 0.815 - 0.254 = 0.541V. Since the oxygen pressure is certainly
less than the 0.2 bar of atmospheric oxygen, and typical mitochondrial matrix pHs are 7.8,
not 7.0, E is probably considerably lower than E⬚⬘ for the second half reaction. The standard free energy difference ⌬ rG⬚⬘ = -v eFE⬚⬘ = -2 * 96485.3 C mol - 1 * 0.561 V =
-109.3 kJ mol - 1.
This reaction is coupled with the transport of two mol protons across the cell membrane
per mol water generated. Since each proton translocated requires approximately 18.3 kJ mol-1,
the reaction is favored by -54.3 kJ mol-1 under standard conditions, although it is probably
somewhat less favored in vivo.
258 Chapter 7 |
Electrochemistry
Mitochondrial Oxidation of NADⴙ
Adding together the overall reactions of complexes I, III, and IV, we obtain the deceptively
simple result
NADH + 1>2 O2 + 11 H +in S NAD + + H2O + 10 H +out .
Four of the protons come from complex I, two from complex III, two from complex IV,
and two because UQ is reduced on the matrix side and oxidized on the intermembrane side.
The standard electrochemical potential for the chemical part of the reaction is
NADH + H +in + 1>2 O2 S NAD + + H2O
E⬚⬘ = +1.139 V.
This corresponds to a ⌬ rG⬚⬘ value of 219.8 kJ mol-1. Approximately 183.0 kJ mol-1
(10 * 18.3) is stored as proton electrochemical potential energy. As we have written it
(without the comparatively small corrections for Q), the efficiency is 83%. By the standards
of most energy transducing systems, this is a marvel of efficiency!
Mitochondrial matrix
δ
α β
b
H+
α
ε
β
γ
a cc c cc
Intermembrane space
FIGURE 7.9 The
mitochondrial> chloroplast F1F0
ATP synthase.
ATP Synthase
While many of the protein complexes involved in oxidative phosphorylation are exquisitely complex, there is surely none so beautiful as the F1F0 ATP synthase, a tiny molecular machine that couples electrochemical potential, mechanical energy, and chemical
synthesis.
The enzyme is depicted in figure 7.9. F1 (short for fraction 1) is the mushroom-shaped,
soluble component, whose subunits are labeled in Greek letters, and has the stoichiometry
a3b3gde. F0 is the membrane-bound component, labeled in Roman letters, and has the
stoichiometry abcn, where n can be anywhere from 10 to 14. F1 binds to F0 on the inner
face of the mitochondrial inner membrane.
The b subunit is catalytically active in ATP synthesis, and research by the Boyer group
showed it has three states: an ‘open state’ (O) in which ATP can dissociate and ADP and
Pi can bind, a ‘loose’ state (L) in which the ADP and Pi are enclosed by the protein but
loosely bound, and a ‘tight’ (T) state in which they have reacted to form a very tightly
bound ATP. Each F1 unit has one b subunit in each of the three states. Boyer proposed
that rotation of the F1 unit switched the b units between the three states cyclically in the
order O S L S T. This was beautifully demonstrated by Yoshida, who attached the F1
unit top down on a glass slide, linked the g or e subunits to a fluorescently labeled actin
filament, and watched the tiny filament rotate by fluorescence microscopy when ATP was
added. Even more elegantly, by attaching a magnetic bead to the particle and applying a
rotating magnetic field, it was possible to drive ATP synthesis by mechanically rotating
the F1 unit. The direction of rotation required for synthesis is the opposite of the direction
in which the unit spins when ATP is applied.
This shows that ATP synthesis and mechanical rotation of the head relative to the
stalk are coupled. Where does the rotation come from? Again, single-molecule microscopy
showed that the membrane embedded ring of 10–15 c subunits rotates with the F1 unit when
F1 is anchored to F0, as it is on the mitochondrial inner membrane. Rotation of this ring is
driven by an electrochemical gradient of H+ ions across the membrane (the protonmotive
force). Protons enter through a pore on the outer face of the a subunit, and protonate a
carboxylic acid side chain on the c subunit immediately adjacent to it. This causes the
c subunit to rotate away from the a subunit, moving the next c subunit into position. This
deposits a proton from the corresponding carboxylic acid to the inner side of the membrane,
and then receives a proton from the outer side, causing it in turn to rotate. For each proton,
the c ring undergoes a 360⬚>n rotation.
Summary | 259
Since each proton induces a 360⬚/n rotation, and it requires one-third of a full rotation,
or 120° rotation, to release an ATP from the T site, the number of protons passing through
the membrane per ATP synthesized is 120⬚>(360⬚>n) = n>3. Depending on the species
and the organelle, n>3 can vary between 31>3 and 5. If we take a minimum value of n = 10,
and our previous ⌬mH + of 18.3 kJ mol-1, this means that 18.3 * 3 1>3 = 61 kJ mol - 1 of
free energy is available to turn the rotor by 120°, and thus synthesize one mol of ATP.
We have previously calculated that at Q = 1000, 49.1 kJ mol-1 of free energy is needed
for ATP synthesis. Thus,
⌬m
n
H +out S H + in
31>3
1
ADP + Pi S ATP + H2O
ADP + Pi + 31>3H + out S ATP + H2O + 31>3H + in
(kJ mol-1)
-18.3
+49.1
⌬ mG
(kJ mol-1)
-61.0
+49.1
-11.9
and the free energy from the proton electrochemical potential difference is more than adequate to drive the synthesis of ATP forward. However, if ΔμH+ drops below 14.7 kJ>mol,
the sign of the net free-energy change will reverse, and ATP hydrolysis will drive protons
out of the mitochondrial interior.
Summary
Galvanic Cells
Electrical work and free energy:
⌬G = QVex = we
(7.5)–(7.6)
⌬G⬚ = - veFE⬚
a
(7.8)
0⌬ rGm
0E
⌬ r S m = v e Fa b
b = - ⌬rSm
0T
0T
p
p
or
⌬ r Hm = ⌬ rGm + T⌬ r S m = v e Fc E + T a
0E
b d
0T p
(7.9)
(7.10)
The Nernst equation:
E = E⬚ E⬚ =
RT
ln Q
veF
RT
ln K
veF
(7.11)
(7.13)
Transmembrane Electrochemical Potential
⌬m = ⌬mE + ⌬mC = ZFV + RT ln Q
(7.17)
V = - (RT/ZF) ln K
(7.18)
Mathematics Needed for Chapter 7
Just algebra and logarithms!
260 Chapter 7 |
Electrochemistry
References
1. Ruma Banerjee et al., 2008. Redox Biochemistry, New York,
Wiley, is an excellent and comprehensive guide to biological oxidation and reduction.
2. The online Electrochemical Encyclopedia at http://
electrochem.cwru.edu/encycl/ is hosted by the Ernest
B. Yeager Center for Electrochemical Sciences (YCES) at
Case Western Reserve university, and has lots of valuable
reference material.
Suggested Reading
Moser, C. C., J. M. Keske, K. Warnke, R. S. Farid, and P. L.
Dutton. 1992. Nature of Biological Electron Transfer.
Nature 355:796–802.
Morth, J. P., Pedersen, B.P., Toustrup-Jensen, M.S., Sørensen, T.
L.M., Petersen, J., Andersen, J. P., Vilsen, B., and Nissen, P.
2007. Crystal structure of the sodium–potassium pump.
Nature (London) 450:1043–1049.
Brandt, U. 2006. Energy Converting NADH:Quinone
Oxidoreductase (Complex I) Annu. Rev. Biochem. 75:69–92.
2003. Architecture of Succinate Dehydrogenase and Reactive
Oxygen Species Generation. Science 299: 700–704.
Dudkinaa, N. V., Kudryashev, M., Stahlberg, H., and Boekemaa,
E. J. 2011. Interaction of complexes I, III, and IV within
the bovine respirasome by single particle cryoelectron
tomography. Proc. Natl. Acad. Sci. USA 37:15196–15200.
Nakamoto, R. K., Scanlon, J. A. B., Al-Shawi, M. K. 2008. The
rotary mechanism of the ATP synthase. Arch. Biochem.
Biophys. 476: 43–50.
Yankovskaya, V., Horsefield, R., Törnroth, S., Luna-Chavez, C.,
Miyoshi, H., Léger, C., Byrne, B., Cecchini, G., Iwata, S.
Problems
1. The cytoplasm of a cell is a (moderately) conducting fluid,
surrounded by an insulating membrane, and embedded in a
solution that is usually at least somewhat conducting. It can
therefore be treated as a capacitor. Assume the cell is spherical with a radius r = 10 mm, the cell membrane has an area
A = 4pr2, and a thickness D ⬃ 5 nm. The capacitance
should then be given by C = eA/D, where the permittivity
e is given by ere0, with er, the dielectric constant, having a
value of approximately 5, and e0 = 8.8542 * 10 - 12 F m - 1.
Calculate the capacitance of the cell membrane, and calculate how many unipositive ions need to cross the cell
membrane to give the interior a potential of 100 mV.
2. This is the half-cell reaction for the reduction of solid
manganese dioxide:
MnO2(s) + 4H + (aq) + 2e - S Mn2 + (aq) + 2H2O (l)
E⬚ = + 1.23 V
Determine the electrochemical potential and the free-energy
change for a system where a solution of 0.1 M Mn2 + at pH 1
is shaken in air at 0.2 bar pressure of O2, resulting in production of MnO2 (s), with the reduction of the oxygen to water.
If the pH is held constant at 1 and the oxygen pressure at
0.2 bar, what will be the equilibrium concentration of Mn2 +
in the presence of excess MnO2 (s)?
3. Ascorbic acid (Asc; Vitamin C) is a monoanion at neutral
pH. It is a powerful antioxidant, which scavenges free
radicals. It is oxidized in a single-electron oxidation to the
monodehydroascorbate radical anion (MDHA-) according
to the reaction
Asc - S MDHA - + H + + e - .
The half-reaction has a standard reduction potential E⬚⬘ of
+0.330 V. Ascorbate also undergoes a two-electron oxidation to dehydroascorbate (DHA):
Asc - S DHA + H + + 2e This half-reaction has a standard reduction potential E⬚⬘ of
+0.08 V. Finally, MDHA– undergoes a disproportionation
reaction:
2 MDHA - + H + S DHA + Asc Determine the biochemical standard free-energy change
⌬G⬚⬘ for this reaction.
4. The standard electrochemical potential for the reduction
of nitrate ion (NO 3- + 3H + + 2e - S HNO2 + H2O) is
E⬚ = + 0.934. The equation reflects the fact that nitric acid
(HNO3) is a strong acid, fully dissociated at pH 0, while
nitrous acid (HNO2 pKa = 3.398) is a weak acid. Calculate
E⬚⬘ for the half-cell reaction at pH 7.
5. Cytochromes are iron–heme proteins in which a porphyrin
ring is coordinated through its central nitrogens to an iron
atom that can undergo a one-electron oxidation–reduction
reaction. Cytochrome f is an example of this class of molecules, and it operates as a redox agent in chloroplast
Problems | 261
photosynthesis. The standard reduction potential E⬚⬘ of
cytochrome f at pH 7 can be determined by coupling it to
an agent of known E⬚⬘, such as ferricyanide> ferrocyanide:
Fe(CN) 63 - + e - S Fe(CN) 64 E⬚ = + 0.440 V
In a typical experiment, carried out spectrophotometrically,
a solution at 25°C and pH 7 containing a ratio
[Fe(CN) 64-]>[Fe(CN) 63 ] = 2
-
is found to have a ratio [Cyt f red]>[ Cyt f ox] = 0.10 at equilibrium.
a. Calculate E⬚⬘ (reduction) for cytochrome f.
b. On the basis of the standard reduction potential E⬚⬘ for
the reduction of O2 to H2O at pH 7 and 25°C, is oxidized
cytochrome f a good enough oxidant to cause the formation of O2 to H2O at pH 7?
6. Fe2 + -myoglobin (Fe2 + -Mb) is spontaneously oxidized
by molecular oxygen in a one-electron process to give
Fe3 + -Mb and superoxide, O2-. The reaction can be written
Fe2 + @Mb + O2 S Fe3 + @Mb + O 2The biochemists’ (pH 7) reduction potential of Fe3 + -Mb is
Fe3 + @Mb + e - S Fe2 + @Mb
E⬚⬘ = + 0.046 V .
O2 can be electrochemically reduced to hydrogen superoxide, a weak acid (pKa ⬃ 4.9):
O2 + H + + e - S HO2
E⬚⬘ = - 1.215 V
a. Calculate the pH 7 reduction potential for oxygen to
superoxide.
b. Calculate the potential for the one-electron oxidation of
myoglobin by oxygen at an oxygen pressure of 0.02 bar
and pH 7.
7. The reaction
glyceraldehyde@3@phosphate + NAD + + Pi S
1,3 diphosphoglycerate + NADH + H +
has ⌬ rG⬚⬘ = 6.3 kJ mol - 1. If the standard reduction potential E⬚⬘ of NAD+ is -0.324 V and the reaction
1,3 diphosphoglycerate + ADP S
3@phosphoglycerate + ATP
has ⌬ rG⬚⬘ = - 18.8 kJ>mol, calculate the standard reduction potential E⬚⬘ for the reaction
3@phosphoglycerate + 2e - + 3H + S
glyceraldehyde@3@phosphate + H2O .
8. Lysozyme (m.w. 14.3 kD) is a rather basic protein; at pH 7,
it has a net positive charge of +18. If we dissolve 5 g of
lysozyme in 100 mL of 0.1 M KCl, and dialyze against
0.1 M KCl, calculate the Donnan potential and the concentration of K + and Cl- inside the membrane.
9. Consider the following reaction, in which two electrons are
transferred from cytochrome-c (reduced):
2cyt c[Fe2+] + pyruvate + 2H+ S
2cyt c[Fe3+] + lactate
a. What is E⬚⬘ for this reaction at pH 7 and 25°C?
b. Calculate the equilibrium constant for the reaction at
pH 7 and 25°C.
c. Calculate the standard Gibbs free-energy change for the
reaction at pH 7 and 25°C.
d. Calculate the Gibbs free-energy change (at pH 7 and
25°C) if the lactate concentration is five times the pyruvate concentration and the cytochrome c (Fe3 + ) is ten
times the cytochrome c (Fe2 + ).
10. The cell
Ag(s), AgI(s)兩KI(10 - 2 M) ‘KCl(10 - 3 M)兩Cl2 (g, 1 bar), Pt(s)
has the voltage 1.5702 V at 298 K.
a. Write the cell reaction.
b. What is ⌬ rG at 298 K?
c. What is ⌬G⬚ at 298 K?
d. Calculate the standard reduction potential for the halfcell on the left.
e. Calculate the solubility product of AgI:
KAgI = aAg + aI f. The cell has a potential of 1.5797 V at 288 K. Estimate
⌬ r S at 298 K for the reaction.
11. Ferredoxins (Fd) are iron- and sulfur-containing proteins
that undergo redox reactions in a variety of microorganisms. A particular ferredoxin is oxidized in a one-electron
reaction, independent of pH, according to the equation
Fdred S Fdox + e - .
To determine the standard potential of Fdred >Fdox a known
amount was placed in a buffer at pH 7.0 and bubbled
with H2 at 1 bar pressure. (Finely divided platinum catalyst was present to ensure reversibility.) At equilibrium,
the ferredoxin was found spectrophotometrically to be
exactly one-third in the reduced form and two-thirds in the
oxidized form.
a. Calculate K⬘, the equilibrium constant, for the system
>2 H2 + Fdox S Fdred + H + .
1
b. Calculate E⬚⬘ for the Fdred/Fdox half-reaction at 25°C.
12. The conversion of b-hydroxybutyrate (b@HB - ) to acetoacetate (AA - ) is an important biochemical redox reaction that
uses molecular oxygen as the ultimate oxidizing agent:
b@HB - + 1>2 O2(g) S AA - + H2O
a. Using the standard reduction potentials given in table 7.1,
calculate ⌬ rG⬚⬘and the equilibrium constant for this system at pH 7 and 25°C.
b. In a solution at pH 7 and 25°C saturated at 1 bar with
respect to dissolved air (which is 20% oxygen), what is
the ratio of AA - to b@HB - at equilibrium?
262 Chapter 7 |
Electrochemistry
13. Consider the oxidation of ethanol to acetaldehyde:
CH3CH2OH + >2 O2(g) S CH3CHO + H2O
1
a. Calculate E⬚⬘ for this reaction at 25°C.
b. Calculate the standard Gibbs free energy (in kJ) for the
reaction at 25°C.
c. Calculate the equilibrium constant at 25°C for the
reaction.
d. Calculate E for the reaction at 25°C when aethanol = 0.1,
pO2 = 4bar, aacetaldehyde = 1, and aH2O = 1.
e. Calculate ⌬ rG for the reaction in part (d).
14. Magnesium ion and other divalent ions form complexes
with adenosine triphosphate, ATP:
ATP + Mg2 + S Mg@ATP
a. Describe an electrochemical cell that would allow you
to measure the activity of Mg2 + at any concentration in
a 0.100 M ATP solution.
b. Describe how you could measure with an electrochemical cell the thermodynamic equilibrium constant for
binding of Mg2 + by ATP.
15. Photosystem 1, in higher plants, converts light into chemical
energy. Energy, in the form of photons, is absorbed by a
chlorophyll complex, P700, which donates an electron to A.
The electron is passed down an electron-transport chain, at
the end of which NADP + is reduced. The reduction potentials of P700 + , A, and NADP + , at pH 7.0 and 25°C are
0.490 V, 0.900 V, and -0.350 V, respectively.
a. Calculate, at pH 7.0 and 25°C, E⬚⬘ of the reaction
P700 + A S P700 + + A - .
b. What is ⌬G⬚⬘ in kJ mol-1, for the same reaction?
c. At pH 7.0 and 25°C find ⌬ rG⬚⬘ in kJ mol-1 for the reaction
NADP + + H2(g) S NADPH + H + .
16. Certain dyes can exist in oxidized or reduced form in solution. The half-reaction for one such dye, methylene blue
(MB), is
MBox(blue) + 2H + + 2e - S MBred(colorless)
E⬚ = + 0.400 V.
As indicated, the oxidized form is blue, and the reduced
form is colorless. From the color of the solution, the relative
amounts of the two forms can be estimated.
a. Write the equation for the half-cell reduction potential
of methylene blue in terms of MBox, [MBred], [H+],
and E⬚⬘.
b. A very small amount of MBox is added to a solution
containing an unknown substance. The pH = 7.0. From
the color of the solution, it was estimated that the ratio of
concentrations [MBox]> [MBred] = 1.00 * 10-3 at equilibrium. Assuming that all activity coefficients are equal
to 1, determine the half-cell potential of the unknown
substance in solution.
17. Consider the following half-cell reactions and their standard
reduction potentials at 298 K and pH 7.0 in aqueous solution:
O2 + 4H + + 4e - S 2H2O
(E⬚⬘ = + 0.815 V)
cystine + 2H + + 2e - S 2 cysteine (E⬚⬘ = - 0.34 V)
a. If you prepare a 0.010 M solution of cysteine at pH 7.0
and let it stand in contact with air at 298 K, what will be
the ratio of cysteine> cysteine at equilibrium? The partial
pressure of oxygen in the air is 0.20 bar. The activity
coefficients may be taken as unity.
b. What is ⌬G for the reaction when the activities of the
reactants and products are the equilibrium values?
18. Typical Mg2 + concentrations in blood plasma are 2.2 mM.
If the membrane potential of erythrocytes is -90 mV
(- inside), calculate the equilibrium concentration of Mg2 +
inside the red blood cell at 37°C.
19. In living biological cells, the concentration of sodium ions
inside the cell is kept at a lower concentration than the concentration outside the cell, because sodium ions are actively
transported from the cell. Consider the following process at
37°C and 1 bar:
1 mol NaCl (0.05M inside) S 1 mol NaCl(0.20M outside)
a. Write an expression for the free-energy change for this
process in terms of activities. Define all symbols used.
b. Calculate ⌬m for the process. You may approximate
the activities by concentrations in M. Will the process
proceed spontaneously?
c. Calculate ⌬G for moving 3 mol of NaCl from inside to
outside under these conditions.
d. Calculate ⌬m for the process if the activity of NaCl
inside is equal to that outside.
e. Calculate ⌬m for the process at equilibrium.
f. The standard free energy for hydrolysis of ATP to
ADP (ATP + H2O S ADP + Pi) in solution is
⌬ rG⬚ = - 31.3 kJ>mol-1 at 37°C and pH 7. The free
energy of this reaction can be used to power the sodiumion pump. For a ratio of ATP to ADP of 10, what must
be the concentration of phosphate to obtain -40 kJ mol-1
for the hydrolysis? Assume that activity coefficients are
1 for the calculation.
g. If the ratio of ATP to ADP is 10, what is the concentration of phosphate at equilibrium? Assume ideal solution
behavior. What do you conclude from your answer?
20. A cell membrane at 37°C is found to be permeable to Ca2 + but
not to anions, and analysis shows the inside concentration to be
0.100 M and the outside concentration to be 0.001 M in Ca2 + .
a. What potential difference in volts would have to exist
across the membrane for Ca2 + to be in equilibrium at the
stated concentrations? Assume that activity coefficients
are equal to 1. Give the sign of the potential inside with
respect to that outside.
b. If the measured inside potential is +100 mV with respect
to the outside, what is the minimum (reversible) work
required to transfer 1 mol of Ca2 + from outside to inside
under these conditions?
Problems | 263
21. When heart muscle is treated with external lithium ions,
the ions pass through the cell membrane and equilibrate. Muscle cells treated with 150 mM lithium chloride solution achieved a steady-state membrane potential
of 40 mV (negative inside). What was the intracellular
lithium concentration?
22. Ascorbate and copper ions have the following electrochemical potentials:
Dehydroascorbate + 2H
Cu2 + + e - S Cu +
+
-
+ 2e S Ascorbate
E⬚⬘ = + 0.08 V
E⬚⬘ = 0.159 V
10 mL of a 0.02 M solution of ascorbate in a buffered
solution at pH 7 at 25°C is mixed with 10 mL of 0.02 M of
Cu2+. Write a balanced chemical equation for the reaction,
and figure out the equilibrium concentration of dehydroascorbate, ascorbate, Cu2+, and Cu+.
23. Mercuric reductase carries out the reaction
NADPH + Hg2 + S NADP + + H + + Hg.
The electrochemical potentials for the two reduction half
reactions are:
E⬚⬘ = - 0.339V
NADP+ + H+ + 2e - S NADPH
Hg2+ + 2e - S Hg
E⬚⬘ = + 0.85 V
Calculate the standard free energy of this reaction, and
the free energy if the NADP + : NADPH ratio is 1, the
Hg2+ concentration is 1 mM, and the Hg concentration
is 0.3 μM.
24. Seawater contains approximately 5 μg of Zn2+ per liter.
Calculate what voltage must be put on a zinc anode,
relative to seawater, to completely prevent zinc from
dissolving in seawater.
Chapter 8
The Motions of Biological
Molecules
“. . . extremely minute particles of solid matter, whether obtained from organic or
inorganic substances, when suspended in pure water, or in some other aqueous fluids,
exhibit motions for which I am unable to account, and which from the irregularity
and seeming independence resemble in a remarkable degree the less rapid motions
of some of the smallest animalcules of infusions”
— Robert Brown, Additional Remarks on Active Molecules, Philosophical
Magazine, Series 2, 6(33), 1829.
Concepts
We have already explored the velocities of gas molecules, both their averages and their
ranges. Even in the gas phase, however, the motion of molecules is not simply governed by
their velocity. Molecules collide with each other, and those collisions dominate the longrange molecular motion. In the liquid phase, because molecules are essentially in contact
with each other, even the short-range motion is dominated by collisions.
In a liquid, the random motion of molecules depends on the temperature, but because
collisions are so important in liquids, molecular motion also depends on the sizes and
shapes of molecules. The larger the volume of a molecule, the slower is its motion; spherical
molecules move faster than those with asymmetric shapes of the same volume. The random
motion of molecules is diffusion; by observing the diffusion, we learn about size and shape.
If a force is applied to a molecule, its motion is no longer random; it is directed. On
Earth’s surface, all molecules experience the force of gravity pulling them toward the
center of Earth. They tend to sediment to the bottom of a container unless their random
motion—their diffusion—keeps the solution mixed. Light molecules in solution remain
uniformly mixed; heavy molecules sediment to the bottom. Earth’s gravitational field is
not very strong, so only large particles, such as viruses or chromosomes, will sediment
significantly. By using a centrifugal field, which can be many orders of magnitude larger
than Earth’s field, any molecule can be sedimented. Sedimentation velocity is the motion
of a particle in a gravitational or centrifugal field; its rate depends on the mass plus the size
and shape of the particle. Sedimentation equilibrium refers to the distribution of molecules
in a gravitational or centrifugal field after equilibrium has been reached. The balance
between the sedimenting force in one direction and the randomizing effect of diffusion
means that a concentration gradient is formed; there is a higher concentration near the
bottom of the container. The concentration gradient at sedimentation equilibrium depends
on the molecular weight of the molecules.
The viscosity of a fluid is its resistance to flow. When a fluid flows, one part of the
fluid moves faster than another part. For example, in a fluid moving through a tube, the
part of the fluid at the wall is stationary, while the part in the middle of the tube is flowing
most rapidly. The molecules in different parts of the fluid thus have different average
264
Molecular Motion and Molecular Collisions | 265
velocities in the direction of flow. The resistance to flow, the viscosity, depends on how
strongly the molecules interact. How much do the stationary molecules at the surface of the
tube slow down the faster-moving molecules toward the middle? For example, the intermolecular interaction provided by the three hydrogen-bond donors and acceptors in glycerol,
HOCH2CHOHCH2OH makes it more than 1000 times more viscous than water. When
macromolecules are added to a solvent, the viscosity increases. The amount of increase
depends on the concentration of the molecules and their size and shape.
Electrophoresis is the motion of charged molecules in an electric field. It depends on
the charge on the molecule, plus its size and shape. For proteins and nucleic acids with many
ionizable groups, the charge on the molecule can be adjusted by changing the pH of the
solution. In aqueous solutions, it is very difficult to quantitate the effective electric field and
effective charge on a macromolecule because of all the other ions present. Thus, electrophoresis is used to measure relative sizes and shapes of molecules rather than absolute values.
All the properties we have described depend on the sizes and shapes of macromolecules; the properties are called transport properties. Diffusion and sedimentation measure
mass transport; viscosity measures momentum transport; and electrophoresis measures
charge transport. We care about transport properties because they tell us what the molecules look like in solution. Are they big or small, compact or extended, rigid or flexible?
Furthermore, diffusion and flow are important methods of transport of molecules in biological cells, animals, and plants. We need to know how fast molecules can move from one
place to another in a living cell, for example, and how much their speed changes when the
temperature or molecular size changes.
Applications
Transport properties are used to analyze, separate, and purify cellular particles, proteins,
and nucleic acids. Sedimentation provides fractionation based on differences in sedimentation coefficients, which depend on the mass of the particle, its shape, and its buoyancy—its
density relative to the solvent. Gel electrophoresis is used to separate native proteins that
differ by as little as one charge or denatured proteins that differ by one peptide unit. Twodimensional gel electrophoresis has the ability to resolve nearly every protein present in
a human cell. The changes in the pattern of proteins produced during differentiation of
cells can be followed. Gel electrophoresis can separate nucleic acids that differ by one
nucleotide, and thus determine their sequence, or can separate DNA fragments that are
millions of base pairs in length.
The absolute molecular weights of macromolecules can be determined by a combination of sedimentation and diffusion or by sedimentation equilibrium. Relative molecular
weights can be determined easily by gel electrophoresis. The molecular weights of macromolecules are a key to their identification. The interactions of macromolecules to form
higher-order structures can be characterized by the molecular weight of the complex.
The shapes of macromolecules and how they change in different environments can be
followed by any of the transport properties. Thus, all sorts of reactions of macromolecules
are studied by measurements of diffusion, sedimentation, viscosity, or electrophoresis.
Molecular Motion and Molecular Collisions
In chapter 2 we derived expressions for the average kinetic energy and the average squared
velocity of molecules in an ideal gas, and in chapter 5 we went on to examine the distribution of molecular velocities at thermal equilibrium. What we have neglected so far are
the collisions between molecules. Molecular collisions are central to the distribution of
266 Chapter 8 |
The Motions of Biological Molecules
(a)
(b)
(c)
d
d
FIGURE 8.1 Two cars passing on
a highway with (a) no collision
(b) collision (c) glancing collision.
t3
t3
t2
t1
t2
t1
v2
v1
FIGURE 8.2 Two gas molecules
passing in close proximity,
showing their velocity vectors
and their locations at times
t1, t2, and t3.
velocities, since only by colliding can molecules redistribute energy throughout the gas.
More importantly, the large scale motions of molecules in a gas undergoing intermolecular
collisions is very different from those in a gas at low pressure, in which the molecules can
travel for macroscopic distances without ever interacting with each other.
Let’s say you’re in class, and someone walks into the classroom on the other side,
wearing heavy perfume or cologne. Using the kinetic theory of gases, it’s easy to estimate
that a ‘typical’ perfume component, say a terpene with M ~ 200 g>mol, should (at ambient temperature) be travelling in excess of 100 m s-1. A naïve application of the kinetic
theory of gases would predict the perfume reaches every point in the room within a
fraction of a second. But, of course, experience tells us that is not what happens. The
scent wafts slowly across the room, first to the people seated next to the wearer, then a
little further, and so on. It might take a substantial fraction of an hour to be detectable
across the room.
Where does the kinetic theory go wrong? It does not account for collisions. The
molecules in the scent do move fast, but they also collide frequently with the nitrogen and
oxygen molecules in the air, knocking them off their track. Instead of a straight trajectory
across the room, they move along a path very similar to the three-dimensional random
walks we discussed in chapter 5. We already know the general properties of these random
walks; what we need to explore now are the details of how far the molecules travel between
collisions and how frequently the collisions occur.
The Collision Tube
The ‘collision tube’ is a useful concept that allows us to slice through the complexities of a
system of huge numbers of molecules, with a wide range of velocities, in ceaseless collision
with each other. Let’s start by considering a very familiar model; two cars approaching
each other on a two-lane highway. Let’s further assume that you, the driver in the light
green car, are travelling on a straight course down the middle of your lane, while the driver
in the darker green car, a car of identical width is talking on a cellphone and distractedly
weaving from side to side. Will you collide?
We start with a very basic but important point; it really doesn’t matter how wildly
driver 2 is weaving, except at the exact time the two of you pass each other. If the car
happens to be in the center of the right hand lane at that instant (figure 8.1 case a), you’re
home free. If the driver is well over in your lane (case b), a collision is inevitable. The
interesting case is case (c), where the left edges of your two cars are exactly in a straight
line, and there is a grazing collision (goodbye wing-mirror!). In a grazing collision such
as case (c), the centers of your two cars are separated by a perpendicular distance equal
width of one car, as can be seen from the figure.
Gas molecules move in three dimensions, but share many of the same features.
Figure 8.2 shows two gas molecules passing near each other; their respective velocities
are v1 and v2, and their positions at times t = t1, t2 and t3 are marked. If we graph the
distance between them, d12 = 兩 v1 - v2 兩 as a function of time, we obtain a graph similar to
that shown in figure 8.3. The intermolecular distance is parabolic; its minimum value we
call the distance of closest approach. Just as with the cars, if d12 is less than the diameter
of one molecule, d, the molecules will collide.
We now do a thought experiment. We take you out of the light green car, shrink you by
a factor of 1010, and seat you on one of the molecules — for simplicity’s sake, let’s say it’s
an atom of argon, and set it going. Other molecules will whiz by. At any instant, any molecule whose distance from you is less than the atomic diameter d will collide. If we draw
a circle around you, perpendicular to the direction of travel, and with a radius equal to
d, the diameter of the argon atom, as your molecule moves, the circle will trace out a
tube of length vt, where v is your speed and t the time of travel. This tube is called the
collision tube, because you will collide with any other atoms that, at their point of closest
approach, fall within the tube. How many atoms is that? Well, we know the volume of the
collision tube; it’s equal to the length times the cross-sectional area. The cross sectional
area is just πd2. So the volume of the tube is V = πd2vt. Because the motion of the molecules
is random, their distribution at the distance of closest approach is just that of an ideal
gas, and so the number of moles of gas within volume V is n = pV>RT. The number of
molecules is just the number of moles multiplied by Avogadro’s number. So the number
of molecules within the tube is
N = NA pV>RT = NA ppd 2 vt>RT .
(8.1)
But, you might (and should) object, once a collision occurs, the velocity will be changed!
Yes indeed, but since the long axis of the tube is pointed along the direction of travel, a
change in the direction of travel is unimportant, from the standpoint of the molecule.
However, the magnitude of the velocity might also change, and so, for a path involving
many collisions, we had better replace v with the average velocity of the molecules in the
gas 8v 9. No problem! We learned how to calculate that in chapter 5!
8v 9 =
8 RT
A pM
d12
Molecular Motion and Molecular Collisions | 267
distance of
closest approach
t
FIGURE 8.3 The intermolecular
distance between two passing
gas molecules, as a function
of time.
(5.16)
A much more subtle objection is that we should not be considering the absolute value of
the average velocity of the gas molecule, but the average velocity relative to other molecules
of the gas. The average relative velocity 8vrel 9 equals the average velocity times √2. We can
now improve Eq. 8.1, inserting the value of 8v 9 from Eq. 5.16:
N =
N Appd2 8v rel9 t
N Appd2t 16RT
16
=
= N Appd2t
RT
RT A pM
A pRTM
(8.2)
The number of collisions per unit time can be obtained by dividing by t:
z =
N
16
= N Appd2
t
A pRTM
(8.3)
This is a simplistic picture. In particular, it treats atoms and molecules as hard
spheres. As we will learn in chapter 11, atomic and molecular orbitals extend out to infinity. Molecules that pass each other at a distance too large to give a ‘hard-sphere’ collision
nonetheless influence each other weakly. Similarly, two atoms colliding at a high relative
velocity can approach closer than the nominal hard-sphere diameter. It is therefore better
center within tube
— collision
d
d
center outside tube
— no collision
FIGURE 8.4 The collision tube.
The green spheres depict atoms
at their distance of closest
approach to the atom indicated
by the grey sphere. Those green
spheres whose centers fall
within the collision tube whose
radius equals the diameter of
the grey sphere, give rise to
collisions.
268 Chapter 8 |
The Motions of Biological Molecules
to replace the collision tube area πd2 with a more empirical quantity called the collision
cross section . This still has the dimensions of area, but is experimentally determined
rather than being a function of a nominal molecular diameter. In the real world, collision
cross sections are somewhat dependent on the molecular velocity:
z = N Aps
16
A pRTM
(8.4)
We can also define a mean free path , which is the average absolute (not relative) distance
a molecule travels between collisions. It is obtained by dividing the average velocity by
the number of collisions per second:
l =
8v9
z
=
8RT
RT
16
n a N Aps
b =
A pM
A pRTM
22N Aps
(8.5)
E X A M P L E 8 .1
For N2 gas at 1 bar and 25°C, calculate (a) the number of collisions each N2 molecule
undergoes in 1 s, (b) the total number of collisions in a volume of 1 m3 in 1 s, and
(c) the mean free path of an N2 molecule. The diameter of an N2 molecule can be taken
as 0.374 nm.
SOLUTION
a. With d = 3.74 : 10-10 m, p = 105 Pa, T = 298.15 K, M = 0.028 kg>mol, and applying Eq. 8.3, z = 7.17 : 109 s-1.
b. For N2 gas at 1 bar and 25 °C, the number of moles per unit volume is p>RT, and the
number of molecules per unit volume is
NAp
(6.022 * 1023 molecules mol-1)(105 Pa)
N
= 2.45 * 1025 molecules m-3 .
=
=
V
RT
(8.314 J K-1 mol-1)(298 K)
From (a.), the number of collisions each molecule encounters per unit time is
z = 7.17 : 109 s-1. Since every collision involves two molecules, the total number of
collisions per m3 is
2.45 : 1025 molecules m-3 : 7.17 : 109 s-1>(2 molecules>collision)
= 8.71 : 1034 collisions m-3 s-1.
c. From Eq. 8.5, the mean free path is
l =
(8.31447 J K - 1 mol - 1)(298 K)
22p(6.022 * 10
23
-1
mol )(10 Pa)(3.74 * 10
5
- 10
2
m)
= 6.62 * 10 - 8 m .
Thus, at 1 bar and 25 °C a typical N2 molecule travels almost 200 times its diameter
between collisions.
Random Walks in a Gas
When we discussed random walks in chapter 5, we discovered that the mean square displacement 8d29 for a random walk of equal steps is given by Nl2, where l is the step length
and N the number of steps (Eq. 5.65). In a gas we do not have equal steps, but if the number
of steps is large enough, we can replace the step length with the mean free path. The number
of steps is just the number of steps per unit time, times the time elapsed, or zt. This gives
8d29 = ztl2 .
(8.6)
Diffusion | 269
Substituting from Eqs. 8.4 and 8.5, we obtain
3
29
8d
= zl t = 8 v 9 lt =
21RT2 2t
2
2MpNA ps
.
(8.7)
This indicates that the mean square displacement should depend linearly on time,
on temperature to the power of 3/2, and on the reciprocal of pressure. The dependence on
molecular size will be complex, since size enters the equation reciprocally with the collision cross-section and with the inverse square root of the molecular mass.
Diffusion
The Diffusion Coefficient and Fick’s First Law
We can calculate the rate of diffusion of molecules in a dilute gas, such as the scent
molecules we discussed above, from the mean speed and mean free path of the molecules.
In a liquid, the mean free path is extremely short, and cannot be meaningfully calculated.
A macroscopic treatment of diffusion is more convenient for liquids, although the random
motion of individual molecules can certainly be measured by single-molecule techniques.
Diffusion occurs whenever there is a concentration difference in a container; diffusion
will eventually make the concentration uniform. Diffusion is a slow process compared to
mixing caused by stirring, for example. But over small distances—as occur in biological cells—diffusion can be an effective method for transporting molecules. The rate of
diffusion depends on the size and shape of the molecule and on the properties—such as
viscosity—of the solvent.
One way to measure diffusion is to measure the movement of molecules caused by a
concentration difference. Think of molecules in solution crossing a 1 m2 cross-sectional
area in the yz-plane (figure 8.5). We define the flux Jx as the net amount of solute that
diffuses through this unit area, per unit time, in the x-direction. The SI units of flux are
mol m-2 s-1. If there is no concentration gradient—no difference in concentration—in the
x-direction, you expect equal numbers of molecules crossing this area from the left and
from the right. Therefore, the net flux Jx is zero. Suppose the concentration increases as x
increases; this means that the concentration gradient dc >dx is positive. The concentration
increases with x; therefore, there are more molecules per unit volume to the right of the
area than to the left. You expect, then, that more molecules per unit time diffuse across
the area from the right than from the left. In other words, there is a net transport of material in the direction opposite to the concentration gradient, as shown in figure 8.5. The
steeper the concentration gradient, the larger is the net flux. These considerations led to
the equation for Fick’s First Law:
Jx = - D a
dc
b
dx
(8.8)
where D, the proportionality constant, is called the diffusion coefficient. The negative
sign indicates that the net transport by diffusion is in a direction opposite to the
concentration gradient. The units of D are m2 s-1, although much of the literature,
particularly the older literature, still uses units of cm2 s-1. The concentration gradient
has units of mol m-4. This corresponds to concentration of solute per distance traveled.
As diffusion proceeds, the solution becomes more uniform and the concentration
gradient dc > dx decreases; eventually, the system approaches homogeneity, and dc > dx
approaches zero. Thus, Eq. 8.8 is given in terms of the instantaneous flux at any
time t. Equation 8.8, Fick’s First Law of Diffusion, has been shown experimentally
to be correct.
z
y
x
Jx
c
FIGURE 8.5 Flux Jx is the net
diffusional transport of material
per unit time in the x-direction,
across a unit cross-sectional
area (enclosed by the dashed
line) perpendicular to x. The
concentration c increases with
increasing x; the flux Jx is in
the direction opposite to the
concentration gradient.
270 Chapter 8 |
The Motions of Biological Molecules
Fick’s Second Law
It is also possible to describe how the concentration in a gradient changes with time.
In a uniform concentration gradient, where dc > dx is constant for all values of x,
Eq. 8.8 tells us that the flux Jx is the same at all positions and, therefore, c will not
change with time. (The flux into every volume element from one side is exactly equal
to that going out the other side.) This describes a steady-state flow of material by
diffusion.
However, if the concentration gradient is not the same everywhere, the concentration will change with time. The change of concentration with time ∂c >∂t at position x is
proportional to the gradient in the flux ∂J>∂x at that point. Intuitively, if more material
is diffusing into the volume element from the right than is diffusing out to the left, then
∂J>∂x < 0 and the concentration inside should increase with time, ∂c >∂t > 0 This logic
leads to Fick’s Second Law:
a
0c
02c
b = Da 2 b
0t x
0x t
(8.9)
We use the notation of partial derivatives because the concentration depends on both time
t and distance x. Fick’s Second Law states that the change in concentration with time at
any position x, (∂c> ∂t)x is proportional to the second derivative of the concentration with
respect to x at time t, (∂2c> ∂x2)t. The proportionality constant is again the diffusion coefficient D, assumed to be independent of concentration. An illustration of the application
of Fick’s Second Law is shown in figure 8.6.
The Einstein-Smoluchowski Relation
In 1905, the year Einstein burst onto the scene with several momentous discoveries in
physics, most famously the theory of special relativity, he published an almost equally
brilliant but less well known paper relating macroscopic diffusion with the microscopic
mean square displacement of molecules undergoing random walks. The proof is simple
and elegant.
Einstein considered material diffusing from a point source, and showed that if x is
the radial distance from the source, then the solution to Fick’s Second Law as a function
of radial distance and time is
1
24pDt
exp a
(8.10)
t = 20
distance, x
t = 400
concentration, c
distance, x
distance, x
-x2
b.
4Dt
concentration, c
t=0
concentration, c
FIGURE 8.6 Simulated
diffusion of a material with a
concentration gradient that
is initially a step, obtained
by numerically propagating
Fick’s Second Law. By t = 20
(units of time are relative) the
step has already smoothed
out; subsequent evolution is
much slower.
concentration, c
c(x,t) =
t=∞
distance, x
Problems | 371
3. The kinetics of the reaction
I - + OCl - m OI - + Cl was studied in basic aqueous media by Chia and Connick
[J. Phys. Chem. 63:1518 (1959)]. The initial rate of I- disappearance is given below for mixtures of various initial
compositions, at 25°C (None of the solutions initially contained OI- or Cl-):
Initial
composition
of Iⴚ (M)
Initial
composition
of OClⴚ (M)
2 * 10 - 3
1.5 * 10 - 3
-3
1.5 * 10 - 3
1.00
3.6 * 10 - 4
2 * 10 - 3
3 * 10 - 3
2.00
1.8 * 10 - 4
-3
-3
1.00
7.2 * 10 - 4
4 * 10
4 * 10
3 * 10
Initial
Initial rate
composition of (mol Iⴚ Lⴚ1sⴚ1)
OHⴚ (M)
1.00
1.8 * 10 - 4
K
b. A + B m AB (fast to equilibrium)
k
Find the values of a, b, and c.
b. Calculate the value of k, including its units.
c. Show whether this rate law is consistent with the
mechanism
K1
OCl - + H2O m HOCl + OH - (fast, equilibrium)
K2
K3
HOI + OH - m H2O + OI -
k2
B + C h D (assuming steady state of B)
d[I - ]
= k[I - ]a[OCl - ]b[OH - ]c .
dt
I - + HOCl h HOI + Cl -
k1
a. A m B
k3
a. The rate law can be expressed in the form
-
a. What is the order of the reaction with respect to A?
b. What is the order of the reaction with respect to B?
c. Use your conclusions in parts (a) and (b) to write a differential equation for the appearance of C.
d. What is the rate coefficient k for the reaction? Do not
omit the units of k.
e. Give a possible mechanism for the reaction and discuss
in words, or give equations, to show how the mechanism is consistent with the experiment.
6. A reaction is zero order in substance S. Starting with the
differential rate law, derive an expression for t½ in terms
of the starting concentration [S]0 and the zero-order rate
coefficient k.
7. Write an expression for the rate of appearance of D from
each of the following mechanisms:
(slow)
(fast, equilibrium)
4. The mechanism for a set of reactions is
k1
A h B
k2
B + C h D.
a.
b.
c.
d.
Write a differential equation for the disappearance of A.
Write a differential equation for d[B]>dt.
Write a differential equation for the appearance of D.
If [A]0 is the concentration of A at zero time, write an
equation that gives [A] at any later time.
5. The stoichiometric equation for a reaction is
AB + C h D
8. The following data were obtained for the concentration vs.
time for a certain chemical reaction. Values were measured
at 1.00 s intervals, beginning at 0.00 and ending at 20.00 s.
Concentrations in mM are:
10.00, 6.91, 4.98, 4.32, 3.55, 3.21, 2.61, 2.50, 2.22, 1.91,
1.80, 1.65, 1.52, 1.36, 1.42, 1.23, 1.20, 1.13, 1.09, 1.00, 0.92
a. Plot concentration (c) vs. time, ln c vs. time, and 1>c
vs. time.
b. Decide whether the data best fit zero-order, first-order,
or second-order kinetics. Calculate the rate coefficient
(with units) for the reaction and write the simplest
mechanism you can for the reaction.
c. Describe an experimental method that you think might
be used to measure the concentrations that are changing
so rapidly. The reaction is nearly over in 20 s.
9. The kinetics of double-strand formation for a DNA oligonucleotide containing a G·T base pair was measured by
temperature-jump kinetics. The reaction is
k1
2CGTGAATTCGCG m DUPLEX .
k-1
The following data were obtained:
A + BSC + D.
The initial rate of formation of C is measured with the following results:
Initial concentration of A (M)
Initial concentration of B (M)
Initial rate
(M sⴚ1)
1.0
1.0
1.0 * 10 - 3
2.0
1.0
4.0 * 10 - 3
1.0
2.0
-3
1.0 * 10
Temperature (°C)
k1(105Mⴚ1 sⴚ1)
kⴚ1(sⴚ1)
31.8
36.8
41.8
46.7
0.8
2.3
3.5
6.0
1.00
3.20
15.4
87.0
a. Determine Ea, ΔH‡ and ΔS‡ for the forward and reverse
reactions; assume that the values are independent of
temperature.
272 Chapter 8 |
The Motions of Biological Molecules
Determination of the Diffusion Coefficient
Classical Flux Measurements
In principle, D can be determined using either of Fick’s laws. A simple way to measure a
diffusion coefficient is measure the amount of material that is transferred through unit area
per unit time—the flux J, analyzing the data using Fick’s First Law. We use a porous glass
disk of thickness Δ x to separate two solutions of two different concentrations, as shown in
figure 8.9. The effective area of the porous disk is determined by calibrating the disk with
a substance of known diffusion coefficient. Dividing the rate of transfer of material by
the effective area of the porous disk gives the flux J. The diffusion coefficient is obtained
from Fick’s First Law (Eq. 8.8):
D = -Ja
porous
disk
solution
solvent
FIGURE 8.9 Measurement of
the diffusion coefficient by the
method of Northrup and Anson.
(Based on data from J.H. Northrop
and M.L. Anson, “A method for
the determination of diffusion
constants and the calculation
of the radius and weight of the
hemoglobin molecule,”Journal of
General Physiology, 12, pp. 543–554.
Copyright (c) 1929 The Rockefeller
University Press.)
x
dx
b _ -Ja b
dc
c
(8.13)
where J is the flux in units of mol m-2 s-1, Δc is the difference in concentrations across
the disk in mol m-3, Δx is its thickness in m, and D the diffusion coefficient in m2 s-1.
D in general varies with concentration. In the experiment shown in figure 8.9, we can
obtain D at any desired concentration, and extrapolate to zero concentration.
To use Fick’s Second Law ( Eq. 8.9) to measure a diffusion coefficient, we measure
the concentration c or the concentration gradient dc > dx as a function of time. For
example, the experiment shown in figure 8.6 can provide a diffusion coefficient. A step
concentration gradient is prepared at the beginning of the experiment, either by carefully removing a slide separating the solutions, or sometimes naturally in the course
of centrifugation (see below). The concentration is usually measured optically, using
the absorbance or refractive index of the solution. The evolution of the concentration
gradient is then followed as a function of time; these days, the evolution of the gradient
is usually fit numerically.
Single Molecule Spectroscopy
In contrast to macroscopic flux measurements, which date back to the 19 th century,
the modern technique of near-field fluorescence microscopy allows us to image
single molecule trajectories in one, two and even three dimensions. These methods
allow us to quantify 8d29 as a function of time, and obtain diffusion coefficients via
the Einstein-Smoluchowski equation ( Eq. 8.12). Fluorescently-tagged receptors have
been observed executing one-dimensional random walks up and down DNA strands;
membrane proteins have been imaged diffusing in two dimensions within membranes;
and free proteins in solution have been recorded in three dimensional Brownian
motion. Because we’re presenting the data in two dimensions on a printed page or
computer screen, it’s convenient to illustrate two dimensional random walks within
membranes.
Figure 8.10 shows a recent example from the research literature. Molecules of
the membrane protein GRP1-PH were tagged with a fluorophore, suspended in a
phospholipid membrane, and their position recorded every 0.05 s using a near-field
microscope. The length of each step was measured, squared, averaged, and divided by
4 : 0.05 (because 8d29 = 4Dt for two-dimensional diffusion), to give a value of D of
2.0 : 10 -12 m 2 s-1.
Diffusion | 273
(b)
(a)
10
n
8
6
1 µm
4
2
0.25 0.50 0.75 1.00 1.25 1.50
r (µm)
EXAMPLE 8.3
The diffusion coefficient of the oxygen-storage protein myoglobin (Mb) is D20,w =
1.13 : 10-10 m2 s-1 (D20,w is the diffusion coefficient at 20°C in water). Estimate a
mean value for the time required for an Mb molecule to diffuse a distance of 10 μm
along a particular direction (say x). 10 μm is the order of the size of a cell.
SOLUTION
The mean-square displacement along the x direction is 8x29 = (10 μm)2. From Eq. 8.12a:
t =
8 x2 9
2D
=
(10 - 5 m)2
= 0.44 s
2(1.13 * 10 - 10 m2 s - 1)
Although the diffusion coefficient of Mb in a cell is smaller than in water because of
the higher viscosity of cytoplasm, nevertheless the diffusion of even relatively large
molecules such as Mb (mol wt 17,000) occurs rapidly across dimensions comparable to
the size of a cell.
Other Methods
The preceding methods start with a non-equilibrium system, and the diffusion coefficient
is evaluated from the way the system moves toward equilibrium. But since the diffusion
coefficient D is related to the random motion of the molecules, one should be able to
measure D for a system at equilibrium by monitoring the random motion of molecules
directly. Two such methods are laser light scattering, sometimes called quasi-elastic light
scattering, and pulsed field-gradient nuclear magnetic resonance (PFG-NMR).
If a beam of monochromatic light of frequency n0 passes through a homogeneous solution, some of the light is deflected and it is no longer monochromatic. Some molecules will
be moving toward the light and some away; this causes a Doppler broadening of the scattered light. The intensity of the scattered light, when plotted as a function of frequency, has a
maximum at n0 and its width at half-height is directly proportional to D (for molecules whose
characteristic dimensions are small compared with the wavelength of the light). For a typical
protein molecule with a D of 10 -10 m2 s-1, the width of the spectrum of the scattered light is
of the order of 10,000 Hz. This is still an extremely sharp line compared with the frequency
FIGURE 8.10 (a) Trajectory of
a single molecule of the general
receptor for phosphoinositides,
PH domain (GRP1-PH) in a
biological membrane, measured
by single-molecule fluorescence
microscopy. Each dot
corresponds to one frame of the
video recording; 20 frames were
collected per second, so the full
trajectory covers 2.1 s. (b) Each
bar represents the number of
frames in the trajectory falling
within a bin of length 0.25 μm.
The green curve is calculated
using Eq. 8.10 from a diffusion
coefficient D computed from the
average squared step length in
(a), using Eq. 8.12.
(Based on data from J.D. Knight,
M.G. Lerner, J.G. Marcano-Velazquez,
R.W. Pastor, J.J. Falke, “Single
Molecule Diffusion of MembraneBound Proteins: Window into Lipid
Contacts and Bilayer Dynamics,”
Biophysical Journal, 99, pp.
2879–2887. Copyright (c) 2010
Elsevier, Inc. (http://linkinghub.
elsevier.com/retrieve/pii/
S0006349510010428).)
274 Chapter 8 |
The Motions of Biological Molecules
of the light (1015 Hz), but lasers can be frequency stabilized to a linewidth of around 1 kHz.
If better resolution is needed, the frequency of the incident photons can be correlated to those
of the scattered photons, allowing even higher resolution. Diffusion coefficients of macromolecules can be measured by this method. Laser light scattering is most efficient when the
wavelength of the light is comparable to the size of the macromolecule.
Diffusion coefficients for smaller molecules can also be measured by pulsed field
gradient NMR. Molecules in an inhomogeneous magnetic field experience a change in
NMR frequency as they diffuse; these changes in NMR frequency cause dephasing of various kinds of spin-echoes, which can be measured by multidimensional NMR techniques.
A more extensive discussion of this method is given in chapter 13.
Gas
D, m2s-1
Ne
5.2 : 10-5
O2
2.3 : 10-5
N2
2.2 : 10-5
Ar
1.8 : 10-5
CO2
1.1 : 10-5
CH4
2.2 : 10-5
C2H6
1.0 : 10-5
Values of the Diffusion and Self-Diffusion Coefficient
C3H8
6.0 : 10-6
n-C4H10
4.3 : 10-6
n-C5H12
3.3 : 10-6
Values of D for low-molecular mass gases at 1 bar pressure and 298 K are given in the sidebar. It is clear that as the molecular mass increases, the self-diffusion coefficient (the diffusion coefficient of a molecule in the pure substance) D decreases more than reciprocally;
M enters Eq. 8.7 both directly as the inverse square root and indirectly through the collisional cross-section. Empirically, within similar classes of compounds, gas phase values
of D are found to be proportional to M-n, where n is somewhat larger than 1.
The short list of self-diffusion coefficients for liquids at 298 K illustrates two principles.
First, for the same substance (here, n-pentane), diffusion coefficients are two to three orders
of magnitude slower in liquids than they are in gases at the same temperature and 1 bar
pressure. This can be attributed primarily to differences in the mean free path, which, as
we have seen, can be hundreds of molecule-lengths in gases at 1 bar, but is a fraction of
the length of a molecule in the liquid state. In addition, it is clear that viscosity has a major
effect on diffusion in the liquid; the highly mobile liquid, pentane, has a diffusion constant
more than 100 times greater than the comparably sized, but highly viscous, glycerol.
When we consider diffusion of molecules in water, changes in viscosity are unimportant, but molecular size plays an important role. As can be seen in the table on the right, as
macromolecular size increases, the diffusion coefficient decreases, but much more slowly
than in gases. Increasing the molecular mass by a factor of over 1000, as we do going from
5-S ribosomal RNA to tobacco mosaic virus (TMV), causes the diffusion coefficient to
decrease by a factor slightly more than 10. This apparent inverse cube root dependence is
justified theoretically in the next section.
D, m2s-1
Liquid
n-C5H12
5.6 : 10-9
CH3OH
2.6 : 10-9
C2H5OH
1.2 : 10-9
glycerol
2.5 : 10-12
M (kD)
D, m2s-1
lysozyme
14.3
1.1 : 10-10
5S-rRNA
40
5.9 : 10-11
hemoglobin
68
7.0 : 10-11
Protein
TMV
40800 4.5 : 10 -12
The Frictional Coefficient
We learned in basic physics that objects in the earth’s gravitational field are subject to an
almost constant acceleration g. We also know that in practice everyday objects in a gravitational field do not continuously accelerate. Once a certain velocity is reached, acceleration
tapers off. That limiting velocity due to friction with the air or for an object sinking in the
water is called the terminal velocity vt.
Since objects subjected to a an external force Fext must accelerate — it’s required by
Newton’s Second Law — it follows that at the terminal velocity the object must experience a retarding force that is exactly equal in magnitude and opposite in direction to the
driving force. Under a limited but quite common set of conditions, the frictional force Ff
is proportional to the velocity, and so we can write
Ff = -Fext = fvt .
(8.14)
The frictional coefficient f (in physics it is called the coefficient of kinetic friction) is a property of the individual object. For example, f of a skydiver is increased by a factor of about
eight when the parachute opens; if it did not, it is unlikely skydiving would be a popular sport!
The Frictional Coefficient | 275
f and D
Consider a solution of lead nitrate in water. Lead ions, like lead metal, are much denser
than water, and are subject to gravitational acceleration. If we put lead shot in a jar of
water, it simply sinks to the bottom. Why don’t lead ions do the same? The difference lies
in the balance between diffusion, which acts by Fick’s First Law to decrease concentration
gradients; and gravity, which pulls the dense objects to the bottom of the jar and creates
concentration gradient. Einstein used the balance between external forces, like gravity
and diffusion, to derive a relation between the frictional coefficient of molecules and the
diffusion coefficient.
Let’s consider the balance between two fluxes; the flux due to diffusion, J D, governed by Fick’s First Law; and the flux Jext due to the external force Fext. The two fluxes,
at equilibrium, must be equal and opposite. Let’s assume further that the external flux
is due to particles that have reached terminal velocity. The force on these particles is
the derivative of the external potential energy and along its direction of maximum
gradient x:
Fext = -dVext >dx
(8.15)
We can then combine Eqs. 8.14 and 8.15:
vt = -(1>f)(dVext>dx)
(8.16)
The external flux is just the concentration times the terminal velocity
Jext = cvt = -(c>f)(dVext >dx) ,
(8.17)
and the total flux, which is zero, is the sum of external and diffusion flux, given by Eq. 8.8:
Jext + JD = 0 = cvt = -(c>f)(dVext >dx) - Ddc>dx
(8.18)
Finally, if we assume the external potential varies purely as a function of x, and dropping
the subscript ext, we can apply Boltzmann statistics (Eq. 5.4, with b = 1> k BT ) to infer that
the concentration as a function of x is proportional to the probability of a molecule being
in that position. This gives
c(x) = N exp (-V(x)>kBT ) ,
(8.19)
where N is an unknown normalization factor. Inserting this in Eq. 8.18, we get
0 = -
-V1x2
dV1x2
dV1x2
-V1x2
N
N
exp a
ba
b + Da
ba
b exp a
b .
f
kBT
dx
kBT
dx
kBT
(8.20)
Moving to the second term to the left, cancelling, and rearranging, we get
kBT
(8.21)
.
f
The frictional coefficient is another essential link between the microscopic and macroscopic worlds. For example in the 19th century, the Irish scientist George Gabriel Stokes
found the frictional coefficient of a spherical object moving slowly through a viscous
medium is
D =
f = 6πhr ,
(8.22)
where h is the viscosity (of which we will have more to say later) and r the radius of
the sphere. This immediately explains the experimental inverse-cube-root dependence of
macromolecular diffusion coefficients on mass. D is proportional to 1> f; f is proportional
to r; and r is proportional to V1>3.
276 Chapter 8 |
The Motions of Biological Molecules
If a macromolecule of known volume is spherical and unsolvated, we can in
principle calculate its radius and then its frictional coefficient from Eq. 8.22. This calculated frictional coefficient can be compared with the measured frictional coefficient
determined from the diffusion coefficient ( Eq. 8.21). The calculated f is usually smaller
than the measured f. The reason for this is usually three-fold. First, the hydrodynamic
volume of macromolecules in water (the effective volume as they diffuse) is increased
by tightly-bound water molecules adhering to their surfaces, the so-called shell of
hydration. Second, macromolecules are often not spherical, and spheroidal or ellipsoidal proteins have larger frictional coefficients than spheres of the same volumes. And
third, it has recently been realized that irregularities and clefts in the protein surface,
which are filled with water, also contribute substantially to retarding its translational
and rotational diffusion.
It is difficult to define precisely how much water is in the shell of hydration, since the
tightness with which this water is bound varies substantially. Some waters are interposed in
salt bridges between cationic and anionic sidechains, and are very tightly held; others are
hydrogen-bonded to hydrophilic groups, and others are hydrogen bonded to the hydrogenbonded waters. There is effectively a gradient of viscosity, beginning with a few waters at
the surface that spend most of the time with the protein, tapering down to waters that have
properties close to bulk water but with slightly altered dynamic properties. A reasonable
but imprecise figure for the hydration is d ~ m H2O>mprotein, where m H2O corresponds to
the water that is well-ordered, has a long average residence time on the protein, and is
substantially (10 - 20%) denser than bulk water.
Shape Factor
For a nonspherical macromolecule, the frictional coefficient is greater than that for a sphere
of the same volume, because of the increased surface area in contact with the solvent.
Computer calculations can be done to calculate the frictional coefficient for a particle of
any shape, but there exist closed-form expressions for simple shapes, such as spheroids
(also called ellipsoids of revolution). A spheroid is the shape formed by rotating an ellipse
about one of its axes, by convention the one labeled a. As a result, one axis is distinct (by
convention the a axis), and all radii are identical and equal to b in the plane perpendicular
to the a axis. A prolate spheroid has a > b, while an oblate spheroid has a < b. Figure 8.11
illustrates these two kinds of spheroids.
Perrin in 1934 calculated the frictional coefficients of spheroids. If we define the axial
ratio p = a > b, and j (the Greek letter xi) by
j =
FIGURE 8.11 Prolate spheroids
(ellipsoids of revolution) are
shaped like footballs; oblate
spheroids are shaped like
pancakes. The axis of revolution
is designated by a for each
molecule.
兩 a2 - b2 兩
B
a2
,
(8.23)
a
a
b
b
oblate spheroid 2a = b
prolate spheroid a = 2b
The Frictional Coefficient | 277
we can define a Perrin S factor for prolate spheroids (Sp) by
2.0
1.8
2tanh j
,
j
with tanh - 1j = 12 3 ln11 + j2 - ln11 - j24 ,
(8.24)
prolate
1.6
f/f0
Sp =
-1
1.4
1.2
and for oblate spheroids by
1.0
2 tan - 1j
So =
.
j
5
(8.25)
The ratio between the frictional factor of the spheroid and the frictional factor of a sphere
with the same volume ( f0) is now
10
15
p = a/b
f/f0
2.0
oblate
1.8
1.6
2
3
f
2p
=
.
f0
S
1.4
(8.26)
1.2
Frictional factors for prolate and oblate spheroids are plotted in figure 8.12. For values of
p = a > b ~ 1, the effect of shape is negligible, and it is evident that one either needs a very
flat oblate spheroid or a prolate spheroid to make a substantial difference to f.
EXAMPLE 8.4
The diffusion coefficient D of bovine pancreatic ribonuclease, an enzyme that digests
RNA, in a dilute buffer at 20°C, has the value is D = 1.31 : 10 -10 m2 s-1.
a. Calculate the frictional coefficient f.
b. Assuming that the protein is spherical, calculate its hydrodynamic radius from the
Stokes equation (Eq. 8.22).
SOLUTION
a. From Eq. 8.21, f = kBT>D = 1.38065 : 10-23 J K-1 : 293 K>(1.31 : 10-10 m2 s-1)
= 3.09 : 10-11 kg s-1.
b. From Eq. 8.22, r = f>(6πh). The viscosity coefficient of a dilute aqueous buffer is
approximately equal to that of water, which, at 20°C, is 1.002 mPa s (see table 2.2).
Thus, r = 3.09 : 10-11 kg s-1>(6 : 3.142 : 1.002 : 10-3 Pa s) = 1.63 : 10-9
m = 1.63 nm
Thus, from a macroscopic measurement of concentration of a solution versus time,
we can measure D and thus learn the radius of the diffusing molecules. Hydration and
a nonspherical shape can affect the calculated radius, but probably not by more than
20 to 30%. Of course, if we knew that the molecule was a long rod, we would not use
the Stokes equation for a sphere to calculate its radius.
Diffusion Coefficients of Random Coils
Very flexible macromolecules, such as DNA molecules with many thousands of base pairs,
are different from the largely rigid, structured proteins discussed above. DNA in solution,
with its length much greater than its diameter, resembles a loose ball of thread. As the DNA
moves through the solution, it carries a large amount of solvent with it. Therefore, the DNA
molecule acts hydrodynamically as a highly hydrated sphere. The radius of this sphere is
expected to be directly proportional to the average three-dimensional size of the molecule.
The root-mean-square end-to-end distance in a random polymer of N units is
8d29½ = n½l .
20
(5.66)
0.2
0.4
0.6
p = a/b
0.8
1.0
FIGURE 8.12 f>f0 as a function
of the axial ratio for prolate and
oblate spheroids. f>f0 represents
the increase in the frictional
coefficient over that of a sphere
of equal volume.
278 Chapter 8 |
The Motions of Biological Molecules
The root-mean-square radius of a random polymer is also proportional to the square
root of the number of monomer units. Thus, the root-mean-square radius is proportional
to the square root of the molecular weight of a macromolecule if the macromolecule is a
very flexible coil. DNA molecules with molecular masses M in the millions of Daltons act
as flexible coils; therefore, their frictional coefficients are expected to be proportional to
M1>2, not M1>3, and their diffusion coefficients are expected to be proportional to M-1>2.
This is found to be approximately true.
Sedimentation
Fg
m Bg
Fb
–mBgvBρA
Ff
–fvt
FIGURE 8.13 The three forces
on a particle in free fall.
Sedimentation is a technique used to separate, purify, and analyze all sorts of cellular
species ranging from individual proteins and nucleic acids to viruses, chromosomes, mitochondria, and so forth. Sedimentation occurs even in the Earth’s gravity, but is usually
done in the much larger field of a centrifuge.
We first consider the free fall of a particle in a viscous medium. As shown in
figure 8.13, there are three forces acting on the particle. The most obvious is the gravitational force due to the Earth’s gravitational acceleration ge or the centrifugal acceleration arot = v 2r, v being the centrifuge’s angular velocity in rad s-1 and r being the
distance from the axis of rotation. The second force is the buoyancy of the object,
famously discovered by Archimedes to be equal to the weight of the solvent it displaces.
And the third is the frictional force on the object as it moves through the medium at
terminal velocity.
We have already discussed the first and third forces; we need briefly to discuss
buoyancy. The buoyant force, according to Archimides, is the weight of the water displaced
by the particle, which is mwater g in the Earth's gravity and mwater arot in a fluid. The mass
of the water displaced equals its volume times the density of the water. The volume displaced, per gram of material added, is simply the partial specific volume, vB — the change
in volume of a solution when a mass mB of solute is added:
Macromolecule vB (mL g-1)
protein
0.69 - 0.75
starch
0.601
DNA
0.54
RNA
0.53
Protein
vB (mL g-1)
vB = a
0V
b
0 mB T,P,mA
(8.27)
The partial specific volume can be measured just as implied by Eq. 8.27. The volume
of a solvent is measured and then a known mass of solute is added.
The partial specific volume can depend on the concentration of the solute, the composition of the solvent, temperature, and pressure. It is, in effect, the equivalent of the
reciprocal density of the solute, dissolved in the solvent. Some representative values of vB
in neutral aqueous solutions are given in the sidebar.
So, now we have for the buoyant force F b:
lysozyme
0.742
serum albumin
0.735
lactalbumin
0.735
We can add the buoyant force to the gravitational force, noting they have opposite signs,
to obtain the effective external force Fext:
ribonuclease
0.700
Fext = Fg + Fb = mBarot - rAvBmBarot = mBarot(1 - rAvB)
hemoglobin
0.748
The effective mass of the particle is effectively multiplied (and reduced) by a factor
(1 -rAvB), which accounts for the buoyancy of the solvent.
The external force, according to Eq. 8.14, is balanced by the frictional force exerted
by the solvent. Substituting this into Eq. 8.29, we obtain
Fb = mdisparot = rAVdisparot = rAvBmB arot
mBarot(1 - rAvB) = fvt .
(8.28)
(8.29)
(8.30)
Sedimentation | 279
And so now we obtain an expression for the terminal velocity of a particle sedimenting
in a fluid:
vt =
mBarot 11 - vBrA 2
(8.31)
f
vt is obviously directly proportional to the external acceleration arot, which in a
centrifuge will depend on experimental parameters (the angular velocity and the distance
from the axis of rotation). We can remove this from the right-hand side by dividing across
by arot; what remains is purely dependent on the system (solvent density, solute partial
specific volume, molecular mass, and frictional coefficient). We give this the designation
s for sedimentation coefficient:
s =
m B 11 - v BrA 2
vt
=
arot
f
(8.32)
The dimensions of the sedimentation coefficient are time, and the usual units are
svedbergs, or S, named after The Svedberg, a pioneer of centrifugation. One svedberg is
10 -13 s. Many cell components are named by their sedimentation coefficient — 5S and
16S RNA, for example. The sedimentation coefficient under standard conditions (20°C,
in pure water) is given a special symbol, s20,w.
Analytical Centrifugation
Two methods can be used to measure the sedimentation coefficients of macromolecules.
In one, a homogeneous solution is spun in an ultracentrifuge. As the macromolecules
move away from the meniscus, they form a boundary with the solvent. By following the
movement of the boundary with time, the sedimentation coefficient can be calculated;
by observing the smoothing out of the boundary with time, we can also determine the
diffusion coefficient. This method is called boundary sedimentation and is illustrated
in figure 8.14.
sample
cell
boundary
FIGURE 8.14 A sector
ultracentrifuge. There are
typically six sectors, each
of which contains a sample
chamber and a reference chamber containing only solvent.
The chambers have transparent
windows top and bottom, to
allow light, flash-synchronized
with the rotor, to pass through
the chamber to an imaging
detector, which provides a
snapshot of the difference in
extinction between reference
and sample as a function of the
position in the chamber.
reference
cell
meniscus
slit
xenon
lamp
rotor
side view
imaging system
+detector
rotor, top view
The Motions of Biological Molecules
The primary apparatus for analytical centrifugation is called a sector ultracentrifuge,
illustrated in figure 8.14. A pancake-shaped rotor is divided into sectors, each of which
holds two transparent walled sample chambers. One contains a blank solution; the other
the sample. As the rotor turns, it triggers a high intensity stroboscopic lamp that shines
through the sample chambers; photosensors on the other side record the extinction of the
solution as a function of the position in the sample chamber.
As the boundary migrates outward as a result of the centrifugal force, diffusion
also causes the sharp step it forms at the beginning (because of the sharpness of the
solution-air boundary) to be smeared out by diffusion. If the diffusion coefficient is
small, the boundary remains sharp. If transport by diffusion is significant, the boundary broadens as it moves downfield. As an approximation, we can assume that the
diffusion process does not affect the rate of movement of the boundary itself, which
simply travels at the terminal velocity of the macromolecules. It is usually sufficiently
accurate to calculate s from the position of the midpoint of the boundary relative to
the center of the rotor, x½:
s =
vt
v2x
=
dx 1>2 >dt
v2x 1>2
=
1 d ln x 1>2
dt
v2
(8.33)
If we plot ln x½ against time, the slope should be s v2. It is also possible to extract D from
the same experiment, by fitting the broadening of the boundary using Fick’s Second Law.
For a more rigorous discussion of the calculation of s and D, a more advanced text should
be consulted. Several suggestions are listed at the end of this chapter.
EXAMPLE 8.5
Escherichia coli DNA ligase, an enzyme that catalyzes the formation of a
phosphodiester bond from a pair of 3'-hydroxyl and 5'-phosphoryl groups in a
double-stranded DNA, has a molecular weight of 74,000 and a vB of 0.737 mL g-1 at
20°C. Boundary sedimentation in a dilute aqueous buffer (0.02 M potassium phosphate, 0.01 M NH4Cl and 0.2 M KCl, pH 6.5) at 20.6°C, rotating at 56,050 rpm gave
the following results (data courtesy of P. Modrich):
Time (min)
x1>2(cm)
1.88
0
5.9110
1.86
20
6.0217
1.84
40
6.1141
60
6.2068
1.80
80
6.3040
1.78
100
6.4047
120
6.5133
140
6.6141
ln x1/2
280 Chapter 8 |
1.82
2
4
6
t (s × 103)
8
a. Calculate s.
b. Calculate the frictional factor f in the dilute aqueous buffer; the density of the buffer
at 20.6°C is 1.01 g mL-1 .
Sedimentation | 281
SOLUTION
a. A plot of ln x½ versus t gives a straight line with a slope of 1.323 : 10-5 s-1. The
angular velocity v is 56050 min-1 : (1 min>60 s) : 2π rad = 5870 rad s-1.
From Eq. 8.33, s = 1.323 : 10-5 s-1> (5870 rad s-1)2 = 3.84 : 10-13 s = 3.84 S.
b. From Eq. 8.32, f =
mB(1 - vBrA)
.
s
mB = 74 kg mol-1> (6.022 : 1023 mol-1) = 1.23 : 10-22 kg .
Therefore, f =
(1.23 * 10 - 22 kg)(1 - 0.737 * 1.010)
3.84 * 10 - 13 s
= 8.24 * 10 - 11 kg s - 1
(radians, the length of an arc divided by the length of a radius, are dimensionless).
Standard Sedimentation Coefficient
To facilitate comparison of sedimentation coefficients measured in solvents with different
values of solvent density r and solvent viscosity h it is common practice to standardize
the measured s. s is directly proportional to the buoyancy factor (1 - rAvB) and inversely
proportional to the frictional coefficient f, which, by Eq. 8.22, is directly proportional to the
solvent viscosity. The buoyancy factor depends on the solvent density. Therefore, we can
calculate a standard value of a sedimentation coefficient, chosen to be the value at 20°C
in water, s20,w, by multiplying the measured s by ratios of viscosities and buoyancy factors:
s20,w = s a
h
h20,w
b
(1 - vBrA)20,w
(1 - vBrA)
(8.34)
To use Eq. 8.34 to convert the measured s to s20,w, we must measure the density and viscosity of the solvent we use for the sedimentation experiments. For the calculation to be
valid for different solvents, it is necessary that the solvent not change the shape or solvation of the macromolecule significantly. In Eq. 8.34, we assume that the only effect on the
frictional coefficient is through the solvent viscosity. If two different solvents give very
different values of s20,w, we know that the macromolecule has different conformations in
the two solvents.
EXAMPLE 8.6
Convert the value of s obtained in example 8.5(a) to s20,w. At 20°C the ratio of the
viscosity of the buffer hb to that of water hw has been measured to be hb > hw = 1.003.
SOLUTION
For a dilute buffer, it is sufficiently accurate to assume that its temperature dependence
of viscosity is the same as that of water.
From standard viscosity tables, we obtain h20.6,w = 0.9906 mPa s-1 and h20,w = 1.0050
mPa s-1. Therefore,
h20.6,b
hb h20.6,w
0.9906
= 0.989 .
=
= 1.003 *
h20,w
hw h20,w
1.0050
We assume for the protein that nB at 20.6°C is the same as nB at 20°C. The density of
water at 20°C is 0.9982 g mL-1 (table 2.2.). Thus,
(1 - vBrA)20,w
(1 - vBrA)
=
1 - 0.737 * 0.9982
= 1.034 .
1 - 0.737 * 1.010
282 Chapter 8 |
The Motions of Biological Molecules
Substituting these values into Eq. 8.34 we obtain
s20,w = 3.84 S * 0.989 * 1.034 = 3.93 S .
Sedimentation Equilibrium
As we have seen, a boundary sedimentation experiment involves a competition between
downward movement of the boundary due to sedimentation and smearing out of the boundary
due to diffusion. At long times, and particularly obviously if D is large and s small, an
equilibrium is reached, as predicted by Eq. 8.18, where transport of the solute downward
by sedimentation exactly balances its movement from upward by diffusion. This situation is
called sedimentation equilibrium. Obviously, one extreme of sedimentation equilibrium exists
if the sedimentation flux is low, as it is in the Earth’s field, and the diffusion rate is high, as it is
for small molecules. In this case, diffusion entirely dominates, and so we don’t expect a sugar
or salt solution on the lab bench to sediment out. Nonetheless a sedimentation equilibrium
will be established over long enough time or if the distances are large enough. For example,
the Earth’s atmosphere is essentially a sedimentation equilibrium on a length scale of about
100 km; while the Earth’s gravitational field pulls the air downward, diffusion causes it flow
upward towards space.
We start with Eq. 8.18, noting that the external flux in this case is due to sedimentation (Js):
Js + JD = 0 = -(c>f)(dVs >dx) - Ddc>dx
(8.35)
In this case, -(dVs > dx) is the force on the molecules due to sedimentation; we will
include in it the buoyant force and replace it with mBarot(1 - rAvB). We can also substitute
Eq. 8.21 into the equation, giving
m Barot(1 - v BrA)c
k BT dc
.
=
f
f dx
(8.36)
The frictional coefficient cancels, as it should; it is a function of the diffusion constant
which should have no role in thermodynamic equilibrium. We can now separate variables:
mBarot(1 - vBrA)
dc
dx =
(8.37)
c
kT
This differential equation can be solved by integration. There are two important cases.
If the gravitational field is independent of x (as when a = ge, the Earth’s field), then
ln c =
mBge(1 - vBrA)
x + C
kBT
(8.38)
and we predict that the log of the concentration should be proportional to distance.
Alternatively, if we do a definite integral between two positions x1 and x2, where the concentrations are c1 and c2, respectively, we obtain
ln a
mBge(1 - vBrA)
c1
b =
1x1 - x2 2 .
c2
kBT
(8.39)
Along the direction of the force (in this case, downward) the concentration increases.
In the second case, in a centrifuge, arot = v2x. The definite integral is
ln a
mBv2(1 - vBrA) 2
c1
b =
1x1 - x22 2 .
c2
2kBT
(8.40)
Sedimentation | 283
Finally, if we multiply above and below on the right-hand side by NA, we replace mB by
MB and kB by R:
ln a
MBv2(1 - vBrA) 2
c1
b =
1x1 - x22 2
c2
2RT
(8.41)
To determine a molecular weight from a sedimentation equilibrium experiment, plot
ln c versus x2. This should give a straight line with slope equal to
MBv2(1 - vBrA)
,
2RT
where MB is the absolute molecular weight. This is important, because with gel electrophoresis methods (see following sections) we can measure only relative molecular weights.
The most accurate molecular weights are obtained from the known amino acid sequence
of a protein or the nucleotide sequence of a nucleic acid, or from mass-spectra. These
provide good standards. However, for proteins containing more than one polypeptide, or
if association or dissociation of macromolecules occurs, then sedimentation-diffusion or
sedimentation-equilibrium can provide the molecular weights of the complexes in their
native state.
Molecular Weights from Sedimentation and Diffusion
Equation 8.32 states that the sedimentation coefficient s depends on the mass m B,
the frictional coefficient f, and the buoyancy factor (1 - r AvB) of a molecule. We
also know f can be obtained from a measured diffusion coefficient, via Eq. 8.21. We
can combine the two equations, with the usual twin substitution of M B for m B and R
for k B, giving
MB =
RTs
.
D(1 - vBrA)
(8.42)
Why do this, when we could solve either Eq. 8.32 or 8.21 directly for the molecular weight by using the Stokes-Einstein equation? As we have discussed, a substantial
inaccuracy is introduced by the difference between the Stokes’ Law radius and the hydrodynamic radius. Factoring f out removes a large part of this inaccuracy. Since in many
cases both s and D can be measured using boundary centrifugation, this gives a singleexperiment method of measuring molecular weights with several sources of systematic
error cancelling out.
EXAMPLE 8.7
The following data have been obtained for ribosomes from a paramecium: s20,w = 82.6 S,
D20,w = 1.52 : 10 -11 m2 s-1, and vB = 0.61 mL g-1 (Reisner, Rowe, and Macindoe,
J. Mol. Biol. 32, 587 [1968]). Calculate the molecular weight of the ribosomes.
SOLUTION
From Eq. 8.42:
MB =
(8.314 J K - 1mol - 1)(293 K)(82.6 * 10 - 13 s)
(1.52 * 10 - 11 m2 s - 1)(1 - 0.61 * 1.00)
= 3.40 * 103 kg mol - 1 = 3.40 * 106 g mol - 1
284 Chapter 8 |
The Motions of Biological Molecules
In a given medium for a family of macromolecules of the same vB, we expect s to be
proportional to MB>f. For random coils such as large DNA molecules, f is proportional to
M1>2, so s is proportional to M>M1>2 = M1>2. Experimentally, for large DNA molecules,
s is found to be proportional to M0.44, close to our expectation. Because for spherical,
unsolvated structured molecules of the same partial specific volume, f is proportional to
M1>3, so s is expected to be proportional to M2>3.
Density-Gradient Centrifugation
ρ (g mL–1)
1.8
1.6
1.4
1.2
1.0
30
32
34 36
x (cm)
38
40
FIGURE 8.15 Equilibrium
density gradient formed by centrifuging a solution of 5.1 M CsCl
(ρ = 1.6344 g>mL) at 10,100 rpm
in a 10 cm cylindrical sample
chamber.
FIGURE 8.16 Density-gradient
centrifugation used in a
Meselson-Stahl type experiment
to prove semi-conservative DNA
replication. (a) 2.0 : 107 g>mol
bacteriophage DNA, grown on
a medium containing 100%
15NH +, buoyant in the CsCl
4
gradient shown in figure 8.15
at a density of 1.763 g mL-1.
(b) After 1 replication cycle in a
medium containing only 14NH4+,
the DNA is now buoyant at a
density of 1.751 g mL-1, consistent with its having one 15N
helix and one 14N helix. (c) After
2 replication cycles in a medium
containing only 14NH4+, 50% of
the DNA molecules are buoyant
at the expected 14N DNA density
of 1.739 g mL-1, while half still
have the intermediate density
of 1.751 g mL-1.
A second application of equilibrium centrifugation involves the spinning of a concentrated salt solution, such as CsCl or sucrose, at high speed. A concentration gradient
is generated, which results in a density gradient, because the density of a CsCl or
sucrose solution increases with increasing concentration. If a macromolecular species
is also present in the solution, it will form a band at a point in the salt gradient
where the macromolecules are of the same density as the solution, or, in other words,
the buoyancy term (1 - rA nB ) is zero. The macromolecules above this point sink;
those below it float. This means that macromolecules, that were uniformly distributed
throughout the cell before sedimentation, concentrate at this position when the density
gradient forms. Alternatively, macromolecules layered on the top of the gradient will
sink to a point where they are buoyant, and then stop. This is illustrated in figure 8.16.
The higher the molecular weight of the macromolecular species, the sharper will be
the band that forms in the density gradient. Many biological macromolecules have
“ buoyant densities” sufficiently different that they can be resolved by the densitygradient centrifugation method.
The original experiment that showed that DNA replicated by making a new
complementary strand for each original strand in the parent DNA was done by using
density-gradient centrifugation. Bacteria with DNA containing 15N were grown in a
medium containing 14N. The DNA was isolated as a function of time and analyzed in a
CsCl density gradient as shown in figure 8.16. The original DNA had both strands labeled
with 15N; this DNA is designated as 15N:15N. After one generation, DNA appears with
one 14N strand and one 15N strand (14N: 15N). It is not until the second generation that
14N: 14N DNA appears.
(a)
(b)
(c)
39.5
39.6
39.7
39.8
x (cm)
39.9
40.0
Viscosity | 285
Viscosity
When an external force F is applied to a solid particle to make it move through a liquid
with a velocity vt, the molecules of the liquid in contact with the particle move at the same
velocity vt because of the attractive forces between the two. Far away from the particle, the
liquid remains stationary. Thus, the movement of the particle through the liquid generates
a velocity gradient in the medium.
Whenever a velocity gradient is maintained in a liquid, momentum is constantly transferred in a direction opposite to the direction of the velocity gradient. This is illustrated in
figure 8.17. Here the liquid is moving in the direction x. The x-components of the average
velocities of molecules in different layers are represented by arrows of different lengths.
The velocity gradient is in the y direction; that is, the x-component of the velocity vx
increases with increasing y. The molecules are also moving in the y- and z-directions.
However, because there is no net flow in the y- and z-directions, the movement of the
molecules in these directions is random. Now consider a unit cross-sectional area in the
xz-plane. In 1 s, a certain number of molecules will move across this area from below, and
an equal number of molecules will move across this area from above. But since molecules
from below have lower vx and consequently lower momentum in the x-direction, there is a
net transfer of momentum in the x-direction from above to below—that is, in a direction
opposite to the velocity gradient. The steeper the velocity gradient, the larger is the net
momentum transfer. Mathematically, we write
Jp = -h
dv x
,
dy
(8.43)
where Jp is the rate of momentum transfer per unit time per unit cross-sectional area. It has
the units of flux, and the quantity h is called the viscosity coefficient, or viscosity. Its SI units
are kg m-1 s-1 or Pa s, but in older tables of viscosity are often given in poise or centipoise:
1 poise = 10 -1 Pa s; 1 cP = 1 mPa s. If h is a constant independent of dvx >dy the fluid is
called a Newtonian fluid. If h is itself dependent on dvx >dy the fluid is non-Newtonian.
Thus, when a particle moves through a stationary fluid under an external force F, a
velocity gradient is generated, and the velocity gradient in turn imposes a viscous drag Fⴕ
on the particle. The final velocity gradient is such that F and Fⴕ balance each other, and
the particle moves at the terminal velocity vt.
Measurement of Viscosity
In our discussion of the free fall of a particle through a viscous medium, we obtained
vt =
m Barot 11 - v BrA 2
.
f
(8.31)
Substituting the Stokes equation for a sphere, f = 6phr (Eq. 8.22) we obtain
h =
m Barot 11 - v BrA 2
6prv t
.
(8.44)
Thus, h for a fluid can be determined by measuring vt for a sphere falling through it. Note
also that, from measurements using the same particle, the relative viscosities of two liquids
h2 and h1 can be calculated from
11 - v BrA,2 2 v t2
h2
=
,
h1
11 - v BrA,1 2 v t1
(8.45)
where the subscripts 1 and 2 refer to the quantities for liquids 1 and 2, respectively.
y
moving wall
velocity
gradient
z x
momentum
transfer
stationary wall
FIGURE 8.17 Uniform velocity
gradient produced in a fluid
placed between a moving wall
and a parallel stationary wall.
The velocity gradient is a vector
in the direction of increasing
velocity y and is perpendicular to the direction of flow x.
Momentum transfer occurs in
the direction -y, opposite to the
velocity gradient.
286 Chapter 8 |
The Motions of Biological Molecules
mark 1
mark 2
A more convenient method is to measure the volume rate of flow through a capillary.
A simple viscometer, called an Ostwald viscometer, is shown in figure 8.18. Here the time
t required for the liquid level to drop from mark 1 to mark 2 is measured. The relative
viscosities of two liquids are measured from the ratio of their flow times, t2 and t1 and the
ratio of their densities, r2 and r1:
h2
r2 t2
=
h1
r1 t1
capillary
(8.46)
Viscosities of Solutions
Adding macromolecules to a solvent increases the viscosity of the solution. The viscosity
of a solution of macromolecules depends on concentration and the size and shape of the
macromolecule. The specific viscosity of a solution is defined as
FIGURE 8.18 Simple Ostwald
viscometer. Liquid is drawn
up initially into the arm on the
right to above mark 1. Upon
release, it begins to flow back
under the influence of gravity,
but it is restricted by viscosity
primarily in the capillary region.
The elapsed time t is measured
from when the meniscus passes
mark 1 until it passes mark 2.
hsp K
h - h
,
h
(8.47)
where h is the viscosity of the solvent and hⴕ is the viscosity of the solution. The
specific viscosity is unitless. To separate the effect of concentration on viscosity
from that of molecular size and shape, the intrinsic viscosity [h] is defined as the limit
of the specific viscosity divided by concentration, as the concentration approaches
zero:
3 h 4 K lim
cS0
hsp
c
(8.48)
The units of [h] are those of reciprocal concentration c-1. If c is in g mL -1, the units of [h]
are mL g-1. The intrinsic viscosity can be related to molecular properties by
3 h 4 = n(vB + dvA) ,
(8.49)
where n (nu) is a unitless shape factor that equals 2.5 for spheres and increases rapidly
for nonspherical shapes. Prolate ellipsoids have higher values than oblate ellipsoids; for
an axial ratio of 10, the value of n is about 15 for prolate ellipsoids and about 8 for oblate
ellipsoids. The intrinsic viscosity depends on the volume of the particle, so it depends
on its partial specific volume vB as well as its hydration d and the specific volume of
the solvent vA.
Measurement of the viscosity of a solution of macromolecules is a convenient method
to study denaturation of a protein or the assembly of several molecules into a large particle.
The change of viscosity as a function of time, pH, or temperature is a measure of the change
in shape or volume of the molecules.
Electrophoresis
Biological macromolecules are usually charged; therefore, they will move in an electric
field. Obviously, molecules with a net positive charge will move toward the negative
electrode, and molecules with a net negative charge will move away from the negative
electrode. The velocity of motion depends on the magnitude of the electric field E, the
net charge on the molecule, and the size and shape of the molecule as characterized by its
frictional coefficient f. The net charge on the molecule is designated by Ze where Z is the
number of electronic charges e.
Electrophoresis | 287
The terminal velocity nt (m s-1) of migration of a particle in a nonconducting solvent
in an electric field is
ZeE
vt =
,
(8.50)
f
where Z is the number of charges, e = 1.60218 : 10 -19 C, E is the electric field in V m-1,
and f the frictional coefficient in kg s-1.
The velocity per unit electric field, vt >E, is called the electrophoretic mobility μ.
The electrophoretic mobility and Eq. 8.50 can be used to measure the charge of a particle in a non-conducting medium. The charge of the electron was originally measured
this way with a charged oil drop in air. However, charged biological macromolecules
are found in aqueous solutions with other ions, buffers, and so on, and are surrounded
by an atmosphere of small ions. This ion atmosphere greatly complicates the interpretation of electrophoretic mobility. The shielding effect of the ion atmosphere reduces the
electric field experienced by the macromolecule. Furthermore, when the macromolecule
moves, it drags its ion atmosphere with it, increasing the frictional coefficient. These
complications make it very difficult to interpret electrophoretic mobility quantitatively as
we have done for other hydrodynamic measurements. Electrophoresis is a powerful analytical tool, however, and macromolecules with very small differences in their properties
can be resolved. One example is the separation of normal hemoglobin and hemoglobin
from patients who suffer from sickle-cell anemia. Sickle-cell hemoglobin S differs from
normal hemoglobin A by a single change of valine for glutamic acid in each of the two
b-chain peptides. As a consequence, the proteins differ by 2 charges per molecule, which
is sufficient to effect a clear separation by electrophoresis.
In the following sections, we describe some applications of electrophoresis to the
characterization of macromolecules.
Gel Electrophoresis
Nearly all electrophoresis of macromolecules is done in gels. We want to separate molecules according to charge or size or shape; the gel greatly reduces mixing of the molecules
by convection and by diffusion.
A gel is a three-dimensional polymer network dispersed in a solvent. A variety of
gels have been used. For example, an agarose gel consists of an aqueous medium and a
polysaccharide obtained from agar. An acrylamide gel consists of an aqueous medium
and a copolymer of acrylamide H2C=CH-CO-NH2 and N,N ⴕ-methylenebisacrylamide
H2C=CH-CO-NHCH2NH-CO-CH=CH2. The water-soluble acrylamide, with its reactive
double bond, polymerizes in the presence of a free-radical initiator into a linear chain:
CH2 CH CH2 CH
CO
CO
NH2
NH2
Adding the bisacrylamide, which has two double bonds, results in the formation of crosslinks between different chains. The degree of cross-linking can be controlled by the ratio
of the bis compound to acrylamide. In a typical gel, over 90% of the space is occupied by
the aqueous medium, but the presence of the three-dimensional polymer network prevents
convectional flow.
The gel contributes additional factors that affect the electrophoretic mobility: (1) The
path traveled by a macromolecule through the porous gel is considerably longer than the
length of the gel; (2) the gel imposes additional frictional resistance to the macromolecules
288 Chapter 8 |
The Motions of Biological Molecules
as they move through pores of comparable size; (3) the macromolecules will hydrogen
bond with polar groups on the polymer network; and (4) some pores in the gel will be too
small for a macromolecule to enter. The combination of these factors further improves the
resolution of electrophoresis.
DNA Sequencing
Gel electrophoresis in a denaturing solvent provides a rapid, effective method to determine the sequence of a single strand of DNA. The mobility of a nucleic acid is determined by the number of nucleotide phosphate groups in the strand when electrophoresis
is done in a gel containing a high concentration of a neutral reagent (such as urea or
formamide) that disrupts base pairing. This simple method has been combined with brilliant logic to invent a quick method for determining the base sequence of a nucleic acid.
The problem was to determine the sequence of four bases in a single strand of nucleic
acid. The solution was to make a break in the strand after only one type of base (such
as guanine) but not at every location of this base. If one end of the strand is labeled, for
example, with 32P we end up with strands of different length, each with 32P on one end
and guanine on the other. Measuring the chain length by gel electrophoresis gives us
the positions of all the guanines. Repeating the procedure for each of the other bases
gives the base sequence. This Nobel Prize-winning method was invented independently
by Maxam and Gilbert [Proc. Natl. Acad. Sci. USA (1977) 74: 560-564] and by Sanger
and colleagues [Proc. Natl. Acad. Sci. USA (1977) 74: 5463-5467]. Figure 8.19 shows
an example of the procedure.
FIGURE 8.19 (a) Schematic
illustration of a sequencing gel
electrophoresis pattern. The
sequence of the first 20 bases of
a single strand of DNA is shown.
The strand is labeled at the
5-hydroxyl group of the first
nucleotide with radioactive 32P.
The strand is broken at adenines
(A), guanines (G), cytosines (C),
or thymines (T) and placed in
four lanes at the top of a denaturing gel of 20% acrylamide,
0.67% methylene bisacrylamide,
and 7 M urea. After electrophoresis (the positive electrode is at
the bottom of the gel), autoradiography for 8 h produced the
pattern shown. The sequence
is now simply read off. (b) An
actual gel pattern.
20 C
19 A
18 G
17 C
16 C
A
G
C
T
15 C
14 T
13 T
12 T
11 G
10 G
9A
8C
7A
6G
5C
4A
3C
2G
1G
(a)
Electrophoresis | 289
Sanger’s method uses an enzyme, DNA polymerase, to make a complement of the DNA
being sequenced. A primer (an oligonucleotide complementary to the DNA sequence) is used
to specify where the DNA polymerase starts synthesis. The enzyme adds deoxynucleoside
triphosphates (dNTPs) to the primer to form the newly synthesized strand; it is the one
actually sequenced. The breaks in the strand, which give the different length fragments,
are made by adding a few percent of a dideoxynucleoside triphosphate (ddNTP) to the
dNTP mixture. Incorporation of a ddNTP into the new strand stops the polymerization at
that point. For example, to obtain the positions of the guanines in the newly synthesized
strand, 10% ddGTP is added. Most of the time, dGTP, which allows the chain to continue,
is incorporated, but 10% of the time, ddGTP, which stops the chain, is incorporated. Thus,
a ladder of chains with increasing lengths is obtained. All strands end in G; all start at
the primer. Electrophoresis as shown in figure 8.19 gives the chain lengths. Repeating
the procedure with each of the other bases completes the sequence. By using fluorescent
derivatives of the ddNTPs, we can do the gel electrophoresis in a single lane instead of four
separate lanes. The lane is scanned by a photodetector. Thus, the position of the band gives
the chain length; its fluorescence characteristics identify the base. This method allows many
different experiments to be done on a single gel and provides a more efficient recording of
the sequence data.
There are currently (mid 2012) around 60 fully sequenced human genomes publicly
available; Apple Computer founder Steve Jobs had his genome sequenced for a mere
$100,000; currently, human genome sequencing is offered commercially for $4,000 (bulk
rate). A draft Neanderthal genome has been published, and the Denisovian genome is
currently being sequenced. In addition, hundreds of plant, animal, fungal and protistal
genomes have been sequenced, including all our closest primate relatives.
Double-Stranded DNA
Double-stranded DNA molecules can be separated according to molecular weight by electrophoresis in either polyacrylamide gels or agarose gels. No denaturant, such as urea,
is added, so the gels are called native gels; the DNA runs in its native form. DNA has a
uniform-charge density because each base pair adds two phosphate charges; therefore, in
free solution the electrophoretic mobility is essentially independent of molecular weight. In
gels the DNA molecules have to wander through the pores of the gel, and good separation
can be obtained in the range of 10 - 100,000 base pairs. The concentration of the agarose
or the cross-linking of the polyacrylamide is optimized to give the best separation for each
size range. Figure 8.20 illustrates the excellent separation of six DNA fragments in the
range of 100 - 1000 base pairs.
DNA Fingerprinting
Although all humans have similar sequences of their DNA—the similarity is what will be
found through the Human Genome Project—we are all unique in our total DNA sequence.
Identical twins start out with identical sequences at conception, but mutations during growth
and development cause slight divergences of their sequences. A DNA sample obtained from
skin, blood, semen, hair, or the like, left at a crime scene, for example, can be compared
with the DNA from a suspect to see if they match.
Only about 2% of human DNA codes for proteins; much of the rest is simple repeating
sequences. The number of repetitions of the sequence and the sequences differ from person
to person. A region of the DNA where this occurs has been called the hypervariable
fingerprint region (Jeffreys, Wilson, and Thein 1986). Restriction enzymes cut doublestranded DNA at specific sequences (such as CAATTG). They produce a distinct set of
different length fragments of the fingerprint DNA; the fragments are separated by native gel
electrophoresis (see figure 8.20). This method is referred to as RFLP analysis: restriction
1100 bp
420 bp
290 bp
160 bp
90 bp
67 bp
FIGURE 8.20 Electrophoresis
of double-stranded DNA fragments on 3.5% acrylamide gel
(acrylamide>bisacrylamide 7:3 by
weight). The mixture originally
started as a thin band at the top
of the picture and was electrophoresed toward the positive
electrode. After electrophoresis, the gel was immersed in a
staining solution containing a fluorescent dye, ethidium, and then
photographed under UV light.
The sizes of the fragments are
given in terms of base pairs (bp).
One base pair corresponds to a
molecular weight of 660 g>mol.
(Unpublished data of T-S. Hsieh
and J. C. Wang.)
290 Chapter 8 |
The Motions of Biological Molecules
fragment length polymorphism analysis; it can distinguish members of a single family,
but not identical twins. DNA fingerprinting, often of short-terminal repeat regions of the
DNA, are used in criminal forensics, but have also been used to identify victims of the
September 11, 2001 terrorist attacks, reunite survivors of the Shoah and their descendants
with their families, and study human historical migration patterns. DNA testing is now a
growing element in family genealogy.
Conformations of Nucleic Acids
We have been describing the separation of nucleic acids based solely on their molecular
weights—the number of nucleotides in single strands on denaturing gels or the number
of base pairs in double strands on native gels. But nucleic acids of the same molecular
weight will have different electrophoretic mobilities when they have different shapes. An
obvious example is linear double-stranded DNA versus circular double-stranded DNA.
The compact circular DNA travels faster than the linear molecule. Many natural DNAs
occur as circles or twisted circles (figure 8’s and so forth). These superhelical DNAs differ
only in their topology—the number of twists in the DNA before the ends are joined; they
are called topological isomers, or topoisomers. The topoisomers can be separated and
quantitated by gel electrophoresis. The concentrations of DNA topoisomers are used to
assay for topoisomerases, enzymes that wind and unwind the DNA during its replication.
These enzymes are targets for anticancer drugs as discussed by Liu (1989), because the
most rapidly dividing cells require the largest amounts of these enzymes.
The DNA double helix is a flexible linear rod that is approximately linear for a few
hundred base pairs; because of its flexibility, however, it acts more like a random coil
when it is thousands of base pairs in length. However, certain sequences of DNA, such
as repeating AT pairs, cause the DNA to bend. Many chemicals that react with DNA,
including carcinogens (such as N-acetylaminofluorene) and anticancer drugs [such as
cis-diamminedichloroplatinum (II)] also cause the DNA to bend. All these bent DNAs
have anomalous electrophoretic mobilities and can be distinguished from linear molecules
of the same size; this is reviewed by Leng (1990).
RNA molecules are synthesized as single strands, which then fold into intramolecular
double-strand regions with single-stranded loops. The simplest folded structure is called a
hairpin—a single stem and loop. More complex structures are formed by further folding
and interactions of different kinds of loops and stems. These structures are vital for the
correct interactions of RNA molecules with proteins and with other RNAs. One example is
transfer RNA, which adds each amino acid to the growing polypeptide chain during protein
synthesis. The folding of RNA molecules can be easily monitored by native gel electrophoresis; mutations can be made in the RNA to determine how the sequence determines
the folding (Jacques and Susskind 1991).
Pulsed-Field Gel Electrophoresis
In conventional gel electrophoresis, the gel pores act like a sieve; different-sized DNA molecules pass through with different mobilities. Unfortunately, standard gel electrophoresis
methods do not work on DNA molecules longer than 50,000 bp. The DNA can no longer
pass through the pores in its randomly coiled form. To travel through the gel, the DNA
is distorted by the field and moves along its long axis through the pores. It travels like a
snake, in a process called reptation. Once the DNA is stretched out, its mobility becomes
nearly independent of length and no separation according to molecular weight is obtained.
The solution to this problem is to make the DNA molecules change direction during electrophoresis. If an electric field is applied perpendicular to the direction of motion of the
DNA, the molecule must change shape to allow it to travel in the new direction. The time
Electrophoresis | 291
Negative electrodes
Positive electrode
Negative electrodes
FIGURE 8.21 A pulsed-field gel
electrophoresis (PFGE) apparatus in which the DNA molecules
move alternately down and
to the right in the figure. The
gel is actually horizontal and is
immersed in a buffer solution.
The net motion of the DNA
is along the diagonal of the
apparatus and is approximately
straight in the gel. The angle
between alternate electric fields
ranges from 100 to 150; this
improves resolution. Each electrical pulse continues from 1 s
to several minutes, depending
on the size of the molecules.
Gel
Positive electrode
necessary for this to occur depends on the size of the DNA. By using a pulsed electric
field that alternates in direction, Schwartz and Cantor (1984) were able to separate DNA
molecules in the million base-pair range. Figure 8.21 shows one configuration of a pulsedfield gel electrophoresis (PFGE) apparatus; many other arrangements of electrodes have
been used. The key variable for obtaining optimum separation is the length of each pulse;
the longer the molecules, the longer are the pulses required. Individual DNA molecules can
be seen moving through a gel by using acridine-labeled DNA and a fluorescence microscope. Computer modeling of this motion allows calculation of the optimum pulse length
to separate a given size range (see Smith, Heller, and Bustamante 1991).
An example of the effect of pulse length in field-inversion gel electrophoresis
(FIGE) is shown in figure 8.22. The sign of the electric field is changed periodically to
Origin
kbp
kbp
170
125
48.5
33.5
170
kbp
125
170
125
48.5
33.5
15.0
48.5
33.5
15.0
15.0
Continuous
field
0.50 s
3s
0.25 s
1s
FIGURE 8.22 Separation
of DNA molecules by fieldinversion gel electrophoresis
(FIGE). The efficiency of
separation is increased by
alternately reversing the
direction of the field. Three
separate experiments are
shown on DNA molecules
ranging in size from 15.0
kilobase pairs (kbp) to 170 kbp.
The DNA molecules are T4
bacteriophage, 170 kbp; T5
bacteriophage, 125 kbp; lambda
bacteriophage, 48.5 kbp; two
restriction enzyme fragments of
lambda bacteriophage, 15.0 and
33.5 kbp. The molecules
start at the top of the figure
(labeled “Origin”) and move
down in a 1% agarose gel. In
a continuous field applied for
4 h all sizes move with similar
mobilities. In a field of 0.50 s
forward and 0.25 s back for 12
h, the two restriction fragments
are well separated. In a field of 3
s forward and 1 s back for 12 h,
the lambda, T4, and T5 DNAs are
well separated.
Based on data from Carle, G. F.,
M. Frank, and M. V. Olson. 1986. Electrophoretic Separation of Large DNA
Molecules by Periodic Inversion of
the Electric Field. Science 232:65–68.
292 Chapter 8 |
The Motions of Biological Molecules
obtain improved separation of the DNA molecules. A field that is positive for 0.50 s and
negative for 0.25 s gives good separation in the size range of 10,000 - 50,000 base pairs.
A field that is positive for 3 s and negative for 1 s gives good separation in the size range
of 50,000 - 200,000 base pairs.
Protein Molecular Weights
Nucleic acids can be separated according to molecular weight by electrophoresis because
there is one charged phosphate group for each nucleotide monomer unit. For proteins, the
number of charges depends on their amino acid compositions and the pH of the buffer.
Furthermore, a polypeptide chain of a specific length can fold into different shapes with
different frictional coefficients. Therefore, to use electrophoresis to determine protein
molecular weight, it is necessary to denature the protein and to introduce a charge on each
peptide. This is done by adding an anionic detergent, sodium dodecylsulfate (SDS), and
2-mercaptoethanol; the latter disrupts sulfur-sulfur linkages in proteins. The combined
action of these reagents causes the proteins to unfold. Furthermore, for most proteins at an
SDS concentration greater than 10 -3 M, a nearly constant amount of SDS is bound per unit
weight of protein (approximately 0.5 detergent molecule is bound per amino acid residue).
Thus, the charge of the protein-dodecylsulfate complex is due primarily to the charges of
the bound dodecylsulfate groups, making the charge per unit weight approximately the
same for most proteins.
Under these conditions, the mobilities of the SDS-treated proteins are determined by
their molecular weights. If M is the molecular weight of a protein and x is the distance
migrated in the gel (proportional to the mobility), the relation
log M = a + bx
(Data collected by Gerard Harbison.)
is usually observed, where a and b are constants for a given gel at a given electric field. A
set of proteins of known molecular weight must be used for calibration in the determination
of a protein of unknown molecular weight. An example is shown in figure 8.23.
5.5
5.0
ln [M/(kg mol–1)]
FIGURE 8.23 The
electrophoretic mobilities of
proteins in SDS-polyacrylamide
gel electrophoresis (SDS-PAGE).
The logarithms of the molecular
weights are linear in the
mobilities of the proteins,
as stated in Eq. 8.51. The
slope and intercept of the
linear function depends on
the amount of cross-linking
and the concentration of
the gel. The acrylamide
concentrations illustrated
are 12%,10%, and 7.5%. The
weight ratio of acrylamide to
methylenebisacrylamide is 37:1.
(8.51)
4.5
7.5% acrylamide
4.0
3.5
10 %
12 %
3.0
0.0
0.2
0.4
0.6
μ
0.8
1.0
Electrophoresis | 293
We emphasize that Eq. 8.51 is dependent on two factors: (1) a constant amount of
bound detergent per unit weight of protein and (2) charges due to bound detergent dominate those carried by the protein itself. Deviations from this type of relation occur when a
protein binds an abnormal amount of dodecylsulfate (such as glycoproteins and membrane
proteins) or carries a large number of charges (such as histones).*
Protein Charge
For native proteins, the mobility depends on the net charge and the frictional coefficient. The
net charge depends on the amino acid composition and the charges of any ligands bound covalently or reversibly. The net charge is always a function of pH; therefore, by judicious choice of
pH, different proteins can be separated. All proteins will be positively charged at low enough
pH because the carboxyl groups of aspartic and glutamic acid will be neutral (-COOH), but
the amino groups of lysine and arginine will be positive (-NH3+). At high enough pH, all
proteins will be negatively charged because now the carboxyls are negative (-COO -) and the
amino groups are neutral (-NH2). This means that at some pH each protein must be neutral.
This pH is the isoelectric point—the pH at which the protein has zero mobility.
Isoelectric focusing (IEF) uses the different isoelectric points of proteins to provide
an effective separation method. Buffers are used to establish a pH gradient in the gel, with
high pH at the negative electrode and low pH at the positive electrode. We can start the
sample on the high-pH side near the negative electrode. The negatively charged proteins
will move toward the positive electrode until each one reaches its isoelectric pH and stops
moving. Similarly, if we start the proteins at the low-pH side near the positive electrode,
each will move to its isoelectric point. Combining IEF in one direction followed with SDS
electrophoresis to separate by molecular weight at right angles to the first separation makes
a very powerful analytical method in modern proteomics. Figure 8.24 shows an example
of the hundreds of different proteins that can be detected from a single cell.
FIGURE 8.24 A twodimensional gel electrophoresis
pattern of urea-denatured
whole cell extract of the
proteobacterium Shewanella
oneidensis, under limited
oxygen conditions, separated
by isoelectric focusing (IEF) in
the first dimension and SDSPAGE in the second dimension.
The proteins are stained with
silver (data from the Argonne
National Laboratory gelbank).
130
120
110
100
90
80
70
60
M (kD)
50
40
30
20
10
4.5
5.0
5.5
6.0
6.5
7.0
7.5
pI
*A glycoprotein is a protein that has oligosaccharide groups attached. Histones are highly positively charged
proteins rich in lysine and arginine residues.
294 Chapter 8 |
The Motions of Biological Molecules
Macromolecular Interactions
Gel electrophoresis can be used to study interactions between macromolecules. To determine the interaction between a protein and a nucleic acid, we do gel electrophoresis of the
mixture. For example, to learn which restriction fragment of DNA binds to a protein, we
compare the gel pattern of the 32P-labeled DNA with and without protein. The fragment
that is retarded in the presence of the protein is the one that is bound. Similar methods can
be used to study the interaction of two proteins or two nucleic acids.
If the kinetics of the macromolecular reactions is slow enough so that re-equilibration
does not occur during electrophoresis, equilibrium constants for reaction can be measured
by gel electrophoresis. Mixtures are equilibrated before the mixture is applied to the gel;
then electrophoresis is done to separate the individual species. If the electrophoresis does
not change the equilibrium concentrations, the concentration of each species determined
after separation determines the equilibrium constant. For example, the binding of an RNA
enzyme—a ribozyme—to its substrate was measured by this method (Pyle, McSwiggen,
and Cech 1990). It is well to emphasize that the gel electrophoresis method for measuring
equilibrium binding is valid only if the kinetics of dissociation is slow. Otherwise, as the
complex is moved away from its substrate, it will dissociate; the apparent equilibrium
constant will be a complicated function of the rate of separation and the rate of reaction.
Size and Shape of Macromolecules
Transport properties—diffusion, sedimentation, viscosity, and electrophoresis—tell us
about sizes and shapes of macromolecules. Molecular weights can be obtained from sedimentation and diffusion, from sedimentation at equilibrium, and from gel electrophoresis.
Properties for some representative macromolecules and particles are given in table 8.1.
From the diffusion coefficient or the sedimentation coefficient, the frictional coefficient f
can be calculated. We can use the Stokes equation f = 6πhr to calculate for comparison the
frictional coefficient f0 for an unsolvated spherical molecule with molecular weight M and
partial specific volume vB that are the same as those of the macromolecule of interest. If f >f0
TABLE 8.1 Transport and Related Properties of Some Proteins
s20,w (S)
D20,w : 1011
(m2 s-1)
vB
(mL g-1)
MB
f>f0
Ferricytochrome c
(bovine heart)
1.91
13.20
0.707
12,744
1.077
Ribonuclease
(bovine pancreas)
2.00
13.10
0.707
13,690
1.066
Myoglobin
(horse heart)
2.04
11.30
0.741
16,951
1.105
Lysozyme
(chicken egg white)
1.91
11.20
0.703
14,314
1.210
Chymotrypsinogen
(bovine pancreas)
2.54
9.50
0.721
25,666
1.193
Immunoglobulin G
(human)
6.6 - 7.2
4.00
0.739
156,000
1.513
Myosin
6.4
1.1
0.728
570,000
3.6
0.44
0.73
40,000,000
2.6
Tobacco mosaic virus
192
Sources: Based on data from Cantor and Schimmel (1980), Andrews, A. T. (1986), Chang, R. (2000), Harding, S. E., A. J. Rowe, and J. C. Horton, eds. (1992),
Schuster, T. M. A., and T. M. Laue, eds. (1994), Van Holde, K. E., W. C. Johnson, and P. S. Ho. (1998) and Celis, J. D., and R. Bravo, eds. (1984).
Summary | 295
is close to 1, we conclude that the macromolecule is approximately spherical and only lightly
solvated. If f > f0 is much greater than 1, the molecule is either highly solvated, asymmetric,
or both. Additional information is provided by the intrinsic viscosity. The viscosity shape
factor v contains information that complements that obtained from the ratio f >f0.
Since the transport properties are sensitive to the sizes and shapes of molecules, they
can be used to study interactions between molecules and conformational changes of molecules. Some examples are interactions between subunits of a macromolecular structure
(such as a multisubunit protein or a ribosome), conformational change of a protein due to
the binding of a substrate, circularization of a linear DNA, unfolding of a protein chain,
and antibody-antigen interactions.
More detailed knowledge on the structural features and functional aspects of macromolecules can be obtained only by other physical-chemical methods, some of which we
will discuss in later chapters. The understanding of how molecules function in biological
processes is usually achieved by a combination of many methods.
Summary
Collisions
Collision Frequency
z = N Appd2
16
16
= N Aps
A pRTM
A pRTM
(8.3, 8.4)
z = number of collisions per molecule per second
d = hard sphere molecular diameter
s = collisional cross section
Mean Free Path
l =
RT
(8.5)
22NA ps
Mean-Square Displacement
3
8 d2 9 = 8 v 9 lt =
2(RT)2t
2MpN A ps
(8.7)
Diffusion
Fick’s First Law
Jx = - D a
dc
b
dx
(8.8)
Jx = net amount of solute diffusing through a 1 m2 area per s, in the x-direction.
D = diffusion coefficient
Fick’s Second Law
a
a
0c
0 2c
b = Da 2 b
0t x
0x t
0c
b = change in concentration c per second at position x.
0t x
(8.9)
296 Chapter 8 |
a
The Motions of Biological Molecules
02c
b = second derivative of concentration c with respect to position x, at fixed time t.
0x 2 t
Diffusion Coefficient D and Mean Square Displacement 8 d2 9
One dimension:
8 d2 9 = 2Dt
(8.12a)
Three dimensions:
8 d2 9 = 6Dt
(8.12b)
Diffusion Coefficient D and Frictional Coefficient f
kBT
f
(8.21)
mB(1 - vBrA)
vt
=
g
f
(8.32)
D =
kB = Boltzmann Constant = 1.38065 : 10-23 J K-1
Sedimentation
s =
s = sedimentation coefficient
mB = mass of macromolecule
nB= partial specific volume of macromolecule
rA = density of solution
f = frictional coefficient
ut = terminal velocity of macromolecule
g = gravitational or centrifugal acceleration
The dimension of s is seconds. The conventional unit for s, the svedberg (S), is 10-13 s.
s20,w = s a
h
h20,w
b
(1-nBrA)20,w
(1 - nBrA)
(8.34)
s20,w = sedimentation coefficient corrected to give expected value in water at 20°C.
Frictional Coefficient and Molecular Parameters
The frictional coefficient f can be obtained from either s or D. For a sphere of radius r,
f = 6phr
h = viscosity of medium.
For an unsolvated, spherical molecule of mass mB and partial specific volume vB, the radius r is
r= a
3V B 1>3
3m Bv B 1>3
b
= a
b .
4p
4p
The frictional coefficient f0 for this unsolvated sphere is
f 0 = 6pha
3m Bv B 1/3
b .
4p
The deviation of f>f0 from unity for a macromolecule can be due to either solvation or asymmetric shape.
(8.22)
Summary | 297
Combination of Diffusion and Sedimentation
MB =
RTs
D(1 - v BrA)
(8.42)
Molecular weight MB can be calculated from s, D, and the partial specific volume vB. rA is the density of the solution.
Equilibrium Centrifugation:
M Bv2(1 - v BrA)
d ln c
=
2RT
d(x 2)
v = 2pn = angular velocity, in radians per second (n is in revolutions per second)
c = concentration
x = distance from center of rotation.
When centrifuged to equilibrium, the molecular weight of a macromolecule can be calculated from the slope of a plot of ln c versus x2.
Viscosity
Jp = -h
dv x
dy
(8.43)
Jp = the rate of momentum transfer per second per meter squared of area. It has the units of flux.
dvx>dy = velocity gradient; rate of change of the x component of the velocity with respect to y.
h = viscosity coefficient.
Newtonian fluid: h is independent of dvx>dy; non-Newtonian fluid: h depends on dvx>dy.
Solutions of macromolecules are often non-Newtonian, and viscosity measurements are usually done at several values of dvx>dy and
extrapolated to dvx>dy = 0.
Specific Viscosity:
hsp K
h - h
h
(8.47)
h = the viscosity coefficient of the solvent
h ⴕ = the viscosity coefficient of the solution of a macromolecule.
Intrinsic viscosity is the limiting value of hsp>c as c approaches 0, where c is the concentration of macromolecules in g mL-1
[h] = lim
cS0
hsp
c
.
(8.48)
Relation between [η] and Molecular Parameters:
[h] = n(v B + dv A)
n = shape factor (2.5 for a sphere)
nB = partial specific volume of the macromolecule
d = mass of solvent hydrodynamically associated with a unit mass of macromolecule.
vA = specific volume of the solvent in dilute solution (1 mL g-1 for dilute aqueous solution).
(8.49)
298 Chapter 8 |
The Motions of Biological Molecules
Very Flexible Coils (Random Colis):
[h] ⬇ M1/2
B
Electrophoresis
Electrophoretic mobility:
m =
vt
E
vt =
ZeE
f
vt = terminal velocity of charged particle, m s-1
E = electric field strength V m-1
(6.64)
Z = number of charges on the macromolecule
f = frictional coefficient
e = fundamental electric charge
μ = electrophoretic mobility = vt >E = terminal velocity per unit electric field.
Gel Electrophoresis
Molecular Weight:
log M = a + bx
(8.51)
M = molecular weight of protein
x = distance traveled in gel
a,b = empirical parameters obtained by referencing proteins of known M.
References
The following books are introductory texts similar in level to this
chapter:
of Data for Biological and Synthetic Polymer Systems. New
York: Springer-Verlag.
1. Andrews, A. T. 1986. Electrophoresis: Theory, Techniques,
and Biochemical and Clinical Applications. Oxford, England:
Oxford University Press.
5. Van Holde, K. E., W. C. Johnson, and P. S. Ho. 1998.
Principles of Physical Biochemistry. Upper Saddle River,
NJ: Prentice Hall.
2. Chang, R., 2000. Physical Chemistry for the Chemical and
Biological Sciences. Sausalito, CA: University Science
Books.
The material in this chapter is covered at a more advanced
level in:
3. Harding, S. E., A. J. Rowe, and J. C. Horton, eds. 1992.
Analytical Ultracentrifugation in Biochemistry and Polymer
Science. Cambridge, England: Royal Society of Chemistry.
4. Schuster, T. M. A., and T. M. Laue, eds. 1994. Modern
Analytical Ultracentrifugation: Acquisition and Interpretation
6. Cantor, C. R., and P. R. Schimmel. 1980. Biophysical
Chemistry. Part II: Techniques for the Study of Biological
Structure and Function. San Francisco: Freeman.
7. Celis, J. D., and R. Bravo, eds. 1984. Two-Dimensional Gel
Electrophoresis of Proteins. Orlando, FL: Academic P
Suggested Reading
Axelrod, D., D. E. Koppel, J. Schlessinger, E. Elson, and W. W.
Webb. 1976. Mobility Measurement by Analysis of Fluorescence Photobleaching Recovery Kinetics. Biophys. J.
16:1055-69.
Bauer, W. R., F. H. C. Crick, and J. H. White. 1980. Supercoiled
DNA. Sci. Am. 243:118.
Carle, G. F., M. Frank, and M. V. Olson. 1986. Electrophoretic
Separation of Large DNA Molecules by Periodic Inversion
of the Electric Field. Science 232:65-68.
Fried, M., and D. M. Crothers. 1981. Equilibria and Kinetics of
the lac Repressor-Operator Interactions by Polyacrylamide
Gel Electrophoresis. Nucleic Acids Research 9:6505-25.
Problems | 299
Hunkapiller, T., R. J. Kaiser, B. F. Koop, and L. Hood. 1991.
Large-Scale and Automated DNA Sequence Determination.
Science 254:59-67.
Jacques, J.-P., and M. M. Susskind. 1991. Use of Electrophoretic
Mobility to Determine the Secondary Structure of a Small
Antisense RNA. Nucleic Acids Res. 19:2971-77.
Jeffreys, A. J., V. Wilson, and S. L. Thein. 1986. IndividualSpecific “Fingerprints” of Human DNA. Nature 316:76-79.
Kheterpal, I., and R. A. Mathies. 1999. Capillary Array
Electrophoresis DNA Sequencing. Analytical Chemistry
71:31A-37A.
Laue, T. M. and W. F. Stafford, III. 1999 Modern applications of
analytical ultracentrifugation. Annu. Rev. Biophys. Biomol.
Struct. 28:75-100
Leng, M. 1990. DNA Bending Induced by Covalently Bound
Drugs: Gel Electrophoresis and Chemical Probe Studies.
Biophys. Chem. 35:155-63.
Liu, L. 1989. DNA Topoisomerase Poisons as Anticancer Drugs.
Annu. Rev. Biochem. 58:351-75.
Pyle, A. M., J. A. McSwiggen, and T. R. Cech. 1990. Direct
Measurement of Oligonucleotide Substrate Binding to WildType and Mutant Ribozymes from Tetrahymena. Proc. Natl.
Acad. Sci. USA 87:8187-91.
Righetti, P. G. 1990. Recent Developments in Electrophoretic
Methods. J. Chromatography 516:3-22.
Schmidt, Th., G. J. Schütz, W. Baumgartner, H. J. Gruber, and
H. Schindler. 1995. Characterization of Photophysics and
Mobility of Single Molecules in a Fluid Lipid Membrane. J.
Phys. Chem. 99:17662-8.
Schmidt, Th., G. J. Schütz, W. Baumgartner, H. J. Gruber, and H.
Schindler. 1996. Imaging of single molecule diffusion. Proc.
Natl. Acd. Sci. USA 93, 292-9.
Schwartz, D. C., and C. R. Cantor. 1984. Separation of Yeast
Chromosome-Sized DNAs by Pulsed-Field Gradient Gel
Electrophoresis. Cell 37:67-75.
Smith, S. B., C. Heller, and C. Bustamante. 1991. Model and
Computer Simulations of the Motion Molecules During Pulse-Field Gel Electrophoresis. Biochemistry 30:
5264-74.
Problems
1. The collisional diameter d of a H2 molecule is about 0.25 nm
= 2.5 : 10-10 m. For H2 gas at 0°C and 1 bar, calculate the
following:
a. The root-mean-square velocity
b. The translational kinetic energy of 1 mol of H2 molecules
c. The number of H2 molecules in 1 mL of the gas
d. The mean free path
e. The number of collisions each H2 molecule undergoes
in 1 s
f. The total number of intermolecular collisions in 1 s in
1 mL of the gas.
2. The following data were reported for human immunoglobulin G (IgG) at 20°C in a dilute aqueous buffer:
M = 156,000 g mol-1
D20,w = 4.0 : 10-11 m2 s-1
vB = 0.739 mL g-1
a. Use the diffusion coefficient to calculate f, the frictional
coefficient.
b. Use the Stokes equation to calculate f0, the frictional
coefficient of an unhydrated sphere of the same M and
vB as IgG.
c. If IgG is not significantly hydrated, use figure 8.12 or
Eqs. 8.23 to 8.26 to estimate the dimensions of a prolate
spheroid that best fit the data.
3. Diffusion, sedimentation, and electrophoresis can be used
to characterize, identify, or separate macromolecules.
a. Describe one experimental method to measure a diffusion
coefficient, one to measure a sedimentation coefficient,
and one to measure an electrophoretic mobility. For each
method, state what is measured and how it is used to
obtain the desired parameter.
b. Describe what you could learn about a protein by
applying the three methods.
c. Describe what you could learn about a nucleic acid by
applying the three methods.
4. The sedimentation coefficient of a certain DNA in 1 M NaCl
at 20 °C was measured by boundary sedimentation at
24,630 rpm. The following data were recorded:
t (min)
Distance of boundary from
center of rotation, x (cm)
16
6.2687
32
6.3507
48
6.4380
64
6.5174
80
6.6047
96
6.6814
a. Plot log x versus t. Calculate the sedimentation
coefficient s.
b. The partial specific volume of the sodium salt of DNA
is 0.556 mL g-1. The viscosity and density of 1 M NaCl
and the viscosity of water are 1.104 mPa s, 1.04 g mL-1
and 1.005 mPa s, respectively. Calculate s20,w for
the DNA.
300 Chapter 8 |
The Motions of Biological Molecules
5. For a bacteriophage T7, the following data at zero
concentration have been obtained [(Dubin et al. 1970,
J. Mol. Biol. 54:547)]:
s20,w = 453 S
D20,w = 6.03 : 10-12 m2 s-1
vB = 0.639 mL g-1
a. Calculate the molecular weight of the bacteriophage.
b. Phosphorus and nitrogen analyses of the bacteriophage show that 51.2% by weight of the bacteriophage is DNA. Calculate the molecular weight of
T7 DNA. Each bacteriophage contains one DNA
molecule.
6. The following questions can be answered by the application of the equations discussed in this chapter.
a. Some lipoproteins sediment in a centrifugal field and
others float. What is the partial specific volume of a
lipoprotein that neither sinks nor floats in a solution of
density 1.125 g mL-1?
b. Calculate the sedimentation coefficient in seconds of a
parachutist who is falling toward Earth at a velocity of
9 m s-1 (about 20 miles per hour).
c. The intrinsic viscosity of a solution of spherical viruses is
1.5 mL g-1. The molecular weight is 5.0 : 106 g mol-1.
What is the volume of each solvated virus particle?
d. A mutant protein with molecular weight 100,000
differs from the normal protein by the replacement of
an alanine amino acid (side chain -CH3) by an asparta
(side chain -CH2COO-). This small change has a negligible effect on the molecular weight and frictional
coefficient. How would you separate the two proteins
from each other?
e. The rate of a gas-phase reaction is directly proportional to the total number of collisions per second
that occur in the container. State by what factor the
rate of the reaction increases when the concentration of molecules doubles; when the collision cross
section s doubles; when the absolute temperature
doubles; when the molecular weight doubles.
7. Nucleosome particles are composed of DNA wrapped
around a “core” of protein. Nucleosome particles were
analyzed using ultracentrifugation and dynamic light scattering. In a solution at 20°C the diffusion constant was
measured to be 4.37 : 10-11 m2 s-1. In the same solution, boundary centrifugation was done at 18,100 rpm to
obtain the following data for the position of the boundary
measured from the center of rotation:
t (min)
Boundary position x (cm)
0
4.460
80
4.593
160
4.713
240
4.844
The density of the solvent was 1.02 g mL-1 and the
partial specific volume of the nucleosome was determined
to be 0.66 mL g-1.
a. What is the molecular weight of a nucleosome particle?
The following gel pattern was found when the DNA in nucleosomes was cut with an enzyme, all protein was removed,
and then the gel electrophoresis of the DNA was done.
We can assume that the shortest fragment found corresponds
to a piece of DNA that had been wrapped around a single
nucleosome core, and longer fragments came from DNA
associated with 2, 3, 4 etc. nucleosome particles. One DNA
base pair has molecular weight 660 g mol-1.
b. How many DNA base pairs are wrapped around each
protein core?
c. What is the molecular weight of a single protein in the
nucleosome if (as other evidence suggests) this “nucleosome core” is made up of eight protein molecules that
bind together, each of about the same molecular weight?
8. Use the data for lysozyme given in table 8.1 to answer the
following questions:
a. Calculate the frictional coefficient f in kg s-1 and the
frictional coefficient f0 that lysozyme would have if it
were an unhydrated sphere.
b. Under certain conditions, lysozyme dimerizes. If the
ratio fdimer >fmonomer = 1.6 calculate s20,w for the lysozyme dimer.
9. Consider a mixture of two proteins with molecular weights
of 20,000 and 200,000. For simplicity of calculation, both
may be approximated as unhydrated spheres with vB =
0.740 mL g-1.
a. Calculate the ratio of the sedimentation coefficients of
the proteins in the same medium (density = 1.05 g mL-1)
at the same temperature.
Problems | 301
pKa
a-COOH
~2
a-NH3+
~9
Side chain
-COOH
~4
Side chain
-NH3+
~ 11
Histidine
~6
A biochemist studying the properties of the enzyme
hexokinase isolates and purifies two different mutants
(I and II) for this enzyme from bacteria. She then
performs isoelectric focusing and runs an SDS-gel
electrophoresis for these mutant enzymes as well as
for the normal enzyme. Her gels after protein
staining are:
SDS-gel
+
Normal
I
II
}
b. Consider the following experiments. A mixture of the
two proteins is placed on top of a centrifuge tube filled
with a dilute aqueous buffer containing a linear gradient of sucrose ranging from 5% at the top to 20%
at the bottom, and the tube is spun in a centrifuge, as
illustrated in the following diagram. The top and bottom of the liquid column are 4.0 and 8.0 cm from the
axis of rotation, respectively. When the larger protein
has sedimented a distance of 3.0 cm, how far has the
smaller protein traveled? [Use your answer to part (a)
in the calculation and neglect the viscosity and density
gradients that are associated with the sucrose concentration gradient.] Based on your calculation, is this an
effective method of separation for the two proteins?
c. As the proteins are sedimenting, the initial concentrated
thin layer of protein molecules will spread out due to diffusion. If the distance over which the protein molecules
spread due to diffusion is much greater than the distance
separating the two proteins, this method will not be useful as a separation approach since the protein layers will
overlap. To see if this might be a problem, calculate the
average distances the two proteins move due to diffusion
at 298 K if the sedimentation experiment lasts 12 h. Take
the average viscosity of the medium as 1.5 mPa s. Is the
“spreading distance” due to diffusion significant compared
to the separation of the proteins due to sedimentation?
d. At sedimentation equilibrium, do you expect to find a
gradual variation in the concentration of protein along r
near the bottom of the tube?
10. Consider a small spherical molecule of radius 0.2 nm.
a. At 298 K, what is the average time required for such a molecule to diffuse across a phospholipid bilayer 4 nm thick?
The viscosity of the bilayer interior is about 0.01 Pa s.
b. At a constant temperature, the average time to diffuse
across a bilayer hydrophobic interior depends only on the
radius of the diffusing molecule. Hence, all molecules
with the same size will cross in the same time. However,
it is well known that for the same concentration gradient
the flux of nonpolar molecules across bilayers is much
greater than that for polar molecules of the same size. Are
these statements contradictory? Explain your answer.
11. The pKa values of various groups in proteins are tabulated
as follows:
Mutants
Isoelectric focusing
pH 8
Mutant II
Normal
Mutant I
+
pH 4
a. On the basis of the gel results, what can you say about
the possible kinds of changes in the amino acid composition of these two mutants? Explain briefly.
b. You have a small unidentified peptide. You determine its
molecular weight by SDS-gel electrophoresis to be 3000
and its isoelectric pH to be 10. With a specific protease,
you cleave a single peptide bond in your protein. You
find that your cleaved peptide still has approximately the
same molecular weight on an SDS-gel electrophoresis,
but its isoelectric point is reduced to about 8. How would
you explain your results?
302 Chapter 8 |
The Motions of Biological Molecules
12. For prolate ellipsoids of large axial ratios—say, with a>b 7
20—it can be shown that the viscosity shape factor
n ~ 0.207(a>b)1.723. A certain rodlike macromolecule
is believed to form an end-to-end dimer. Show that the
intrinsic viscosity of the dimer is expected to be 3.32 times
that of the monomer.
13. The DNA from an animal virus, polyoma, has a sedimentation coefficient s20,w of 20 S. Digesting the DNA very
briefly with pancreatic DNase I, an enzyme that introduces
single-chain breaks into a double-stranded DNA, converts
it to a species sedimenting at 16 S. This reduction in s
could be due to either a reduction in molecular weight or
a conformational change of the DNA (so that its frictional
coefficient is increased). How would you design an experiment to decide between these possibilities?
14. Two spherical viruses of molecular weights M1 and M2
respectively, happen to have the same partial specific
volume vB. Neglecting hydration, what are the expected
values for the ratios s2>s1, D2>D1, and [h]2>[h]1?
15. The following data have been obtained for human serum
albumin:
s20,w = 4.6 S
D20,w = 6.1 * 10-11 m2 s-1
[h] = 4.2 mL g-1
vB = 0.733 mL g-1
Calculate the molecular weight of this protein.
16. a. Proteins A and B with molecular weight of 16,500 and
35,400 move 4.60 cm and 1.30 cm, respectively, during
electrophoresis and in SDS polyacrylamide gel. What
is the molecular weight of protein C that moves 2.80
cm under the same conditions in the same gel?
b. The sedimentation coefficient of a DNA molecule
is 22.0 S. A dilute solution of the DNA is spun in a
centrifuge at 40,000 rpm starting from a distance of
6.0 cm from the axis of rotation. How far will the DNA
move in 20 min?
17. In 6 M guanidine hydrochloride and in the presence of
2-mercaptoethanol, it is generally believed that complete
unfolding of proteins occurs. Could you test whether this is
true by viscosity measurements? Give a brief and concise
discussion. Some experimental data are listed below in
the following table (taken from C. Tanford, 1968, Adv.
Protein Chem. 23:121):
[H], mL g-1
M
normal
unfolded
Ribonuclease (cow)
13,690
3.3
16.6
Myoglobin (horse)
17,568
3.1
20.9
Chymotrypsinogen(cow)
25,666
2.5
26.8
Serum albumin (cow)
66,296
3.7
52.2
Protein
18. The O2-carrying protein, hemoglobin, contains a total of
four polypeptide chains: two a chains and two b chains
per molecule. The hemoglobin of a certain person from
Boston, designated hemoglobin M Boston, differs from
normal hemoglobin (hemoglobin A) in that a histidine
residue in each of the a chains of hemoglobin A is substituted by a tyrosine residue. From the ionization constants
given in table 4.2, do you expect the electrophoretic
mobilities at pH 7 of the two proteins, hemoglobin M
Boston and hemoglobin A, to differ? Give your reasons.
What would be a reasonable pH to use to separate the two
by electrophoresis? Which is more negatively charged
at this pH?
19. The protein b-lactalbumin (M = 14,000 g mol-1)
has been studied under different solution conditions
by dynamic light scattering. At pH 7 and 40°C in a
dilute solution, the diffusion constant was found to be
14.25 : 10-11 m2 s-1.
a. If the viscosity of the solvent is 1.01 mPa s, estimate
the diameter of the protein, assuming that it is spherical in shape.
b. At pH < 2 this protein becomes inactive and changes
spectral characteristics. Under these conditions, the
diffusion constant was found to be 12.80 : 10-11 m2
s-1. Calculate the change in volume of the protein
(assume all else remained the same).
c. At intermediate pH, there is an equilibrium between
these two forms of the protein. Calculate the ratio
of sedimentation coefficients (shigh pH >slow pH) for
these two forms. Assume the solvent has a density
of 1.04 g mL-1.
20. A protein involved in light harvesting for photosynthesis in bacteria was investigated using sedimentation
equilibrium and SDS-gel electrophoresis. Two centrifuge experiments were done, the first in 1 M NaCl solution and the second in 1 M NaSCN. The samples were
spun at 20°C at 21,380 rpm until an equilibrium concentration gradient was established. The partial specific
volume of the protein vB was found to be 0.709 mL g-1;
the density r for the salt solutions was 1.20 g mL-1.
Ignore any possible effects of density gradients that are
established due to the salt in the solutions. The concentration profile was determined and used to generate the
following graph; c = protein concentration at position x
in the centrifuge cell. Below is a drawing of the denaturing SDS-gel electrophoresis.
Problems | 303
log c
−3
Equilibrium sedimentation
−4
NaSCN
NaCl
c. From X-ray diffraction, it was shown that the presence
of ligands causes a contraction of the enzyme of about
1.2 nm along one axis. Why did the radius not decrease
by 0.6 nm? Explain briefly.
22. T4 is a large bacterial virus. The virus has a symmetric
(approximately spherical) head group that contains DNA.
These head particles were studied and were found to have
the following parameters:
s20,w = 1025 S
D20,w = 3.60 : 10-12 m2 s-1
−5
25
30
35
40
x2 (cm2)
SDS-gel electrophoresis
Reference
Protein
28 kD
18 kD
10 kD
6 kD
+
a. From the centrifuge data, calculate the effective molecular weight of the light-harvesting protein in each of the
two salt solutions.
b. From the SDS–gel electrophoresis data, estimate the
molecular weights of the two polypeptides that make up
the light-harvesting protein.
c. How can you explain the two sets of data? Give a concise
explanation of what each observation indicates and how
the observations fit together.
21. The enzyme aspartate transcarbamylase (ATCase) has a
molecular weight of 310 kD and undergoes a change
in shape upon binding of substrate or inhibitors. This
change has been characterized by ultracentrifugation. In
the absence of any ligands, a sedimentation coefficient
of s20,w = 11.70 S was measured. The solvent density
was 1.00 g mL-1; the viscosity was 1.005 mPa s, and the
partial specific volume of the protein was 0.732 mL g-1.
a. Find the radius of the protein in nm, assuming that it is
a sphere.
b. Upon adding a ligand that bound to the enzyme, the
sedimentation coefficient increased by 3.5%. What is
the radius of the ligated enzyme?
vB = 0.605 mL g-1
Calculate the following:
a. The molecular weight of the head group
b. The volume in mL of the head group from vB (this assumes
no hydration)
c. The frictional coefficient of the head group from the
diffusion coefficient
d. The volume of the head group from the frictional
coefficient in part (c) and the Stokes equation. The difference between the calculated volume in part (b) and
the measured volume in part (d) is caused by hydration
of the head group.
23. A small spherical virus has a molecular weight of
1.25 : 106 g mol-1 and a diameter of 13.5 nm; it is not
significantly hydrated. Calculate the intrinsic viscosity in
units of mL g-1 of an aqueous solution of this virus.
24. For each of the following changes, state whether the sedimentation coefficient of the particles will increase, decrease,
remain the same, or whether it is impossible to tell. Also
give an equation or a one- or two-sentence explanation that
supports your answer.
a. The temperature of the solvent is increased from 20°C
to 30°C.
b. The length of the long axis of the particle (which is a
prolate ellipsoid) is halved.
c. 15N is substituted for 14N in the particle.
d. Provide answers for the diffusion coefficient of the
particle in parts (a)-(c).
25. An important step in the blood coagulation process is the
enzymatic conversion of the protein prothrombin to the
lower-molecular-weight clotting agent, thrombin, via cleavage of a portion of the prothrombin polypeptide.
Prothrombin S thrombin + cleaved peptide
The values of s20,w and D20,w for prothrombin have
been found to be 4.85 S and 6.24 : 10-11 m2 s-1
respectively. The diffusion coefficient for thrombin is
8.76 : 10-11 m2 s-1. Assuming that both thrombin and
prothrombin are unhydrated spheres with vB = 0.70 ml g-1,
calculate the molecular weights of prothrombin, thrombin,
and the cleaved peptide.
304 Chapter 8 |
The Motions of Biological Molecules
26. a. Gel electrophoresis is used to determine the sequences
of DNA molecules. Describe how this is done. In particular, describe how the positions of bands on a gel are
related to the sequence of the DNA.
b. Gel electrophoresis is used to determine molecular
weights of proteins. Describe how this is done and
mention the limitations to this method.
c. Describe one other method that could be used to determine the molecular weight of a protein. State what is
measured and how this is related to molecular weight.
d. Describe how an equilibrium-binding constant can be
measured by gel electrophoresis.
27. Three transport properties of proteins are easy to measure:
s (sedimentation coefficient), D (diffusion coefficient),
and μ (electrophoretic mobility). State what happens to
each of these three quantities for the following changes.
Be quantitative and give equations to justify your answers.
a. The molecular weight of the protein doubles, but the
frictional coefficient is constant.
b. The frictional coefficient of the protein doubles, but the
molecular weight is constant.
c. The charge on the protein doubles, but nothing else
changes.
d. The protein changes from a prolate ellipsoid to an
oblate ellipsoid of the same axial ratio; nothing else
changes.
e. The amount of hydration of the protein doubles.
28. Bacteriophage PL25 has s20,w = 485 S; D20,w =
6.8 : 10-12m2 sec-1, and a partial specific volume of
0.68 mL> g. Compute its molar mass.
Electron microscopy showed the length of its double
stranded DNA to be 12.6 μm. If each base pair of DNA
has a length of 0.3 nm, and each base pair has a molar
mass of 660 g mol-1, compute the molar mass of the
phage DNA, and by subtracting this from the total mass,
estimate the molar mass of PL25 protein.
29. At acid pH, yeast hexokinase has a dimer molecular
weight of 107210 Daltons. It is approximately spherical
and has a partial specific volume of 0.74 mL g-1. Predict
its radius and its Stokes’ Law sedimentation and diffusion
coefficients in water at 20 °C (h = 1.003 : 10-3 Pa s).
30. RNA with a sedimentation coefficient of 5S is one of
the best-known ribosomal components. What would be
the (a) radius (b) molecular weight of 5S RNA if it were
spherical and unhydrated? You can assume the partial
specific volume of RNA to be 0.56 mL g-1. Assume the
viscosity of water at 20°C is 0.0010 Pa s.
Chapter 9
Kinetics: Rates of Chemical
Reactions
We have explored how fast molecules move and how often they collide; now we can
begin to learn how fast they react. Thermodynamics tells us whether a reaction can
occur; kinetics tells us when it will occur—in a hundred years or in a millisecond. Kinetics helps us to influence rates—to speed them up and to slow them.
Concepts
A chemical reaction may occur when two molecules collide. For the reaction to occur,
the molecules must have enough energy to break the covalent bonds of the reactants.
Increasing concentrations of reactants and increasing the temperature nearly always
increase the rate of a chemical reaction. The higher concentrations mean more frequent
collisions; the higher temperature provides more energy per collision and also more collisions per second. However, not all reaction rates increase with increasing temperature
and concentration.
Chemical kinetics includes two distinct parts: experiments and theory. First, we must
do experiments to measure the dependence of the rate on concentrations of reactants and,
possibly, of products. (There is often product inhibition of the rate, and there may be
catalysis by the product.) We change the solvent environment. Concentrations of molecules
not directly involved in the reaction can have a large effect on the rate. Added salts, metal
ions, H+ and so forth can act as catalysts or inhibitors of the reaction. We determine the
effect of temperature and any other pertinent variables. The rate may depend on whether
the reaction is done in a glass container or a plastic container. It may depend on whether
it is done in a dark room or in sunlight. Joseph Priestley accidentally learned that plants
produce oxygen only in sunlight, not in the dark. We have since learned much more about
photosynthesis and photochemical reactions in general. The important lesson is that many
things may have large effects on the rate of a reaction. A spark introduced into a container
of H2 and O2 gas changes the rate of formation of H2O by more than 10 orders of magnitude: a stable gas mixture is converted into a bomb.
Once we have determined the effects of many variables on the rate, the next step is to
explain the effects in terms of what the molecules are doing. The theoretical explanation
is called a mechanism of the reaction. A mechanism is one or more equations showing which molecules react and which bonds break or form. The equations represent
elementary reactions. Elementary reactions involve one or two molecules, at most three
molecules, that collide and react. The mechanism is distinct from the stoichiometric
reaction. The stoichiometry tells us the number of moles of reactants that produce
the number of moles of products. The stoichiometry tells us nothing about the dependence of rate on concentration; the mechanism is an explanation of the experimental
rate data. For a chemical reaction, there is only one stoichiometric reaction; however,
305
306 Chapter 9 |
Kinetics: Rates of Chemical Reactions
many mechanisms are consistent with the rate data. Each mechanism is a hypothesis
to explain the kinetic data; the mechanism is a guide to design further experiments to
better understand the results.
An important concept in understanding the mechanism of a reaction and the temperature dependence of a rate is the transition state. For a reaction in which one species
is converted to another, the structure of the transition state is assumed to be intermediate
between the two species. It is an unstable structure (unstable means that it exists only for
the time of a molecular vibration) that quickly transforms to either stable species. The
transition state corresponds to an energy maximum between two stable species that exist
in energy minima. The transition state energy is the energy difference between the transition state and a reactant; its magnitude characterizes the temperature dependence of the
rate of the reaction. A small transition state energy means that little energy is required to
transform reactant to product and the rate does not depend much on temperature. A large
transition state energy produces a large temperature dependence of the rate.
Applications
Kinetics—the study and understanding of the rate of change of anything—is useful in all
areas of medicine, biology, and biochemistry. The effect of food supply, predators, and climate on the number of animals in a population follows the same differential equations used
for treating molecular reactions in the atmosphere or in a single biological cell. Bacterial
growth rates, radioactive decay, biosynthesis of deoxynucleotides, viral infectivity, and
antibiotic cures all depend on the rates of reactions. We will learn how to measure rates
of reaction and to study the variables that affect the rates. For some purposes, this may be
enough. For example, if we use an enzyme as a catalyst, it may be sufficient to assay its
activity every day. When the activity gets too low, we throw out the old solution and make
a new batch. The rate of denaturation of the enzyme is important, but the mechanism may
be too complicated to understand. We are satisfied with the empirical rate data.
For most reactions, we do want to understand what is happening. Why do some reaction rates increase by a factor of 2 for a 10-degree rise in temperature, but some do not
change or even decrease? When should a salt not directly involved in the reaction affect
the rate? The energy and structure of the transition state is important in answering these
questions. It may be difficult to generalize the concept of a mechanism and a transition state
to the complex rates involved in predator–prey dependence. However, the idea is to propose
a hypothesis that explains the data. The test of the hypothesis is how well it explains other
data for different animals under different conditions.
Kinetics
Chemical kinetics is the study of rates of reactions. Some reactions, such as that between
hydrogen gas and oxygen gas in an undisturbed clean flask, occur so slowly as to be
unmeasurable. Radioisotopes of some nuclei have very long lifetimes; the carbon isotope
14C decays so slowly that half of the initial amount is still present after 5770 years. Other
radionuclei have half-lives orders of magnitude longer than this. Processes such as the growth
of bacterial cells are slow but easily measurable. The rate of reaction between H+ and OHto form water is so fast that its study requires special techniques, such as temperature-jump
kinetics. The frontier is expanding to include still faster processes, and reaction times of fractions of a picosecond (1 ps = 10 -12 s) are currently being studied. Nuclear reactions involving
species with lifetimes shorter than 10 -20 s are known. Clearly, the methods of observation
are very different to include processes over such an enormous range of time.
The first chemical reaction rate studied quantitatively involved a compound of
biological origin. In 1850 L. Wilhelmy reported that the hydrolysis of a solution of sucrose
Kinetics | 307
to glucose and fructose occurred at a rate that decreased steadily with time but always
remained proportional to the concentration of sucrose remaining in the solution. He followed
the reaction indirectly by measuring the change with time of the optical rotation—the rotation of the plane of polarization of light passing through the solution. The phenomenon of
optical rotation results from the molecular chirality of sugars such as sucrose, glucose, and
fructose; chirality (handedness) refers to the fact that these molecules are structurally (and
chemically) distinct from their mirror-image molecules (see chapter 13). Subsequent work
demonstrated that systems at equilibrium are not static but are undergoing transformations
between reactants and products in both directions and at equal rates.
The role of catalysts, substances that increase the rates of reactions without themselves
being consumed, was recognized early from the influence of hydrogen ion on the rate of
sucrose hydrolysis. The early “ferments” used to convert sugars from grain or grapes into
beverages contained enzyme catalysts that greatly sped up the rates of these processes.
Biological organisms contain thousands of different enzymes—protein molecules that
selectively catalyze all the reactions in living cells. In the hydrogen–oxygen gas mixture,
the introduction of a trace of finely divided platinum catalyst leads to a violent explosion.
Because catalysts have no effect on the position of equilibrium, we can conclude that the
hydrogen–oxygen mixture in the absence of a catalyst is not at equilibrium.
The ability of increasing temperature to speed most chemical reactions was put on
a quantitative basis by Arrhenius (1889). The mixture of hydrogen and oxygen, which is
stable indefinitely at room temperature, explodes on heating it to temperatures in excess
of 400 ⬚C. In other cases, such as the processes that occur in biological cells, an increase
in temperature causes the enzyme-catalyzed reactions to cease altogether.
Light or electromagnetic radiation serves as a “reagent” in some biological processes
that are essential to our survival. Photosynthesis and vision are just two of the most obvious
examples. Other processes that are not chemical in nature, such as the nuclear reactions
that provide the energy source of the Sun and have been the major source of heat within
Earth, can nevertheless be described using the methods of chemical kinetics. Population
dynamics, ecological changes and balance, atmospheric pollution, and biological waste
disposal are just a few of the relatively new applications of this powerful approach.
The methods of chemical kinetics or reaction-rate analysis were developed for the
resolution and understanding of the relatively simple systems encountered by chemists.
These approaches are also valuable in analyzing the much more complex processes of
biology. The reason that the methods work is often because one or a few steps control the
rate of an extensive chain of reactions. All the steps involved in metabolism—cell division
and replication, muscular contraction, and so on—are subject to the same basic principles,
as are the elementary reactions of the chemist.
The rate or velocity, v, of a reaction or process describes how fast it occurs. Usually,
the rate is expressed as a change in concentration per unit time,
dc
= rate of reaction ,
v =
dt
but it may alternatively express the change of a population of cells with time, the increase
or decrease in the pressure of a gas with time, or a change in the absorption of light by a
colored solution with time. In general, the rate of a process depends in some way on the
concentrations or amounts involved; the rate is a function of the concentrations. This relation is known as the rate law:
v = f(concentrations)
The rate law may be simple (v = constant, for example) or complex, but it gives important
information about the mechanism of the process. One of the main objectives of research
in kinetics is the determination of the rate law.
308 Chapter 9 |
Kinetics: Rates of Chemical Reactions
Rate Law
Substances that influence the rate of a reaction can be grouped into two categories:
1. Those whose concentration changes with time during the course of the reaction:
Reactants—decrease with time.
Products—increase with time.
Intermediates—increase and then decrease during the course of the reaction. An
example is substance C in the following two-step reaction:
A h C h B
2. Those whose concentrations do not of necessity change with time:
Catalysts (both promoters and inhibitors), including enzymes and active surfaces.
Intermediates in a steady-state process, including reactions under flowing conditions.
Components that are buffered by means of equilibrium with large reservoirs.
Solvents and the environment in general.
These influences do not change during a single run, but they can be changed from one
experiment to the next. The concentrations of these components frequently do influence
the rates of reactions.
Order of a Reaction
It is important in kinetics to learn immediately the vocabulary that kineticists use. We need
to distinguish the stoichiometric reaction, the order of the reaction, and the mechanism of
the reaction. It is essential to understand these terms. The stoichiometry of the reaction
describes how many moles of each reactant are needed to form each mole of products.
Only ratios of moles are significant. For example,
H2 + 12 O2 = H2O
2H2 + O2 = 2H2O
are both correct stoichiometric reactions. The mechanism of a reaction describes how
the molecules react to form products. The mechanism is, in general, a set of elementary
reactions consistent with the stoichiometric reaction. For the reaction of H2 and O2 in the
gas phase, the reaction is probably a chain involving H, O, and OH radicals:
H#
H2 h 2H #
+ O2 h # OH + O
# OH + H2 h H2O + H #
O + H2 h # OH + H #
Each step in the mechanism describes an elementary reaction; the four reactions constitute
the proposed mechanism.
The kinetic order of a reaction describes the way in which the rate of the reaction
depends on the concentration. Consider a reaction whose stoichiometry is
A + B h P.
For many such reactions, the rate law is of the form
v = k[A]m[B]n[P]q
(9.1)
where the concentrations [A], [B], [P] are raised to powers m, n, and q, that are usually
integers or zero but may be nonintegral as well. The order of the reaction with respect
to a particular component A, B, and P, is just the exponent of its concentration. Because
Kinetics | 309
the rate may depend on the concentrations of several species, we need to distinguish
between the order with respect to a particular component and the overall order, which
is the sum of the exponents of all components. Some representative examples are listed
in table 9.1.
If the concentration of a component is unchanged during the course of the reaction,
it is frequently omitted in the rate-law expression. A more complete rate law for the first
reaction listed in table 9.1 is
v = k ⬘[sucrose][H + ][H2O] .
TABLE 9.1 Rate Laws and Kinetic Order for Some Reactions
Stoichiometric reaction
Rate law
Kinetic order
sucrose + H2O h fructose + glucose
v = k[sucrose]
1
L-isoleucine h D-isoleucine
v = k[L-isoleucine]
1
v = k[14C]
1
v=
k[proflavin]2
2
h
p-nitrophenylacetate + 2
p-nitrophenolate + acetate + H2O (pH 9)
v=
k[p-nitrophenylacetate][OH-]
2 (overall)
hemoglobin·3O2 + O2 h
hemoglobin·4O2
v = k[Hb·3O2][O2]
2 (overall)
H2 + I2 h 2 HI
v = k[H2][ I2]
2 (overall)
14C
h
14N
+ b-
2 proflavin h proflavin dimer
OH-
H2 + Br2 h 2 HBr
v =
k 3 H2 4 3 Br2
4 1>2
Complex
k⬘ + 3 HBr]>[Br2 4
CH3CHO h CH4 + CO
v ⬵ [CH3CHO]3>2
3>2 (approx.)
C2H5OH h CH3CHO (liver enzymes)
v constant
0
However, H+ is a catalyst, and its concentration is constant during a run; the concentration
of H2O, the solvent, is also little changed because it is present in vast excess. Therefore,
the terms [H+] and [H2O] are omitted in the rate law (called pseudo-first order) given in
the table. When the reaction is carried out in the presence of different concentrations of
[H+] or with an added inert solvent, the first-order dependence of the reaction on [H+] and
on [H2O] is seen.
Table 9.1 illustrates that there is no simple relation between the stoichiometry and
the rate law. It is never possible to deduce the order of the reaction by inspection of the
stoichiometric equation. Kinetic experiments must be done to measure the order of the
reaction. The reaction of H 2 and I2 is first order with respect to each reactant over a
wide range of conditions, whereas the similar reaction of H 2 + Br2 exhibits a complex
rate law that cannot be described by a single “order” under all conditions. Note that in
this case the rate depends on the concentration of a product as well as on reactants. This
comes about because of a reverse step in the mechanism that becomes important as the
product concentration builds up. Reaction orders may be nonintegral, as in the case of
the thermal decomposition of acetaldehyde, and they may be significantly different during the initial stages, when the reaction is getting under way, or at the end, when other
complications set in. The significance of a zero-order reaction is that the rate is constant
and independent of the concentration of the reactants. This is characteristic of reactions
catalyzed by enzymes, such as liver alcohol dehydrogenase, under the special condition
where the enzyme is saturated with reactants (called substrates in enzyme reactions).
The role of enzymes and other catalysts, such as H+ in the sucrose hydrolysis reaction,
310 Chapter 9 |
Kinetics: Rates of Chemical Reactions
is detected by observing that a change in the catalyst concentration produces change in
the experimental rate coefficient for the reaction, even though the catalyst concentration does not vary with time during the course of any single experiment. Even these
preliminary comments indicate that the rate law contains important information about
the mechanism of the reaction.
Experimental Rate Data
The rate law for a reaction must be determined from experimental data. We may simply
want to know how the rate depends on concentrations for practical reasons, or we may
want to understand the mechanism of the reaction. For example, if we are inactivating a
virus by reaction with formaldehyde, it is vital for us to know how the rate depends on the
concentrations of virus and formaldehyde. If the rate is first order in virus concentration,
the rate will be one-tenth as fast for 105 viruses per milliliter as for 106 viruses per
milliliter. If the rate is second order in virus concentration (unlikely), the rate will be only
one-hundredth as fast. Knowledge of the rate law is necessary to determine the time
of treatment with formaldehyde needed to ensure that all the viruses are inactivated before
they are used to immunize a population. We also must know how temperature, pH, and
solvent affect the rate.
There are many possible ways to obtain the rate data. A usual method is to obtain
concentrations of reactants and products at different times during the reaction. For the virus
inactivation, we would presumably measure live virus versus time by an infectivity assay.
In general, any analytical method that determines concentration can be used. If the time
required to perform the analysis is long relative to the rate of the reaction, quenching, or
sudden stopping of the reaction, is necessary. An enzyme-catalyzed reaction, for example,
can be quenched by cooling the reaction mixture quickly, by the addition of an agent that
denatures the enzyme or by the addition of a chelating compound if a multivalent metal ion
is necessary for catalytic activity. In a quenched-flow apparatus, reactants in two syringes
are forced through a mixing chamber to initiate the reaction. The mixture flows through a
tube into a second mixing chamber, where it is mixed with a stopping reagent. The reaction
time in such an experiment is the time it takes for the solution to flow from the first to the
second mixing chamber and can be as short as several milliseconds.
If there is a physical property that changes significantly as the reaction proceeds, it can
be used to follow the reaction. If the reactants and products absorb light at characteristic
wavelengths, the absorbance can be related to the extent of the reaction (see chapter 13).
The concentration of H+ can be easily monitored with a glass electrode, but different
techniques are needed when the pH changes rapidly with time. Reactions between charged
species can be followed by the electrical conductivity of the solution.
Very fast reactions require special experimental approaches. The relaxation methods,
such as temperature-jump methods, will be described later in this chapter.
Zero-Order Reactions
A zero-order reaction corresponds to the rate law:
dc
= -k ,
dt
(9.2)
where c is the concentration of a reactant and k is a constant called the rate coefficient or
rate constant. The units of the rate coefficient k for a zero-order reaction are obviously
concentration per time, such as M s−1. This expression can readily be integrated by writing
it in the differential form with the variables c and t separated on each side of the equality:
dc = -k dt
Kinetics | 311
Integrating both sides, we obtain
3 dc = -k 3 dt + C ,
where the rate coefficient k is placed outside the integral sign because it does not depend
on time. It can therefore easily be integrated, giving
c = -kt + C .
(9.3)
Alternatively, if the initial and final conditions are specified, the equation can be written
as a definite integral. If the concentration is c1 at time t1 and c2 at time t2, then
c2
t2
3 dc = -k 3 dt
c1
t1
c2 - c1 = -k(t2 - t1) .
More commonly, we can determine the value of C in Eq. 9.3 by considering the value of
C at the initial condition (t = 0) , where c = C = c0 . Substituting this in Eq. 9.3:
c = -kt + c0 or c - c0 = -kt
(9.4)
This is the equation for a straight line giving the dependence of c on t; the slope is the rate
coefficient k .
This behavior is illustrated by the conversion of ethanol to acetaldehyde by the enzyme
liver alcohol dehydrogenase (LADH). The oxidizing agent is nicotinamide adenine dinucleotide (NAD+) and the reaction can be written
LADH
CH3CH2OH + NAD + h CH3CHO + NADH + H + .
In the presence of an excess of alcohol over the enzyme and with the NAD+ buffered via
metabolic reactions that rapidly restore it, the rate of this reaction in the liver is zero order
over most of its course:
v = -
d[CH3CH2OH]
d[CH3CHO]
=
= k
dt
dt
(9.5)
The negative sign is used with the reactant, ethanol, because its concentration decreases
with time; the concentration of the product, acetaldehyde, increases with time. This
behavior is illustrated in figure 9.1.
The reaction cannot be of zero order for all times; because obviously the reactant
concentration cannot become less than zero, and the product concentration cannot become
larger than the initial reactant concentration. For the oxidation of alcohol by LADH the
reaction is zero order only while the concentration of alcohol is significantly in excess of
that of the enzyme.
First-Order Reactions
A first-order reaction corresponds to the rate law:
dc
= -kc
dt
(9.6)
The units of k are reciprocal time, usually s-1. There are no concentration units in k for a
first order reaction so it is clear that we do not need to know absolute concentrations; only
relative concentrations are needed. An elementary step in a reaction of the form
A h B
c0
CH3CHO
c
CH3CH2OH
t
FIGURE 9.1 Plot of concentration versus time for a zeroorder reaction. For ethanol, the
equation is c = -kt + c0 . For
acetaldehyde, the equation is
c = kt . The magnitude of the
slope of each straight line is
equal to the rate coefficient k.
The order must eventually
change from being zero order as
the concentration of CH3CH2OH
approaches zero.
312 Chapter 9 |
Kinetics: Rates of Chemical Reactions
has a rate law of the form
v = -
d[A]
d[B]
=
= k[A] ,
dt
dt
(9.7)
where k is the rate coefficient for the reaction and [A] and [B] are concentrations.
The rate of the reaction can be expressed in terms of either the rate of disappearance
of reactant, -d[A]>dt , or the rate of formation of product, d[B]>dt . The stoichiometric equation assures us that these two quantities will always be equal to one another.
To solve the rate-law expression, we choose the form involving the smallest number
of variables:
d[A]
= k[A]
(9.8)
dt
Here, time is one variable, and the concentration of A is the other We can separate the
variables; [A] on the left hand side, t on the right:
d[A]
= -k dt
[A]
Once the variables are separated, the equation can be integrated, separately, on each side:
d[A]
3 [A] = 3 -k dt + C = -k 3 dt + C
ln [A] = -kt + C ,
(9.9)
where C is a constant of integration. This states that for a first-order reaction, the
logarithm of the concentration will be a linear function of time, as shown in figure 9.2.
To evaluate the constant C, we need to know one concentration at one time. For example,
if [A]0 is the value of the concentration initially when t = 0, it too must satisfy Eq. 9.9.
Substitution gives
ln [A]0 = C ,
which provides an alternative form of Eq. 9.9:
ln
[A]
= -kt
[A]0
(9.10)
A more general form involving any two points during the course of a first-order
reaction is
ln
ln c0
[A]2
= -k(t2 - t1) .
[A]1
(9.11)
By taking the exponential of each side of Eq. 9.11, we obtain (remember that eln x = x)
[A] = [A]0e - kt = [A]010 - kt>2.303 .
ln c
t
FIGURE 9.2 Semilogarithmic
straight-line plot for a first-order
reaction.
(9.12)
This says that the concentration of A decreases exponentially with time for a firstorder reaction. It starts at an initial value [A]0 since e0 = 1 , and reaches zero only after
infinite time! Strictly speaking, the reaction is never “finished.” Before worrying about the
philosophical implications of this, be assured that the inability to detect any remaining [A]
will occur in finite time even using the most sensitive analytical methods.
Let’s consider the stability of the antibiotic penicillin as a practical example. Assume
that our job is to learn how long penicillin remains active when it is stored at room
Kinetics | 313
temperature. We want to determine its activity versus time. The general structure of
penicillin is
H H
N
R
H
S
CH3
O
N
CH3
O
H
COOH
Ultraviolet absorbance, infrared absorbance, or nuclear magnetic resonance could be used
to measure its concentration, but we want to know if it will kill bacteria; therefore, we must
measure its antibiotic properties. We use a standard assay to count the number of bacterial
colonies that survive after treatment with successive dilutions of the penicillin solution
under standardized conditions. We thus obtain the number of standard units of penicillin
in the sample. The penicillin is in a well-buffered solution at pH 7 kept at 25 ⬚C. The mean
values for data measured in triplicate are given in table 9.2.
Table 9.2 The Number of Units of Penicillin Present After Storage at 25 °C.
Penicillin Units Are Proportional to the Concentration of Penicillin; They
Are a Measure of the Antibiotic Effectiveness of the Penicillin Solution
Time (weeks)
0
1.00
2.00
3.00
4.00
5.00
7.00
8.00
9.00
10.00
11.00
12.00
13.00
14.00
15.00
16.00
17.00
18.00
19.00
20.00
*The
Penicillin units*
10,100
8180
6900
5380
4320
3870
2190
2000
1790
1330
1040
898
750
572
403
403
314
279
181
167
ln (penicillin units)
9.220
9.009
8.839
8.590
8.371
8.261
7.692
7.601
7.490
7.193
6.947
6.800
6.620
6.349
5.999
5.999
5.749
5.631
5.198
5.118
precision of the data produces only three significant figures for this column.
A plot of penicillin units versus time is shown in figure 9.3. The amount of penicillin
decreases rapidly with time; an exponential decrease is a good guess:
[penicillin] = [penicillin]0 e - kt
Kinetics: Rates of Chemical Reactions
FIGURE 9.3 The concentration
(in units of antibiotic activity) of
penicillin plotted versus time (in
weeks). The concentration is not
linear in time; thus, the data are
not consistent with a zero-order
reaction.
10000
Penicillin units
314 Chapter 9 |
8000
6000
4000
2000
5
10
t (weeks)
15
20
We note that after three to four weeks, the penicillin is only half as active as it was
originally. The data clearly are not consistent with zero-order kinetics; the concentration
is not linear in time. To test for first-order kinetics, the natural logarithm (or the base 10
logarithm) of the penicillin units is plotted versus time, as shown in figure 9.4. A linear
plot is found; this indicates that the data are consistent with first-order kinetics. A best
least-squares fit to the data gives the equation
ln (penicillin units) = 9.220 - 0.2036t (weeks) .
Comparison with Eq. 9.10 shows that the slope of the line is the first-order rate coefficient;
k = 0.2036 week-1. The intercept is ln (penicillin units) extrapolated to zero time; therefore,
ln [penicillin]0 = 9.220 and [penicillin]0 = 10100, equal to the measured value of 10,100.
A common parameter used to describe the kinetics of first-order rate processes is the
half-life (t½) This is simply the time required for half the initial concentration to react; we
saw that for the data in figure 9.3 this was about 3.5 weeks. For a quantitative determination,
we put [A] = ½[A]0 into Eq. 9.12:
1
2 [A]0
1
2
= e - kt1>2
5
10
9
ln(Penicillin units)
FIGURE 9.4 The ln (concentration) of penicillin plotted versus
time. The data are consistent
with first-order kinetics; the
slope of the line is equal to
minus the rate coefficient k for
the reaction.
= [A]0 e - kt1>2
8
7
6
5
t (weeks)
15
20
Kinetics | 315
Taking logarithms of both sides, we have
ln 12 = -kt1>2 .
Since ln ½ = -0.6931 and therefore ln 2 = 0.6931 we see that
t1>2 =
ln 2
0.6931
=
.
k
k
(9.13)
For the penicillin data, k = 0.2036 week-1 and the half-life is t½ = 3.404 weeks.
E X E R C I S E 9 .1
From the definition of half-life show that
[A]
= 2 - t>t1>2
(9.14)
[A]0
and that t½ is independent of the initial concentration of reactant. Just combine Eqs. 9.12
and 9.13 to obtain Eq. 9.14; then show that the time for the concentration to change from
[A]0 to ½[A]0 is the same as the time for the concentration to change from ½[A]0 to ¼[A]0.
Another parameter used to characterize first-order kinetic data is the relaxation time t;
it is the reciprocal of the rate coefficient k. From Eq. 9.12,
1
t =
(9.15a)
k
[A]
= e - t>t .
(9.15b)
[A]0
The relaxation time t is the time required for the concentration to decrease to 1>e = 0.3679
times its initial value (at t = t, e-t>t = e-1).
Note that the half-life, the relaxation time, and the first-order rate coefficient
depend only on ratios of concentrations. We do not need to know actual concentrations to characterize the kinetics of first-order reactions. We can measure any property
proportional to concentration, such as the optical absorbance or the mutagenic ability.
It is only for first-order reactions that the rate coefficient, with units of t-1, does not
contain units of concentration. A familiar example is radioactive decay, which is always
first order. It does not matter whether there is a large or small amount of a radioactive
isotope—the half-life is the same. For all other rate processes, the rate coefficient does
contain units of concentration, and half-lives and relaxation times depend on initial
concentrations.
We can now characterize the loss of activity of the penicillin at one temperature
and one buffer solution by a single number—the half-life of the first-order reaction is
3.40 weeks. It is useful to be able to characterize the data taken over a 20-week period by
one number instead of a table of many numbers, such as table 9.2. How can knowledge
of kinetics help further? The fact that most reactions are faster at higher temperatures
suggests that penicillin will remain active longer when stored in a refrigerator at 4 ⬚C or
in liquid nitrogen at 77 K. Because it might take years to measure the stability at low temperatures, it will be quicker to measure the rate of decomposition at higher temperatures
and use the Arrhenius equation, which we will discuss shortly, to extrapolate to 4 ⬚C. We
would need to make measurements in frozen solutions at low temperatures to extrapolate
to 77 K.
A proposed mechanism can help us understand the data and may help us formulate a
better buffer solution for the penicillin. Penicillin has a lactam structure—an amide ring.
316 Chapter 9 |
Kinetics: Rates of Chemical Reactions
The strained four-member ring is easily hydrolyzed to give an inactive compound. The
stoichiometric reaction is
H 2O
R
N
O–
O
R
O
N
H2+
By analogy with other ester and amide hydrolysis mechanisms, we propose
OH–
slow
N
O
R
OH–
H2O
fast
N
–O
R
N
H
R
R
fast
OH
O
O
O–
N
H2+
OH
The reaction starts with a hydroxide ion adding to the carbonyl carbon of the lactam.
This is proposed to be the slowest step in the reaction; therefore, we expect that the
rate of the reaction will be directly proportional to the concentration of OH-. The
next two steps are much faster than the first step, so they do not affect the kinetics.
We say that the first step is rate determining. In a buffer with a pH of 6 instead of 7,
the proposed mechanism predicts that the rate of decomposition should be one-tenth as
fast. The half-life would be 33.7 weeks instead of 3.37 weeks. Of course, the rate of
penicillin decomposition at pH 6 would have to be measured, and the effect of injection
into patients of pH 6 solution would need to be determined, but clearly knowledge of
kinetics has been helpful.
We will discuss reaction mechanisms in much more detail later in this chapter. Here
we just want to give an idea of how kinetic measurements can be used.
E X A M P L E 9 .1
Carbon dioxide in the atmosphere contains a small but readily detectable amount of the
radioactive isotope 14C. This isotope is produced by high-energy neutrons (in cosmic
rays) that transform nuclei of nitrogen atoms by the process
14N
+ 1n h
14C
+ 1H .
The 14C nucleus is unstable and decays by the first-order process
14C
h
14N
+ b,
with a half-life of 5770 years. The production and decay of 14C leads to a nearly
constant steady-state concentration. The amount of 14C present in a carbon-containing
sample can be determined by measuring its radioactivity (the rate of production of
high-energy electrons, b particles).
In the atmosphere, CO2 contains a nearly constant amount of 14C. Once CO2 is
“fixed” by photosynthesis, however, it is taken out of the atmosphere and new 14C
is no longer added to it. The level of radioactivity then decreases by a first-order
process with a 5770-year half-life.
A sample of wood from the core of an ancient bristlecone pine in the White Mountains
of California shows a 14C content that is 54.9% as great as that of atmospheric CO2. What
is the approximate age of the tree?
Kinetics | 317
SOLUTION
Since the process is first order, we need know only the ratio of the present radioactivity to
the original value. Assume that the level of 14C in the atmosphere has not changed over the
life of the tree. (This is not strictly true. Corrections can be made, but they are small.) Thus,
3 14C 4
= 0.549 .
3 14C 4 0
Using Eqs. 9.9 and 9.12, we obtain
lna
3 14C 4
0.693
b = t,
14
t1>2
3 C40
where t is the age of the wood. Therefore,
(0.600)(5770 yr)
0.693
ln (0.549) = t or t =
= 4990 yr .
5770 yr
0.693
We conclude that the carbon in the tree was taken from the atmosphere approximately
4990 years ago.
Second-Order Reactions
A second-order reaction corresponds to the rate law v = kc2 or v = kcAcB. The units of k
are those of c-1t-1, typically M-1s-1.
It is useful to separate the treatment of second-order reactions into two major classifications, depending on whether the rate law depends on (I) the second power of a single
reactant species or (II) the product of the concentrations of two different reagents.
Class 1
v = k[A]2
Although the stoichiometric equation may involve either one or several components,
the rate law for many reactions depends only on the second power of a single component.
Some examples are:
2 proflavin h [proflavin]2 ; v = k[proflavin]2
NH4OCN h NH2CONH2 ; v = k[NH4OCN]2
ammonium cyanate
A–A–G–C–U–U
k1
·
·
·
·
·
·
2A–A–G–C–U–U
urea
U–U–C–G–A–A
hexanucleotide*
; v=k2[A2GCU2]
2
hexanucleotide dimer
In each case, the rate law is of the form
v = -
d[A]
= k[A]2 .
dt
The variables can be separated,
-d[A]
= k dt
[A]2
and each side integrated, to give
1
= kt + C .
[A]
*See the appendix for the structure of the nucleotides.
(9.16)
318 Chapter 9 |
Kinetics: Rates of Chemical Reactions
2
c (M)
B
1
A
C
t
FIGURE 9.5 Reaction
A + B h C, [A]0= 1 M,
[B]0= 2 M, via a second order
rate equation v = k[A][B].
Since [A] = [A]0 when t = 0, the constant of integration is 1>[A]0, and we obtain an integrated form of the second-order rate equation:
1
1
= kt
(9.17)
[A]
[A]0
Note that for second-order kinetics one expects a linear relation between the reciprocal
of the reactant concentration and time. This is in contrast to first-order kinetics where the
logarithm of concentration is linear in time.
Class II
v = k[A][B]
(9.18)
A reaction that is second order overall may be first order with respect to each of the
two reactants. Some examples are:
CH3COOC2H5 + OH - h CH3COO - + C2H5OH; v = k[CH3COOC2H5][OH - ]
NO(g) + O3(g) h NO2(g) + O2(g); v = k[NO][O3]
H2O2 + 2Fe2 + + 2H + (excess) h 2H2O + 2Fe3 + ; v = k[H2O2][Fe2 + ]
Again we see that the overall stoichiometric equation is not a valid indicator of the rate.
We may expect that there are significant underlying differences in the mechanisms of
these reactions.
In general, for reactions of the type
A + B h products
exhibiting class II kinetics, the initial concentrations of the reactants, a and b, need not be
the stoichiometric ratio. For this reason, we rewrite the rate law
v = k[A][B]
(9.19)
ln([A]/[B])
in terms of the reaction variable x, where x = the concentration of each species reacted.
For a stoichiometric reaction of A + B h products:
[A] = a - x;
[A]0 = a
[B] = b - x;
[B]0 = b
Following substitution into Eq. 9.19 and separation of variables, we obtain
t
FIGURE 9.6 Plot of ln([A]>[B])
vs. t for the reaction shown in
figure 9.5.
dx
= kdt.
(9.20)
(a - x)(b - x)
A table of integrals or symbolic algebra program gives the result
b(a - x)
1
ln
= kt (a ⬆ b)
(9.21)
a - b a(b - x)
or, alternatively,
[B]0[A]
1
ln
= kt .
(9.22)
[A]0 - [B]0 [A]0[B]
After separating the constant terms involving only initial concentrations, we see that class II
second-order reactions exhibit a linear relation between ln ([A]>[B]) and time.
Note that when a = b (when the initial concentrations are stoichiometric), then Eqs. 9.22
and 9.23 do not apply. In this case, however, the method of class I is appropriate because
the values of [A] and [B] will be equal throughout the entire course of the reaction. That
is, if [A] = [B] at all times then
v = k[A][B] = k[A]2 ,
which can be integrated exactly as in class I to give Eq. 9.17.
Kinetics | 319
E X AMPLE 9. 2
Hydrogen peroxide reacts with ferrous ion in acidic aqueous solution according to
the reaction
H2O2 + 2Fe2+ + 2H+ h 2H2O + 2Fe3+ .
From the following data, obtained at 25 ⬚C, determine the order of the reaction and the
rate coefficient:
[H2O2]0 = 1.00 * 10-5 M
[Fe2+]0 = 1.00 * 10-5 M
[H+]0 = 1.00 M
Time (min)
105
*
[Fe2+],
M
0
5.3
8.7
11.3
16.2
18.5
24.6
34.1
0
0.309
0.417
0.507
0.588
0.632
0.741
0.814
SOLUTION
Because the H+ concentration is in huge excess, it will not change significantly during
the course of the reaction. As a consequence, we can write the rate expression as
v = k[H2O2]i[Fe2 + ] j ,
recognizing that this is the simplest general form. Note also that the initial
concentration of H2O2 is twice the amount needed to react with all of the Fe2+ present.
1. Test for first order in [Fe2+], where i = 0 and j = 1. In this case, we expect from
Eq. 9.10 that
ln
[Fe2 + ]0
[Fe2 + ]
= kt or
1 [Fe2 + ]0
ln
= k.
t [Fe2 + ]
2. Test for first order in [H2O2], where i = 1 and j = 0. Then
1 [H2O2]0
ln
= k.
t [H2O2]
The data for testing proposals 1 and 2 are as follows:
Time (min)
105
*
[Fe2+],
M
2+
ln
(1)
[Fe ]0
0
5.3
8.7
11.3
16.2
18.5
24.6
34.1
1.00
0.691
0.583
0.493
0.412
0.368
0.259
0.186
0
0.370
0.540
0.707
0.887
1.000
1.351
1.682
__
69.8
62.0
62.6
54.7
54.0
54.9
49.3
1.00
0.845
0.791
0.741
0.706
0.684
0.630
0.593
__
31.8
26.9
26.5
21.5
20.5
18.8
15.3
2+
[Fe ]
103 [Fe2 + ]0
ln
t
[Fe2 + ]
105 * [H2O2], M
(2) 103 ln
[H2O2]0
[H2O2]
The values in the rows designated (1) and (2) are not constant, but show a trend,
indicating that neither of these predictions is valid for the reaction.
320 Chapter 9 |
Kinetics: Rates of Chemical Reactions
3. Test for second order overall, where i = 1 and j = 1. Because the stoichiometric
coefficients are not unity, we need to modify the development of Eq. 9.22. For
a reaction with stoichiometry A + 2B S product, we define
v = -
d[A]
1 d[B]
= k[A][B] = ,
dt
2 dt
where [A] = a - x, [A]0 = a and [B] = b - 2x, [B]0 = b. The equation analogous
to Eq. 9.22 is now
dx
= k dt
(a - x)(b - 2x)
or multiplying the equation by 2,
dx
= 2kdt .
(a - x)(b>2 - x)
This is readily solved by noting that with respect to Eq. 9.21 we have just replaced b
by b>2 and k by 2k. We simply make the same replacements in the solution, Eq. 9.22
to obtain the result
b
(a - x)
1
2
ln
= 2kt
b
b
a a a - xb
2
2
or
b(a - x)
1
ln
= kt
2a - b a(b - 2x)
or
[Fe2 + ]0[H2O2]
1
ln
= kt .
2[H2O2]0 - [Fe2 + ]0 [H2O2]0[Fe2 + ]
Since [Fe2+]0>[H2O2]0 = 1 from the initial conditions, we expect the relation
1 [H2O2]
ln
= constant
t [Fe2 + ]
to hold if the rate law is first order with respect to each reactant.
Proposal 3 is tested by the following tabulation:
Time (min)
[H2O2]
0
5.3
8.7
11.3
16.2
18.5
24.6
34.1
1.00
1.223
1.357
1.503
1.714
1.859
2.432
3.188
[Fe2 + ]
(3)
1 [H2O2]
(min - 1)
ln
t
[Fe2 + ]
__
0.0380 0.0351 0.0361 0.0332 0.0335 0.0361 0.0340
The entries in the bottom row are approximately constant and show that the results
do correspond to this rate law. Therefore,
v = k[H2O2][Fe2 + ] .
Kinetics | 321
The value of k can be determined from
k =
1
1 [H2O2]
d.
c ln
2[H2O2]0 - [Fe2 + ]0 t [Fe2 + ]
The average value of (1>t)ln{[H2O2]>[Fe2+]} from the table is 0.035 min-1.
Therefore,
k2 =
0.035 min - 1
= 3.5 * 103 M - 1 min - 1 .
(2 * 10 - 5 - 1 * 10 - 5) M
Renaturation of DNA as an Example of a Second-Order Reaction
A double-stranded DNA is made of two antiparallel chains with nucleotide sequences that
are complementary to each other. If a linear DNA fragment is heated or subjected to high
pH, the two chains separate, and this disordered product is termed the denatured, or coiled,
form. If the temperature or the pH is then lowered so that the double helix is again the
stable form, pairing of the bases between chains of complementary sequences occurs, and
the chains will reassociate, or renature. If we designate the strands with complementary
sequences as A and A⬘, the renaturation reaction can be written as
k
A + A⬘ h AA⬘,
which has been found experimentally to be second order. The most striking feature of
renaturation kinetics is that when DNA from different sources is first broken down to about
the same size (by sonication, for example) and then the rates of renaturation are measured
at the same DNA concentrations in moles of nucleotides per liter, the rates are found to
span a range of several orders of magnitude (figure 9.7). The following analysis shows that
the rate is expected to be inversely proportional to the sequence complexity of the DNA.
Figure 9.8 illustrates that the rate of renaturation depends on the repetitiveness of the
DNA sequence. When a long DNA molecule with no repetitive sequences is fragmented
into pieces longer than 20 base pairs (bp), each piece should be a unique sequence. For
example, a DNA 106 bp long can be broken into essentially 106 (actually 106 - 19) pieces
of length 20. The number of possible DNA sequences that are 20 bp long is 420 (any of
the 4 bases can be in each position). Because 420 is slightly greater than 1012, which is
much larger than 106, each piece is likely to be unique. When the pieces are denatured,
the concentration of each unique strand is very small, and the probability of finding
its Watson–Crick complementary strand is small. The rate of renaturation will be very
slow [figure 9.6(a)]. However, if there are repetitive sequences, some of the fragmented
pieces can be identical, and it will be easy for a denatured strand to find its complement
[figure 9.6(b)]. The analysis can be done quantitatively.
Let c0 be the total concentration (in moles of nucleotide per liter) of all the single
strands before any renaturation occurs. This total concentration can be measured easily
from the UV absorbance of the solution. The rate of renaturation will depend on [A]0, the
initial concentration (in moles of nucleotide per liter) of fragment A that is complementary
to fragment A⬘. We see from figure 9.7 that [A]0 can vary from c0 >2 for poly dA · poly dT
[figure 9.6(b)] to c0 >2N for a DNA of N bp of a nonrepeating sequence.
Therefore, we define N as the sequence complexity—the total number of base pairs in
the smallest repeating sequence in the DNA. The concentration [A]0 is proportional to c0 >N.
If the DNA is that of the bacterium E. coli, for example, there is very little repetition of the
sequence, and N is the same as the number of base pairs per genome, or about 3 * 106. If the
DNA is polydeoxyadenylate (poly dA) in one strand and polydeoxythymidylate (poly dT)
in the other, then N is 1, since the smallest repeating sequence is just a single base pair.
322 Chapter 9 |
Kinetics: Rates of Chemical Reactions
Complexity (nucleotide pairs)
1
10
102
103
104
105
106
107
108
109
1010
0
0
Calf
nonrepetitive
fraction
Fraction reassociated
T4
E. coli
Poly U
+ Poly A
0.5
0.5
MS-2
Mouse
satellite
.
1.0
10−6
10−5
10−4
10−3
10−2
1.0
0.1
C0t (M s)
1
10
100
1000
10,000
FIGURE 9.7 Reassociation rates in solutions containing 0.18 M Na+ for several double-stranded DNAs (mouse
satellite—part of the mouse genome that has highly repetitive sequences, T4 bacteriophage, E. coli bacteria, and
part of the calf genome). Data for two double-stranded RNAs (MS-2 viral RNA and a synthetic RNA, poly A·poly
U) are also shown. The large nucleic acids were first broken into double-stranded fragments of about 400 base
pairs by sonication. The fragments were then denatured into single strands of about 400 nucleotides by heating
the solution to 90°C for a few minutes. The solution was cooled quickly and the fraction reassociated to double
strands was measured versus time t. The rate of reassociation (renaturation) to double strands depends on
the concentration of the single strands c0 and on the number of different sequences in the nucleic acids—the
sequence complexity. c0 is expressed in units of moles of nucleotides per liter. The complexity of the samples,
indicated by arrows over the upper scale, range from 1 for poly A·poly U to over 109 for the nonrepetitive fraction of calf DNA. (Based on data from R. Britten and D. Kohne, 1968, Science 161:529.)
The rate of renaturation is
-d[A]
-d[A⬘]
=
= k[A][A⬘] = k[A]2
dt
dt
The DNA is originally double stranded, so the concentrations of A and A⬘ must be equal.
The equation is the same as Eq. 9.16, and the time to go half way to completion is, from
Eq. 9.17:
1
t1>2 =
.
k[A]0
But [A]0 is proportional to c0>N and so substituting into the equation above gives
N
t1>2 ⬀
kc0
or
c0t1>2 ⬀ N .
Thus, the half-time of renaturation times the total initial denatured strand concentration is
proportional to the sequence complexity of the DNA. A long half-time means that each
Kinetics | 323
single strand will make many collisions before it finds its complement; the sequence is
complex. The quantity c0t½ can thus be used to measure the complexity of the nucleotide
sequence of the genome of an organism. The method is obviously applicable to doublestranded RNA as well. Figure 9.5 shows that the sequence complexity varies from 1 for
synthetic poly A·poly U to 109 for the nonrepetitive fraction of calf thymus DNA.
DNA with nonrepetitive sequence of 106 base pairs
Sonicate
Sonication produces ≈106 different fragments, each
containing about 400 base pairs.
Denature
F
A
I′
L
M
N′
M′
I
K′
C
F′
B′
D
E
J′
K
H
J
L′
A′
E′
G
C′
H′
G′
N
B
D′
Renature
F
A
G
L
E
A′
G′
L′
E′
J
J′
D
D′
K
H
C
N
C′
I′
M
K′
M′
H′
F′
I
B′
N′
B
Renaturation is very slow because strands A and A′
are in very low concentrations; many collisions are
needed to find complementary partners.
(a)
FIGURE 9.8 Renaturation (or reassociation) rates depend on the sequence complexity of the DNA. In (a) a
complex DNA, such as E. coli DNA with a nonrepeating sequence, is fragmented to give essentially all different
fragments. The concentration of each fragment is very small compared with the total concentration of all
fragments. After the double-stranded fragments are denatured, the low concentrations of the complementary
single strands lead to a slow rate of renaturation. In (b) all the fragments of poly dA·poly dT are the same. The
concentrations of the single strands are high, and the rate of renaturation is much faster.
324 Chapter 9 |
Kinetics: Rates of Chemical Reactions
DNA with repeating sequence of 106 base pairs—poly dA · poly dT
Sonicate
Sonication produces ≈106 identical fragments, each
containing about 400 base pairs.
Denature
A
T
T
T
A
A
A
A
A
A
A
T
A
T
T
T
A
T
A
T
A
T
A
T
A
A
A
T
T
T
A
T
T
A
T
T
Renature
T
A
A
T
A
T
A
A
T
A
T
T
T
A
A
T
A
A
T
T
T
A
T
A
T
T
A
T
A
A
Renaturation is very fast because strands
A and T are in high concentrations.
(b)
FIGURE 9.8 (continued)
Reactions of Other Orders
It is easy to generalize the treatment of class I reactions of order n, where n is any positive
or negative number except +1. The rate law is then
v = k[A]n ,
(9.23)
1
1
1
c
d = kt (n ⬆ 1) .
n
1
n - 1 [A]
[A]n0 - 1
(9.24)
which can be integrated to give
For example, the thermal decomposition of acetaldehyde is 3/2 order over most of the
course of the reaction. A linear plot should be obtained, according to Eq. 9.24, if [A]-½ is
plotted versus time.
Kinetics | 325
Determining the Order and Rate Coefficient of a Reaction
We have discussed various simple orders that can represent reaction kinetics, and we have
given specific illustrations. Here we mention some general methods that can be used to
determine the order and rate coefficient for a reaction. We need to know the stoichiometry of the reaction, and we need to be able to measure the concentration of a product or
reactant. To determine the order of the reaction with respect to each reactant, we actually
need only to measure a property that is directly proportional to concentration. Except for
first-order reactions, we do need to know concentrations to obtain rate coefficients. The
characteristics of the different order reactions are given in table 9.3.
TABLE 9.3 The Characteristics of Simple-Order Reactions
Reaction order
Linear plot
Rate proportional to
Units of k
Zero-order
c vs t
dc>dt = -k
M s-1
First-order
ln c vs. t
dc>dt = -kc
s-1
Second-order (I)
1>c vs t
dc>dt = -kc2
M-1 s-1
nth-order
1>cn-1
dc>dt =
M-(n-1) s-1
vs t
-kcn
Direct Data Plots
A plot of concentration versus time gives a clue to the order. If the plot is linear, a zero-order
reaction is indicated. If a curved plot results, other plots, such as ln c versus time or 1>c
versus time, can be tried. Look for a linear plot that indicates the order and whose slope
is proportional to the rate coefficient.
Rate Versus Concentration Plots (Differentiation of the Data)
From measurements of the slopes of plots of concentration versus time, you can determine the rate of the reaction at different times. These rates can then be plotted versus the
corresponding reactant concentration to try to deduce the form of the rate law. This
approach requires many experimental points or a continuous curve.
Method of Initial Rates
The rate is measured during the earliest stages of the reaction. The concentrations do not
change much with time, and they can be replaced in the rate law by the initial concentrations. This method is particularly useful when the rate law is not of the simplest form. For
example, in many enzyme-catalyzed reactions, the method of initial rates has been used
to find that
k[S]
v =
,
K + [S]
where k and K are constants and [S] is the substrate concentration. (We discuss this
Michaelis–Menten type of kinetics in chapter 10.)
Changes in Initial Concentrations
The influence of different initial concentrations on the initial rate, the half-life, or some
other parameter can be determined. The method measures the order of the reaction with
respect to a particular component directly. To find the order with respect to component
A in the expression
v = k[A]a[B]b[C]c
we measure the initial rate v1 with initial concentrations [A]0, [B]0, [C]0:
v 1 = k[A0]a[B0]b[C0]c
326 Chapter 9 |
Kinetics: Rates of Chemical Reactions
We then double the concentration of A0 keeping the others constant, and measure the new
initial rate v2:
v 2 = k(2[A0])a[B0]b[C0]c
The ratio of these two rates leads to the value of a,
v2
= 2a ,
v1
because all other factors cancel.
Methods of Reagents in Excess
We can decrease the kinetic order of a reaction by choosing one or more of the reactant
concentrations to be in large excess. Normally, the concentration of a reactant decreases
steadily during the course of a reaction. This complicates the kinetic analysis, especially if the rate law is not a simple one. By using a large excess of one component,
its concentration is maintained nearly constant, and the corresponding term in the rate
law now hardly changes during the course of the reaction. In the example above, we
could choose B and C to be 1 M and A to be 0.01 M . Concentrations of B and C would
change only slightly (depending on the stoichiometry) during the complete reaction of A.
We could essentially study the kinetics with respect to A independently of the B and C
kinetics.
Once the order of the reaction is known with respect to each component, the value of
the rate coefficient can be calculated. It is important to specify what rate and what species
are being described. Uncertainty can occur when the stoichiometric coefficients for the
reaction are not all the same. For example, the reaction
N2O5 h 2NO2 +
1
2 O2
exhibits first-order kinetics. If we express concentrations of these substances in mol L-1
the rate of disappearance of reactant N2O5 is half as great as the rate of formation of the
product NO2 and twice as great as the rate of formation of O2. Expressed mathematically, this is
-
d[N2O5]
d[O2]
1 d[NO2]
=
= 2
.
dt
2 dt
dt
The coefficients in this rate equation are simply the reciprocals (apart from the signs) of
the stoichiometric coefficients. The standard convention is that for the following stoichiometric equation
mM + nN h pP + qQ
the rate of the reaction is written as
v = -
1 d[N]
1 d[P]
1 d[Q]
1 d[M]
= =
=
.
q dt
m dt
n dt
p dt
The rate-law expression
v = k[M]i[N]j #
##
will serve to define the relation of the rate coefficient to the rate of the reaction. (Remember
that the exponents i and j in the rate law are not related to the stoichiometric coefficients.)
Reaction Mechanisms and Rate Laws | 327
Reaction Mechanisms and Rate Laws
The form of the rate law determined from the kinetics of a reaction does not tell us the actual
mechanism, although it gives important clues. A very complex process, such as the growth
and multiplication of bacterial cells, exhibits simple first-order kinetics. Growth, as measured
by the increase in the number of cells, occurs in proportion to the number of cells present at
any given time and with a rate coefficient determined by the turnaround time for the complex events in the cell cycle. We cannot hope for this single parameter, the rate coefficient,
to give us detailed information about the individual steps of DNA replication, transcription
into RNA, synthesis of protein, mitotic division, and so on that are involved in the cell cycle.
The rate law does, however, make a quantitative statement about how fast the cells grow.
Typically, a complex reaction sequence or mechanism is made up of a set of elementary
reactions. Consider, for example, the process by which the ozone layer of the upper atmosphere is maintained. This ozone layer acts as an important shield, absorbing potentially
harmful UV radiation from the Sun and protecting organisms on the surface of Earth
from dangerous biological consequences. Ozone is formed from atmospheric oxygen by a
process that can be described by the stoichiometric equation
3O2 h 2O3 .
Ozone formation starts with a photochemical step resulting from the absorption of shortwavelength UV radiation by O2, followed by a second elementary reaction in which the
oxygen atoms produced in the first step react with O2 to form ozone:
hv(below 242 nm)
h O + O
O + O2 + M h O3 + M
O2
where M is a third body (any other gas molecule) whose function is to remove excess
energy from the excited ozone initially formed. The protective role of ozone results from its
absorption of somewhat longer wavelength UV light not absorbed by O2 and this radiation
leads to the decomposition of ozone by the elementary reaction
O3
hv (190 to 300 nm)
h O2 + O .
The balance in the ozone concentration is achieved by these three elementary steps
operating together, supplemented by lesser contributions of additional reactions involving
other elements.
Johnston (1975) pointed out that the flights of supersonic transports in the upper
atmosphere posed a threat to the stability of the protective ozone layer. This comes about
because of the role of oxides of nitrogen (NOx) in the exhaust gases of the jets. The oxides
catalyze the decomposition of ozone:
NOx
O + O3 h 2O2
This process can be understood in terms of the elementary reactions involved, which are:
NO + O3 h NO2 + O2
NO2 + O h NO + O2
Not only does this show in molecular detail how NOx produces the ozone decomposition,
but it also shows that NO and NO2 are not used up in the process. Together they serve a
true catalytic function.
Another class of catalyst for the destruction of ozone is chlorofluorocarbons (Molina
and Rowland 1974; Anderson, Toohey, and Brune 1991). Chlorine atoms are produced by
the photochemical decomposition of, for example, dichlorodifluoromethane (Freon 12):
CCl2F2 h CClF2 + Clⴢ
328 Chapter 9 |
Kinetics: Rates of Chemical Reactions
The chlorine atoms catalyze the decomposition of ozone molecules by oxygen atoms:
Clⴢ + O3 h ClOⴢ + O2
O + ClOⴢ h Clⴢ + O2
The net reaction is that ozone is converted to oxygen and the chlorine atoms are not used
up. Because of the environmental dangers caused by the loss of ozone, the use of chlorofluorocarbons in aerosol cans has been banned in the United States.
It often happens that more than one mechanism can be written consistent with a given
rate law. The best one can be chosen only if we learn more about the intermediates, the
energetics of the process, and the rates of elementary steps determined from other reactions. The concept of elementary reactions is important to chemical kinetics for the same
reason that the study of chemical bonds or of functional groups is important in simplifying and organizing the chemistry of millions of different organic compounds. The kinetic
parameters of an elementary step are the same regardless of the overall reaction in which
it is participating.
Elementary reactions are described in terms of their molecularity: the number of
reactant particles involved in the elementary reaction. Reactions between two particles
are called bimolecular. They may be thought of in terms of a collision between the two
reactants to form a transition state. A transition state is an unstable arrangement of atoms
intermediate between reactants and products. The transition state exists for no longer than
the time it takes for a molecular vibration (of order 10 femtoseconds). For example, the
reaction of ozone with nitric oxide in the gas phase occurs as a simple bimolecular process.
The transition state is the transient species in which an O=O bond is being broken as an
O=N bond forms:
collision
O3 + NO
O
transition state
O
O
N
O
O2 + NO2
Since the process is bimolecular, it follows that it must exhibit second-order kinetics.
The converse is not true, however; many second-order reactions do not have simple bimolecular mechanisms but are more complex. The form of the rate law alone cannot predict
the mechanism but the mechanism directly provides the rate law. This is an important
distinction to learn.
From the example given above, we can define molecularity as the number of particles that come together to form the transition state in an elementary reaction. Note that the
term molecularity applies only to elementary reactions and not to the overall mechanism.
Unimolecular reactions are those that involve only a single reactant particle—for example,
radioactive decay. True unimolecular processes are rare because most reactions require some
activation energy in order to proceed. This energy is commonly picked up by collisions with
the surroundings (other molecules, walls of the vessel). Termolecular reactions involve collisions of three molecules and are more likely to occur at high pressures or in liquid solution.
The simplest examples of reaction mechanisms are those that involve only a single
elementary reaction. In this case, the rate law can be written by inspection. The rate is
proportional to the concentration of each of the species involved in forming the transition state, and the exponents are determined by the numbers of each kind of particle. For
example, if the reaction
A + B h products
is elementary, its rate will be given by
v = k[A][B] .
Reaction Mechanisms and Rate Laws | 329
The real problem occurs when the overall reaction is complex and consists of several
connected elementary reactions. The overwhelming majority of reactions and processes
in nature are complex.
A few examples will illustrate some important methods of dealing with complex
reactions. One general method is to see whether one elementary reaction controls the
overall kinetics—that is, to look for the rate-determining step. For certain mechanisms,
one elementary reaction is found to dominate the kinetics of the complex reaction. When
this is so, the analysis of the mechanism is greatly simplified. We can write the rate for
the rate-determining step and equate it to the overall rate. In the following examples, we
discuss the general case first and then indicate the simplification that occurs if there is a
rate-determining step.
Parallel Reactions
Parallel reactions are of the following type:
k1
B
k2
C
A
The substance A can decompose by either of two paths, giving rise to different products,
B and C. The rate expressions are
d[A]
= -k 1[A] - k 2[A] = -(k 1 + k 2)[A]
(9.25)
dt
d[B]
= k1[A]
(9.26)
dt
d[C]
= k2[A] .
(9.27)
dt
The solution to the first equation, which has the form of the first-order rate law, Eq. 9.7, is
[A]
ln
= -(k 1 + k 2)t
[A]0
[A] = [A]0exp[ -(k 1 + k 2)t] .
(9.28)
Thus, [A] decays exponentially, as illustrated in figure 9.9. To find out how [B] and [C]
increase with time, we need to solve Eqs. 9.26 and 9.27. Because each of these equations
contains three variables, it cannot be solved as it stands. However, we can substitute for
A using Eq. 9.28. Thus,
d[B]
= k 1[A] = k 1[A]0exp[ -(k 1 + k 2)t]
dt
and
d[C]
= k 2[A] = k 2[A]0exp[ -(k 1 + k 2)t] .
dt
A
B
We now separate variables and integrate. If [B]0 = [C]0 = 0, the solutions are
c
[B] =
k 1[A]0
5 1 - exp[ -(k 1 + k 2)t] 6 ,
k1 + k2
(9.29)
[C] =
k 2[A]0
5 1 - exp[ -(k 1 + k 2)t] 6 .
k1 + k2
(9.30)
Note that B and C are formed in a constant ratio, [B]>[C] = k1>k2, throughout the course of the
reaction. The extension of this method to three or more parallel reactions is straightforward.
C
t
FIGURE 9.9 Kinetic curves for
reactant and products for two
first-order reactions in parallel.
330 Chapter 9 |
Kinetics: Rates of Chemical Reactions
E X AMPLE 9. 3
Radioactive decay frequently occurs simultaneously by two separate first-order paths.
For example 64Cu is unstable against decay, emitting either an electron or a positron.
The scheme can be written as follows:
64Cu
k1
64Zn
+ β– (62%)
k2
64Ni
+ β+ (38%)
The decay of 64Cu occurs exponentially with a single half-life of 12.80 h. Calculate the
rate coefficients for the two paths, k1 and k2.
SOLUTION
From Eq. 9.25 we see that 64Cu decays as a first-order process with rate coefficient
equal to k1 + k2 Thus,
ln 2
ln 2
=
k1 + k2 =
= 0.05415 h - 1 .
t1/2
12.80 h
But
38
k 2 >k 1 =
.
62
Combining the two equations for k1 and k2,
k1 =
0.05415
k2 =
0.05415
1 +
1 +
38
62
62
38
= 0.0336 h - 1 ,
= 0.0206 h-1 .
In parallel reactions, if one step is much faster than the others, this fast step primarily
determines the rate of reaction. This is very different from the case of series reactions,
in which the slow step determines the rate, as we will see in the next section. In the
example above, if 99% of the reaction goes through the b- path, we can ignore the b+
path in calculating the decay. We can see this quantitatively in Eqs. 9.29 and 9.30.
When k1 >> k2 the decay of [A] depends only on k1 and [A] ⬵ [A]0exp(-k1t).
Series Reactions (First Order)
Series reactions are of the type
k1
k2
A h B h C.
Compound A reacts to form B, which then goes on to form C. The rate expressions are
v1 = -
d[A]
= k 1[A] ,
dt
[A] = [A]0exp(-k 1t) ,
d[B]
= k 1[A] - k 2[B] = k 1[A]0exp(-k 1t) - k 2[B] . (9.31)
dt
This differential equation is slightly more difficult to solve, but we show the result. If [B]0 = 0:
[B] =
k 1[A]0
[exp(-k 1t) - exp(-k 2t)] .
k2 - k1
(9.32)
Reaction Mechanisms and Rate Laws | 331
Similarly,
v2 =
=
d[C]
= k2[B]
dt
k 1k 2[A]0
[exp(-k 1t) - exp(-k 2t)] .
k2 - k1
Variables [C] and t can be separated here to give, for [C]0 = 0,
[C] = [A]0 e 1 -
1
[k exp(-k 1t) - k 1exp(-k 2t)] f .
k2 - k1 2
(9.33)
Where such complex formulas are involved, it is important to get some intuitive feeling for
how these reactions behave. Figure 9.10 illustrates the pattern for series first-order reactions.
As the concentration of A decreases, it does so in a simple exponential fashion. If there is no
B present initially, the rate of the second reaction will be negligible at the beginning, (v2)0 =
0. As B is formed by the first step, its concentration rises, reaches a maximum, and then
declines to zero as the second step removes it from the reaction mixture. In accordance with
the rate law, the curve for [C] starts out with zero slope, where v2 = 0. The rising portion
of curve [C] then undergoes an inflection (the point where d2[C]>dt2 = 0) when [B] has its
maximum value. Finally, [C] approaches a limiting value as the reaction nears completion.
These properties are useful diagnostics for series reactions.
When two or more reactions occur in series, the slow reaction is the rate-determining
step. It dominates in the kinetic control of the overall process. The analogy with a city waterdistribution system is appropriate. It does not matter how large the reservoir, water mains,
and distribution pipes are. If the smallest cross section of the whole system is the ½ inch
pipe coming into your house, which will limit the maximum possible flow, and the only
practical way to increase the flow is to replace the small pipe with one of larger diameter.
For the series reactions
k1
k2
A h B h C
(a)
B
C
c
we can consider the two cases k1 W k2. and k1 V k2. The first situation means that the
first step is much faster than the second, and the second is the rate-determining step. Once
the reaction is started, A will be converted to B rapidly in comparison with the subsequent
reaction of B to C. During most of the course of the reaction, B undergoes a first-order
conversion to C, which is controlled by the rate coefficient k2.If the formation of C is our
measure of the rate of the reaction, its appearance is nearly identical to that of a single-step
(elementary) first-order reaction. This is illustrated schematically in figure 9.10(a). We
arrive at the same conclusion from examination of Eq. 9.33 in the limit of k1 W k2. Under
those conditions, Eq. 9.35 reduces to
A
t
(b)
A
C
c
[C] = [A]0[1 - exp(-k2t)] ,
which has the same form as the simple first-order formation kinetics.
The second situation, where k1 V k2 starts out with a slow conversion of A to B and
is followed by a very rapid reaction for B going on to C. In this case, the concentration
of B remains small throughout the course of the reaction, and C appears essentially as A
disappears. This is illustrated in figure 9.10(b). The rate of the reaction can be measured
either by the formation of C or by the disappearance of A. Thus, Eq. 9.33 applies to the
overall reaction.
The chief reason for emphasizing rate-limiting steps in series reactions is because
complex processes in biology frequently have rate-limiting steps. For our analysis, it does
B
t
FIGURE 9.10 Concentrations
for reactants and product
in the series reaction
A h B h C (a) k1 = 5,
k2 = 0.5 (b) k1 = 0.5, k2 = 5.
332 Chapter 9 |
Kinetics: Rates of Chemical Reactions
not matter how many steps are involved in the series mechanism; all that is required is
that one step should be appreciably slower than any of the others. If the rate-limiting step
is not first order, as in the example treated here, it must be treated using the kinetic equations for the appropriate order.
E X AMPLE 9. 4
A complex but illustrative example of the kinetic behavior of series reactions
occurs in the clotting of blood. The fi nal step in the process is the conversion of
fibrinogen to fibrin (the clotting material) catalyzed by the proteolytic (protein
cutting) enzyme thrombin. Thrombin is not normally present in the blood but
is converted from a precursor, prothrombin, by the action of another activated
proteolytic enzyme and calcium ion. The overall process involves a cascade of at
least eight such steps:
Tissue injury activates
Factor XII
(Hageman factor)
XIIa (active)
Factor XI
XIa (active)
Factor IX
(Christmas factor)
IXa (active)
Factor VIII
(antihemophilic factor)
VIIIa (active)
Factor X
(Stuart factor)
Factor V
(proaccelerin)
Xa (active)
Va
Prothrombin
Fibrinogen
Thrombin
Fibrin
(blood clot)
antithrombin Inactive
thrombin
Reaction Mechanisms and Rate Laws | 333
A consequence of this cascade of steps is the fact that the thrombin concentration,
as measured by the decrease in clotting time for blood samples, increases dramatically
following wounding and then decreases again once the clot is formed. The behavior,
shown in figure 9.11, is very much like that of an intermediate in a series of reactions.
In the proposed mechanism, the injury sets off a cascade of reactions involving a series
of proteolytic enzymes, each of which activates a target protein. This cascade phenomenon serves to accelerate the mobilization of the clotting mechanism so that it can be
switched from fully off to fully on in less than 1 min. The conversion of prothrombin
to thrombin, catalyzed by active proaccelerin and Ca2+, is the immediate cause of
clotting. The level of thrombin in the blood must not remain high, however, or it will
cause fibrin to form in the circulatory system and block the normal flow of blood.
Antithrombin is an additional factor that inactivates thrombin, usually within a few
minutes after it has reached its maximum level. In this way, the action of thrombin is
restricted to the period when it is critically needed.
Although this overall process is much more elaborate than the simple example of
series reactions considered later, a similarity is seen when we inspect the formation
and disappearance of thrombin as rate-limiting steps, shown in green. For this purpose,
prothrombin plays the role of A, thrombin is B, and inactive thrombin is C. The
appearance of active proaccelerin turns on the first step (A S B) and the subsequent
formation of antithrombin turns on the second step (B S C) The rate of clotting of
fibrinogen serves (kinetically speaking) as a simple indicator of the level of thrombin
present at any time in the course of this process.
FIGURE 9.11 Concentration of
active thrombin after wounding.
The clotting rate for the conversion of fibrinogen to fibrin is a
direct measure of the thrombin
concentration.
7
thrombin units
6
5
4
3
2
1
1
2
3
t (min)
4
5
Reproduced with permission of The Royal Society of Chemistry.
Equilibrium and Kinetics
All reactions approach equilibrium. Therefore, in principle, for every forward reaction
step, there is a backward step. In practice, we can sometimes ignore the backward step
because either the equilibrium constant is very large or the concentrations of products
are kept very small. Nevertheless, it is important to know the general relation between
kinetic rate coefficients (k) and thermodynamic equilibrium constants (K ).
Let’s consider an elementary first-order reversible reaction:
k1
A m B
k-1
The rate of disappearance of A is
-
d[A]
= k 1[A] - k - 1[B] .
dt
334 Chapter 9 |
Kinetics: Rates of Chemical Reactions
At equilibrium, -d[A]>dt = 0; therefore,
[B]eq
=
[A]eq
k1
= K.
k-1
E XE RCISE 9. 2
Show that for the elementary first-order reversible reaction
k1
A m B
k-1
the approach to equilibrium concentrations starting from any concentrations of A and
B is a first-order reaction with a rate coefficient equal to the sum of the rate coefficients
for the forward and backward reactions
ln
[A] - [A]eq
= ln
[A]0 - [A]eq
[B] - [B]eq
[B]0 - [B]eq
= -(k1 + k - 1)t ,
where [A]0 and [B]0 are the concentrations of A and B, respectively, at t = 0.
Hint: Let x = [A] - [A]eq = [B] - [B]eq, then
-
d[A]
= k 1[A] - k - 1[B]
dt
= k 1([A]eq + x) + k - 1(x - [B]eq) .
But because
-
d[A]eq
dt
= k 1[A]eq - k - 1[B]eq = 0 ,
we therefore obtain
-
d[A]
dx
= = (k 1 + k - 1)x
dt
dt
ln
x
= -(k1 + k - 1)t .
x0
From the definition of x, we obtain the equation we set out to prove
ln
[A] - [A]eq
[A]0 - [A]eq
= ln
[B] - [B]eq
[B]0 - [B]eq
= -(k 1 + k - 1)t.
Often there is more than one path for the reaction of A to form B. To be consistent with
the principles of equilibrium thermodynamics, we must apply the principle of microscopic
reversibility. The principle requires that at equilibrium the reactions between any pair of
reactant, intermediate, or product species must occur with equal frequency in both directions, that is, the following mechanism is not possible:
A
B
C
Reaction Mechanisms and Rate Laws | 335
Each elementary step in the reaction must be reversible to be consistent with the principle
of microscopic reversibility. Thus, we must formulate the mechanism as follows:
k1
A
k3
k–1
k–3 k–2
B
k2
C
The relation to thermodynamics requires further that
K =
[B]eq
[A]eq
=
[B]eq [C]eq
[C]eq [A]eq
and thus
k2 k3
k1
=
.
k -1
k -2 k -3
K =
(9.34)
The six rate coefficients are therefore not independent.
Complex Reactions
For many reactions, particularly those involving enzymes as catalysts, a series of reversible
steps is involved. An example is the following set of coupled elementary reactions:
k1
A + B m X
k-1
k2
X m P + Q
k-2
In this case, the exact solution to the rate equations is complex because two of the elementary reactions are bimolecular and a total of five molecular species are involved. It is useful
to learn some approximations that can be applied to these complex reactions.
Initial-Rate Approximation
At the beginning of a reaction, the product concentrations are usually very small or zero.
For the set of the above opposed reactions, the step designated -2 can be neglected in an
analysis of initial velocities. This simplifies the kinetic expressions considerably. As the
extent of reaction increases, the experimental results will begin to differ from the prediction of the approximate theory.
Prior-Equilibrium Approximation
This approach is valuable when steps 1 and −1 are rapid in comparison with (irreversible)
step 2:
k1
A + B m X (fast, equilibrium)
k-1
k2
X h P + Q (slow)
A, B, and X rapidly attain a state of quasi-equilibrium, such that
v1 = v - 1
336 Chapter 9 |
Kinetics: Rates of Chemical Reactions
and
k 1[A][B] = k - 1[X] .
We can write this as an equilibrium expression:
K =
k1
[X]
=
k -1
[A][B]
Step 2 is the rate-limiting step, and the rate of formation of product is given by
v =
d[Q]
d[P]
=
= k2[X] .
dt
dt
Substituting from above, we see that the rate
v =
k2k1
[A][B]
k-1
(9.35)
can be expressed in terms of reactant concentrations only. This is particularly valuable if
the relative concentration of X remains very small (k1 W k-1), but is just as valid when X
is fairly large (k1 ~ k-1). The important criterion for the prior-equilibrium approximation
to apply can be written
v ⬵ v2 V v1 ⬵ v-1 .
This should be read: “The overall rate of the reaction is limited by the slow step 2, and the
rate is much slower than the forward and reverse reactions of step 1, which are essentially
in equilibrium.”
Steady-State Approximation
It frequently happens that an intermediate is formed that is very reactive. As a consequence,
it never builds up to any significant concentration during the course of the reaction:
k1
A + B h X
k2
X + D h P
(slow)
(fast)
To a first approximation, X reacts as rapidly as it is formed:
v1 ⬵ v2
(9.36)
k1[A][B] ⬵ k2[X][D]
Hence,
v =
d[P]
= k1[A][B]
dt
in terms of reactant concentrations only. A more general way of formulating the steadystate approximation is to consider all steps involving formation and disappearance of the
reactive intermediate and to set the sum of their rates equal to zero. In this example, X is
formed in step 1, and it disappears in step 2. Thus,
d[X]
= k1[A][B] - k2[X][D] ⬵ 0 .
dt
(9.37)
Reaction Mechanisms and Rate Laws | 337
This gives the same result (Eq. 9.36). However, this second approach is more general.
The significance of this equation is not that [X] is constant during the course of the
reaction. Such is never the case. It is true, however, that the slope, d[X]>dt, of the curve
of [X] versus time is much smaller than those typical of reactants or products. For example,
this can be seen for component B for the scheme shown in figure 9.10(b). It is this nearly
zero slope throughout the course of the reaction that is the characteristic of the steadystate condition.
The steady-state approximation can be applied in many chemical and biochemical
reactions. Care must be taken that the intermediate satisfies the criterion that d[X]>dt ⬵ 0.
Note that this is not necessarily the case where prior equilibrium is involved. It is also not
true for the intermediate B illustrated in figure 9.10(a).
Deducing a Mechanism from Kinetic Data
There is no straightforward way to obtain a mechanism from kinetic data. There is always
more than one mechanism that will be consistent with the kinetic data. Therefore, why
propose a mechanism at all? We like to think that we understand a reaction and that we can
guess what the molecules are doing. A mechanism gives us some basis for predicting what
should happen for other reactions.
The best way to illustrate how to hypothesize a mechanism is to discuss specific
examples. The only general advice that can be given is to think of a simple and plausible
mechanism and then calculate the kinetics and see if they are consistent with the data. A
unique rate equation can always be obtained from a proposed mechanism. If the rate law is
simple first or second order, assume a unimolecular or bimolecular rate-determining step.
Then, if necessary, add other steps to agree with the stoichiometry. For example, suppose
for the stoichiometric reaction
A + B h C
the rate law is found to be
-
d[A]
= k[A][OH - ] .
dt
We can assume the mechanism
k1
A + OH - h M k2
M - + B h C + OH - .
The first elementary reaction postulates that A + OH- can react to form an intermediate, M-; this gives the correct rate law. The second elementary reaction must be added to
account for the stoichiometry. From the proposed mechanism, we can now predict rate laws
for [B] and [C] and experimentally check the predictions. Another mechanism consistent
with the data is:
A + OH - N N K
k1
N- h P-
k2
P + B S C + OH
(slow, rate determining)
-
The rate-determining step is
-
(fast to equilibrium)
d[N - ]
= k 1[N - ]
dt
(fast)
338 Chapter 9 |
Kinetics: Rates of Chemical Reactions
but
[N - ]
= K.
[A][OH - ]
Therefore,
-
d[N - ]
= k1K[A][OH - ] .
dt
Because all other elementary reactions are fast, each time N- reacts to form P-, A and B
react to form C. The rate laws are then
-
d[A]
d[B]
d[C]
= =
= k1K[A][OH - ] .
dt
dt
dt
We can do other kinetic experiments to decide between the two mechanisms and can
make attempts to detect the postulated intermediates. Many other mechanisms could also
be written. It should be clear that proposing reasonable mechanisms requires practice and
experience.
Temperature Dependence
It seems almost intuitive that if you want a reaction to proceed faster, you should heat it.
It came as a surprise when some chemical reactions were found to have negative temperature coefficients; that is, the reaction was slower at higher temperature. In biology this is a
common occurrence. Few organisms or biochemical processes can withstand temperatures
above 50 ⬚C for very long because of irreversible structural changes that alter the mechanisms or destroy the organism.* Processes such as coagulation of the protein egg albumin
(egg white) occur during cooking, and they occur more rapidly at high temperatures. This
seems normal. However, if a protein is first carefully denatured, it can be renatured more
rapidly at a lower temperature than at a higher one! There is a kinetic effect in addition to
the shift in equilibrium constant for this process.
Observations by Arrhenius and others showed that for most chemical reactions there
is a logarithmic relation between the rate coefficient and the reciprocal of the temperature
in kelvins. We can therefore write the Arrhenius result as
ln k = - a
Ea
b + ln A
RT
(9.38)
or
k = A expa-
Ea
b.
RT
(9.39)
The activation energy for the reaction is Ea and A is called the pre-exponential factor.
This is an empirical equation that represents the behavior of many chemical systems and
even some complex biological processes (figure 9.12). The usual way of obtaining the
activation energy is to plot ln k versus 1/T . The slope of the line is equal to -Ea>R. Once
Ea is known, Eq. 9.39 can be solved for A. An alternative method of obtaining Ea and A
is illustrated in example 9.5.
are, however, thermophiles—organisms that live at temperatures of 100 ⬚C or higher in hot pools such as
those in Yellowstone Park and in hot deep-sea vents.
*There
Temperature Dependence | 339
FIGURE 9.12 Arrhenius plot of
the chirping rate of tree crickets
at different temperatures. The
linearity of this plot suggests that
the complex biological process
is governed by a rate-limiting
step with an activation energy
of 50.2 kJ mol-1. The chirping
frequency changes by a factor
of over 3 in the temperature
range from 282 K to 297 K. (Data
from Edes, R.T., The American
Naturalist, vol. 33, p. 935, 1899.)
5.25
ln (chirps/minute)
5.00
4.75
4.50
4.25
4.00
3.75
3.35
3.40
3.45
1000/T (1000
3.50
3.55
K–1)
E X AMPLE 9. 5
The decomposition of urea in 0.1 M HCl occurs according to the reaction
NH2CONH2 + 2 H2O h 2 NH4+ + CO32 - .
The first-order rate coefficient for this reaction was measured as a function of
temperature, with the following results:
Expt.
Temperature (°C)
k (minⴚ1)
1
61.0
0.713 * 10-5
2
71.2
2.77 * 10-5
Calculate the activation energy Ea and the A factor in Eq. 9.39 for this reaction.
SOLUTION
This problem can be solved either numerically or graphically. The numerical solution
is given here. Since ln k = -Ea>RT + ln A, we will first calculate ln k and 1>T for each
experiment.
Expt.
ln k
1>T (Kⴚ1)
1
-11.85
2.922 * 10-3
2
-10.49
2.904 * 10-3
We can write
ln k2 - ln k1 = -
Ea 1
1
a
b.
R T2
T1
Combining experiments 1 and 2 in this way, we obtain
Ea = = -
R(ln k2 - ln k1)
1>T2 - 1>T1
8.314 J K - 1mol - 1 (-10.49 + 11.85)
(2.904-2.992) * 10 - 3 K - 1
= 128500 J mol - 1 = 128.5 kJ mol - 1 .
340 Chapter 9 |
Kinetics: Rates of Chemical Reactions
Using the data from either experiment, we can calculate a value for A from
ln A = ln k +
Ea
.
RT
Using the data of experiment 2,
128500
(8.314)(344.4)
= -10.49 + 44.84 = 34.35 .
ln A = -10.49 +
Therefore,
A = exp(ln A) = 8.28 * 1014 min - 1
= 1.38 * 1013 s - 1 .
Note that the value for A is in the range of infrared vibrational frequencies. This
means that with enough energy—at very high temperatures—the reaction occurs about
as rapidly as the atoms can move.
To understand the meaning of the activation energy, consider the elementary reaction
M + N h P with an energy barrier between reactants and products, as illustrated in
figure 9.13. In this figure, we plot the energy versus the reaction coordinate. The reaction
coordinate is a convenient way to represent the change of the reactants into products as
a reaction takes place. It is not a simple coordinate, such as the x-coordinate for a point
in space; instead, it can represent the positions of all the significant atoms in the reacting
molecules. For example, in the reaction
F- + CH3OH h CH3F + OH the reaction coordinate represents the concerted motions of the OH- approaching the
central carbon, while the F- is leaving and the three hydrogen atoms are also moving.
We can plot any property versus the reaction coordinate, but energy or free energy is
particularly useful.
Ea –1
80
Energy (kJ mol–1)
FIGURE 9.13 Energy changes
that occur throughout the
course of a representative reaction (The SN2 displacement of
OH- from CH3OH by F-). The x
coordinate is the C- O distance,
which is an appropriate reaction
coordinate for this reaction. Ea
is the activation energy for the
energy for the forward reaction
(CH3OH + F- h CH3F) while
Ea-1 is the activation energy for
the reverse reaction.
60
Ea
40
ΔE
20
0
reactants
140
products
180
rC–O (pm)
220
260
Transition-State Theory | 341
The molecules M and N must have sufficient energy to react. Because the process
must be reversible at the microscopic level, the top of the barrier will be the same for the
reverse reaction. The energy of the products of a reaction is usually different from that of
the reactants; the energy difference, ΔE can be positive or negative:
⌬E = EP - (EM + EN)
Therefore, the activation energy for the reverse reaction is usually not the same as for the
forward reaction. The relation
(Ea)forward - (Ea)reverse = ⌬E
(9.40)
does hold, however, at least for elementary reactions.
As the temperature is increased, the fraction of molecules having energy greater than
Ea increases, and the rate of reaction correspondingly increases. In some cases, correlations
can be made between the activation energy and the energy of a bond that must be broken
in order for a reaction to proceed.
Very detailed theories of chemical kinetics exist that can predict rate coefficients for
simple, gas-phase reactions. We discuss more general theories that are useful mainly in
helping us understand the factors that affect the kinetics.
In the collision theory of chemical kinetics, we assume that, to react, molecules must
collide (including collisions at the walls of the vessel in many cases). The rate of a reaction
whose mechanism is
M+N h P
is given by
v = k[M][N] = A [M][N]expa -
Ea
b .
RT
(9.41)
At high temperatures, the exponential approaches unity, and k ⬵ A:
lim
c A expa S
T
⬁
Ea
bd = A
RT
The maximum rate of this reaction at high temperature would be A[M][N], which is therefore
interpretable as a collision frequency for the molecules M and N. The exponential factor
resembles a Boltzmann probability distribution (see chapter 5). It can be considered to
represent the fraction of molecules with proper orientation that have sufficient energy to react.
Examined in detail, the collision theory has serious shortcomings. Collision frequencies
of molecules in the gas phase can be calculated quite well using kinetic theory of gases (see
chapter 8). From Eq. 9.41, theoretical values for the rate coefficient of a bimolecular gas-phase
reaction can be calculated. These are on the order of 1011 mol-1 L s-1 for small molecules,
somewhat dependent on the molecular radii. Experimentally, values of A range widely, from
102 to 1013 mol-1 L s-1 for reactions that are thought to be relatively simple. Although a few
reactions proceed much faster than the rates calculated from simple collision theory, many
are orders of magnitude slower. Explaining these results by collision theory is difficult.
Transition-State Theory
An alternative model was introduced in 1935 by Eyring and others. Its significant new
feature was that it considered the reactive intermediate, or transition state* (an unstable
species at a free-energy maximum), to be just like a stable molecule, except for motion of
atoms along the reaction coordinate between reactants and products. Thus, the transition
*The
transition state was originally called an activated complex.
342 Chapter 9 |
Kinetics: Rates of Chemical Reactions
state is a molecule lasting during only a few molecular vibrations. When the transition-state
theory was proposed, it was not possible to study a transition state because of its short
lifetime. However, transition states in the gas phase now have been characterized spectroscopically (Zewail and Bernstein 1988; Gruebele and Zewail 1990) using laser pulses of
femtoseconds (1 fs = 10 -15 s) to picoseconds (1 ps = 10 -12 s) in length.
In transition-state theory, every elementary reaction
k
M + N h P
is rewritten as
K[
k[
M + N N MN[ h P .
It is assumed that reactants are in rapid equilibrium with the transition state, MN‡. Here K‡
is the equilibrium constant between the transition state and reactants, and k‡ is a universal
rate coefficient:
[MN[]
(9.42)
K[ =
[M][N]
d[P]
= k[[MN[]
(9.43)
dt
Neither K‡ nor k‡ are experimentally measurable from Eqs. 9.42 and 9.43 because the
concentration of the transition state is not measurable. However, K‡and k‡ can be related to
concentrations of reactants and to measurable rates. Combining Eqs. 9.42 and 9.43 we obtain
d[P]
= k[K[[M][N] .
(9.44)
dt
The experimental rate expression is
d[P]
= k[M][N] .
dt
So we have related an experimental rate coefficient k to properties of the transition state:
k = k‡K‡
(9.45)
But every equilibrium constant is related to the standard free energy of the reaction
⌬G[ = -RT ln K [
and transition-state theory shows that the universal rate coefficient k[ is equal to
k[ =
k BT
,
h
(9.46)
where k B is the Boltzmann constant and h is the Planck constant. Combining Eqs. 9.45
and 9.46, we obtain
kBT [
kBT
⌬G[
K =
expa b.
k =
(9.47)
h
h
RT
This equation permits a calculation of the free energy of activation, ΔG‡, from a measured rate
coefficient. The rate coefficient must have units (s-1) consistent with the Boltzmann and Planck
constants; if the reaction is second order, the concentration units (usually M) are defined by the
standard state for the free energy. The same derivation could be given for the reverse reaction.
So for any reaction, the picture shown in figure 9.14 holds. The free-energy difference between
products and reactants is equal to the difference in activation free energies:
[
[
- ⌬Greverse
⌬ rG = ⌬Gforward
(9.48)
Transition-State Theory | 343
FIGURE 9.14 Progress of
a reaction according to the
transition-state theory.
MN‡
ΔG‡f
M+N
ΔG‡r
ΔG
ΔrG
P+Q
reaction coordinate
MN‡
M+N
P+Q
Additional insight comes if we consider the distinction between energy (enthalpy) and
entropy contributions. At constant temperature,
⌬G[ = ⌬H[ - T⌬S[ ,
[
(9.49)
[
where ⌬H and ⌬S are the activation enthalpy and entropy, respectively. Substituting
into Eq. 9.47,
kBT
⌬S[
⌬H[
(9.50)
k =
expa
b expa b.
h
R
RT
From this equation we can see that transition-state theory predicts a slightly different
temperature dependence from that of Arrhenius, although the main temperature dependence for both are in the exponential terms. According to Arrhenius (Eq. 9.38), a plot of
ln k versus 1>T gives a straight line with slope equal to -Ea>R. In transition-state theory, a
plot of ln k versus 1>T has a slope of -(ΔH‡ + RT)>R. This is seen by taking the derivative
of ln k with respect to 1>T. For most reactions, ΔH‡ is much larger than RT, so the slope is
nearly equal to -ΔH‡>R analogous to the Arrhenius theory. Thus,
ΔH‡ ⬵ Ea .
ΔS‡
(9.51)
ΔH‡
The value of
can be obtained from Eq. 9.50 and
it can also be calculated from the
Arrhenius pre-exponential term A. Substituting Eq. 9.51 into Eq. 9.50 and comparing the
result with Eq. 9.39, we obtain
Ah
b.
(9.52)
⌬S[ ⬇ Ra ln
kBT
In Eq. 9.52 we use an average T in the temperature range of interest.
The chief advantage of the transition-state theory is that it relates kinetic rates to
thermodynamic properties of reactants and a transition intermediate. This can help us
understand some of many factors that influence the rates. Entropies of activation are
typically negative for simple gas-phase reactions. They reflect the loss in randomness in
the transition state relative to the reactants.
Consider the bimolecular dimerization of butadiene in the gas phase:
2 C4H6
h (C4H6 cC4H6)‡ h C8H12
- ⌬S‡ = + 68 JK-1 mol-1
344 Chapter 9 |
Kinetics: Rates of Chemical Reactions
The fairly large negative entropy of activation corresponds to the loss of three translational
degrees of freedom in going from two independent reactant particles to the single one in
the associated transition state.
E X AMPLE 9.6
Calculate the entropy of activation at 71.2 ⬚C for the reaction in example 9.5 involving
the decomposition of urea.
SOLUTION
Using the approximation that ΔH‡ ⬵ Ea we use Eq. 9.54, remembering that A must
have units of s-1. Therefore, for A = 1.38 * 10-13 s-1,
⌬S[ = Ra ln A - ln
kBT
b
h
= (8.314 J K - 1 mol - 1) c30.256 - ln
(1.318 * 10 - 23)(344.4)
d
6.626 * 10 - 34
= (8.314 J K - 1 mol - 1)(30.256 - 29.602)
= 5.44 J K - 1 mol - 1 .
The entropy of activation for this reaction is small. This is consistent with our conclusion
in example 9.5 that, in the absence of an energy barrier, this reaction occurs as rapidly
as the atoms can move. For reactions in which the solvent water is able to interact less
favorably with the transition state than with the reactant, a positive entropy of activation
is often observed.
Electron Transfer Reactions: Marcus Theory
Theoretical calculations for the energies along possible reaction paths have been done for
reactions of diatomic or triatomic molecules in the gas phase. This leads to an energy surface
for the reaction and predictions of the reactivity of each vibrational and rotational state of
the molecules. The predictions can be tested by molecular beam experiments, which provide
kinetic data for molecules colliding with different amounts of kinetic energy (translational,
vibrational, and rotational) and different orientations relative to each other (Lee 1987).
Reaction cross sections as a function of energy and angle of collision are measured. Values
of the rate coefficient, its temperature dependence, and thus ΔH‡ and ΔS‡ have been calculated for reactions in the gas phase. These parameters represent combinations of the detailed
reaction cross-sections that have been obtained by molecular beam experiments.
Theoretical calculations for solution reactions are in general much more difficult, but
success has been obtained for charge-transfer reactions such as an electron transfer from
Fe2+ to Ce4+ to produce Fe3+ and Ce3+ in aqueous solution. Electron-transfer reactions
do not involve bond breakage or formation, but they are crucial in photosynthesis and in
biochemical oxidation–reduction reactions. In 1956 Marcus derived a theory for the rates
of electron-transfer reactions based on some simple ideas (see Nobel lecture, Marcus 1993).
He thought about the simplest electron-transfer reaction—one in which the products and
reactants are the same but one of the reactants is radioactively labeled (designated by *):
Fe2 + + *Fe3 + h Fe3 + + *Fe2 +
or
Ce4 + + *Ce3 + h Ce3 + + *Ce4 +
Ionic Reactions and Salt Effects | 345
What controls the rates of these reactions? Clearly, the free-energy change for both
reactions is effectively zero. What about the free energies of activation that characterize
the rates? The motions of electrons are very fast compared with motions of atoms, so
Marcus reasoned that the reorientation of water molecules around the ions must be the
rate-limiting process. The radius of Fe3+ is smaller than that of Fe2+, and the larger charge
on Fe3+ orients the surrounding water molecules more tightly. Therefore, the orientations
of the water molecules must be different for reactant and product ions, and the transition
state (where the electron transfer occurs) must correspond to an unstable configuration of
water molecules that is in between the stable configurations for products and reactants.
Starting from these simple reactions, Marcus derived an expression for the free energy of
activation ΔG‡ for electron-transfer reactions in general:
⌬G[ =
l
⌬G⬚ 2
a1 +
b
4
l
(9.53)
ΔG° is the standard free energy of the reaction, and the factor l is a reorganization
energy with contributions from the solvent and from the reactants (when the reactants
are molecules, not spherical metal ions). Note that for the ionic reactions above where
the standard free-energy change ΔG° = 0 the free energy of activation is simply l>4. If
the solvent is treated as a continuum with refractive index n and dielectric constant e, the
Marcus solvent reorientation energy has a simple form:
lsolv = (⌬e)2 a
1
1
1
1
1
+
- ba 2 - b
er
2a1
2a2
R
n
(9.54)
Δe = charge transferred
a1, a2 = radii of ions
R = center-to-center distance between the reactants
n = refractive index of solvent
er = dielectric constant (relative permittivity) of the solvent
Equation 9.54 has the form of the Coulomb energy of interaction of two charges q
at distance r in a dielectric medium (q2>er). To calculate l it is convenient to measure
charge in electrons transferred (Δe = 1,2,…) and distances in Å (1 Å = 10 -10 m). To obtain
l in kJ mol-1, the results must be multiplied by 1389. For water, the refractive index
is about 1.33, and the dielectric constant is about 80; both are dimensionless. Marcus’
Nobel Prize address (1993) describes what led him to the problem and its solution; he also
discusses applications to biochemical reactions.
Ionic Reactions and Salt Effects
The effect of the nonideal behavior of reactants in solution can be treated by extending the
transition-state theory to include activities and activity coefficients explicitly. Equation 9.43
is written as
v = k [c [ =
kBT [
c
h
(9.55)
and Eq. 9.42 can be rewritten as
K[ =
c [g [
[MN[]
=
,
cMgMcNgN
[M][N]
(9.56)
346 Chapter 9 |
Kinetics: Rates of Chemical Reactions
where c is the respective concentration and g the activity coefficient. Combining Eqs. 9.55
and 9.56 and substituting into Eq. 9.44, we obtain the result
v =
kBT [ gMgN
K
cMcN .
h
g[
(9.57)
In this case, the rate coefficient can be written as
k =
gMgN
kBT [ gMgN
K
= k0 [ ,
[
h
g
g
(9.58)
where k0 is the rate coefficient extrapolated to infinite dilution of all species in solution;
at infinite dilution, all activity coefficients are unity.
For reactions involving ionic species, we can use the Debye–Hückel Limiting Law
(chapter 4) to express the dependence of the rate coefficient k on ionic strength. For each
ionic species, we use
log gi = -0.51Z 2i 2I
I =
1 all species 2
ciZ i ,
2 a
i
(9.59)
where the constant 0.51 applies to water solutions at 25°C, Zi is the charge on species i,
and I is the ionic strength. The charge relation for the reaction M + N h MN‡ is clearly
ZM + ZN = Z‡ and we can write the logarithm of Eq. 9.58 as
log k = log k0 - 0.51[Z2M + Z2N - (Z2M + Z2N)2]2I
= log k0 + 2(0.51)ZMZN 2I .
(9.60)
Equation 9.60 predicts that a plot of log k versus √I for dilute ionic solutions should give a
straight line with a slope 2(0.51)ZM ZN ⬵ ZM ZN. The slope then gives direct information
about the charges of the reacting species. This has been verified for a variety of ionic reactions in dilute aqueous solutions. For example, when the rate of decomposition of H2O2 in
the presence of HBr was studied as a function of ionic strength, it was found that the data
fit Eq. 9.60 with a slope of -1. The stoichiometric reaction
H2O2 S H2O + ½ O2
does not involve ionic species, but the kinetics imply that H+ and Br- are involved in the
formation of the transition-state species.
E X AMPLE 9.7
The acid denaturation of CO-hemoglobin has been studied at pH 3.42 in a formic acid–
sodium formate buffer as a function of sodium formate (Na+HCOO-) concentration
(Zaiser and Steinhardt 1951). The half-lives for the first-order denaturation are as follows:
NaOOCH (M)
0.007
0.010
0.015
0.020
t½
20.2
13.6
8.1
5.9
Determine whether these results follow the Debye–Hückel Limiting Law in the
dependence of the rate coefficient on ionic strength and, if so, what is the apparent
charge on the CO-hemoglobin?
Isotopes and Stereochemical Properties | 347
–0.8
–1.0
log k
SOLUTION
Since NaCOOH is a 1:1 electrolyte, the ionic strength I is equal to the molar
concentration c of sodium formate (we ignore the small H+ concentration). For a
first-order reaction,
ln 2
0.693
k =
=
.
t1/2
t1.2
–1.4
Therefore,
0.08
√I
k
0.084
0.100
0.122
0.141
(min-1)
0.0343
0.0510
0.0856
0.117
log k
-1.465
-1.293
-1.068
-0.930
A plot of these data is shown in figure 9.15. Although the data show some curvature,
there is a reasonably good fit to a straight line with a slope of 9. Since the denaturation
occurs by reaction with monovalent H+ this suggests a charge of approximately +9 for
the CO-hemoglobin at pH 3.42.
Isotopes and Stereochemical Properties
The rate law, the activation energy, the ionic effects, and so on, all provide important clues
to the mechanism of a reaction. A number of other experimental and theoretical approaches
have been devised; the use of isotopes and the examination of the stereochemical properties
of reactants and products are two of the most powerful experimental methods to provide
the structural details of a reaction. In the hydrolysis of an ester,
O
CH3
–1.2
C
H
O
O
R
+ H2O
C
CH3
H
C
OH + HO
R′
R
C
R′
the rate law cannot tell us at which of the two bonds indicated by arrows is cleaved. When
the reaction is carried out in 18O (a stable isotope of the more abundant 16O) labeled H2O,
18O is found only in the acid and not in the alcohol produced. Thus, the cleavage must be
occurring at the position indicated by the green arrow. In the reaction between acetaldehyde
bound to an enzyme,
O
CH3
C
H
and a nucleophilic (electron-donating) reagent X- the stereospecificity of the product is
different depending on the direction of attack by X- The nucleophile can approach from
above or below the plane containing the four atoms:
O
C
C
H
The two products are depicted as follows:
O–
CH3
C
H
X
X
CH3
C
H
O–
They are not superimposable but are mirror images (chapter 13). Therefore, from the
stereospecificity of the product, we can deduce from which side of the plane the attacking
0.1
0.12
√I
0.14
FIGURE 9.15 Data for the rate
of acid denaturation of CO hemoglobin. (Based on data
from Zaiser, E.M. and Steinhardt,
J.S., Journal of the American
Chemical Society, 73, p. 5568.
Copyright © 1951 American
Chemical Society.)
348 Chapter 9 |
Kinetics: Rates of Chemical Reactions
group approaches. Kineticists have made extensive use of isotopic labeling and stereospecificity in deducing mechanisms; the two examples given provide a glimpse of how
this might be achieved.
Very Fast Reactions
The rates of many reactions can be measured by mixing two solutions and measuring the
change of concentrations with time. The concentrations are measured continuously by
absorbance or other spectroscopic method, or the reaction is quenched and an appropriate
analytical method is used. However, some reactions are too fast for this technique. Consider
the very simple and very important reactions
H+ + OH- S H2O
CH3COO - + H+ S CH3COOH .
Many reactions in solution involve neutralization as part of the mechanism, and this
affects the kinetics. Yet the kinetics of these reactions cannot be studied by mixing sodium
hydroxide and hydrochloric acid or sodium acetate and hydrochloric acid. The reaction
is over long before the solutions can be mixed thoroughly; therefore, the apparent rate of
reaction is actually a measure of the rate of mixing. Methods that do not require mixing two
solutions are needed for these very fast reactions. Relaxation methods allow you to measure
a rate by perturbing an equilibrium and following the return of the system to equilibrium.
Relaxation Methods
Consider a reversible system:
k1
A m
B
k
-1
At equilibrium, the rates of the forward and reverse reactions are the same. The position
of equilibrium is represented in terms of an equilibrium constant,
K =
[B]eq
k1
=
,
k-1
[A]eq
which is the ratio of the rate coefficients k1 and k-1. The usual measurements of the
“average” composition of the system at equilibrium can provide a value for only the ratio,
not for the individual rate coefficients. Clearly, k1 and k-1 can range from very large to
very small and still give the same ratio (the equilibrium constant K). If we change the
temperature or pressure of the system so as to attain a new equilibrium state, the new state
will, in general, have a different value for the equilibrium constant. We are still completely
ignorant of the individual rate coefficients, however.
Several approaches are possible to extract the desired rate coefficients and mechanistic
information. One category involves the use of relaxation methods, in which the system at
equilibrium is suddenly perturbed in such a way that it finds itself out of equilibrium under
the new conditions. It then relaxes to a state of equilibrium under the new conditions, and
the rate of relaxation is governed by the rate coefficients and mechanism of the process.
Examples of relaxation processes include:
1. Temperature jump, which can be achieved by the discharge of an electric capacitor,
the use of a rapid laser pulse, or by plunging the sample into hot water to produce a
sudden increase in temperature of a system
2. Pressure jump, where a restraining diaphragm is suddenly ruptured to release pressurization of the system
Very Fast Reactions | 349
3. Flash- or laser-pulse photolysis, where a pulse of light produces a sudden change in
the population of electronically excited states of absorbing molecules. In cases where
the excited state is more acidic than the ground state, this results in a jump in [H+]
In each case, the system suddenly finds itself out of equilibrium at the instant of the perturbation, and the experimenter monitors the approach to a new equilibrium state under
the altered conditions.
A second major category of methods for dealing with systems at equilibrium involves
the study of fluctuations about the equilibrium state. Because of both the dynamic and
statistical nature of systems at equilibrium, the system and many of its properties undergo
constant and rapid fluctuations about some average or median state. Any normal (longterm) measurements always give the same result if the system is truly at equilibrium, and
these results serve to characterize the average equilibrium state. On a short-enough time
scale, however, we observe that the properties of the system exhibit a kind of Brownian
motion, in the sense that they are constantly moving one way or the other with respect
to their average equilibrium value. This can be true for such properties as the following:
1. Refractive index, where fluctuations result in variations in scattering of light from laser
beams
2. Concentration, which can be detected as fluctuations in the absorption or fluorescence
of particular species
3. Position, involved in scattering of fine laser beams off single particles
4. Pressure, where fluctuations can produce “noise” in an acoustic detector
In each case, there is no need to perturb the system artificially; we simply take advantage
of the fluctuations that occur naturally because of the particulate and statistical nature of
matter.
Relaxation Kinetics
As an example of relaxation kinetics, consider an equilibrium of the type
k1
A + B m P.
k -1
The rate expression is
d[P]
= k1[A][B] - k - 1[P] .
dt
(9.61)
At equilibrium, d[P]>dt is zero, and
K =
[P]eq
k1
=
,
k-1
[A]eq[B]eq
(9.62)
where the bar over the concentration terms emphasizes the equilibrium values. Now generate a nearly discontinuous change (typically within 10-6 s but it can be as short as 10-8 s)
of temperature or pressure. Changing temperature or pressure will in general cause the
equilibrium concentrations to change; we can use thermodynamics to tell us how large a
change to expect. From chapter 3, we remember that
a
0⌬G
b = - ⌬S and
0T p
a
0⌬G
b = ⌬V .
0p T
(9.63)
Combining these equations with ΔG° = -RT ln K, we obtain
a
0 ln K
⌬H⬚
b =
0T p
RT 2
and
a
0 ln K
⌬V⬚
b = .
0p T
RT
(9.64)
350 Chapter 9 |
Kinetics: Rates of Chemical Reactions
Equation 9.64 tells us that the equilibrium constant K will be different at the new
temperature or pressure immediately after the pulse. How much different will depend on
the enthalpy change of the reaction for a T jump or the volume change of the reaction for
a p jump. The system finds itself suddenly out of equilibrium under the new conditions,
and we can follow the concentration changes during the subsequent relaxation to the new
equilibrium state.
Conditions are usually chosen so that the T jump or p jump produces relatively small
displacements from equilibrium; that is, a jump of 5 to 10 degrees or 100 to 1000 bar is
used. This simplifies not only the mathematics but also the thermodynamic analysis of the
results. Suppose, following the sudden perturbation, the system relaxes so as to increase
the concentration of A by a small amount Δ[A], [B] by Δ[B], and to decrease [P] by Δ[P],
as shown in figure 9.16. The perturbation is small enough so that (Δ[A])2 is negligible
compared with Δ[A] (and likewise for B and P). The values of Δ[A], Δ[B], and Δ[P] are all
defined as positive, and because of the stoichiometry, they are all equal. It is then convenient to rewrite the instantaneous, time-dependent concentrations as
[P] = [P]eq + ⌬[P]
[A] = [A]eq - ⌬[A] = [A]eq - ⌬[P]
[B] = [B]eq - ⌬[B] = [B]eq - ⌬[P] .
The values of [P]eq, [A]eq, and [B]eq refer to the equilibrium concentrations at the final
temperature or pressure. Therefore,
d([P]eq + ⌬[P])
d(⌬[P])
d[P]
=
=
dt
dt
dt
because d[P]eq>dt = 0. Combining this result with Eq. 9.61, we obtain
d(⌬[P])
d[P]
=
= k 1[A][B] - k - 1[P]
dt
dt
= k 1([A]eq - ⌬[P])([B]eq - ⌬[P]) - k - 1([P]eq + ⌬[P])
(9.65)
= k 1[A]eq[B]eq - k -1[P]eq - k 1 1[A]eq ⌬[P] + [B]eq ⌬[P] - (⌬[P])2 2 - k -1 ⌬[P] .
FIGURE 9.16 Effect of
temperature or pressure jump
on concentrations of reactants
and products for a system
initially at equilibrium or in a
steady state. The reaction is
A + B S P, and the perturbation causes the concentration
of P to decrease slightly while
A and B increase slightly. After
a short time, a new equilibrium
is reached with concentrations
[A]eq, [B]eq, and [P]eq at the
higher temperature or pressure.
Δ[A]0
[A]eq
Δ[A]
[B]eq
Δ[B]0
Δ[B]
Δ[P]0
Δ[P]
[P]eq
T or p jump
t
Very Fast Reactions | 351
-
d(⌬[P])
= 5 k 1([A]eq + [B]eq) + k - 1 6 ⌬[P]
dt
(9.66)
To achieve Eq. 9.68, we dropped the term involving (Δ[P])2 because it is small for
small displacements from equilibrium, and we have used Eq. 9.61 to show that the first
two terms of Eq. 9.65 sum to zero. The resulting Eq. 9.66 is a simple first-order equation
because {k1([A]eq + [B]eq) + k-1} is independent of time. If we rewrite Eq. 9.68 as
-
d(⌬[P])
⌬[P]
=
t
dt
it can be integrated to give
⌬[P] = ⌬[P]0e - t/t ,
where t is the relaxation time for the process. For this example,
1
t =
.
k - 1 + k1([A]eq + [B]eq)
(9.67)
(9.68)
Results for other examples are presented in table 9.4.
A very important and at first surprising result of analyses of each of the examples
listed in table 9.4 is that the relaxation kinetics is always simple first order (exponential in
time), regardless of the number of molecules involved as reactants or products. This is true
because the perturbations produce only small changes in the equilibrium concentrations,
and we can ignore squared terms. The behavior is illustrated in figure 9.16 for the case
where the perturbation shifts the equilibrium in favor of reactants. The reverse situation is
also found, depending on the sign of ΔH° and ΔV °.
Curves similar to those depicted in figure 9.16 were collected for the dimerization
reaction of proflavin, according to the equation
k1
2P m P2 .
k -1
Because of the fast kinetics involved, it was necessary to use a pulsed laser to provide the
temperature jump. For dimerizations, it is possible to simplify the relaxation expression
TABLE 9.4 Analysis of Relaxation Times for Single-Step Equilibria
Relaxation time*
Mechanism
k1
A m P
k -1
k1
2A m P
k -1
k1
A + B m P
k -1
k1
A + B m P + Q
k -1
k1
2A + B m P
k -1
k1
A + B + C m P
k -1
Slope
Intercept
t =
1
k1 + k - 1
t =
1
4k 1[A]eq + k -1
t =
1
k - 1 + k 1([A]eq + [B]eq)
Plot 1>t (y-axis) vs. ([A]eq + [B]eq) (x-axis)
t =
1
k 1([A]eq + [B]eq) + k - 1([P]eq + [Q]eq)
Plot 1>t (y-axis) vs. ([A]eq + [B]eq) (x-axis)
t =
1
k 1(4[A]eq[Beq] + [A]2eq) + k - 1[P]eq
t =
1
Plot 1>t vs. ([A]eq[B]eq + [B]eq[C]eq + [A]eq[C]eq)
k -1 + k 1([A]eq[B]eq + [B]eq[C]eq + [A]eq[C]eq)
k
k
Plot 1>t (y-axis) vs. [A]eq (x-axis)
4k1
k1
k1
k-1
k-1
k-1([P]eq + [Q]eq)
Plot 1>t (y-axis) vs. [P]eq (x-axis)
k-1
1
Source: Based on data from M. Eigen and L. De Maeyer, Investigation of Rates and Mechanisms of Reactions, 3d ed., vol. 6,
part 2, ed. G. G. Hammes (New York: Wiley-Interscience, 1974), chapter 3.
k1(4[A]eq[B]eq+ [A]2eq)
-1
Kinetics: Rates of Chemical Reactions
by writing it in terms of the total concentration [P]t of monomers and dimers. Turner and
colleagues (1972) transformed the last equation in table 9.4 to give
1
= k 2- 1 + 8k 1k - 1[P]t .
t2
(9.69)
The analysis according to Eq. 9.69 does not require a prior knowledge of the equilibrium
constant to determine both rate coefficients. The data should fall on a straight line when
plotted according to Eq. 9.69. The plot shown in figure 9.17 demonstrates not only that
this is observed but also that the rate coefficients are independent of pH between 4 and 7.8.
From the slope and intercept of the line shown in figure 9.17, k1 = 8 * 108 M-1 s-1 and
k-1 = 2.0 * 106 s-1 at 25 °C were obtained. The dimerization rate coefficient corresponds
to a process that is almost at the limit placed by the diffusion of the monomeric species in
aqueous solution, as discussed in the next section.
80
60
τ–2, µs–2
352 Chapter 9 |
40
20
0
0
1
2
3
[P]t, mM
4
5
6
FIGURE 9.17 Plot of relaxation data for proflavin dimerization as a function of total proflavin
concentration, [P]t.Green points were collected at pH 7.8; black at pH 4.0; T = 25 °C. (Based on data from
Turner DH, Flynn GW, Lundberg SK, Faller LD, Sutin N. Dimerization of proflavin by the laser raman
temperature-jump method, Nature, 239(5369), p. 215–217. Copyright © 1972 Nature Publishing Group.)
E XE RCISE 9. 3
Use the results of this kinetic analysis to determine the equilibrium constant for
proflavin dimerization at 25 ⬚C.
E X AMPLE 9. 8
The dimerization of the decanucleotide A4GCU4 has been studied by Pörschke,
Uhlenbeck, and Martin (1973). The letters A, G, C, and U represent the bases adenine,
guanine, cytosine, and uracil, and this oligonucleotide forms base-paired doublestranded helices, which are models for nucleic acids. The two strands are antiparallel;
therefore, the oligonucleotide is self-complementary.
a. A temperature jump was applied to a solution containing 7.45 * 10-6 M
oligonucleotide at pH 7.0 to give a final temperature of 32.4 °C.
2 A4GCU4
k1
k–1
A-A-A-A-G-C-U-U-U-U
U-U-U-U-C-G-A-A-A-A
Very Fast Reactions | 353
ΔA265
From the change in the UV absorption at 265 nm wavelength, shown in the figure, the
following data can be extracted:
0
200
t (ms)
400
Time (ms)
0
50
100
150
200
250
300
∞
100 : ⌬ A265
0
2.0
3.2
3.9
4.28
4.47
4.57
4.70
The difference in absorbance at 265 nm ΔA265 is directly proportional to the concentration of product at time t minus the concentration at zero time. Use the results to
calculate the relaxation time constant under these conditions.
b. Similar experiments were carried out at 23.3°C and pH 7.0 for a set of concentrations of single-stranded oligomer. The observed relaxation times were as follows:
t (ms)
455.
370.
323.
244.
[M]eq (μM)
1.63
2.45
3.45
5.90
Determine k1 and k-1 for double-strand (dimer) formation and dissociation,
respectively, under these conditions.
SOLUTION
a. First, test the data for first-order relaxation. The absorbance should approach
its value at infinite time (its equilibrium value) exponentially:
⌬A ⬁ - ⌬A t
= e - t/t
⌬A ⬁
This is equivalent to Eq. 9.67.
0
50
100
150
200
250
300
100 : (⌬ A∞ - ⌬At)
4.7
2.7
1.5
0.8
0.42
0.23
0.13
0
100
200
t (ms)
1n(⌬At– ⌬A⬁ )
Time (ms)
1
0
–1
–2
300
The data fit a linear first-order plot with a slope of 12.1 s-1. The relaxation time is
83 ms (if we do a regression fit to the log plot) or 86 ms (based on a direct fit to the
exponential function; better).
Kinetics: Rates of Chemical Reactions
b. We use the appropriate expression from table 9.4,
1
t =
,
4k1[M]eq + k - 1
where [M]eq is the concentration of oligonucleotide. Thus, a plot of 1>t versus [M]eq
should give a straight line.
[M]eq (μM)
1.63
2.45
3.45
5.90
1>t (s-1)
2.20
2.70
3.10
4.10
4
1/τ (s–1)
354 Chapter 9 |
k1 = 0.166 μM–1 s–1
3
2
1
k–1 = 1.48 s–1
0
0
2
4
c (μM)
6
The straight line corresponds to the equation
1
= 4k1[M]eq + k - 1 .
t
Therefore, from the plot shown, k-1 = 1.48 s-1 and k1 = .116 μM-1 s-1.
In this case, the dimerization step is several powers of 10 slower than the diffusion
limit.
Diffusion-Controlled Reactions
The concept of a diffusion-controlled reaction is useful for interpreting the kinetics of
fast reactions. Imagine a simplified system where reactant species M and N initially move
about in the system independently of one another and far enough apart so that they do not
interact. In the gas phase, the intervening space is mostly empty, and the molecules would
come together, react, and the product(s) depart without interference from other collisions.
The situation in the liquid (or solid) phase is quite different. As we saw in chapter 8, the
reactant species are always in contact with the solvent or other solutes, and they are constantly
being bumped by their neighbors. The motion of each reactant molecule is a random-walk
process as described earlier. Because the spaces between molecules in the liquid phase are
small (typically 10% or so of the molecular diameters), there results a cage effect in which
the molecule makes many collisions with its neighbors before it moves away to a new site,
where it encounters new neighbors. The motion is similar to that of people in a crowded
room. The time required to move across the room is very great when the room is crowded,
but a nearly empty room is easily traversed. The mobility of a molecule in a liquid medium is
characterized by its diffusion coefficient D. Hydrogen ions have an especially large diffusion
coefficient in water because they can “hop” from one water molecule to another.
The effect of diffusion on kinetics can be developed mathematically so as to calculate
an encounter frequency for two solute species M and N. Putting the encounter frequency
into the rate expression, we obtain a value for the A factor in the Arrhenius expression, in
units of m3 mol-1 s-1:
A