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Thermodynamics,
Statistical Thermodynamics,
and Kinetics
FOURTH EDITION
Thomas Engel • Philip Reid
PHYSICAL CHEMISTRY
Thermodynamics,
Statistical Thermodynamics,
and Kinetics
FOURTH EDITION
GLOBAL EDITION
Thomas Engel
University of Washington
Philip Reid
University of Washington
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A01_ENGE7707_04_GE_FM.indd 2
22/07/2020 17:56
To Walter and Juliane,
my first teachers,
and to Gloria,
Alex, Gabrielle,
and Amelie.
THOMAS ENGEL
To my family.
PHILIP REID
Brief Contents
THERMODYNAMICS, STATISTICAL THERMODYNAMICS,
AND KINETICS
1
Fundamental Concepts of
Thermodynamics 21
12
Probability 337
2
13
The Boltzmann Distribution 365
Heat, Work, Internal Energy, Enthalpy, and the
First Law of Thermodynamics 45
14
3
Ensemble and Molecular Partition
Functions 389
The Importance of State Functions:
Internal Energy and Enthalpy 81
15
Statistical Thermodynamics 423
4
Thermochemistry 103
16
Kinetic Theory of Gases 457
5
Entropy and the Second and Third Laws
of Thermodynamics 123
17
Transport Phenomena 479
18
Elementary Chemical Kinetics 509
6
Chemical Equilibrium 163
19
Complex Reaction Mechanisms 557
7
The Properties of Real Gases 205
20
Macromolecules 609
8
Phase Diagrams and the Relative Stability
of Solids, Liquids, and Gases 223
APPENDIX
9
Ideal and Real Solutions 253
10
Electrolyte Solutions 289
11
Electrochemical Cells, Batteries,
and Fuel Cells 307
4
A
Credits 659
Index 660
Data Tables 641
Detailed Contents
THERMODYNAMICS, STATISTICAL THERMODYNAMICS,
AND KINETICS
Preface 10
Math Essential 1
Math Essential 3
Units, Significant Figures, and
Solving End of Chapter Problems
1 Fundamental Concepts
of Thermodynamics 21
1.1
1.2
1.3
1.4
1.5
What Is Thermodynamics and Why Is It
Useful? 21
The Macroscopic Variables Volume, Pressure,
and Temperature 22
Basic Definitions Needed to Describe
Thermodynamic Systems 26
Equations of State and the Ideal Gas Law 28
A Brief Introduction to Real Gases 30
Math Essential 2
Differentiation and Integration
2 Heat, Work, Internal Energy,
Enthalpy, and the First Law
of Thermodynamics 45
2.1
Internal Energy and the First Law
of Thermodynamics 45
2.2 Heat 46
2.3 Work 47
2.4 Equilibrium, Change, and Reversibility 49
2.5 The Work of Reversible Compression or
Expansion of an Ideal Gas 50
2.6 The Work of Irreversible Compression or
Expansion of an Ideal Gas 52
2.7 Other Examples of Work 53
2.8 State Functions and Path Functions 55
2.9 Comparing Work for Reversible and Irreversible
Processes 57
2.10 Changing the System Energy from a MolecularLevel Perspective 61
2.11 Heat Capacity 63
2.12 Determining ∆U and Introducing the State
Function Enthalpy 66
2.13 Calculating q, w, ∆U, and ∆H for Processes
Involving Ideal Gases 67
2.14 Reversible Adiabatic Expansion and Compression
of an Ideal Gas 71
Partial Derivatives
3 The Importance of State
Functions: Internal Energy
and Enthalpy 81
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Mathematical Properties of State Functions 81
Dependence of U on V and T 84
Does the Internal Energy Depend More Strongly
on V or T? 86
Variation of Enthalpy with Temperature
at Constant Pressure 90
How are CP and CV Related? 92
Variation of Enthalpy with Pressure at Constant
Temperature 93
The Joule–Thomson Experiment 95
Liquefying Gases Using an Isenthalpic
Expansion 97
4 Thermochemistry
4.1
4.2
4.3
4.4
4.5
4.6
103
Energy Stored in Chemical Bonds Is Released
or Absorbed in Chemical Reactions 103
Internal Energy and Enthalpy Changes Associated
with Chemical Reactions 104
Hess’s Law Is Based on Enthalpy Being a State
Function 107
Temperature Dependence of Reaction
Enthalpies 109
Experimental Determination of ∆U and ∆H
for Chemical Reactions 111
Differential Scanning Calorimetry 113
5 Entropy and the Second and
Third Laws of Thermodynamics
5.1
5.2
5.3
5.4
5.5
123
What Determines the Direction of Spontaneous
Change in a Process? 123
The Second Law of Thermodynamics,
Spontaneity, and the Sign of ∆S 125
Calculating Changes in Entropy as T, P, or V
Change 126
Understanding Changes in Entropy at the
Molecular Level 130
The Clausius Inequality 132
5
6
CONTENTS
5.6
5.7
5.8
5.9
5.10
5.11
5.12
5.13
The Change of Entropy in the Surroundings and
∆Stot = ∆S + ∆Ssur 133
Absolute Entropies and the Third Law of
Thermodynamics 135
Standard States in Entropy Calculations 139
Entropy Changes in Chemical Reactions 139
Heat Engines and the Carnot Cycle 141
How Does S Depend on V and T ? 146
Dependence of S on T and P 147
Energy Efficiency, Heat Pumps, Refrigerators,
and Real Engines 148
6 Chemical Equilibrium
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
163
Gibbs Energy and Helmholtz Energy 163
Differential Forms of U, H, A, and G 167
Dependence of Gibbs and Helmholtz Energies
on P, V, and T 169
Gibbs Energy of a Reaction Mixture 171
Calculating the Gibbs Energy of Mixing for Ideal
Gases 173
Calculating the Equilibrium Position for a
Gas-Phase Chemical Reaction 175
Introducing the Equilibrium Constant for a
Mixture of Ideal Gases 178
Calculating the Equilibrium Partial Pressures
in a Mixture of Ideal Gases 182
Variation of KP with Temperature 183
Equilibria Involving Ideal Gases and Solid
or Liquid Phases 185
Expressing the Equilibrium Constant in Terms
of Mole Fraction or Molarity 187
Expressing U, H, and Heat Capacities Solely
in Terms of Measurable Quantities 188
A Case Study: The Synthesis of Ammonia 192
Measuring ∆G for the Unfolding of Single RNA
Molecules 196
7 The Properties of Real Gases
7.1
7.2
7.3
7.4
7.5
205
Real Gases and Ideal Gases 205
Equations of State for Real Gases and Their
Range of Applicability 206
The Compression Factor 210
The Law of Corresponding States 213
Fugacity and the Equilibrium Constant
for Real Gases 216
8 Phase Diagrams and the Relative
Stability of Solids, Liquids,
and Gases 223
8.1
What Determines the Relative Stability of the
Solid, Liquid, and Gas Phases? 223
8.2 The Pressure–Temperature Phase Diagram 226
8.3 The Phase Rule 233
8.4 Pressure–Volume and Pressure–Volume–
Temperature Phase Diagrams 233
8.5 Providing a Theoretical Basis for the P–T
Phase Diagram 235
8.6 Using the Clausius–Clapeyron Equation
to Calculate Vapor Pressure as a Function
of T 237
8.7 Dependence of Vapor Pressure of a Pure
Substance on Applied Pressure 239
8.8 Surface Tension 240
8.9 Chemistry in Supercritical Fluids 243
8.10 Liquid Crystal Displays 244
9 Ideal and Real Solutions
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
9.10
9.11
9.12
9.13
9.14
9.15
253
Defining the Ideal Solution 253
The Chemical Potential of a Component in the
Gas and Solution Phases 255
Applying the Ideal Solution Model to Binary
Solutions 256
The Temperature–Composition Diagram
and Fractional Distillation 260
The Gibbs–Duhem Equation 262
Colligative Properties 263
Freezing Point Depression and Boiling Point
Elevation 264
Osmotic Pressure 266
Deviations from Raoult’s Law in Real
Solutions 268
The Ideal Dilute Solution 270
Activities are Defined with Respect to Standard
States 272
Henry’s Law and the Solubility of Gases
in a Solvent 276
Chemical Equilibrium in Solutions 277
Solutions Formed from Partially Miscible
Liquids 280
Solid–Solution Equilibrium 282
CONTENTS
10 Electrolyte Solutions
289
10.1 Enthalpy, Entropy, and Gibbs Energy of Ion
Formation in Solutions 289
10.2 Understanding the Thermodynamics of Ion
Formation and Solvation 291
10.3 Activities and Activity Coefficients for
Electrolyte Solutions 294
10.4 Calculating g{ Using the Debye–Hückel
Theory 296
10.5 Chemical Equilibrium in Electrolyte
Solutions 300
11 Electrochemical Cells, Batteries,
and Fuel Cells 307
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
11.10
11.11
11.12
11.13
11.14
11.15
The Effect of an Electrical Potential on the
Chemical Potential of Charged Species 307
Conventions and Standard States
in Electrochemistry 309
Measurement of the Reversible Cell
Potential 312
Chemical Reactions in Electrochemical Cells
and the Nernst Equation 312
Combining Standard Electrode Potentials
to Determine the Cell Potential 314
Obtaining Reaction Gibbs Energies and
Reaction Entropies from Cell Potentials 316
Relationship Between the Cell EMF and the
Equilibrium Constant 316
Determination of E ~ and Activity Coefficients
Using an Electrochemical Cell 318
Cell Nomenclature and Types of
Electrochemical Cells 319
The Electrochemical Series 320
Thermodynamics of Batteries and Fuel Cells 321
Electrochemistry of Commonly Used
Batteries 322
Fuel Cells 326
Electrochemistry at the Atomic Scale 328
Using Electrochemistry for Nanoscale
Machining 331
12 Probability
12.1
12.2
12.3
12.4
12.5
337
Why Probability? 337
Basic Probability Theory 338
Stirling’s Approximation 346
Probability Distribution Functions 347
Probability Distributions Involving Discrete
and Continuous Variables 349
12.6 Characterizing Distribution Functions 352
Math Essential 4
7
Lagrange Multipliers
13 The Boltzmann Distribution
365
13.1
13.2
13.3
13.4
Microstates and Configurations 365
Derivation of the Boltzmann Distribution 371
Dominance of the Boltzmann Distribution 376
Physical Meaning of the Boltzmann
Distribution Law 378
13.5 The Definition of b 379
14 Ensemble and Molecular Partition
Functions 389
14.1
14.2
14.3
14.4
14.5
The Canonical Ensemble 389
Relating Q to q for an Ideal Gas 391
Molecular Energy Levels 393
Translational Partition Function 394
Rotational Partition Function:
Diatomic Molecules 396
14.6 Rotational Partition Function:
Polyatomic Molecules 404
14.7 Vibrational Partition Function 406
14.8 The Equipartition Theorem 411
14.9 Electronic Partition Function 412
14.10 Review 416
15 Statistical Thermodynamics
423
15.1 Energy 423
15.2 Energy and Molecular Energetic Degrees
of Freedom 427
15.3 Heat Capacity 432
15.4 Entropy 437
15.5 Residual Entropy 442
15.6 Other Thermodynamic Functions 443
15.7 Chemical Equilibrium 447
16 Kinetic Theory of Gases 457
16.1 Kinetic Theory of Gas Motion and
Pressure 457
16.2 Velocity Distribution in One
Dimension 458
16.3 The Maxwell Distribution of Molecular
Speeds 462
16.4 Comparative Values for Speed
Distributions 465
16.5 Gas Effusion 467
16.6 Molecular Collisions 469
16.7 The Mean Free Path 473
8
CONTENTS
17 Transport Phenomena
17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8
17.9
What Is Transport? 479
Mass Transport: Diffusion 481
Time Evolution of a Concentration Gradient 485
Statistical View of Diffusion 487
Thermal Conduction 489
Viscosity of Gases 492
Measuring Viscosity 495
Diffusion and Viscosity of Liquids 496
Ionic Conduction 498
18 Elementary Chemical Kinetics
18.1
18.2
18.3
18.4
18.5
18.6
18.7
18.8
18.9
18.10
18.11
18.12
18.13
18.14
18.15
19 Complex Reaction Mechanisms
479
509
Introduction to Kinetics 509
Reaction Rates 510
Rate Laws 512
Reaction Mechanisms 517
Integrated Rate Law Expressions 518
Numerical Approaches 523
Sequential First-Order Reactions 524
Parallel Reactions 529
Temperature Dependence of Rate
Constants 531
Reversible Reactions and Equilibrium 533
Perturbation-Relaxation Methods 537
The Autoionization of Water: A TemperatureJump Example 539
Potential Energy Surfaces 540
Activated Complex Theory 542
Diffusion-Controlled Reactions 546
19.1
19.2
19.3
19.4
19.5
19.6
19.7
19.8
Reaction Mechanisms and Rate Laws 557
The Preequilibrium Approximation 559
The Lindemann Mechanism 561
Catalysis 563
Radical-Chain Reactions 574
Radical-Chain Polymerization 577
Explosions 578
Feedback, Nonlinearity, and Oscillating
Reactions 580
19.9 Photochemistry 583
19.10 Electron Transfer 595
20 Macromolecules
20.1
20.2
20.3
20.4
20.5
20.6
20.7
609
What Are Macromolecules? 609
Macromolecular Structure 610
Random-Coil Model 612
Biological Polymers 615
Synthetic Polymers 623
Characterizing Macromolecules 626
Self-Assembly, Micelles, and Biological
Membranes 633
APPENDIX
A
Credits 659
Index 660
Data Tables 641
557
About the Authors
THOMAS ENGEL taught chemistry at the University of Washington for more than 20
years, where he is currently professor emeritus of chemistry. Professor Engel received
his bachelor’s and master’s degrees in chemistry from the Johns Hopkins University
and his Ph.D. in chemistry from the University of Chicago. He then spent 11 years as
a researcher in Germany and Switzerland, during which time he received the Dr. rer.
nat. habil. degree from the Ludwig Maximilians University in Munich. In 1980, he left
the IBM research laboratory in Zurich to become a faculty member at the University
of Washington.
Professor Engel has published more than 80 articles and book chapters in the area
of surface chemistry. He has received the Surface Chemistry or Colloids Award from
the American Chemical Society and a Senior Humboldt Research Award from the Alexander von Humboldt Foundation. Other than this textbook, his current primary science
interests are in energy policy and energy conservation. He serves on the citizen’s advisory board of his local electrical utility, and his energy-efficient house could be heated in
winter using only a hand-held hair dryer. He currently drives a hybrid vehicle and plans
to transition to an electric vehicle soon to further reduce his carbon footprint.
PHILIP REID has taught chemistry at the University of Washington since 1995.
Professor Reid received his bachelor’s degree from the University of Puget Sound in
1986 and his Ph.D. from the University of California, Berkeley in 1992. He performed
postdoctoral research at the University of Minnesota-Twin Cities before moving to
Washington.
Professor Reid’s research interests are in the areas of atmospheric chemistry, ultrafast condensed-phase reaction dynamics, and organic electronics. He has published
more than 140 articles in these fields. Professor Reid is the recipient of a CAREER
Award from the National Science Foundation, is a Cottrell Scholar of the Research Corporation, and is a Sloan Fellow. He received the University of Washington Distinguished
Teaching Award in 2005.
9
Preface
The fourth edition of Thermodynamics, Statistical Thermodynamics, and Kinetics includes many changes to the presentation
and content at both a global and chapter level. These updates
have been made to enhance the student learning experience
and update the discussion of research areas. At the global level,
changes that readers will see throughout the textbook include:
p­roblems from the fourth edition, new self-guided,
adaptive Dynamic Study Modules with wrong answer
feedback and remediation. Students who solve homework problems using MasteringTM Chemistry obtain
immediate feedback, which greatly enhances learning associated with solving homework problems. This
platform can also be used for pre-class reading quizzes
that are useful in ensuring students remain current in
their studies.
• Review of relevant mathematics skills. One of the
•
•
•
•
•
•
•
primary reasons that students experience physical chemistry as a challenging course is that they find it difficult to
transfer skills previously acquired in a mathematics course
to their physical chemistry course. To address this issue,
contents of the third edition’s Math Supplement have been
expanded and split into 4 two- to seven-page Math Essentials, which are inserted at appropriate places throughout
this book, just before the math skills are required. Our
intent in doing so is to provide “just-in-time” math help
and to enable students to refresh math skills specifically
needed in the following chapter.
Concept and Connection. New Concept and Connection features have been added to each chapter to present students with a quick visual summary of the most
important ideas within the chapter. In each chapter, approximately 10–15 of the most important concepts and/or
connections are highlighted in the margins.
End-of-Chapter Problems. Numerical Problems are now
organized by section number within chapters to make it
easier for instructors to create assignments for specific
parts of each chapter. Furthermore, a number of new
Conceptual Questions and Numerical Problems have
been added to the book. Numerical Problems from the
previous edition have been revised.
Introductory chapter materials. Introductory paragraphs
of all chapters have been replaced by a set of three questions plus responses to those questions. This new feature
makes the importance of the chapter clear to students at
the outset.
Figures. All figures have been revised to improve clarity.
Also, for many figures, additional annotation has been
included to help tie concepts to the visual program.
Key Equations. An end-of-chapter table that summarizes Key Equations has been added to allow students to
focus on the most important of the many equations in each
chapter. Equations in this table are set in red type where
they appear in the body of the chapter.
Further Reading. A Further Reading section has been
added to each chapter to provide references for students
and instructors who would like a deeper understanding of
various aspects of the chapter material.
Guided Practice and Interactivity
TM
° Mastering Chemistry has been significantly ex-
panded to include a wealth of new end-of-chapter
10
° NEW! Pearson eText gives students access to their
textbook anytime, anywhere.
■
■
Pearson eText mobile app offers offline access
and can be downloaded for most iOS and Android
phones/tablets from the Apple App Store or
Google Play Store.
Instructor and student note-taking, highlighting,
bookmarking, and search functionalities
° NEW! 66 Dynamic Study Modules help students
°
°
study effectively on their own by continuously assessing their activity and performance in real time.
Students complete a set of questions with a unique
answer format that also asks them to indicate their
confidence level. Questions repeat until the student can answer them all correctly and confidently.
These are available as graded assignments prior to
class and are accessible on smartphones, tablets, and
computers.
Topics include key math skills, as well as a refresher
of general chemistry concepts, such as understanding
matter, chemical reactions, and the periodic table and
atomic structure. Topics can be added or removed to
match your coverage.
In terms of chapter and section content, many changes were
made. The most significant of these changes are as follows:
• A new chapter entitled Macromolecules (Chapter 20) has
•
•
been added. The motivation for this chapter is that assemblies of smaller molecules form large molecules, such as
proteins or polymers. The resulting macromolecules can
exhibit new structures and functions that are not reflected
by the individual molecular components. Understanding
the factors that influence macromolecular structure is
critical in understanding the chemical behavior of these
important molecules.
A more detailed discussion of system-based and
­surroundings-based work has been added in Chapter 2
to help clarify the confusion that has appeared in the
chemical education literature about using the system or
­surroundings pressure in calculating work.
The discussion on entropy and the second law of thermodynamics in Chapter 5 has been substantially revised. As a
PREFACE
•
result, calculations of entropy changes now appear earlier
in the chapter, and the material on the reversible Carnot
cycle has been shifted to a later section.
The approach to chemical equilibrium in Chapter 6 has
been substantially revised to present a formulation in
terms of the extent of reaction. This change has been made
to focus more clearly on changes in chemical potential as
the driving force in reaching equilibrium.
For those not familiar with the third edition of Thermodynamics,
Statistical Thermodynamics, and Kinetics, our approach to
teaching physical chemistry begins with our target audience—
undergraduate students majoring in chemistry, biochemistry,
and chemical engineering, as well as many students majoring
in the atmospheric sciences and the biological sciences. The
following objectives outline our approach to teaching physical chemistry.
• Focus on teaching core concepts. The central principles
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•
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•
of physical chemistry are explored by focusing on core ideas
and then extending these ideas to a variety of problems. The
goal is to build a solid foundation of student understanding
in a limited number of areas rather than to provide a condensed encyclopedia of physical chemistry. We believe this
approach teaches students how to learn and enables them
to apply their newly acquired skills to master related fields.
Illustrate the relevance of physical chemistry to the
world around us. Physical chemistry becomes more
relevant to a student if it is connected to the world around
us. Therefore, example problems and specific topics are
tied together to help the student develop this connection.
For example, fuel cells, refrigerators, heat pumps, and real
engines are discussed in connection with the second law of
thermodynamics. Every attempt is made to connect fundamental ideas to applications that could be of interest to
the student.
Link the macroscopic and atomic-level worlds. One
of the strengths of thermodynamics is that it is not dependent on a microscopic description of matter. However,
students benefit from a discussion of issues such as how
pressure originates from the random motion of molecules.
Present exciting new science in the field of physical
chemistry. Physical chemistry lies at the forefront of
many emerging areas of modern chemical research.
Heterogeneous catalysis has benefited greatly from mechanistic studies carried out using the techniques of modern
surface science. Atomic-scale electrochemistry has become
possible through scanning tunneling microscopy. The role
of physical chemistry in these and other emerging areas
is highlighted throughout the text.
Provide a versatile online homework program with
tutorials. Students who submit homework problems
using MasteringTM Chemistry obtain immediate feedback,
•
11
a feature that greatly enhances learning. Also, tutorials
with wrong answer feedback offer students a self-paced
learning environment.
Use web-based simulations to illustrate the concepts
being explored and avoid math overload. Mathematics
is central to physical chemistry; however, the mathematics
can distract the student from “seeing” the underlying concepts. To circumvent this problem, web-based simulations
have been incorporated as end-of-chapter problems in several chapters so that the student can focus on the science
and avoid a math overload. These web-based simulations
can also be used by instructors during lecture. Simulations, animations, and homework problem worksheets can
be accessed on Mastering™ Chemistry.
Effective use of Thermodynamics, Statistical Thermodynamics,
and Kinetics does not require proceeding sequentially through
the chapters or including all sections. Some topics are discussed in supplemental sections, which can be omitted if they
are not viewed as essential to the course. Also, many sections
are sufficiently self-contained that they can be readily omitted
if they do not serve the needs of the instructor and students.
This textbook is constructed to be flexible to your needs.
We welcome the comments of both students and instructors
on how the material was used and how the presentation can
be improved.
Thomas Engel and Philip Reid
University of Washington
ACKNOWLEDGMENTS
Many individuals have helped us to bring the text into its current form. Students have provided us with feedback directly
and through the questions they have asked, which has helped
us to understand how they learn. Many of our colleagues,
including Peter Armentrout, Doug Doren, Gary Drobny, Eric
Gislason, Graeme Henkelman, Lewis Johnson, Tom Pratum,
Bill Reinhardt, Peter Rosky, George Schatz, Michael Schick,
Gabrielle Varani, and especially Bruce Robinson, have been
invaluable in advising us. We are also fortunate to have
access to some end-of-chapter problems that were originally
presented in Physical Chemistry, 3rd edition, by Joseph H.
Noggle and in Physical Chemistry, 3rd edition, by Gilbert
W. Castellan. The reviewers, who are listed separately, have
made many suggestions for improvement, for which we are
very grateful. All those involved in the production process
have helped to make this book a reality through their efforts.
Special thanks are due to Jim Smith, who guided us through
the first edition, to our current editor Jeanne Zalesky, to
our developmental editor Spencer Cotkin, and to Jennifer
Hart and Beth Sweeten at Pearson, who have led the production process.
12
PREFACE
4TH EDITION MANUSCRIPT REVIEWERS
David Coker,
Boston University
Yingbin Ge,
Central Washington University
Eric Gislason,
University of Illinois, Chicago
Nathan Hammer,
University of Mississippi
George Papadantonakis,
University of Illinois, Chicago
Stefan Stoll,
University of Washington
Liliya Yatsunyk,
Swarthmore College
4TH EDITION ACCURACY REVIEWERS
Garry Crosson,
University of Dayton
Benjamin Huddle,
Roanoke College
Andrea Munro,
Pacific Lutheran University
4TH EDITION PRESCRIPTIVE REVIEWERS
Joseph Alia,
University of Minnesota, Morris
Herbert Axelrod,
California State University, Fullerton
Timothy Brewer,
Eastern Michigan University
Paul Cooper,
George Mason University
Bridget DePrince,
Florida State University
Patrick Fleming,
California State University, East Bay
Richard Mabbs,
Washington University, St. Louis
Vicki Moravec,
Trine University
Andrew Petit,
California State University, Fullerton
Richard Schwenz,
University of Northern Colorado
Ronald Terry,
Western Illinois University
Dunwei Wang,
Boston College
Gerard Harbison,
University of Nebraska, Lincoln
Sophya Garashchuk,
University of South Carolina
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PREVIOUS EDITION REVIEWERS
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ACKNOWLEDGMENTS FOR THE GLOBAL EDITION
Contributor
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Reviewers
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A Visual, Conceptual, and Contemporary
Approach to Physical Chemistry
NEW! Math Essentials provide a review of
MATH ESSENTIAL 2:
relevant math skills, offer “just in time” math
help, and enable students to refresh math skills
specifically needed in the chapter that follows.
Differentiation and Integration
UPDATED! Introductory paragraphs of all chapters
Differential and integral calculus is used extensively in physical chemistry. In this unit
we review the most relevant aspects of calculus needed to understand the chapter discussions and to solve the end-of-chapter problems.
ME2.1
The Definition and Properties
of a Function
ME2.2
The First Derivative
of a Function
ME2.1 THE DEFINITION AND PROPERTIES
ME2.3
The Chain Rule
ME2.4
The Sum and Product Rules
ME2.5
The Reciprocal Rule and the
Quotient Rule
ME2.6
Higher-Order Derivatives:
Maxima, Minima, and
Inflection Points
ME2.7
Definite and Indefinite
Integrals
OF A FUNCTION
have been replaced by a set of three questions plus
responses to those questions making the relevance
of the chapter clear at the outset.
A function ƒ is a rule that generates a value y from the value of a variable x. Mathematically, we write this as y = ƒ1x2. The set of values x over which ƒ is defined is the domain of the function. Single-valued functions have a single value of y for a given value
of x. Most functions that we will deal with in physical chemistry are single valued.
However, inverse trigonometric functions and 1 are examples of common functions
that are multivalued. A function is continuous if it satisfies these three conditions:
ƒ1x2 is defined at a
lim ƒ1x2 exists
xSa
(ME2.1)
lim ƒ1x2 = ƒ1a2
xSa
ME2.2 THE FIRST DERIVATIVE OF A FUNCTION
C H A P T E R
5
What Determines the Direction
of Spontaneous Change
in a Process?
5.2
The Second Law of
Thermodynamics, Spontaneity,
and the Sign of ∆S
5.3
Calculating Changes in Entropy
as T, P, or V Change
5.4
Understanding Changes in
Entropy at the Molecular Level
WHY is this material important?
5.5
The Clausius Inequality
Mixtures of chemically reactive species can evolve toward reactants or toward products, defining the direction of spontaneous change. How do we know which of these
reactions is spontaneous? Entropy, designated by S, is the state function that predicts
the direction of natural, or spontaneous, change. In this chapter, you will learn how to
carry out entropy calculations.
5.6
The Change of Entropy
in the Surroundings and
∆Stot = ∆S + ∆Ssur
5.7
Absolute Entropies and the
Third Law of Thermodynamics
5.8
Standard States in Entropy
Calculations
5.9
Entropy Changes in Chemical
Reactions
WHAT are the most important concepts and results?
Heat flows from hotter bodies to colder bodies, and gases mix rather than separate.
Entropy always increases for a spontaneous change in an isolated system. For a spontaneous change in a system interacting with its environment, the sum of the entropy
of the system and that of the surroundings always increases. Thermodynamics can be
used to devise ways to reduce our energy use and to transition away from fossil fuels.
WHAT would be helpful for you to review for this chapter?
Because differential and integral calculus and partial derivatives are used extensively in
this chapter, it would be helpful to review the material on calculus in Math Essential 2
and 3.
20
10
24
2
22
4 x
210
The symbol ƒ′1x2 is often used in place of dƒ1x2>dx. For the function of interest,
dƒ1x2
1x + h22 - 1x22
2hx + h2
= lim
= lim
= lim 2x + h = 2x
hS0
hS0
hS0
dx
h
h
Figure ME2.1
(ME2.3)
In order for dƒ1x2>dx to be defined over an interval in x, ƒ1x2 must be continuous over
the interval. Next, we present rules for differentiating simple functions. Some of these
functions and their derivatives are as follows:
d1ax n2
= anx n - 1, where a is a constant and n is any real number
dx
d1aex2
= aex,
dx
The function y = x2 plotted as a function of x. The dashed line is the tangent to
the curve at x = 1.5.
(ME2.4)
(ME2.5)
where a is a constant
d ln x
1
=
x
dx
(ME2.6)
37
5.10 Heat Engines and the
Carnot Cycle
5.11 (Supplemental Section) How
M03_ENGE7707_04_GE_ME2.indd 37
Does S Depend on V and T ?
23/05/20 7:22 PM
5.12 (Supplemental Section)
Dependence of S on T and P
5.13 (Supplemental Section) Energy
Efficiency, Heat Pumps,
Refrigerators, and Real Engines
5.1 WHAT DETERMINES THE DIRECTION OF
NEW! Concept and Connection features
in each chapter present students with quick
visual summaries of the core concepts within
the chapter, highlighting key take aways and
providing students with an easy way to review
the material.
SPONTANEOUS CHANGE IN A PROCESS?
Concept
A process is defined to be
spontaneous even if it occurs only
for some conditions because of an
energy barrier.
Figure 8.15
P
m
P
Solid
Critical
point
Solid
Solid–Liquid
Critical
point
T
Gas
m
l
q
a
id
Solid
b
c
a
p
lid
lum
e
Critical
point
f
–G
L
iqu
i Gas id e
h
le
as
Gas
d
V
G
as
Lin
e
So
Vo
Liquid
Tri
p
Liq
Ga uid
s
So
lid
–G
as
0
k
g
Triple
point
A P–V–T phase diagram for a substance
that expands upon melting. The indicated processes are discussed in the text.
Liquid
f
Pressure
revised to improve clarity and for
many figures, additional annotation
has been included to help tie concepts
to the visual program.
28/05/20 3:28 PM
uid
M08_ENGE7707_04_GE_C05.indd 123
UPDATED! All figures have been
123
Liq
The first law of thermodynamics constrains the range of possible processes to those
that conserve energy and rules out machines that work indefinitely without an energy
source (perpetual motion machines) that have been the dream of inventors over centuries. Any system that is not at equilibrium will evolve with time. We refer to the
direction of evolution as the spontaneous direction. Spontaneous does not mean that
the process occurs immediately, but rather that it will occur with high probability if any
barrier to the change is overcome. For example, the transformation of a piece of wood
to CO2 and H2O in the presence of oxygen is spontaneous, but it only occurs at elevated
temperatures because an activation energy barrier must be overcome for the reaction to
proceed. In this chapter, we discuss the second law of thermodynamics, which allows
us to predict if a process is spontaneous. Before stating the second law, we discuss two
processes that satisfy the first law because they conserve energy and attempt to identify
criteria that can predict the spontaneous direction of evolution.
f(x)
f(x)5x2
hS0
Solid–Liqu
Entropy and the
Second and Third Laws
of Thermodynamics
5.1
The first derivative of a function has as its physical interpretation the slope of the function evaluated at the point of interest. In order for the first derivative to exist at a
point a, the function must be continuous at x = a, and the slope of the function at
x = a must be the same when approaching a from x 6 a and x 7 a. For example, the
slope of the function y = x 2 at the point x = 1.5 is indicated by the line tangent to the
curve shown in Figure ME2.1.
Mathematically, the first derivative of a function ƒ1x2 is denoted dƒ1x2>dx. It is
defined by
dƒ1x2
ƒ1x + h2 - ƒ1x2
= lim
(ME2.2)
dx
h
g
o
n
T2
T1
T4
e
ur
rat
e
mp
Te
T3Tc
Continuous Learning Before, During,
and After Class
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NEW! 66 Dynamic Study Modules
help students study effectively on their own
by continuously assessing their activity and
performance in real time.
Students complete a set of questions with
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and confidently. These are available as graded
assignments prior to class and are accessible on
smartphones, tablets, and computers.
Topics include key math skills as well as a
refresher of general chemistry concepts such
as understanding matter, chemical reactions,
and understanding the periodic table & atomic
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214
CHAPTER 7 The Properties of Real Gases
Concept
1.0
Tr 5 2.00
Tr 5 1.50
Compression factor (z)
Values of compression factor plotted
as a function of the reduced pressure Pr
for seven gases. Compression factors calculated for the six values of reduced temperatures are indicated in the figure. The
solid curves are drawn to guide the eye.
CHAPTER 7 The Properties of Real Gases
Figure 7.7
Values of compression factor plotted
as a function of the reduced pressure Pr
for seven gases. Compression factors calculated for the six values of reduced temperatures are indicated in the figure. The
solid curves are drawn to guide the eye.
0.8
Concept
Tr 5 1.30
Tr 5 1.20
Methane
Tr 5 1.10
Ethene
Tr 5 1.00
0.2
4
5
3
Reduced pressure (Pr)
2
6
7
The law of corresponding states implicitly assumes that two parameters are sufficient to describe an intermolecular potential. This assumption is best for molecules that
are nearly spherical because for such molecules the potential is independent of the molecular orientation. It is not nearly as good for dipolar molecules such as HF, for which
the potential is orientation dependent.
The results shown in Figure 7.7 demonstrate the validity of the law of corresponding states. How can this law be applied to a specific gas? The goal is to calculate z for specific values of Pr and Tr and to use these z values to estimate the error in
using the ideal gas law. A convenient way to display these results is in the form of a
graph. Calculated results for z using the van der Waals equation of state as a function
of Pr for different values of Tr are shown in Figure 7.8. For a given gas and specific
P and T values, Pr and Tr can be calculated. A value of z can then be read from the
curves in Figure 7.8.
From Figure 7.8, we see that for Pr 6 5.5, z 6 1, as long as Tr 6 2. This means
that the real gas exerts a smaller pressure than an ideal gas in this range of Tr and Pr.
We conclude that for these values of Tr and Pr, the molecules are more influenced by
the attractive part of the potential than by the repulsive part that arises from the finite
molecular volume. However, z 7 1 for Pr 7 7 for all values of Tr as well as for all
The compression factor plotted as a
function of Pr for selected values of Tr.
Values of Tr are indicated adjacent to the
curves. The curves were calculated using
the van der Waals equation of state.
1.3
0.6
6
8
10
Reduced pressure (Pr)
1.2
214
CHAPTER 7 The Properties of Real Gases
1.1
0.4
Figure 7.7
Values of compression factor plotted
as a function of the reduced pressure Pr
for seven gases. Compression factors calculated for the six values of reduced temperatures are indicated in the figure. The
solid curves are drawn to guide the eye.
1.0
27/05/2020 17:53
Concept
1.0
Tr 5 2.00
Tr 5 1.50
Compression factor (z)
M10_ENGE7707_04_GE_C07.indd 214
4.0
1.0
0.8
Tr 5 1.30
Nitrogen
Tr 5 1.20
0.6
Methane
Propane
Tr 5 1.10
Ethane
Butane
0.4
Isopentane
The law of corresponding states
allows the compression factors for
very different gases to be estimated.
Ethene
Tr 5 1.00
0.2
1
4
5
3
Reduced pressure (Pr)
2
6
7
The law of corresponding states implicitly assumes that two parameters are sufficient to describe an intermolecular potential. This assumption is best for molecules that
are nearly spherical because for such molecules the potential is independent of the molecular orientation. It is not nearly as good for dipolar molecules such as HF, for which
the potential is orientation dependent.
The results shown in Figure 7.7 demonstrate the validity of the law of corresponding states. How can this law be applied to a specific gas? The goal is to calculate z for specific values of Pr and Tr and to use these z values to estimate the error in
using the ideal gas law. A convenient way to display these results is in the form of a
graph. Calculated results for z using the van der Waals equation of state as a function
of Pr for different values of Tr are shown in Figure 7.8. For a given gas and specific
P and T values, Pr and Tr can be calculated. A value of z can then be read from the
curves in Figure 7.8.
From Figure 7.8, we see that for Pr 6 5.5, z 6 1, as long as Tr 6 2. This means
that the real gas exerts a smaller pressure than an ideal gas in this range of Tr and Pr.
We conclude that for these values of Tr and Pr, the molecules are more influenced by
the attractive part of the potential than by the repulsive part that arises from the finite
molecular volume. However, z 7 1 for Pr 7 7 for all values of Tr as well as for all
1.6
1.4
Compression factor (z)
Compression factor (z)
Figure 7.8
The compression factor plotted as a
function of Pr for selected values of Tr.
Values of Tr are indicated adjacent to the
curves. The curves were calculated using
the van der Waals equation of state.
1.2
4
Ethane
Butane
Isopentane
Ethene
0.2
Compression factor (z)
Figure 7.8
M10_ENGE7707_04_GE_C07.indd 214
1.7
1.5
1.4
Propane
Tr 5 1.00
4
5
3
Reduced pressure (Pr)
2
6
7
1.4
1.4
0.8
Methane
Tr 5 1.10
0.4
1.6
1.6
2.0
Nitrogen
Tr 5 1.20
0.6
The law of corresponding states implicitly assumes that two parameters are sufficient to describe an intermolecular potential. This assumption is best for molecules that
are nearly spherical because for such molecules the potential is independent of the molecular orientation. It is not nearly as good for dipolar molecules such as HF, for which
the potential is orientation dependent.
The results shown in Figure 7.7 demonstrate the validity of the law of corresponding states. How can this law be applied to a specific gas? The goal is to calculate z for specific values of Pr and Tr and to use these z values to estimate the error in
using the ideal gas law. A convenient way to display these results is in the form of a
graph. Calculated results for z using the van der Waals equation of state as a function
of Pr for different values of Tr are shown in Figure 7.8. For a given gas and specific
P and T values, Pr and Tr can be calculated. A value of z can then be read from the
curves in Figure 7.8.
From Figure 7.8, we see that for Pr 6 5.5, z 6 1, as long as Tr 6 2. This means
that the real gas exerts a smaller pressure than an ideal gas in this range of Tr and Pr.
We conclude that for these values of Tr and Pr, the molecules are more influenced by
the attractive part of the potential than by the repulsive part that arises from the finite
molecular volume. However, z 7 1 for Pr 7 7 for all values of Tr as well as for all
Butane
2
Tr 5 1.30
1
Ethane
0.4
1
0.8
Propane
Isopentane
The law of corresponding states
allows the compression factors for
very different gases to be estimated.
Tr 5 2.00
Tr 5 1.50
The law of corresponding states
allows the compression factors for
very different gases to be estimated.
Nitrogen
0.6
1.0
Compression factor (z)
214
Figure 7.7
Figure 7.8
The compression factor plotted as a
function of Pr for selected values of Tr.
Values of Tr are indicated adjacent to the
curves. The curves were calculated using
the van der Waals equation of state.
M10_ENGE7707_04_GE_C07.indd 214
1.2
4.0
1.0
2
0.8
1.3
0.6
2.0
1.7
1.5
1.4
4
6
8
10
Reduced pressure (Pr)
1.2
1.1
0.4
1.0
27/05/2020 17:53
1.2
4.0
1.0
2
0.8
1.3
0.6
2.0
1.7
1.5
1.4
4
6
8
10
Reduced pressure (Pr)
1.2
1.1
0.4
1.0
27/05/2020 17:53
This page intentionally left blank
MATH ESSENTIAL 1:
Units, Significant Figures, and
Solving End of Chapter Problems
ME1.1 UNITS
Quantities of interest in physical chemistry such as pressure, volume, or temperature
are characterized by their magnitude and their units. In this textbook, we use the SI
(from the French Le Système international d'unités) system of units. All physical quantities can be defined in terms of the seven base units listed in Table ME1.1. For more
details, see http://physics.nist.gov/cuu/Units/units.html. The definition of temperature
is based on the coexistence of the solid, gaseous, and liquid phases of water at a pressure of 1 bar.
ME1.1
Units
ME1.2
Uncertainty and Significant
Figures
ME1.3
Solving End-of-Chapter
Problems
TABLE ME1.1 Base SI Units
Base Unit
Unit
Definition of Unit
Unit of length
meter (m)
The meter is the length of the path traveled by light in vacuum during a time
interval of 1>299,792,458 of a second.
Unit of mass
kilogram (kg)
The kilogram is the unit of mass; it is defined by taking the value of Planck’s
constant h to be exactly 6.62607015 * 10-34 kg . m2/s.
Unit of time
second (s)
The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state
of the cesium 133 atom.
Unit of electric current
ampere (A)
The ampere is the constant current that, if maintained in two straight parallel
conductors of infinite length, is of negligible circular cross section, and if placed
1 meter apart in a vacuum would produce between these conductors a force
equal to 2 * 10-7 kg m s-2 per meter of length. In this definition, 2 is an
exact number.
Unit of thermodynamic
temperature
kelvin (K)
The kelvin, unit of thermodynamic temperature, is defined by taking the value
of the Boltzmann constant k to be exactly 1.380649 * 10-23 J/K.
Unit of amount of substance
mole (mol)
The mole is the SI unit of substance. One mole contains exactly
6.02214076 * 1023 elementary entities. When the mole is used, the elementary
entities must be specified and may be atoms, molecules, ions, electrons, other
particles, or specified groups of such particles.
Unit of luminous intensity
candela (cd)
The candela is the luminous intensity, in a given direction, of a source that
emits monochromatic radiation of frequency 540. * 1012 hertz and that has a
radiant intensity in that direction of 1>683 watt per steradian.
Quantities of interest other than the seven base quantities can be expressed in terms
of the units meter, kilogram, second, ampere, kelvin, mole, and candela. The most important of these derived units, some of which have special names as indicated, are listed
in Table ME1.2. A more inclusive list of derived units can be found at http://physics
.nist.gov/cuu/Units/units.html.
17
18
MATH ESSENTIAL 1 Units, Significant Figures, and Solving End of Chapter Problems
TABLE ME1.2 Derived Units
Unit
Definition
Relation to Base Units
Special Name
Abbreviation
Area
Size of a surface
m2
m2
3
Volume
Amount of three-dimensional space an object
occupies
m
m3
Velocity
Measure of the rate of motion
m s-1
m s-1
Acceleration
Rate of change of velocity
m s-2
m s-2
Linear
momentum
Product of mass and linear velocity of an object
kg m s-1
kg m s-1
Angular
momentum
Product of the moment of inertia of a body
about an axis and its angular velocity with
respect to the same axis
kg m2 s-1
kg m2 s-1
Force
Any interaction that, when unopposed, will
change the motion of an object
kg m s-2
newton
N
Pressure
Force acting per unit area
kg m-1 s-2
N m-2
pascal
Pa
Work
Product of force on an object and movement
along the direction of the force
kg m2 s-2
joule
J
Kinetic energy
Energy an object possesses because of its
motion
kg m2 s-2
joule
J
Potential energy
Energy an object possesses because of its
position or condition
kg m2 s-2
joule
J
Power
Rate at which energy is produced or
consumed
kg m2 s-3
watt
W
Mass density
Mass per unit volume
kg m-3
kg m-3
Radian
Angle at the center of a circle whose arc is
equal in length to the radius
m>m = 1
m>m = 1
Steradian
Angle at the center of a sphere subtended by
a part of the surface equal in area to the square
of the radius
m2 >m2 = 1
m2 >m2 = 1
Frequency
Number of repeat units of a wave per unit time
s-1
hertz
Hz
Electrical charge
Physical property of matter that causes it to
experience an electrostatic force
As
coulomb
C
Electrical potential
Work done in moving a unit positive charge
from infinity to that point
volt
V
Electrical resistance
Ratio of the voltage to the electric current that
flows through a conductive material
kg m2 s-3 >A
W>A
ohm
Ω
kg m2 s-3 >A2 W>A2
If SI units are used throughout the calculation of a quantity, the result will have
SI units. For example, consider a unit analysis of the electrostatic force between two
charges:
F =
=
q1q2
8pe0r 2
=
C2
A2 s2
=
8p * kg -1s4A2 m-3 * m2
8p * kg -1s4A2 m-3 * m2
1
1
kg m s-2 =
N
8p
8p
Therefore, in carrying out a calculation, it is only necessary to make sure that all quantities are expressed in SI units rather than carrying out a detailed unit analysis of the
entire calculation.
ME1.3 Solving End-of-Chapter Problems
ME1.2 UNCERTAINTY AND SIGNIFICANT
FIGURES
In carrying out a calculation, it is important to take into account the uncertainty of
the individual quantities that go into the calculation. The uncertainty is indicated by
the number of significant figures. For example, the mass 1.356 g has four significant
figures. The mass 0.003 g has one significant figure, and the mass 0.01200 g has four
significant figures. By convention, the uncertainty of a number is {1 in the rightmost
digit. A zero at the end of a number that is not to the right of a decimal point is not
significant. For example, 150 has two significant figures, but 150. has three significant
figures. Some numbers are exact and have no uncertainty. For example, 1.00 * 106
has three significant figures because the 10 and 6 are exact numbers. By definition, the
mass of one atom of 12C is exactly 12 atomic mass units.
If a calculation involves quantities with a different number of significant figures,
the following rules regarding the number of significant figures in the result apply:
• In addition and subtraction, the result has the number of digits to the right of the
decimal point corresponding to the number that has the smallest number of digits to the right of the decimal point. For example 101 + 24.56 = 126 and
0.523 + 0.10 = 0.62.
In multiplication or division, the result has the number of significant figures corresponding to the number with the smallest number of significant figures. For
example, 3.0 * 16.00 = 48 and 0.05 * 100. = 5.
•
It is good practice to carry forward a sufficiently large number of significant figures in
different parts of the calculation and to round off to the appropriate number of significant figures at the end.
ME1.3 SOLVING END-OF-CHAPTER PROBLEMS
Because calculations in physical chemistry often involve multiple inputs, it is useful to
carry out calculations in a manner that they can be reviewed and easily corrected. For
example, the input and output for the calculation of the pressure exerted by gaseous
benzene with a molar volume of 2.00 L at a temperature of 595 K using the Redlich–
RT
a
1
Kwong equation of state P =
in Mathematica is shown
Vm - b
V
1V
2T m m + b2
below. The statement in the first line clears the previous values of all listed quantities,
and the semicolon after each input value suppresses its appearance in the output.
In[36]:=
out[42]=
Clear[r, t, vm, a, b, prk]
r = 8.314 * 10^ -2;
t = 595;
vm = 2.00;
a = 452;
b = .08271;
rt
a
1
prk =
vm - b
2t vm(vm + b)
21.3526
Invoking the rules for significant figures, the final answer is P = 21.4 bar.
The same problem can be solved using Microsoft Excel as shown in the following
table.
A
B
C
D
E
F
1
R
T
Vm
a
b
=((A2*B2)/(C2-E2))-(D2/SQRT(B2))*(1/(C2*(C2+E2)))
2
0.08314
595
2
452
0.08271
21.35257941
19
This page intentionally left blank
C H A P T E R
1
Fundamental Concepts
of Thermodynamics
WHY is this material important?
Thermodynamics is a powerful science that allows predictions to be made about chemical reactions, the efficiency of engines, and the potential of new energy sources. It is
a macroscopic science and does not depend on a description of matter at the molecular
scale. In this chapter, we introduce basic concepts such as the system variables pressure, temperature, and volume, and equations of state that relate these variables with
one another.
1.1
What Is Thermodynamics
and Why Is It Useful?
1.2
The Macroscopic Variables
Volume, Pressure, and
Temperature
1.3
Basic Definitions Needed to
Describe Thermodynamic
Systems
1.4
Equations of State and the Ideal
Gas Law
1.5
A Brief Introduction
to Real Gases
WHAT are the most important concepts and results?
Processes such as chemical reactions occur in an apparatus whose contents we call the
system. The rest of the universe is the surroundings. The exchange of energy and matter between the system and surroundings is central to thermodynamics. We will show
that the macroscopic gas property pressure arises through the random thermal motion
of atoms and molecules. Equations of state such as the ideal gas law allow us to calculate how one system variable changes when another variable is increased or decreased.
WHAT would be helpful for you to review for this chapter?
It would be useful to review the material on units and problem solving discussed in
Math Essential 1.
1.1 WHAT IS THERMODYNAMICS
AND WHY IS IT USEFUL?
Thermodynamics is the branch of science that describes the behavior of matter and
the transformation between different forms of energy on a macroscopic scale, which
is the scale of phenomena experienced by humans, as well as larger-scale phenomena
(e.g., astronomical scale). Thermodynamics describes a system of interest in terms of
its bulk properties. Only a few variables are needed to describe such a system, and the
variables are generally directly accessible through measurements. A thermodynamic
description of matter does not make reference to its structure and behavior at the microscopic level. For example, 1 mol of gaseous water at a sufficiently low density is completely described by two of the three macroscopic variables of pressure, volume, and
temperature. By contrast, the microscopic scale refers to dimensions on the order of the
size of molecules. At the microscopic level, water is described as a dipolar triatomic
molecule, H2O, with a bond angle of 104.5° that forms a network of hydrogen bonds.
In the first part of this book (Chapters 1–11), we will discuss thermodynamics. Later
in the book, we will turn to statistical thermodynamics. Statistical thermodynamics
Concept
Because thermodynamics does not
make reference to a description
of matter at the microscopic level,
it is equally applicable to a liter of
garbage and a liter of pure water.
21
22
CHAPTER 1 Fundamental Concepts of Thermodynamics
uses atomic and molecular properties to calculate the macroscopic properties of matter.
For example, statistical thermodynamic analysis shows that liquid water is the stable
form of aggregation at a pressure of 1 bar and a temperature of 90°C, whereas gaseous
water is the stable form at 1 bar and 110°C. Using statistical thermodynamics, we can
calculate the macroscopic properties of matter from underlying molecular properties.
Given that the microscopic nature of matter is becoming increasingly well understood using theories such as quantum mechanics, why is the macroscopic science
thermodynamics relevant today? The usefulness of thermodynamics can be illustrated
by describing four applications of thermodynamics that you will have mastered after
working through this book:
• You have built an industrial plant to synthesize NH3(g) gas from N2(g) and H2(g).
•
•
•
You find that the yield is insufficient to make the process profitable, and you decide
to try to improve the NH3 output by changing either the temperature or pressure of
synthesis, or both. However, you do not know whether to increase or decrease the
values of these variables. As will be shown in Chapter 6, the ammonia yield will be
higher at equilibrium if the temperature is decreased and the pressure is increased.
You wish to use methanol to power a car. One engineer provides a design for an
internal combustion engine that will burn methanol efficiently according to the reaction CH3OH(l) + 3>2O2(g) S CO2(g) + 2H2O(l). A second engineer designs
an electrochemical fuel cell that carries out the same reaction. He claims that the
vehicle will travel much farther if it is powered by the fuel cell rather than by the
internal combustion engine. As will be shown in Chapter 5, this assertion is correct,
and an estimate of the relative efficiencies of the two propulsion systems can be
made.
You are asked to design a new battery that will be used to power a hybrid car.
Because the voltage required by the driving motors is much higher than can be
generated in a single electrochemical cell, many cells must be connected in series.
Because the space for the battery is limited, as few cells as possible should be used.
You are given a list of possible cell reactions and told to determine the number of
cells needed to generate the required voltage. As you will learn in Chapter 11, this
problem can be solved using tabulated values of thermodynamic functions.
Your attempts to synthesize a new and potentially very marketable compound have
consistently led to yields that make it unprofitable to begin production. A supervisor suggests a major effort to make the compound by first synthesizing a catalyst
that promotes the reaction. How can you decide if this effort is worth the required
investment? As will be shown in Chapter 6, the maximum yield expected under
equilibrium conditions should be calculated first. If this yield is insufficient, a catalyst is useless.
1.2 THE MACROSCOPIC VARIABLES VOLUME,
PRESSURE, AND TEMPERATURE
Concept
The origin of pressure in a gas is the
random thermally induced motion of
individual molecules.
We begin our discussion of thermodynamics by considering a bottle of a gas such as
He or CH4. At a macroscopic level, the sample of known chemical composition is completely described by the measurable quantities volume, pressure, and temperature for
which we use the symbols V, P, and T. The volume V is just that of the bottle. What
physical association do we have with P and T?
Pressure is the force exerted by the gas per unit area of the container. It is most
easily understood by considering a microscopic model of the gas known as the kinetic
theory of gases. The gas is described by two assumptions: first, the atoms or molecules
of an ideal gas do not interact with one another, and second, the atoms or molecules
can be treated as point masses. The pressure exerted by a gas on the container that confines the gas arises from collisions of randomly moving gas molecules with the container walls. Because the number of molecules in a small volume of the gas is on the
order of Avogadro’s number NA, the number of collisions between molecules is also
23
1.2 The Macroscopic Variables Volume, Pressure, and Temperature
z
large. To describe pressure, a molecule is envisioned as traveling through space with a
velocity vector v that can be resolved into three Cartesian components: vx, vy, and vz, as
illustrated in Figure 1.1.
The square of the magnitude of the velocity v2 in terms of the three velocity components is
v2 = v # v = v2x + v2y + v2z
vz
(1.1)
v
The particle kinetic energy is 1>2 mv2 such that
1
1
1
1
etr = mv2 = mv2x + mv2y + mv2z = etrx + etry + etrz
2
2
2
2
mai
F
m dvi
1 dmvi
1 dpi
P =
=
= a
b = a
b = a
b
A
A
A dt
A
dt
A dt
∆p
* (number of molecules)
molecule
x
Figure 1.1
Cartesian components of velocity. The
particle velocity v can be resolved into
three velocity components: vx, vy, and vz.
2mvx
mvx
(1.3)
In Equation (1.3), F is the force of the collision, A is the area of the wall with which
the particle has collided, m is the mass of the particle, vi is the velocity component along
the i direction (i = x, y, or z), and pi is the particle linear momentum in the i direction.
Equation (1.3) illustrates that pressure is related to the change in linear momentum
with respect to time that occurs during a collision. Due to conservation of momentum,
any change in particle linear momentum must result in an equal and opposite change
in momentum of the container wall. A single collision is depicted in Figure 1.2. This
figure illustrates that the particle linear momentum change in the x direction is -2mvx
(note that there is no change in momentum in the y or z direction). Accordingly, a corresponding momentum change of 2mvx must occur for the wall.
The pressure measured at the container wall corresponds to the sum of collisions
involving a large number of particles that occur per unit time. Therefore, the total momentum change that gives rise to the pressure is equal to the product of the momentum
change from a single-particle collision and the total number of particles that collide
with the wall:
∆ptotal =
y
vx
(1.2)
where e is kinetic energy and the subscript tr indicates that the energy corresponds
to translational motion of the particle. Furthermore, this equation states that the
total translational energy is the sum of translational energy along each Cartesian
dimension.
Pressure arises from the collisions of gas particles with the walls of the container;
therefore, to describe pressure, we must consider what occurs when a gas particle collides
with the wall. First, we assume that the collisions with the wall are elastic collisions,
meaning that translational energy of the particle is conserved. Although the collision is
elastic, this does not mean that nothing happens. As a result of the collision, linear momentum is imparted to the wall, which results in pressure. The definition of pressure is
force per unit area, and, by Newton’s second law, force is equal to the product of mass
and acceleration. Using these two definitions, we find that the pressure arising from the
collision of a single molecule with the wall is expressed as
vy
x
Figure 1.2
Collision between a gas particle and a
wall. Before the collision, the particle has
a momentum of mvx in the x direction,
whereas after the collision the momentum
is -mvx. Therefore, the change in particle
momentum resulting from the collision is
- 2mvx. By conservation of momentum,
the change in momentum of the wall must
be 2mvx. The incoming and outgoing trajectories are offset to show the individual
momentum components.
(1.4)
How many molecules strike the side of the container in a given period of time? To
answer this question, the time over which collisions are counted must be considered.
Consider a volume element defined by the area of the wall A multiplied by length ∆x,
as illustrated in Figure 1.3. The collisional volume element depicted in Figure 1.3 is
given by
V = A∆x
(1.5)
The length of the box ∆x is related to the time period over which collisions will be
counted ∆t and the component of particle velocity parallel to the side of the box (taken
to be the x direction):
∆x = vx ∆t
(1.6)
Area 5 A
Dx 5 vxDt
Figure 1.3
Volume element used to determine the
number of collisions with a wall per
unit time.
24
CHAPTER 1 Fundamental Concepts of Thermodynamics
In this expression, vx is for a single particle; however, an average of this quantity
will be used when describing the collisions from a collection of particles. Finally, the
number of particles that will collide with the container wall Ncoll in the time interval
∼
∆t is equal to the number density N. This quantity is equal to the number of particles
in the container N divided by the container volume V and multiplied by the collisional
volume element depicted in Figure 1.3:
nNA
1
1
∼
Ncoll = N * 1Avx ∆t2 a b =
1Avx ∆t2 a b
2
V
2
(1.7)
We have used the equality N = n NA where NA is Avogadro’s number and n is the
number of moles of gas in the second part of Equation (1.7). Because particles travel
in either the +x or -x direction with equal probability, only those molecules traveling
in the +x direction will strike the area of interest. Therefore, the total number of collisions is divided by two to take the direction of particle motion into account. Employing
Equation (1.7), we see that the total change in linear momentum of the container wall
imparted by particle collisions is given by
∆ptotal = 12mvx21Ncoll2
= 12mvx2 a
=
nNA Avx ∆t
b
V
2
nNA
A∆t m 8 v2x 9
V
(1.8)
In Equation (1.8), angle brackets appear around v2x to indicate that this quantity
represents an average value, given that the particles will demonstrate a distribution of
velocities. This distribution is considered in detail later in Chapter 13. With the total
change in linear momentum provided in Equation (1.8), the force and corresponding
pressure exerted by the gas on the container wall [Equation (1.3)] are as follows:
F =
P =
∆ptotal
nNA
=
Am 8 v2x 9
∆t
V
nNA
F
=
m 8 v2x 9
A
V
(1.9)
Equation (1.9) can be converted into a more familiar expression once 1>2 m 8v2x 9 is
recognized as the translational energy in the x direction. In Chapter 14, it will be shown
that the average translational energy for an individual particle in one dimension is
m 8v2x 9
k BT
=
2
2
(1.10)
where T is the gas temperature.
Substituting this result into Equation (1.9) results in the following expression for
pressure:
P =
Concept
The ideal gas law assumes that
individual molecules are point masses
that do not interact.
Concept
Temperature can only be measured
indirectly through a physical property
such as the volume of a gas or the
voltage across a thermocouple.
nNA
nNA
nRT
m 8 v2x 9 =
k T =
V
V B
V
(1.11)
We have used the equality NA kB = R where kB is the Boltzmann constant and R is the
ideal gas constant in the last part of Equation (1.11). The Boltzmann constant relates
the average kinetic energy of molecules to the temperature of the gas, whereas the ideal
gas constant relates average kinetic energy per mole to temperature. Equation (1.11) is
the ideal gas law. Although this relationship is familiar, we have derived it by employing a classical description of a single molecular collision with the container wall and
then scaling this result up to macroscopic proportions. We see that the origin of the
pressure exerted by a gas on its container is the momentum exchange of the randomly
moving gas molecules with the container walls.
What physical association can we make with the temperature T? At the microscopic
level, temperature is related to the mean kinetic energy of molecules as shown by
1.2 The Macroscopic Variables Volume, Pressure, and Temperature
T1x2 = a + bx
(1.12)
Equation (1.12) defines a temperature scale in terms of a specific thermometric property, once the constants a and b are determined. The constant a determines the zero of
the temperature scale because T102 = a and the constant b determines the size of a
unit of temperature, called a degree.
One of the first practical thermometers was the mercury-in-glass thermometer. This
thermometer utilizes the thermometric property that the volume of mercury increases
monotonically over the temperature range -38.8°C to 356.7°C in which Hg is a liquid.
In 1745, Carolus Linnaeus gave this thermometer a standardized scale by arbitrarily
assigning the values 0 and 100 to the freezing and boiling points of water, respectively.
Because there are 100 degrees between the two calibration points, this scale is known
as the centigrade scale.
The centigrade scale has been superseded by the Celsius scale. The Celsius scale
(denoted in units of °C) is similar to the centigrade scale. However, rather than being
determined by two fixed points, the Celsius scale is determined by one fixed reference
point at which ice, liquid water, and gaseous water are in equilibrium. This point is
called the triple point (see Section 8.2) and is assigned the value 0.01°C. On the Celsius
scale, the boiling point of water at a pressure of 1 atmosphere is 99.975°C. The size of
the degree is chosen to be the same as on the centigrade scale.
Although the Celsius scale is used widely throughout the world today, the
numerical values for this temperature scale are completely arbitrary because a liquid
other than water could have been chosen as a reference. It would be preferable to have
a temperature scale derived directly from physical principles. There is such a scale,
called the thermodynamic temperature scale or absolute temperature scale. For
such a scale, the temperature is independent of the substance used in the thermometer,
and the constant a in Equation (1.12) is zero. The gas thermometer is a practical thermometer with which the absolute temperature can be measured. A gas thermometer
contains a dilute gas under conditions in which the ideal gas law of Equation (1.11)
describes the relationship among P, T, and the molar density rm = n>V with sufficient accuracy:
P = rm RT
(1.13)
P
rm R
(1.14)
Ambient temperature
1024 s after Big Bang
1012 K
1010 K
Core of red giant star
108 K
Core of sun
Solar corona
106 K
Surface of sun
104 K
Mercury boils
H2O is liquid
102 K
Oxygen boils
Helium boils
1K
Average temperature
of universe
1022 K
3He
1024 K
superfluid
Figure 1.4
Absolute temperature displayed on a logarithmic scale. The temperature of a number
of physical phenomena is also shown.
0.1 L
2.5
2.0
Pressure (bar)
Equation (1.10). We defer the discussion of temperature at the microscopic level until
Chapter 13 and focus on a macroscopic level discussion here. Although each of us has
a sense of a “temperature scale” based on the qualitative descriptors hot and cold, a
more quantitative and transferable measure of temperature that is not grounded in individual experience is needed. The quantitative measurement of temperature is accomplished using a thermometer. For any useful thermometer, the measured temperature,
T, must be a single-valued, continuous, and monotonic function of some thermometric
system property such as the volume of mercury confined to a narrow capillary, the
electromotive force generated at the junction of two dissimilar metals, or the electrical
resistance of a platinum wire.
The simplest case that one can imagine is when T is linearly related to the value of
the thermometric property x:
25
1.5
0.2 L
1.0
0.3 L
0.4 L
0.5 L
0.6 L
0.5
Equation (1.13) can be rewritten as
T =
showing that for a gas thermometer, the thermometric property is the temperature
dependence of P for a dilute gas at constant V. The gas thermometer provides the
international standard for thermometry at very low temperatures. At intermediate
temperatures, the electrical resistance of platinum wire is the standard, and at higher
temperatures the radiated energy emitted from glowing silver is the standard. The
absolute temperature is shown in Figure 1.4 on a logarithmic scale together with
associated physical phenomena.
Equation (1.14) implies that as T S 0, P S 0. Measurements carried out by
Guillaume Amontons in the 17th century demonstrated that the pressure exerted by a fixed
amount of gas at constant V varies linearly with temperature as shown in Figure 1.5.
2200 2100 0
100 200 300 400
Temperature (Celsius)
Figure 1.5
Relationship between temperature and
pressure for a dilute gas. The pressure
exerted by 5.00 * 10-3 mol of a dilute gas
is shown as a function of the temperature
measured on the Celsius scale for different
fixed volumes. The dashed segments of lines
indicate that the data are extrapolated to
lower temperatures than could be achieved
experimentally by early investigators.
26
CHAPTER 1 Fundamental Concepts of Thermodynamics
At the time of these experiments, temperatures below -30°C were not attainable in the
laboratory. However, the P versus TC (temperature on the Celsius scale) data can be
extrapolated to the limiting TC value at which P S 0. It is found that these straight lines
obtained for different values of V intersect at a common point on the TC axis that lies
near -273°C.
The data in Figure 1.5 show that at constant V the thermometric property P varies
with temperature as
P = a + bTC
Nucleus
where a and b are experimentally obtained proportionality constants. Figure 1.5 shows
that all lines intersect at a single point, even for different gases. This suggests a unique
reference point for temperature, rather than the two reference points used in constructing the centigrade scale. The value zero is given to the temperature at which P S 0,
so that a = 0. However, this choice is not sufficient to define the temperature scale
because the size of the degree is undefined. By convention, the size of the degree on
the absolute temperature scale is set equal to the size of the degree on the Celsius scale.
With these two choices, the absolute and Celsius temperature scales are related by
Equation (1.16). The scale measured by the ideal gas thermometer is the absolute temperature scale used in thermodynamics. The unit of temperature on this scale is called
the kelvin, abbreviated K (without a degree sign):
T>K = TC >°C + 273.15
Mitochondrion
Plasma membrane
Animal cell
(a)
Nucleus
Vacuole
Cell wall Chloroplast
Mitochondrion
Plant cell
(b)
Figure 1.6
Animal and plant cells are open
systems. Sketch of (a) an animal cell
and (b) a plant cell shown with selected
organelles. The contents of the animal cell
include the cytosol fluid and the numerous
organelles (e.g., nucleus, mitochondria)
that are separated from the surroundings
by a lipid-rich plasma membrane. The
plasma membrane acts as a boundary
layer that can transmit energy and is
selectively permeable to ions and various
metabolites. A plant cell is surrounded by
a cell wall that similarly encases the cytosol and organelles, including chloroplasts
(in which photosynthesis occurs).
(1.15)
(1.16)
1.3 BASIC DEFINITIONS NEEDED TO DESCRIBE
THERMODYNAMIC SYSTEMS
In the first two sections of this chapter, we discussed the macroscopic variables pressure, volume, and temperature. With this background information, we are now ready to
introduce some important concepts used in thermodynamics. A thermodynamic system
consists of all the materials involved in the process under study. This material could be
the contents of an open beaker containing reagents, the electrolyte solution within an
electrochemical cell, or the contents of a cylinder and movable piston assembly in an
engine. In thermodynamics, the rest of the universe is referred to as the surroundings.
If a system can exchange matter with the surroundings, it is called an open system; if
not, it is a closed system. Living cells are open systems (see Figure 1.6). Both open and
closed systems can exchange energy with the surroundings. Systems that can exchange
neither matter nor energy with the surroundings are called isolated systems.
The interface between the system and its surroundings is called the boundary.
Boundaries determine if energy and mass can be transferred between the system and
the surroundings and lead to the distinction between open, closed, and isolated systems.
Consider Earth’s oceans as a system, with the rest of the universe being the surroundings. The system–surroundings boundary consists of the solid–liquid interface between
the continents and the ocean floor and the water–air interface at the ocean surface. For
an open beaker in which the system is the contents, the boundary surface is just inside
the inner wall of the beaker, and it passes across the open top of the beaker. In this case,
energy can be exchanged freely between the system and surroundings through the side
and bottom walls, and both matter and energy can be exchanged between the system
and surroundings through the open top boundary. The portion of the boundary formed
by the beaker in the previous example is called a wall. Walls can be rigid or movable
and permeable or nonpermeable. An example of a movable wall is the surface of a balloon. An example of a selectively permeable wall is the fabric used in raingear, which
is permeable to water vapor but not to liquid water.
The exchange of energy and matter across the boundary between system and
surroundings is central to the important concept of equilibrium. The system and surroundings can be in equilibrium with respect to one or more of several different system
variables such as pressure (P), temperature (T ), and concentration. Thermodynamic
equilibrium refers to a condition in which equilibrium exists with respect to P, T, and
1.3 Basic Definitions Needed to Describe Thermodynamic Systems
concentration. What conditions are necessary for a system to come to equilibrium with
its surroundings? Equilibrium is established with respect to a given variable only if that
variable does not change with time, and if it has the same value in all parts of the system and surroundings. For example, the interior of a soap bubble1 (the system) and the
surroundings (the room) are in equilibrium with respect to P because the movable wall
(the bubble) can reach a position where P on both sides of the wall is the same, and
because P has the same value throughout the system and surroundings. Equilibrium
with respect to concentration exists only if transport of all species across the boundary
in both directions is possible. If the boundary is a movable wall that is not permeable
to all species, equilibrium can exist with respect to P but not to concentration. Because
N21g2 and O21g2 cannot diffuse through the (idealized) bubble, the system and surroundings are in equilibrium with respect to P but not to concentration. Equilibrium
with respect to temperature is a special case that is discussed next.
Two systems that have the same temperature are in thermal equilibrium. We use
the concepts of temperature and thermal equilibrium to characterize the walls between
a system and its surroundings. Consider the two systems with rigid walls shown in
Figure 1.7a. Each system has the same molar density and is equipped with a pressure
gauge. If we bring the two systems into direct contact, two limiting behaviors are observed. If neither pressure gauge changes, as in Figure 1.7b, the wall has not allowed
the transfer of energy, and we refer to the wall as being adiabatic. Because P1 ≠ P2,
the systems are not in thermal equilibrium and, therefore, have different temperatures.
An example of a system surrounded by adiabatic walls is coffee in a Styrofoam cup
with a Styrofoam lid.2 Experience shows that it is not possible to bring two systems
enclosed by adiabatic walls into thermal equilibrium by bringing them into contact
because adiabatic walls insulate against the transfer of “heat.” If we push a Styrofoam
cup containing hot coffee against one containing ice water, they will not reach the same
temperature. Rely on experience at this point regarding the meaning of heat; a thermodynamic definition will be given in Chapter 2.
The second limiting case is shown in Figure 1.7c. In bringing the systems into
intimate contact, both pressures change and reach the same value after some time. We
conclude that the systems have the same temperature, T1 = T2, and say that they are in
thermal equilibrium. We refer to the walls as being diathermal. Two systems in contact separated by diathermal walls reach thermal equilibrium because diathermal walls
conduct heat. Hot coffee stored in a copper cup is an example of a system surrounded
by diathermal walls. Because the walls are diathermal, the coffee will quickly reach
room temperature.
The zeroth law of thermodynamics generalizes the experiments illustrated in
Figure 1.7 and asserts the existence of an objective temperature that can be used to
define the condition of thermal equilibrium. The formal statement of this law is as
follows:
Two systems that are separately in thermal equilibrium with a third system are
also in thermal equilibrium with one another.
The unfortunate name was assigned to the “zeroth” law because it was formulated
after the first law of thermodynamics, but logically it precedes it. The zeroth law tells
us that we can determine if two systems are in thermal equilibrium without bringing
them into contact. Imagine the third system to be a thermometer, which is defined more
precisely in the next section. The third system can be used to compare the temperatures
of the other two systems; if they have the same temperature, they will be in thermal
equilibrium if placed in contact.
1
For this example, the surface tension of the bubble is assumed to be so small that it can be set equal to zero.
This is in keeping with the thermodynamic tradition of weightless pistons and frictionless pulleys.
2
In this discussion, Styrofoam is assumed to be a perfect insulator.
27
Concept
Two systems are in thermodynamic
equilibrium if pressure, temperature,
and all concentrations are the same in
both systems.
(a)
Adiabatic wall does not
allow thermal equilibrium
(b)
(c)
Diathermal wall allows
thermal equilibrium
Figure 1.7
Comparison of adiabatic and diathermal walls. (a) Two separated systems
with rigid walls and the same molar
density have different temperatures.
(b) Two systems are brought together so
that their adiabatic walls are in intimate
contact. The pressure in each system will
not change unless heat transfer is possible.
(c) As in part (b), two systems are brought
together so that their diathermal walls are
in intimate contact. The pressures become
equal.
28
CHAPTER 1 Fundamental Concepts of Thermodynamics
1.4 EQUATIONS OF STATE
AND THE IDEAL GAS LAW
Concept
An equation of state links the
variables that define the system.
Macroscopic models in which the system is described by a set of variables are based
on experience. It is particularly useful to formulate an equation of state that relates the
state variables. A dilute gas can be modeled as consisting of point masses that do not
interact with one another; we call this an ideal gas. The equation of state for an ideal
gas was first determined from experiments by the English chemist Robert Boyle in the
17th century. If the pressure of He is measured as a function of the volume for different
values of temperature, the set of nonintersecting hyperbolas as shown in Figure 1.8 is
obtained. The curves in this figure can be quantitatively fit by the functional form
PV = aT
Pressure (106 Pa)
4
(1.17)
3
where T is the absolute temperature as defined by Equation (1.16), allowing a to be
determined. The constant a is found to be directly proportional to the mass of gas used.
It is useful to separate this dependence by writing a = nR, where n is the number of
moles of the gas and R is a constant that is independent of the size of the system. The
result is the ideal gas equation of state
2
PV = NkBT = nRT
700 K
1
200 K
0.2 0.4 0.6 0.8 1.0 1.2 1.4
Volume (1022 m3)
as derived in Equation (1.11). The equation of state given in Equation (1.18) is familiar as
the ideal gas law. Because the four variables P, V, T, and n are related through the equation
of state, any three of these variables are sufficient to completely describe the ideal gas.
Of these four variables, P and T are independent of the amount of gas, whereas V
and n are proportional to the amount of gas. A variable that is independent of the size of
the system (for example, P and T ) is referred to as an intensive variable, and one that
is proportional to the size of the system (for example, V ) is referred to as an extensive
variable. Equation (1.18) can be written in terms of intensive variables exclusively:
P = rmRT
Figure 1.8
Plot showing the relationship between
pressure and volume. Red curves
represent fixed values of temperature
for 1.00 mol of He. The interval of curves
is 100 K.
(1.18)
(1.13)
For a fixed number of moles, the ideal gas equation of state has only two independent
intensive variables: any two of P, T, and rm.
For an ideal gas mixture
PV = a ni RT
i
(1.19)
because ideal gas molecules do not interact with one another. Equation (1.19) can be
rewritten in the form
P = a
i
ni RT
= a Pi = P1 + P2 + P3 + c
V
i
(1.20)
In Equation (1.20), Pi is the partial pressure of each gas. This equation states that
each ideal gas exerts a pressure that is independent of the other gases in the mixture.
We also have
ni RT
ni RT
Pi
ni
V
V
= xi
=
=
=
(1.21)
n
ni RT
P
nRT
a V
V
i
Concept
In an ideal gas mixture, the total
pressure is the sum of the partial
pressures.
which relates the partial pressure of a component in the mixture Pi with its mole fraction,
xi = ni >n, and the total pressure P.
In the SI system of units, pressure is measured in Pascal (Pa) units, where
1 Pa = 1 N>m2. Volume is measured in cubic meters, and temperature is measured
in kelvin. However, other units of pressure are frequently used; these other units are
related to the Pascal, as indicated in Table 1.1. In this table, numbers that are not exact
have been given to five significant figures. The other commonly used unit of volume is
the liter (L), where 1 m3 = 103 L and 1 L = 1 dm3 = 10-3 m3.
1.4 Equations of State and the Ideal Gas Law
TABLE 1.1 Units of Pressure and Conversion Factors
Unit of Pressure
Symbol
Numerical Value
Pascal
Pa
1 N m-2 = 1 kg m-1 s-2
Atmosphere
atm
1 atm = 101,325 Pa (exactly)
Bar
bar
1 bar = 105 Pa
Torr or millimeters of Hg
Torr
1 Torr = 101,325>760 = 133.32 Pa
Pounds per square inch
psi
1 psi = 6,894.8 Pa
EXAMPLE PROBLEM 1.1
Before you begin a day trip, you inflate the tires on your automobile to a recommended
pressure of 3.21 * 105 Pa on a day when the temperature is -5.00°C. On your trip you
drive to the beach, where the temperature is 28.0°C. (a) What is the final pressure in the
tires, assuming constant volume? (b) Derive a formula for the final pressure, assuming
more realistically that the volume of the tires increases with increasing pressure as
Vƒ = Vi11 + g 3 Pƒ - Pi 4 2 where g is an experimentally determined constant.
Solution
a. Because the number of moles is constant,
PƒVƒ
Pi Vi Tƒ
Pi Vi
=
; Pƒ =
;
Ti
Tƒ
VƒTi
Pƒ =
b.
Pi Vi Tƒ
VƒTi
= 3.21 * 105 Pa *
PƒVi11 + g 3 Pƒ - Pi 4 2
Pi Vi
=
;
Ti
Tƒ
1273.15 + 28.02
Vi
*
= 3.61 * 105 Pa
Vi
1273.15 - 5.002
Pi Tƒ = PƒTi11 + g 3 Pƒ - Pi 4 2
P 2ƒ Ti g + PƒTi11 - Pi g2 - Pi Tƒ = 0
Pƒ =
-Ti11 - Pi g2 { 2T 2i 11 - Pi g22 + 4Ti Tƒ gPi
2Ti g
We leave it to the end-of-chapter problems to show that this expression for Pƒ
has the correct limit as g S 0.
In the SI system, the constant R that appears in the ideal gas law has the value
8.314 J K-1 mol-1, where the joule (J) is the unit of energy in the SI system. To simplify
calculations for other units of pressure and volume, the values of the constant R with
different combinations of units are given in Table 1.2.
TABLE 1.2 The Ideal Gas Constant, R,
in Various Units
R = 8.314 J K-1 mol-1
R = 8.314 Pa m3 K-1 mol-1
R = 8.314 * 10-2 L bar K-1 mol-1
R = 8.206 * 10-2 L atm K-1 mol-1
R = 62.36 L Torr K-1 mol-1
29
30
CHAPTER 1 Fundamental Concepts of Thermodynamics
EXAMPLE PROBLEM 1.2
Consider the composite system, which is held at 298 K, shown in the following figure.
Assuming ideal gas behavior, calculate the total pressure and the partial pressure of
each component if the barriers separating the compartments are removed. Assume that
the volume of the barriers is negligible.
He
2.00 L
1.50 bar
Ne
3.00 L
2.50 bar
Xe
1.00 L
1.00 bar
Solution
The number of moles of He, Ne, and Xe is given by
PV
RT
PV
nNe =
RT
PV
nXe =
RT
n = nHe
nHe =
=
=
8.314
8.314
=
8.314
+ nNe +
1.50 bar * 2.00 L
= 0.121 mol
* 10-2 L bar K-1 mol-1 * 298 K
2.50 bar * 3.00 L
= 0.303 mol
* 10-2 L bar K-1 mol-1 * 298 K
1.00 bar * 1.00 L
= 0.0403 mol
* 10-2 L bar K-1 mol-1 * 298 K
nXe = 0.464
The mole fractions are
nHe
0.121
=
= 0.261
n
0.464
nNe
0.303
=
=
= 0.653
n
0.464
nXe
0.0403
=
=
= 0.0860
n
0.464
xHe =
xNe
xXe
The total pressure is given by
P =
1nHe + nNe + nXe2RT
V
0.464 mol * 8.314 * 10-2 L bar K-1 mol-1 * 298 K
6.00 L
= 1.92 bar
=
The partial pressures are given by
PHe = xHeP = 0.261 * 1.92 bar = 0.501 bar
PNe = xNeP = 0.653 * 1.92 bar = 1.25 bar
PXe = xXeP = 0.0860 * 1.92 bar = 0.165 bar
1.5 A BRIEF INTRODUCTION TO REAL GASES
The ideal gas law provides a useful means of describing a system in terms of macroscopic parameters. However, we should also emphasize the downside of not taking the
microscopic nature of the system into account. For example, the ideal gas law only
holds for gases at low densities. In practice, deviations from the ideal gas law that occur
for real gases must be taken into account in such applications as a gas thermometer.
31
1.5 A Brief Introduction to Real Gases
P =
nRT
n2a
- 2
V - nb
V
(1.22)
This equation of state has two parameters that are substance dependent and must be
experimentally determined. The parameters b and a take the finite size of the molecules
and the strength of the attractive interaction into account, respectively. (Values of a and
b for selected gases are listed in Table 7.4.) The van der Waals equation of state is more
accurate in calculating the relationship between P, V, and T for gases than the ideal gas
law because a and b have been optimized using experimental results. However, there
are other more accurate equations of state that are valid over a wider range than the van
der Waals equation, as will be discussed in Chapter 7.
EXAMPLE PROBLEM 1.3
Van der Waals parameters are generally tabulated with either of two sets of units:
a. Pa m6 mol-2 or bar dm6 mol-2
b. m3 mol-1 or dm3 mol-1
Determine the conversion factor to convert one system of units to the other. Note that
1 dm3 = 10-3 m3 = 1 L.
Solution
Pa m6 mol-2 *
m3 mol-1 *
bar
105 Pa
*
106 dm6
= 10 bar dm6 mol-2
m6
103 dm3
= 103 dm3 mol-1
m3
Real gas
Ideal gas
rtransition
rV50
0
V(r)
For most applications, calculations based on the ideal gas law are valid to pressures
on the order of 1 bar. Real gases will be discussed in detail in Chapter 7. However, because we need to take nonideal gas behavior into account in Chapters 2 through 6, we
introduce an equation of state that is valid to greater densities in this section.
The assumptions of the ideal gas law—that atoms or molecules do not interact
with one another and can be treated as point masses—limit the range of validity of
the model. Consider the potential energy function typical for a real gas, as shown in
Figure 1.9. This figure shows the potential energy of interaction of two gas molecules
as a function of the distance between them. The intermolecular potential can be divided
into regions in which the potential energy is essentially zero 1r 7 rtransition2, negative
(attractive interaction) 1rtransition 7 r 7 rV = 02, and positive (repulsive interaction)
1r 6 rV = 02. The distance rtransition is not uniquely defined and depends on the energy
of the molecule. The value of rtransition is on the order of the molecular size.
As density is increased from very low values, molecules approach one another to
within a few molecular diameters and experience a long-range attractive van der Waals
force due to time-fluctuating dipole moments in each molecule. This strength of the
attractive interaction is proportional to the polarizability of the electron charge in a
molecule and is, therefore, substance dependent. In the attractive region, P is lower
than that calculated using the ideal gas law. This is the case because the attractive interaction brings the atoms or molecules closer than they would be if they did not interact.
At sufficiently high densities, the atoms or molecules experience a short-range repulsive interaction due to the overlap of the electron charge distributions. Because of this
interaction, P is higher than that calculated using the ideal gas law. We see that for a
real gas P can be either greater or less than the ideal gas value. Note that the potential
becomes repulsive for a value of r greater than zero. As a consequence, the volume of
a gas even well above its boiling temperature approaches a finite limiting value as P
increases. By contrast, the ideal gas law predicts that V S 0 as P S ∞ .
Given the potential energy function depicted in Figure 1.9 under what conditions
is the ideal gas equation of state valid? A real gas behaves ideally only at low densities for which r 7 rtransition, and the value of rtransition is substance dependent. The
van der Waals equation of state takes both the finite size of molecules and the attractive potential into account. It has the form
r
0
Figure 1.9
The potential energy of interaction of
two molecules or atoms as a function of
their separation r. The red curve shows
the potential energy function for an ideal
gas. At values of r less than that indicated
by the dashed line, an equation of state
more accurate than the ideal gas law
should be used. V1r2 = 0 at r = rV = 0
and as r S ∞ .
32
CHAPTER 1 Fundamental Concepts of Thermodynamics
In Example Problem 1.4, a comparison is made of the molar volume for N2 calculated at low and high pressures, using the ideal gas and van der Waals equations of state.
EXAMPLE PROBLEM 1.4
Connection
The pressure exerted by a real gas
can be larger or smaller than the
pressure of an ideal gas under the
same conditions.
a. Calculate the pressure exerted by N2 at 300. K for molar volumes of 250. L mol-1
and 0.100 L mol-1 using the ideal gas and the van der Waals equations of state. The
values of parameters a and b for N2 are 1.370 bar dm6 mol-2 and 0.0387 dm3 mol-1,
respectively.
b. Compare the results of your calculations at the two pressures. If P calculated
using the van der Waals equation of state is greater than those calculated with the
ideal gas law, we can conclude that the repulsive interaction of the N2 molecules
outweighs the attractive interaction for the calculated value of the density. A similar statement can be made regarding the attractive interaction. Is the attractive or
repulsive interaction greater for N2 at 300. K and Vm = 0.100 L?
Solution
a. The pressures calculated from the ideal gas equation of state are
P =
nRT
1 mol * 8.314 * 10-2 L bar mol-1K-1 * 300. K
=
= 9.98 * 10-2 bar
V
250. L
P =
nRT
1 mol * 8.314 * 10-2 L bar mol-1K-1 * 300. K
=
= 249 bar
V
0.100 L
The pressures calculated from the van der Waals equation of state are
P =
=
nRT
n2a
- 2
V - nb
V
1 mol * 8.314 * 10-2 L bar mol-1 K-1 * 300. K
250. L - 1 mol * 0.0387 dm3 mol-1
-
11 mol22 * 1.370 bar dm6 mol-2
1250. L22
= 9.98 * 10-2 bar
P =
1 mol * 8.314 * 10-2 L bar mol-1 K-1 * 300 K
0.100 L - 1 mol * 0.0387 dm3 mol-1
-
11 mol22 * 1.370 bar dm6 mol-2
= 270. bar
10.100 L22
b. Note that the result is identical to the result for the ideal gas law for Vm = 250. L
and that the result calculated for Vm = 0.100 L deviates from the ideal gas law
result. Because Preal 7 Pideal, we conclude that the repulsive interaction is more
important than the attractive interaction for this specific value of molar volume
and temperature.
VOCABULARY
absolute temperature scale
adiabatic
Boltzmann constant
boundary
Celsius scale
centigrade scale
closed system
diathermal
elastic collision
equation of state
equilibrium
extensive variable
gas thermometer
ideal gas
ideal gas constant
ideal gas law
intensive variable
isolated system
Conceptual Problems
kelvin
macroscopic scale
macroscopic variables
mole fraction
open system
partial pressure
surroundings
system
system variables
temperature
temperature scale
thermal equilibrium
33
thermodynamic equilibrium
thermodynamic temperature scale
thermometer
van der Waals equation of state
zeroth law of thermodynamics
KEY EQUATIONS
Equation
P =
P =
nNA
nNA
m 8v2x 9 =
kT
V
V
nRT
V
T>K = TC >°C + 273.15
P = a
i
ni RT
= a Pi = P1 + P2 + P3 + c
V
i
Significance of Equation
Equation Number
Molecular level origin of pressure
1.11
Ideal gas equation of state
1.11
Definition of absolute temperature scale relative
to Celsius scale
1.16
Relation between total and partial pressure
1.20
Pi
=
P
niRT
niRT
ni
V
V
=
=
= xi
niRT
n
nRT
a V
V
i
Relation between partial pressure, total pressure,
and mole fraction
1.21
P =
nRT
n2a
- 2
V - nb
V
Van der Waals equation of state
1.22
CONCEPTUAL PROBLEMS
Q1.1 The location of the boundary between the system and
the surroundings is a choice that must be made by the thermodynamicist. Consider a beaker of boiling water in an airtight
room. Is the system open or closed if you place the boundary
just outside the liquid water? Is the system open or closed if
you place the boundary just inside the walls of the room?
Q1.2 Real walls are never totally adiabatic. Order the following walls in increasing order with respect to being diathermal:
1-cm-thick concrete, 1-cm-thick vacuum, 1-cm-thick copper,
1-cm-thick cork.
Q1.3 Why is the possibility of exchange of matter or energy
a necessary condition for equilibrium between two systems?
Q1.4 At sufficiently high temperatures, the van der Waals
RT
equation has the form P ≈
. Note that the attractive
Vm - b
part of the potential has no influence in this expression. Justify
this behavior using the potential energy diagram in Figure 1.9.
Q1.5 The parameter a in the van der Waals equation is
greater for H2O than for He. What does this say about the
form of the potential function in Figure 1.9 for the two gases?
Q1.6 Can temperature be measured directly? Explain your
answer.
Q1.7 Give an example of two systems that are in equilibrium
with respect to only one of two state variables.
Q1.8 Explain how the ideal gas law can be deduced for the
measurements shown in Figures 1.5 and 1.8.
Q1.9 Give an example based on molecule–molecule
interactions illustrating how the total pressure upon mixing
two real gases could be different from the sum of the partial
pressures.
Q1.10 Which of the following systems are open? (a) a dog,
(b) an incandescent light bulb, (c) a tomato plant, (d) a can of
tomatoes. Explain your answers.
Q1.11 Which of the following systems are isolated? (a) a
bottle of wine, (b) a tightly sealed, perfectly insulated
thermos bottle, (c) a closed tube of toothpaste, (d) our solar
system. Explain your answers.
Q1.12 Why do the z and y components of the velocity vector
not change in the collision depicted in Figure 1.2?
34
CHAPTER 1 Fundamental Concepts of Thermodynamics
Q1.13 If the wall depicted in Figure 1.2 were a movable
piston, under what conditions would it move as a result of the
molecular collisions?
Q1.14 The mass of an He atom is less than that of an Ar atom.
Does that mean that because of its larger mass, Argon exerts
a higher pressure on the container walls than He at the same
molar density, volume, and temperature? Explain your answer.
Q1.15 Explain why attractive interactions between molecules
in gas make the pressure less than that predicted by the ideal
gas equation of state.
Q1.16 State whether the following are intensive or extensive
variables. (a) temperature, (b) pressure, (c) mass. Explain
your answers.
Q1.17 State whether the following are intensive or extensive
variables. (a) density, (b) mean square speed 8 v2x 9 , (c) volume.
Explain your answers.
NUMERICAL PROBLEMS
Section 1.2
P1.1 Devise a temperature scale, abbreviated G, for which
the magnitude of the ideal gas constant is 9.34 J G-1 mol-1.
P1.2 Suppose that you measured the product PV of 1 mol
of a dilute gas and found that PV = 23.60 L atm at 0.00°C
and 32.35 L atm at 100.°C. Assume that the ideal gas law is
valid, with T = t1°C2 + a, and that the values of R and a
are not known. Determine R and a from the measurements
provided.
Section 1.4
P1.3 A sample of propane 1C3H82 is placed in a closed
vessel together with an amount of O2 that is 2.60 times the
amount needed to completely oxidize the propane to CO2
and H2O at constant temperature. Calculate the mole fraction
of each component in the resulting mixture after oxidation
assuming that the H2O is present as a gas.
P1.4 A gas sample is known to be a mixture of ethane and
butane. A bulb having a 250.0 cm3 capacity is filled with the
gas to a pressure of 101 * 103 Pa at 20.0° C. If the weight of
the gas in the bulb is 0.3800 g, what is the mole percent of
butane in the mixture?
P1.5 One liter of fully oxygenated blood can contain and
transport 0.210 liters of O2 measured at T = 298 K and
P = 1.00 atm. Calculate the number of moles of O2 carried
per liter of blood. Hemoglobin, the oxygen transport protein
in blood, has four oxygen-binding sites. How many hemoglobin molecules are required to transport the O2 in 2.50 L
of fully oxygenated blood?
P1.6 Yeast and other organisms can convert glucose
1C6H12O62 to ethanol 1CH3CH2OH2 by a process called alcoholic fermentation. The net reaction is
C6H12O61s2 S 2C2H5OH1l2 + 2CO21g2
Calculate the mass of glucose required to produce 4.50 L of
CO2 measured at P = 1.00 atm and T = 303 K.
P1.7 A vessel contains 4.15 g H2O1 l2 in equilibrium with
water vapor at 30.°C. At this temperature, the vapor pressure
of H2O is 31.82 Torr. What volume increase is necessary for
all the water to evaporate?
P1.8 Consider a 25.5-L sample of moist air at 60.°C and
1.00 atm in which the partial pressure of water vapor is
0.131 atm. Assume that dry air has the composition 78.0 mole
percent N2, 21.0 mole percent O2, and 1.00 mole percent Ar.
a. What are the mole percentages of each of the gases in the
sample?
b. The percent relative humidity is defined as %RH =
PH2O >P *H2O, where PH2O is the partial pressure of water in
the sample and P *H2O = 0.197 atm is the equilibrium vapor
pressure of water at 60.°C. The gas is compressed at 60.°C
until the relative humidity is 100.%. What volume does the
mixture contain now?
c. What fraction of the water will be condensed if the total pressure of the mixture is isothermally increased to 75.0 atm?
P1.9 A typical diver inhales 0.450 L of air per breath and
carries a 25 L breathing tank containing air at a pressure of
275. bar. As she dives deeper, the pressure increases by 1 bar
for every 10.08 m. How many breaths can the diver take from
this tank at a depth of 40. m? Assume that the temperature
remains constant.
P1.10 Consider two connected rigid vessels separated by
a valve. The first rigid vessel has a volume of 0.600 m3 and
contains H2 at 18.0°C and a pressure of 400. * 103 Pa. The
second rigid vessel has a volume of 0.750 m3 and contains Ar
at 25.0°C at a pressure of 250. * 103 Pa. The valve separating the two vessels is opened, and both are cooled to a temperature of 17.0°C. What is the final pressure in the vessels?
P1.11 A mixture of oxygen and hydrogen is analyzed
by passing it over hot copper oxide and through a drying
tube. Hydrogen reduces the CuO according to the reaction
CuO1s2 + H21g2 S Cu1s2 + H2O1l2, and oxygen reoxidizes the copper formed according to Cu1s2 + 1>2 O21g2 S
CuO1s2. At 25°C and 750. Torr, 155.0 cm3 of the mixture
yields 68.0 cm3 of dry oxygen measured at 25°C and 750. Torr
after passage over CuO and the drying agent. What is the
original composition of the mixture?
P1.12 An athlete at high performance inhales ∼4.00 L of air
at 1.00 atm and 298 K. The inhaled air and the exhaled air
contain 0.45 and 5.9% by volume of water, respectively.
Numerical Problems
For a respiration rate of 29 breaths per minute, how many
moles of water per minute are expelled from the body
through the lungs?
P1.13 Aerobic cells metabolize glucose in the respiratory
system. This reaction proceeds according to the overall
reaction
6O21g2 + C6H12O61s2 S 6CO21g2 + 6H2O1l2
Calculate the volume of oxygen required at STP to metabolize
0.0375 kg of glucose 1C6H12O62. STP refers to standard temperature and pressure; that is, T = 273 K and P = 1.00 atm.
Assume oxygen behaves ideally at STP.
P1.14 An athlete at high performance inhales ∼4.00 L of air
at 1.0 atm and 298 K at a respiration rate of 35 breaths per
minute. If the exhaled and inhaled air contain 14.0 and 20.9%
by volume of oxygen, respectively, how many moles of oxygen per minute are absorbed by the athlete’s body?
P1.15 A mixture of 1.75 * 10-3 g of O2, 4.10 * 10-3 mol
of N2, and 6.50 * 1020 molecules of CO are placed into a
vessel of volume 4.75 L at 15.0°C.
a. Calculate the total pressure in the vessel.
b. Calculate the mole fractions and partial pressures of
each gas.
P1.16 In the absence of turbulent mixing, the partial pressure of each constituent of air would fall off with height
above sea level in the Earth’s atmosphere as Pi = P 0i e-Migz>RT
where Pi is the partial pressure at the height z, P 0i is the
partial pressure of component i at sea level, g is the acceleration of gravity, R is the gas constant, T is the absolute
temperature, and Mi is the molecular mass of the gas. As a
result of turbulent mixing, the composition of the Earth’s
atmosphere is constant below an altitude of 100 km, but the
total pressure decreases with altitude as P = P 0e-Mavegz>RT ,
where Mave is the mean molecular weight of air. At sea level,
xN2 = 0.78084, xHe = 0.00000524, and T = 300 K.
a. Calculate the total pressure at 10.5 km assuming a mean
molecular mass of 28.9 g mol-1 and that T = 300. K
throughout this altitude range.
b. Calculate the value that xN2 >xHe would have at 10.5 km
in the absence of turbulent mixing. Compare your answer
with the correct value.
P1.17 An initial step in the biosynthesis of glucose C6H12O6
is the carboxylation of pyruvic acid CH3COCOOH to form
oxaloacetic acid HOOCCOCH2COOH
CH3COCOOH1s2 + CO21g2 S HOOCCOCH2COOH1s2
If you knew nothing else about the intervening reactions
involved in glucose biosynthesis other than no further carboxylations occur, what volume of CO2 is required to produce
1.00 g of glucose? Assume P = 1 atm and T = 298. K.
P1.18 Consider the oxidation of the amino acid glycine
NH2CH2COOH to produce water, carbon dioxide, and urea
NH2CONH2:
NH2CH2COOH1s2 + 3O21g2 S
NH2CONH21s2 + 3CO21g2 + 3H2O1l2
35
Calculate the volume of carbon dioxide evolved at P = 1.00
atm and T = 305. K from the oxidation of 0.050 g of glycine.
P1.19 Assume that air has a mean molar mass of 28.9 g mol-1
and that the atmosphere has a uniform temperature of 23.0°C.
Calculate the barometric pressure in Pa in Madrid, for which
z = 667. m. Assume pressure at sea level is 105 Pa. Use the
information contained in Problem P1.16.
P1.20 When Julius Caesar expired, his last exhalation had a
volume of 375 cm3 and contained 1.00 mole percent argon.
Assume that T = 300. K and P = 1.00 atm at the location of
his demise. Assume further that T has the same value throughout the Earth’s atmosphere. If all of his exhaled Ar atoms are
now uniformly distributed throughout the atmosphere, how
many inhalations of 375 cm3 must we make to inhale one
of the Ar atoms exhaled in Caesar’s last breath? Assume
the radius of the Earth to be 6.37 * 106 m. [Hint: Calculate
the number of Ar atoms in the atmosphere in the simplified
geometry of a plane of area equal to that of the Earth’s surface.
See Problem P1.16 for the dependence of the barometric pressure on the height above the Earth’s surface.]
P1.21 Calculate the number of molecules per m3 in an ideal
gas at the standard temperature and pressure conditions of
0.00°C and 1.00 atm. Repeat the calculation for a temperature
of 0.00°C and pressure of 1.00 Pa.
P1.22 Consider a gas mixture in a 1.50@dm3 flask at 15.0°C.
For each of the following mixtures, calculate the partial pressure of each gas, the total pressure, and the composition of the
mixture in mole percent:
a. 4.50 g H2 and 3.25 g O2
b. 3.25 g N2 and 2.00 g O2
c. 4.50 g CH4 and 2.25 g NH3
P1.23 A mixture of H21g2 and NH31g2 has a volume of
250 cm3 at 20.00°C and 1 atm. The mixture is cooled to the
temperature of liquid nitrogen, at which point ammonia freezes
out and is removed, leaving in the remaining gas. Upon warming the vessel to 20.00°C and 1 atm, the volume is 115.0 cm3.
Calculate the mole fraction of NH3 in the original mixture.
P1.24 A sealed flask with a capacity of 1.15 dm3 contains
14.0 g of carbon dioxide. The flask is so weak that it will
burst if the pressure exceeds 1.00 * 106 Pa. At what
temperature will the pressure of the gas exceed the bursting
pressure?
P1.25 A balloon filled with 55.5 L of He at 15.0°C and
1.0 atm rises to a height in the atmosphere of 10 km, where
the pressure is 0.17 atm and the temperature is -43.15°C.
What is the final volume of the balloon? Assume that the
pressure inside and outside the balloon have the same value.
P1.26 Carbon monoxide competes with oxygen for binding
sites on the transport protein hemoglobin. CO can be poisonous if inhaled in large quantities. A safe level of CO in air is
50. parts per million (ppm). When the CO level increases to
800. ppm, dizziness, nausea, and unconsciousness occur,
followed by death. Assuming the partial pressure of oxygen
in air at sea level is 0.20 atm, what proportion of CO to O2
is fatal?
36
CHAPTER 1 Fundamental Concepts of Thermodynamics
P1.27 The total pressure of a mixture of oxygen and hydrogen
is 1.80 atm. The mixture is ignited and the water is removed.
The remaining gas is pure hydrogen and exerts a pressure of
0.230 atm when measured at the same values of T and V as
the original mixture. What was the composition of the original
mixture in mole percent?
P1.28 Liquid N2 has a density of 875.4 kg m-3 at its normal
boiling point. What volume does a balloon occupy at 298 K
and a pressure of 1.00 atm if 5.25 * 10-3 L of liquid N2 is
injected into it? Assume that there is no pressure difference
between the inside and outside of the balloon.
P1.29 Calculate the volume of all gases evolved by the
complete oxidation of 0.500 g of the amino acid alanine
NH2CHCH3COOH if the products are liquid water, nitrogen
gas, and carbon dioxide gas, the total pressure is 1.00 atm,
and T = 305 K.
P1.30 As a result of photosynthesis, an acre of forest
(1 acre = 4047 square meter) can take up 1000. kg of CO2.
Assuming air is 0.0412% CO2 by volume, what volume of air
is required to provide 500. kg of CO2? Assume T = 298 K
and P = 1.00 atm.
P1.31 A gas thermometer of volume 0.205 L contains 0.325 g
of gas at 759.0 Torr and 15.0°C. What is the molar mass of
the gas?
P1.32 A 550 cm3 vessel contains a mixture of Ar and Xe.
If the mass of the gas mixture is 2.150 g at 20.0°C and the
pressure is 760. Torr, calculate the mole fraction of Xe in the
mixture.
P1.33 Use L’Hôpital’s rule, lim 3 ƒ1x2>g1x24 x S 0 =
dƒ1x2>dx
lim c
d
to show that the expression derived for
dg1x2>dx x S 0
Pƒ in part (b) of Example Problem 1.1 has the correct limit as
g S 0.
P1.34 Many processes such as the fabrication of integrated
circuits are carried out in a vacuum chamber to avoid reaction
of the material with oxygen in the atmosphere. It is difficult
to routinely lower the pressure in a vacuum chamber below
8.0 * 10-11 Torr. Calculate the molar density at this pressure at
300. K. What fraction of the gas-phase molecules initially present for 1.0 atm in the chamber is present at 1.0 * 10-10 Torr?
P1.35 Approximately how many oxygen molecules arrive
each second at the mitochondrion of an active person with
a mass of 71 kg? The following data are available: Oxygen
consumption is approximately 38 mL of O2 per minute
per kilogram of body weight, measured at T = 300. K and
P = 1.00 atm. In an adult there are about 1.6 * 1010 cells
per kg body mass. Each cell contains approximately 800.
mitochondria.
P1.36 A cylinder of compressed gas contains 3.21 * 103 g
of N2 gas at a pressure of 3.75 * 107 Pa and a temperature
of 15.0°C. What volume of gas has been released into the
atmosphere if the final pressure in the cylinder is 2.35 * 105
Pa? Assume ideal behavior and that the gas temperature is
unchanged.
Section 1.5
P1.37 A real gas is in a regime in which it is accurately
described by the ideal gas equation of state. 19.0 g of the gas
occupies 50.0 L at a pressure of 0.500 atm and a temperature
of 449 K. Identify the gas.
P1.38 Calculate the pressure exerted by Ar for a molar
volume of 1.54 L mol-1 at 375 K using the van der Waals
equation of state. The van der Waals parameters a and b for
Ar are 1.355 bar dm6 mol-2 and 0.0320 dm3 mol-1, respectively. Is the attractive or repulsive portion of the potential
dominant under these conditions?
P1.39 Use the ideal gas and van der Waals equations to
calculate the pressure when 1.75 mol H2 are confined to a
volume of 2.05 L at 298 K. Is the gas in the repulsive or
attractive region of the molecule–molecule potential?
P1.40 Calculate the pressure exerted by benzene for a molar
volume of 1.75 L at 605 K using the Redlich–Kwong equation of state:
P =
=
1
RT
a
V
1V
Vm - b
2T m m + b2
1
nRT
n2a
V - nb
2T V1V + nb2
The Redlich–Kwong parameters a and b for benzene are
452.0 bar dm6 mol-2 K1>2 and 0.08271 dm3 mol-1, respectively. Is the attractive or repulsive portion of the potential
dominant under these conditions?
P1.41 Rewrite the van der Waals equation using the molar
volume rather than V and n.
FURTHER READING
Laugier, A., and Garai, J. “Derivation of the Ideal Gas Law.”
Journal of Chemical Education (2007): 1832–1833.
Eberhart, J. “The Many Faces of van der Waals’s Equation
of State.” Journal of Chemical Education 66 (1989):
906–909.
Rigby, M. “A van der Waals Equation for Nonspherical
Molecules.” Journal of Physical Chemistry 76 (1972):
2014–2016.
Soave, G. “Improvement of the van der Waals Equation of
State.” Chemical Engineering Science 39 (1984): 357–369.
MATH ESSENTIAL 2:
Differentiation and Integration
Differential and integral calculus is used extensively in physical chemistry. In this unit
we review the most relevant aspects of calculus needed to understand the chapter discussions and to solve the end-of-chapter problems.
ME2.1
The Definition and Properties
of a Function
ME2.2
The First Derivative
of a Function
ME2.1 THE DEFINITION AND PROPERTIES
ME2.3
The Chain Rule
ME2.4
The Sum and Product Rules
ME2.5
The Reciprocal Rule and the
Quotient Rule
ME2.6
Higher-Order Derivatives:
Maxima, Minima, and
Inflection Points
ME2.7
Definite and Indefinite
Integrals
OF A FUNCTION
A function ƒ is a rule that generates a value y from the value of a variable x. Mathematically, we write this as y = ƒ1x2. The set of values x over which ƒ is defined is the domain of the function. Single-valued functions have a single value of y for a given value
of x. Most functions that we will deal with in physical chemistry are single valued.
However, inverse trigonometric functions and 1 are examples of common functions
that are multivalued. A function is continuous if it satisfies these three conditions:
ƒ1x2 is defined at a
lim ƒ1x2 exists
xSa
lim ƒ1x2 = ƒ1a2
xSa
(ME2.1)
ME2.2 THE FIRST DERIVATIVE OF A FUNCTION
The first derivative of a function has as its physical interpretation the slope of the function evaluated at the point of interest. In order for the first derivative to exist at a
point a, the function must be continuous at x = a, and the slope of the function at
x = a must be the same when approaching a from x 6 a and x 7 a. For example, the
slope of the function y = x 2 at the point x = 1.5 is indicated by the line tangent to the
curve shown in Figure ME2.1.
Mathematically, the first derivative of a function ƒ1x2 is denoted dƒ1x2>dx. It is
defined by
dƒ1x2
ƒ1x + h2 - ƒ1x2
= lim
(ME2.2)
hS0
dx
h
f(x)
f(x)5x2
20
10
24
2
22
4 x
210
The symbol ƒ′1x2 is often used in place of dƒ1x2>dx. For the function of interest,
dƒ1x2
1x + h22 - 1x22
2hx + h2
= lim
=
lim
= lim
2x + h = 2x
hS0
hS0
hS0
dx
h
h
Figure ME2.1
(ME2.3)
In order for dƒ1x2>dx to be defined over an interval in x, ƒ1x2 must be continuous over
the interval. Next, we present rules for differentiating simple functions. Some of these
functions and their derivatives are as follows:
d1ax n2
= anx n - 1, where a is a constant and n is any real number
dx
d1aex2
= aex,
dx
d ln x
1
=
x
dx
where a is a constant
The function y = x2 plotted as a function of x. The dashed line is the tangent to
the curve at x = 1.5.
(ME2.4)
(ME2.5)
(ME2.6)
37
38
MATH ESSENTIAL 2 Differentiation and Integration
d1a sin x2
= a cos x,
dx
where a is a constant
d1a cos x2
= -a sin x,
dx
where a is a constant
(ME2.7)
(ME2.8)
ME2.3 THE CHAIN RULE
In this section, we deal with the differentiation of more complicated functions. Suppose
that y = ƒ1u2 and u = g1x2. From the previous section, we know how to calculate
dƒ1u2>du. But how do we calculate dƒ1u2>dx? The answer to this question is stated as
the chain rule:
dƒ1u2
dƒ1u2 du
=
dx
du dx
(ME2.9)
Several examples illustrating the chain rule follow:
d sin13x2
d sin13x2 d13x2
=
= 3 cos13x2
dx
d13x2
dx
d ax +
dx
1 -4
b
x
d ln1x 22
d ln1x 22 d1x 22
2x
2
=
= 2 =
x
dx
dx
d1x 22
x
1 -4
1
b d ax + b
x
x
1 -5
1
=
= -4 ax + b a1 - 2 b
x
dx
1
x
d ax + b
x
d ax +
(ME2.10)
(ME2.11)
(ME2.12)
d exp1ax 22
d exp1ax 22 d1ax 22
=
= 2ax exp1ax 22, where a is a constant (ME2.13)
dx
dx
d1ax 22
Additional examples of use of the chain rule include:
d 1 23x 4>3 2 >dx = 14>3223x 1>3
d 1 5e322x 2 >dx = 1522e322x
(ME2.14)
(ME2.15)
d14 sin kx2
= 4k cos x, where k is a constant
dx
d 1 23 cos 2px 2
dx
= -223p sin 2px
(ME2.16)
ME2.4 THE SUM AND PRODUCT RULES
Two useful rules in evaluating the derivative of a function that is itself the sum or product of two functions are as follows:
d3 ƒ1x2 + g1x24
dƒ1x2
dg1x2
=
+
dx
dx
dx
(ME2.17)
d1x 3 + sin x2
dx 3
d sin x
=
+
= 3x 2 + cos x
dx
dx
dx
(ME2.18)
For example,
ME2.6 Higher-Order Derivatives: Maxima, Minima, and Inflection Points
d3 ƒ1x2g1x24
dƒ1x2
dg1x2
= g1x2
+ ƒ1x2
dx
dx
dx
(ME2.19)
For example,
d3 sin1x2 cos1x24
d sin1x2
d cos1x2
= cos1x2
+ sin1x2
dx
dx
dx
= cos2 x - sin2 x
(ME2.20)
ME2.5 THE RECIPROCAL RULE
AND THE QUOTIENT RULE
How is the first derivative calculated if the function to be differentiated does not have a
simple form such as those listed in the preceding section? In many cases, the derivative
is found by using the product rule and the quotient rule given by
da
For example,
da
dc
For example,
1
b
dƒ1x2
ƒ1x2
1
= 2
dx
dx
3 ƒ1x24
1
b
sin x
1 d sin x
-cos x
= - 2
=
dx
sin x dx
sin2 x
ƒ1x2
dƒ1x2
dg1x2
d
g1x2
- ƒ1x2
g1x2
dx
dx
=
2
dx
3 g1x24
da
x2
b
sin x
2x sin x - x 2 cos x
=
dx
sin2 x
(ME2.21)
(ME2.22)
(ME2.23)
(ME2.24)
ME2.6 HIGHER-ORDER DERIVATIVES: MAXIMA,
MINIMA, AND INFLECTION POINTS
A function ƒ1x2 can have higher-order derivatives in addition to the first derivative.
The second derivative of a function is the slope of a graph of the slope of the function
versus the variable. In order for the second derivative to exist, the first derivative must
be continuous at the point of interest. Mathematically,
d 2ƒ1x2
dx
For example,
d 2 exp1ax 22
dx 2
=
2
=
d dƒ1x2
a
b
dx
dx
d3 2ax exp1ax 224
d d exp1ax 22
c
d =
dx
dx
dx
= 2a exp1ax 22 + 4a2x 2 exp1ax 22, where a is a constant
(ME2.25)
(ME2.26)
The symbol ƒ″1x2 is often used in place of d 2ƒ1x2>dx 2. If a function ƒ1x2 has a
concave upward shape 1∪2 at the point of interest, its first derivative is increasing with
x and, therefore, ƒ″1x2 7 0. If a function ƒ1x2 has a concave downward shape 1¨2 at
the point of interest, ƒ″1x2 6 0.
39
40
MATH ESSENTIAL 2 Differentiation and Integration
4
f(x)5x225x
22
f(x)
2
1
21
x
2
22
d1x 3 - 5x2
5
= 3x 2 - 5 = 0, which has the solutions x = {
= 1.291
dx
A3
24
Figure ME2.2
The maxima and minima can also be determined by graphing the derivative and finding
the zero crossings, as shown in Figure ME2.3.
Graphing the function clearly shows that the function has one maximum and one
minimum in the range specified. Which criterion can be used to distinguish between
these extrema if the function is not graphed? The sign of the second derivative, evaluated at the point for which the first derivative is zero, can be used to distinguish
between a maximum and a minimum:
ƒ1x2 = x3 − 5x plotted as a function
of x. Note that it has a maximum and a
minimum in the range shown.
f(x)
f(x)53x225
The second derivative is useful in identifying where a function has its minimum or
maximum value within a range of the variable, as shown next. Because the first derivative is zero at a local maximum or minimum, dƒ1x2>dx = 0 at the values xmax and xmin.
Consider the function ƒ1x2 = x 3 - 5x shown in Figure ME2.2 over the range
-2.5 … x … 2.5.
By taking the derivative of this function and setting it equal to zero, we find the
minima and maxima of this function in the range
10
d 2ƒ1x2
dx
5
2
d 2ƒ1x2
22
1
21
dx
x
2
2
=
=
d dƒ1x2
c
d 6 0 for a maximum
dx dx
d dƒ1x2
c
d 7 0 for a minimum
dx dx
(ME2.27)
We return to the function graphed earlier and calculate the second derivative:
25
d 21x 3 - 5x2
Figure ME2.3
dx 2
The first derivative of the function
shown in the previous figure as a
function of x.
7.5
By evaluating
d 2ƒ1x2
dx
f(x)
2.5
x
1
21
2
22.5
f(x)5x3
25.0
27.5
Figure ME2.4
2
d13x 2 - 52
d d1x 3 - 5x2
c
d =
= 6x
dx
dx
dx
at x = {
5
= {1.291
A3
(ME2.28)
(ME2.29)
we see that x = 1.291 corresponds to the minimum, and x = -1.291 corresponds to
the maximum.
If a function has an inflection point in the interval of interest, then
5.0
22
=
ƒ1x2 = x3 plotted as a function of x.
The value of x at which the tangent to the
curve is horizontal is called an inflection
point.
dƒ1x2
= 0 and
dx
d 2ƒ1x2
= 0
dx 2
(ME2.30)
An example for an inflection point is x = 0 for ƒ1x2 = x 3. A graph of this function in
the interval -2 … x … 2 is shown in Figure ME2.4. In this case,
dx 3
= 3x 2 = 0 at x = 0 and
dx
d 21x 32
= 6x = 0 at x = 0
dx 2
(ME2.31)
ME2.7 DEFINITE AND INDEFINITE INTEGRALS
In many areas of physical chemistry, the property of interest is the integral of a function
over an interval in the variable of interest. For example, the work done in expanding an
ideal gas from the initial volume Vi to the final volume Vƒ is the integral of the external
pressure Pext over the volume
xƒ
w = -
Lxi
Vƒ
Pexternal Adx = -
LVi
Pexternal dV
(ME2.32)
Equation ME2.13 defines a definite integral in which the lower and upper limits of integration are given. Geometrically, the integral of a function over an interval is the area
41
ME2.7 Definite and Indefinite Integrals
under the curve describing the function. For example, the integral 1-2.31x 3 - 5x2dx
is the sum of the areas of the individual rectangles in Figure ME2.5 in the limit within
which the width of the rectangles approaches zero. If the rectangles lie below the zero
line, the incremental area is negative; if the rectangles lie above the zero line, the incremental area is positive. In this case, the total area is zero because the total negative area
equals the total positive area.
The integral can also be understood as an antiderivative. From this point of view,
the integral symbol is defined by the relation
2.3
dƒ1x2
ƒ1x2 =
dx
L dx
L
x4
5x 2
+ C
4
2
(ME2.33)
(ME2.34)
Note the constant that appears in the evaluation of every indefinite integral. By
differentiating the function obtained upon integration, you should convince yourself
that any constant will lead to the same integrand. In contrast, a definite integral has no
constant of integration. If we evaluate the definite integral
2.3
L
-2.3
1x 3 - 5x2dx = a
x4
5x 2
x4
5x 2
+ Cb
- a + Cb
(ME2.35)
4
2
4
2
x = 2.3
x = -2.3
we see that the constant of integration cancels. The function obtained upon integration
2.3
is an even function of x, and 1-2.31x 3 - 5x2dx = 0, just as we saw in the geometric
interpretation of the integral.
Some indefinite integrals are encountered so often by students of physical chemistry that they become second nature and are recalled at will. These integrals are
directly related to the derivatives discussed in Sections ME2.2–ME2.5 and include the
following:
L
dƒ1x2 = ƒ1x2 + C
L
x ndx =
xn + 1
+ C
n + 1
dx
= ln x + C
L x
eax
+ C, where a is a constant
a
f(x)5x325x
22
f(x)
2
1
21
2
22
24
and the function that appears under the integral sign is called the integrand. Interpreting
the integral in terms of area, we evaluate a definite integral, and the interval over which
the integration occurs is specified. The interval is not specified for an indefinite integral.
The geometrical interpretation is often useful in obtaining the value of a definite integral from experimental data when the functional form of the integrand is not known.
For our purposes, the interpretation of the integral as an antiderivative is more useful.
The value of the indefinite integral 1 1x 3 - 5x2dx is that function which, when differentiated, gives the integrand. Using the rules for differentiation discussed earlier, you
can verify that
1x 3 - 5x2dx =
4
(ME2.36)
L
eax =
L
sin x dx = -cos x + C
(ME2.38)
L
cos x dx = sin x + C
(ME2.39)
(ME2.37)
Figure ME2.5
The integral of a function over a given
range corresponds to the area under
the curve. The area under the curve
is shown approximately by the green
rectangles.
x
42
MATH ESSENTIAL 2 Differentiation and Integration
Although students will no doubt be able to recall the most commonly used integrals, the primary tool for the physical chemist in evaluating integrals is a good set of
integral tables. Some commonly encountered integrals are listed below. The first group
presents indefinite integrals.
L
L
L
L
L
L
L
L
L
L
L
L
1sin ax2dx = -
(ME2.40)
1
sin ax + C
a
1cos ax2dx =
1sin 2 ax2dx =
1cos2 ax2dx =
1
cos ax + C
a
(ME2.41)
1
1
x sin 2ax + C
2
4a
(ME2.42)
1
1
x +
sin 2ax + C
2
4a
(ME2.43)
x sin2 bxdx =
x2
cos 2bx
x sin 2bx
+ C
2
4
4b
8b
(ME2.44)
x cos2 bxdx =
x2
cos 2bx
x sin 2bx
+
+
+ C
2
4
4b
8b
(ME2.45)
1x 2 sin2 ax2dx =
1 3
1
1
1
x - a x2 b sin 2ax x cos 2ax + C
3
6
4a
4a2
8a
1x 2 cos2 ax2dx =
1 3
1
1
1
x + a x2 b sin 2ax +
x cos 2ax + C
3
6
4a
4a2
8a
1sin3 ax2dx = 1x 2 sin ax2dx =
1x 2 cos ax2dx =
x meaxdx =
(ME2.48)
1a2x 2 - 22cos ax
+
2x sin ax
+ C
a2
(ME2.49)
1a2x 2 - 22sin ax
+
2x cos ax
+ C
a2
(ME2.50)
a
3
a
3
eax
1
eax
a
eax
dx
=
+
dx + C
m
m - 1 xm - 1
m - 1 L xm - 1
Lx
L
(ME2.47)
3 cos ax
cos 3ax
+
+ C
4a
12a
x meax
m
x m - 1eaxdx + C
a
aL
r 2e-ardr = -
(ME2.46)
e-ar 2 2
1a r + 2ar + 22 + C
a3
(ME2.51)
(ME2.52)
(ME2.53)
The following group includes definite integrals.
a
a
0
0
npx
mpx
npx
mpx
a
sin a
b * sin a
b dx =
cos a
b * cos a
b dx = dmn
a
a
a
a
2
L
L
where dmn is one if m = n and 0 if m ≠ n
a
L
0
c sin a
npx
npx
b d * c cos a
b d dx = 0
a
a
(ME2.54)
(ME2.55)
ME2.7 Definite and Indefinite Integrals
p
L
p
2
sin mx dx =
0
∞
0
L 2x
0
sin x
∞
dx =
∞
L
L 2x
0
x ne-axdx =
0
2
cos x
n!
a
∞
L
L
cos2 mx dx =
x 2ne-ax dx =
n+1
1#3#5#
∞
L
x 2n + 1e-ax dx =
0
∞
L
0
2
e-ax dx = a
(ME2.56)
p
A2
(ME2.57)
1a 7 0, n positive integer2
# # 12n - 12
n+1 n
2
0
2
dx =
p
2
a
p
1a 7 0, n positive integer2
Aa
n!
1a 7 0, n positive integer2
2 an + 1
p 1>2
b
4a
(ME2.58)
(ME2.59)
(ME2.60)
(ME2.61)
Commercially available software such as Mathematica™, Maple™, Matlab™, and
MathCad™ can evaluate both definite and indefinite integrals.
43
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C H A P T E R
2
Heat, Work, Internal Energy,
Enthalpy, and the First Law
of Thermodynamics
2.1
Internal Energy and the First Law
of Thermodynamics
2.2
Heat
2.3
Work
2.4
Equilibrium, Change,
and Reversibility
2.5
The Work of Reversible
Compression or Expansion
of an Ideal Gas
WHY is this material important?
2.6
The Work of Irreversible
Compression or Expansion
of an Ideal Gas
2.7
Other Examples of Work
2.8
State Functions and Path
Functions
2.9
Comparing Work for Reversible
and Irreversible Processes
Energy transformations are an important feature of many processes such as chemical
reactions and phase transformations. In this chapter, you will learn how to calculate
changes in the internal energy and enthalpy of a system and its surroundings as system
variables are changed. You will also learn how macroscopic processes such as heat and
work flow can be understood at the molecular level.
WHAT are the most important concepts and results?
Changes in the internal energy and enthalpy depend only on the initial and final states
of a system. By contrast, work and heat flow, which are responsible for changing the
energy of a system, depend on the path between the initial and final states. The heat
capacity of gases, liquids, and solids depends strongly on temperature, which can be
understood by considering the translational, rotational, and vibrational degrees of
freedom of molecules.
WHAT would be helpful for you to review for this chapter?
Because differential and integral calculus is used extensively in this chapter, it would
be helpful to review the material on calculus in Math Essential 2.
2.10 Changing the System Energy
from a Molecular-Level
Perspective
2.11 Heat Capacity
2.12 Determining ∆U and Introducing
the Sate Function Enthalpy
2.13 Calculating q, w, ∆U, and ∆H for
Processes Involving Ideal Gases
2.14 Reversible Adiabatic Expansion
and Compression of an Ideal Gas
2.1 INTERNAL ENERGY AND THE FIRST
LAW OF THERMODYNAMICS
This section focuses on the change in energy of the system and surroundings during a
thermodynamic process such as an expansion or compression of a gas. In thermodynamics, we are interested in the internal energy of a system, as opposed to the energy
associated with a system relative to a particular frame of reference. For example, a container of gas in an airplane has a kinetic energy relative to an observer on the ground.
However, the internal energy of the gas is defined relative to a coordinate system
fixed on the container that moves as the airplane moves. Viewed at a molecular level,
the internal energy can take on a number of forms such as
• the translational energy of the molecules.
• the potential energy of the constituents of the system. For example, a crystal con•
sisting of polarizable molecules will experience a change in its potential energy as
an electric field is applied to the system.
the internal energy stored in the form of molecular vibrations and rotations.
45
46
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
• the internal energy stored in the form of chemical bonds that can be released
through a chemical reaction.
• the potential energy of interaction between molecules.
The total of all these forms of energy for the system of interest is given the symbol
U and is called the internal energy.1
The first law of thermodynamics is based on the results of many experiments
showing that energy can be neither created nor destroyed, if the energies of both the
system and the surroundings are considered. This law can be formulated in a number of
equivalent forms. Our initial formulation of this law is as follows:
The internal energy U of an isolated system is constant.
This form of the first law looks uninteresting because it suggests that nothing happens
in an isolated system when viewed from outside the system. How can the first law tell us
anything about thermodynamic processes such as chemical reactions? Consider separating an isolated system into two subsystems, the system and the surroundings. When
changes in U occur in a system in contact with its surroundings, the change in total
energy∆Utot is given by
∆Utot = ∆Usys + ∆Usur = 0
(2.1)
Therefore, the first law becomes
∆Usys = - ∆Usur
Concept
The energy of a system can only be
changed by transfer of heat or work
between system and surroundings.
(2.2)
For any decrease of Usys, Usur must increase by exactly the same amount. For example,
if a gas (the system) is cooled, the temperature of the surroundings must increase.
How can the energy of a system be changed? There are many ways to alter U,
several of which are discussed in this chapter. All changes in a closed system in which
no chemical reactions or phase changes occur can be classified only as heat, work, or
a combination of both. Therefore, the internal energy of such a system can only be
changed by the flow of heat or work across the boundary between the system and surroundings. For example, U for a gas can be increased by putting it in an oven or by
compressing it. In both cases, the temperature of the system increases. This important
recognition leads to a second and more useful formulation of the first law:
∆U = q + w
(2.3)
where q and w designate heat and work, respectively. Throughout this textbook, quantities without a subscript such as ∆U, q, and w refer to the system. What do we mean by
heat and work? In the following two sections, we define these important concepts and
discuss how they differ.
The symbol ∆ is used to indicate a change that occurs as a result of an arbitrary
process. The simplest processes are those in which one of P, V, or T remains constant.
A constant temperature process is referred to as isothermal, and the corresponding
terms for constants P and V are isobaric and isochoric, respectively.
2.2 HEAT
In this and the next section, we discuss the two ways in which the internal energy of
a system can be changed. Heat,2 in the context of thermodynamics, is defined as the
quantity of energy that flows across the boundary between the system and surroundings
1
We could include other terms such as the binding energy of the atomic nuclei, but we choose to include only
the forms of energy that are likely to change in simple chemical reactions.
2
Heat is perhaps the most misused term in thermodynamics, as discussed in the article by Robert Romer (2001)
cited in Further Reading. In common usage, it is incorrectly referred to as a substance as in the phrase “Close the
door; you’re letting the heat out!” An equally inappropriate term is heat capacity (discussed in Section 2.11),
because it implies that materials have the capacity to hold heat rather than to store energy. We use the terms
heat flow or heat transfer to emphasize the transitory nature of heat. However, you should not think of heat
as a fluid or a substance.
2.3 Work
because of a temperature difference between the system and the surroundings. Heat
always flows spontaneously from regions of high temperature to regions of low temperature. Heat has several important characteristics, notably:
• Heat is transitory. That is, it only appears during a change in state of the system and
•
•
surroundings. Heat is not associated with the initial and final states of the system
and the surroundings.
The net effect of heat is to change the internal energy of the system and surroundings in accordance with the first law. If the only change in the surroundings is a
change in temperature of a reservoir, heat has flowed between the system and the
surroundings. The quantity of heat that has flowed is directly proportional to the
change in temperature of the reservoir.
The sign convention for heat is as follows. If the temperature of the system is raised,
q is positive; if it is lowered, q is negative. It is common usage to state that if q is
positive, heat is withdrawn from the surroundings and deposited in the system. If q
is negative, heat is withdrawn from the system and deposited in the surroundings.
Defining the surroundings as the rest of the universe is impractical because it is not
realistic to search through the whole universe to see if a mass has been raised or lowered and if the temperature of a reservoir has changed. Experience shows that, in general, only those parts of the universe close to the system interact with it. Experiments
can be constructed to ensure that this is the case, as shown in Figure 2.1.
Imagine that we are interested in an exothermic chemical reaction that is carried out
in a rigid sealed container with diathermal walls. We define the system as consisting
solely of the reactant and product mixture. The vessel containing the system is immersed in an inner water bath separated from an outer water bath by a container with
rigid diathermal walls. During the reaction, heat flows out of the system 1q 6 02, and
the temperature of the inner water bath increases to Tƒ. Using an electrical heater, we
can increase the temperature of the outer water bath so that at all times, Touter = Tinner .
Therefore, no heat flows across the boundary between the two water baths, and because
the container enclosing the inner water bath is rigid, no work flows across this boundary. Therefore, ∆U = q + w = 0 + 0 = 0 for the composite system made up of the
inner water bath and everything within it. This composite system is an isolated system
that does not interact with the rest of the universe. To determine q and w for the reactant and product mixture, we need to examine only the composite system and we can
disregard the rest of the universe.
2.3 WORK
Work, in the context of thermodynamics, is defined as any action that transfers energy
across the boundary between the system and surroundings; this transfer results when
a force in the surroundings causes a displacement in the system along the direction of
the force. For example, adding a mass to a spring (the system) tethered to the ceiling
causes the spring to increase in length. This process involves work. Other examples of
work are charging a battery using a current flow across an electrical potential or compressing a gas.
In each of these examples, there is a motion along the direction of the force. The
discussion throughout this chapter is limited to processes in which the system and
surroundings have no kinetic energy at the beginning and end of the process. We also
assume that there are no frictional forces between moving components in the system
or surroundings.
Just as for heat, work has several important characteristics:
• Work is transitory: it only appears during a change in state of the system and sur•
roundings. Only energy, and not work, is associated with the initial and final states
of the system.
The net effect of work is to change U of the system and surroundings in accordance
with the first law, according to Equation (2.3).
47
Concept
If heat is withdrawn from the
surroundings and deposited in the
system, q is positive.
Rest of universe
Thermometers
Outer water bath
Inner water bath
Sealed
reaction
vessel
Tinner
Touter 5 Tinner
Heating coil
Figure 2.1
Isolating a system of interest. A sealed
reaction vessel with rigid walls is
immersed in an inner water bath, which
in turn is immersed in an outer water bath.
A heater in the outer water bath is used
to always keep the temperatures of the
outer and inner water baths equal. This
allows an isolated composite system to be
created in which the surroundings to the
system of interest are limited in extent.
48
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Concept
• In principle, work can always be carried out so that a mass in the surroundings is
If in doing work, a mass in the
surroundings is lowered, w is positive.
•
•
raised or lowered in the Earth’s gravitational field; this is the signature of work as
opposed to heat. If the mass in the surroundings has been raised, w is defined to be
negative. In contrast, if the mass in the surroundings has been lowered, w is defined
to be positive. If w 7 0, it is common usage to say that work has been carried out
on the system by the surroundings; if w 6 0, we say that work has been carried out
by the system on the surroundings.
Work can be defined from a reference point in either the system or the surroundings. In the absence of friction, wsur = -w.
Work can be measured and calculated by following changes either in the system or
in the surroundings.3 Work by or on the system results in the change of height of a
mass in the surroundings, and the amount of work done is given by
w = -mg∆h
(2.4)
where ∆h is the change in height of a mass in the surroundings and g is the gravitational constant. Alternately, the work for the same process can be calculated from
changes in the system using the following equation
xƒ
w =
•
Lxi
F # dx.
(2.5)
where F is the force in the surroundings acting on the system and dx is the displacement. Note that because of the scalar product in the integral, the work will be zero
unless the force has a component along the displacement direction. The sign of
the work follows from evaluating the integral in Equation (2.5). If the vectors F
and dx point in the same direction, w is positive. In contrast, if the vectors point in
the opposite direction, w is negative. Example Problem 2.2 illustrates both ways to
calculate work.
At a molecular level, work is associated with a coordinated motion. For example, in
doing electrical work, the electrons move from a lower to a higher electrical potential. In compressing a gas, there is a concerted motion of the molecules.
Common forms of work are listed in Table 2.1 and shown in Figure 2.2.
Next, we will distinguish between two types of processes: reversible and irreversible.
After introducing these two types of processes, we will discuss how to calculate the
work of compressing or expanding an ideal gas.
TABLE 2.1 Types of Work
Equation for
System-Based Work
Types of Work
Variables, System definition
Gas expansion
and compression
Pressure in the surroundings at the system–
surroundings boundary 1Pext2, volume 1V2
The gas is the system.
-
Force 1F2, distance 1x2
The spring is the system.
w =
Surface tension 1g2, surface area 1s2
The content of the bubble is the system.
w = -
Electrical potential difference 1f2,
electrical charge 1Q2
The conductor is the system.
w =
Spring stretching and
compression
Bubble expansion and
contraction
Current passes
through conductor
3
Vƒ
LVi
Pext dV
xƒ
Lxi
F # dx
sƒ
Lsi
L0
gds
Q
f dQ′
SI Units
Pa m3 = J
Nm = J
1N m-12 1m22 = J
VC = J
The relative merits of calculating system- and surroundings-based work are discussed in the articles by
Gislason and Craig (2005, 2007) and Schmidt-Rohr (2014) cited in Further Reading.
2.4 Equilibrium, Change, and Reversibility
49
Mass
Mass
Mechanical
stops
Piston
Potential energy
of a spring
increases in both
compression
and expansion.
Mass
Pi ,Vi
Mass
Initial state
Mass
Piston
Mass
Mass
Force
vector
Pf ,Vf
Mass
Final state
Increase in energy of the
system: Temperature increase
(a) Compression work done
on a gas (the system).
The initial and final
volumes are defined by
the mechanical stops
indicated in the figure.
Increase in the surface
area of water
(b) The surface of water
(the system) is deformed
when a cylindrical mass
with a hydrophobic
coating is placed on it.
Increase in potential energy stored in the spring
(c) A spring (the system),
anchored at the top is
stretched in response
to attachment of a
spherical mass to its
lower end.
Figure 2.2
Four processes that involve work. The initial state is shown in the top row, and the final state
is shown in the bottom row. (a) Compression work done on a gas (the system). (b) The surface of
water (the system) is deformed by a mass placed on it. (c) A stretched spring (the system). (d) A
compressed spring (the system). In each process, the mass in the surroundings is lowered, showing that work has been done on the system; thus, w 7 0. The increase in energy of the system
manifests as a temperature increase in the gas (a), an increase in the surface area of the water
(b), and an increase in potential energy stored in the spring (c and d). Note that unlike a gas, the
potential energy of a spring increases in both compression and expansion. The darker red color in
the bottom diagram in (a) relative to that in the top diagram implies a higher temperature.
2.4 EQUILIBRIUM, CHANGE, AND REVERSIBILITY
Thermodynamics can only be applied to systems in internal equilibrium, and a requirement for equilibrium is that the overall rate of change of all processes such as diffusion
or chemical reaction be zero. How do we reconcile these statements with our calculations of q, w, and ∆U associated with processes in which there is a macroscopic change
in the system? To answer this question, it is important to distinguish between the system and surroundings each being in internal equilibrium, and between the system and
surroundings each being in equilibrium with one another.
(d) A spring (the system),
anchored at the bottom is
compressed as a mass is
attached to its upper end.
In parts (c) and (d), the
mass is moved horizontally
and then attached to the
spring in the inital state.
50
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
First, there is the issue of internal equilibrium. Consider a system composed of an ideal gas, which satisfies the equation of state,
P = nRT>V. All combinations of P, V, and T consistent with this
B: Pressure and
equation of state form a surface in P–V–T space, as shown in
3
temperature consistent
Figure 2.3. All points on the surface correspond to states of interwith a V = 4.0 L
nal equilibrium; that is, the system is uniform on all length scales
2
i
and is characterized by single values of T, P, and concentration.
Nonequilibrium situations cannot be represented on such a plot
1
f
because T, P, and concentration do not have unique values. An
C: Neither constant
temperature nor
0
example of a system that is not in internal equilibrium is a gas
constant volume
12.5 10 7.5 5.0 2.5
that is expanding so rapidly that the exchange of energy between
V (L)
molecules through collisions is slow compared with the rate of
expansion. In this case, different regions of the gas may have difFigure 2.3
ferent values for the density, pressure, and temperature.
A pressure–volume–temperature diagram for an ideal gas.
Next, consider a process in which the system changes from an
The surface shown satisfies the equation PV = RT; therefore,
all possible states for one mole of an ideal gas lie on this surinitial state characterized by Pi, Vi, and Ti to a final state characterface. All combinations of pressure and volume consistent with
ized by Pƒ, Vƒ, and Tƒ, as shown in Figure 2.3. If the rate of change
T = 800 K (curve A) and all combinations of pressure and
of the macroscopic variables is small, the system passes through
temperature consistent with a volume of 4.0 L (curve B) are
a succession of states that are close to and essentially indistinshown as black curves that lie in the P–V–T surface. The third
guishable from equilibrium states as it goes from the initial to the
curve corresponds to a path between an initial state i and a
final state. Such a process is called a quasi-static process. We
final state ƒ that is neither a constant temperature nor a
now visualize a process in which the system undergoes a major
constant volume path (curve C).
change in terms of a directed path consisting of a sequence of
quasi-static processes, and we distinguish between two important classes of quasistatic processes, namely, reversible and irreversible processes. The term path as used
Concept
in thermodynamics refers to a curve in the space defined by the changing variables
In a quasi-static process, the system
that represents a process.
passes through a continuous
Reversibility in a chemical system can be illustrated by a system consisting of liqsuccession of equilibrium states.
uid water in equilibrium with gaseous water that is surrounded by a thermal reservoir.
The system and surroundings are both at temperature T. An infinitesimally small increase in T results in a small increase of the amount of water in the gaseous phase and a
small decrease in the liquid phase. An equally small decrease in the temperature has the
opposite effect. Therefore, fluctuations in T give rise to corresponding fluctuations in
the composition of the system. Because an infinitesimal change in the direction of the
variable that drives the process (T in this case) causes a reversal in the direction of the
process, we say that the process is reversible.
If an infinitesimal change in the direction of the driving variable does not change
the direction of the process, we say that the process is irreversible. For example, if
a stepwise temperature increase is induced in the system discussed here using a heat
pulse, the amount of water in the gas phase increases rapidly. In this case, the composition of the system cannot be returned to its initial value by an infinitesimal temperature
decrease. Although any process that takes place at a rapid rate in the real world is irreversible, real processes can approach reversibility in the appropriate limit. For example, a slow increase in the electrical potential in an electrochemical cell can convert
Piston
reactants to products in a nearly reversible process.
1000
800
600
400
200
P (10
26
Pa)
T (K)
A: Pressure and
volume consistent
with T = 800 K
Ti ,Vi
Piston
Ti ,1/2 Vi
Initial state
Final state
Figure 2.4
An ideal gas confined in a piston and
cylinder assembly. Reversible isothermal
compression is shown in which Vƒ = Vi >2.
Reversibility is achieved by adjusting
the rate at which the beaker on top of the
piston is filled with water relative to the
evaporation rate.
2.5 THE WORK OF REVERSIBLE COMPRESSION
OR EXPANSION OF AN IDEAL GAS
Imagine reversibly compressing an ideal gas using the setup shown in Figure 2.4. In
this reversible process, the upward force on the inner face of the piston is given by
PA, where P is the gas pressure and A is the area of the piston. The downward force
in the surroundings is given by mg = Pext A, where m is the mass of the piston and
water beaker and Pext is the pressure in the surroundings at the system–surroundings
boundary. P and Pext differ only by an infinitesimal amount, and both pressures change
continuously in the course of this quasi-static reversible process. The work done on the
2.5 The Work of Reversible Compression or Expansion of an Ideal Gas
gas in moving the piston and mass from an initial height hi to a final height hƒ where
hƒ 6 hi is
hƒ
wrev compression =
Lhi
F # dx =
hƒ
Lhi
hƒ
mgdx =
Vƒ
= -
LVi
Lhi
Pext Adx
Vƒ
Pext dV = -
LVi
PdV
(2.6)
In the third integral, we have converted force to pressure, and in the fourth integral, we
have changed the integration variable to volume rather than distance. The minus sign in
the fourth integral arises because, for a compression, dV = -Adx. Because P = Pext in
a reversible process, either of these variables can be used in Equation (2.6); however,
it is convenient to use P because it can be related to other measurable variables by the
equation of state for the gas.
In order to evaluate the integral, we need to know the functional dependence of P
on V. For the example of an isothermal compression of an ideal gas,
Vƒ
wisothermal rev compression = -
LVi
Vƒ
PdV = -nRT
Vƒ
dV
= -nRT ln 7 0
Vi
LVi V
(2.7)
Because compression and expansion by the interchanges of the limits of integration
and the sign of dV, wcompression = -wexpansion. Note that because a compression process in the system is an expansion process in the surroundings, w = -wsur.
We can also calculate the work done in this process by observing changes in the
surroundings rather than the system. In this case w = -mg∆h, where m is the mass of
the piston and water beaker and ∆h is the distance that the mass moves in the process.
Both expressions for w give the same answer, as is shown in Example Problem 2.1.
EXAMPLE PROBLEM 2.1
1.00 mol of an ideal gas (the system) is confined in a piston and cylinder assembly
similar to that shown in Figure 2.2. The initial temperature is 300. K, and the initial
pressure is 0.200 bar. The surroundings are a water bath at 300. K.
a. Write an expression in terms of the system variables for the work in a reversible
process in which the volume of the gas is doubled. Can you obtain a numerical
value for w?
b. Assume that the reversible process is isothermal. Calculate w by evaluating the
integral in part (a).
Solution
Vƒ
a. wrev expansion = -
LVi
Vƒ
Pext dV = -
LVi
P1V2dV because P = Pext.
The integral can be evaluated only if the functional dependence P1V2 is known.
nRT
b. P1V2 =
For an isothermal expansion, the product nRT can be taken outside
V
the integral.
Vƒ
wrev expansion = -
LVi
Vƒ
PdV = -nRT
Vƒ
dV
= -nRT ln
Vi
LVi V
= -8.314 J mol-1 K-1 * 1.00 mol * 300. K ln 2 = -1.73 kJ
In end-of-chapter Problem P2.18, you will be asked to calculate the work for
this process by considering the movement of masses in the surroundings.
51
Concept
In a reversible process, an infinitesimal
change in the driving force can reverse
the direction of change.
52
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
2.6 THE WORK OF IRREVERSIBLE
COMPRESSION OR EXPANSION
OF AN IDEAL GAS
Imagine the irreversible compression of an ideal gas in the apparatus shown in
Figure 2.2 in which both stops are removed abruptly. The frictionless piston falls
quickly and oscillates around its final height until coming to rest.4 The height of the
mass on the piston decreases from hi to hƒ. In this case, the gas pressure will not be
uniform or well defined if the piston falls quickly enough. In addition, we cannot
calculate the work from the system variables because they do not have the same
values throughout the system. However, if we wait until the piston has come to rest,
we can calculate the work in the system from the change in potential energy in the
surroundings using
w = -wsur = -mg∆h
(2.8)
as shown in Example Problem 2.2. The change in potential energy in the surroundings
has two components: the height of the mass on the piston has changed, and part of the
surroundings at constant pressure Psur has been displaced.
hƒ
wsur =
Lhi
hƒ
mgdx +
Lhi
hƒ
Psur Adx =
Lhi
a
Vƒ
mg
+ Psur b Adx =
Pext dVsur
A
LVi
(2.9)
Note that Pext also has two components: one from the gas pressure in the surroundings and the second from the force exerted by the piston. Because dVsur = -dV,
V
w = -wsur = - 1Vi ƒPext dV for an irreversible expansion or compression. Note that
we obtained the same formula for the reversible process in which P = Pext.
EXAMPLE PROBLEM 2.2
Concept
Work for both reversible and
irreversible processes can always be
V
calculated from w = - 1V ƒPext dV .
i
1.00 mol of an ideal gas (the system) is confined in a piston and cylinder assembly
similar to that shown in Figure 2.2. The piston radius is 10.0 cm, and the total mass on
the weightless piston is 64.0 kg. The initial temperature is 300. K. The gas is irreversibly expanded at constant temperature by removing half of the mass. The surroundings
are evacuated.
a. Describe in words the evolution of the system when the mass on the piston is
reduced.
b. Calculate w for this irreversible process using Equation (2.6).
c. Calculate w for this process using Equation (2.8). Compare your result with your
results for part (b).
Solution
a. The piston moves rapidly upward and because of its inertia overshoots its equilibrium position. It oscillates back and forth around the equilibrium position until the
oscillations are damped out and the piston comes to rest with P = Pext = 1.00 bar.
In this case, Psur = 0. Because the expansion may be too rapid to allow internal
equilibrium to be maintained, the system-based work cannot in general be calculated for an irreversible expansion.
4
Even for an ideal gas in a frictionless piston and cylinder assembly, the oscillations are damped out. See the
references to Gislason and Craig (2005, 2007) in Further Reading for an explanation.
2.7 Other Examples of Work
53
b. To use Equation (2.6), we must calculate the external pressure and the initial and
final volumes.
Pi =
64.0 kg * 9.81 m s-2
mg
Force
=
=
= 2.00 * 104 Pa
Area
pr 2
p * 110.0 * 10-2 m22
Pƒ = 0.500Pi
Vi =
nRT
8.314 J mol-1 K-1 * 1.00 mol * 300. K
=
= 0.125 m3
Pi
2.00 * 104 Pa
Vƒ =
PiVi
= 2Vi = 0.250 m3
Pƒ
Mass
Pext = Pƒ = 1.00 * 104 Pa
Pi ,Vi
Vƒ
w = -
LVi
PextdV = Pext1Vƒ - Vi2
O2
H2
1
H2O
2
= -1.00 * 104 Pa * 10.250 m3 - 0.125 m32 = -1.25 * 103 J
c. To calculate w using Equation (2.8), we must first calculate the distance by which
the piston is raised in the expansion. The initial height is
hi =
Vi
pr 2
=
0.125 m3
= 3.97 m
p * 110.0 * 10-2 m22
Electrical
generator
Mass
Because the volume is doubled, ∆h = hi
Initial state
w = -mg∆h = -32.0 kg * 9.81 m s-2 * 0.125 m = -1.25 * 103 J
The results to parts (b) and (c) are the same as they must be if friction is neglected.
Note that although the initial and final states of the system are the same in
Example Problems 2.1 and 2.2, w is not the same. We will discuss the origin of
this difference in Section 2.9.
We summarize this discussion of the reversible or irreversible compression or expansion
of an ideal gas in a frictionless apparatus by stating that the work for both processes
V
can be calculated from the equation w = - 1Vi ƒPext dV. Evaluation of the integral requires that the functional dependence of Pext on V be known. Note that unlike the case of
the reversible expansion discussed in the previous section, only the surroundings-based
calculation can be carried out for an irreversible expansion or compression.
Mass
Pi,Vf
H2
1
O2
H2O
2
Electrical
generator
2.7 OTHER EXAMPLES OF WORK
An example of another important kind of work, electrical work, is shown in Figure 2.5
in which the contents of the cylinder is the system. Electrical current flows through a
conductive aqueous solution, and water undergoes electrolysis to produce H2 and O2
gas. The current is produced by a generator. As current flows, the mass that drives
the generator is lowered. In this case, the surroundings do electrical work on the
system. As a result, some of the liquid water is transformed to H2 and O2. From electrostatics, the work done in transporting a charge, Q, through an electrical potential
difference, f, is
welectrical = Qf
(2.10)
For a constant current, I, that flows for a time, t, Q = It. Therefore,
welectrical = Ift
(2.11)
Mass
Final state
Figure 2.5
A piston and cylinder assembly that
contains liquid water. The water is
electrolyzed to produce H21g2 and O21g2
by passing current between the two electrodes shown in the figure. In doing so,
work is done on the system, as shown by
the lowered mass linked to the generator.
54
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
The system also does work on the surroundings through the increase in the volume of
the gas phase at the constant external pressure Pi , as shown by the raised mass on the
piston. The total work done is
Vƒ
w = wPV + welectrical = Ift -
LVi
Pext dV = Ift - Pext1Vƒ - Vi2
(2.12)
Several additional examples of work calculations are given in Example Problems 2.3
and 2.4.
EXAMPLE PROBLEM 2.3
a. Calculate the work to expand 20.0 L of an ideal gas to a final volume of 85.0 L
against a constant external pressure of 2.50 bar. The system is the gas.
b. An air bubble in liquid water is expanded from a radius of 0.100 cm to a radius of
0.325 cm. The surface tension of water is 71.99 N m-1. Calculate the work for this
process. The system is the contents of the bubble.
c. A current of 3.20 A is passed through a heating coil for 30.0 s. The electrical
potential across the resistor is 14.5 V. Calculate the work for this process. The
system is the coil.
d. The force that a H2 molecule exerts on whatever is stretching or compressing it
through a distance x is given by F = -kx with k = 575 N m-1. The minus sign
indicates that the force opposes the motion. Calculate the work to compress the
bond 1.0 pm. The system is the molecule.
Solution
Vƒ
a. w = -
LVi
Pext dV = -Pext1Vƒ - Vi2
= -2.50 bar *
sƒ
b. w = -
Lsi
105 Pa
10-3 m3
* 185.0 L - 20.0 L2 *
= -16.3 kJ
bar
L
gds = -g4p1r 2ƒ - r 2i 2
= -4p * 71.99 N m-110.3252 cm2 - 0.1002 cm22 *
10-4 m2
cm2
= -8.65 * 10-3 J
c. w =
L0
Q
f dQ′ = fQ = fIt = 14.5 V * 3.20 A * 30.0 s = 1.39 kJ
d. In calculating the work done on the molecule, we must consider the force exerted
by the surroundings on the molecule, which is the negative of the force exerted by
the molecule on the surroundings.
w =
L
F # dx =
xƒ
L
x0
= 2.9 * 10-22 J
kxdx = c
kx 2 xƒ
575 N m-1 * x 2 -1.0 * 10
d = c
d
2 x0
2
0
-12
EXAMPLE PROBLEM 2.4
A heating coil is immersed in a 100. g sample of H2O liquid at the standard boiling
temperature of 99.61°C in an open insulated beaker on a laboratory bench at 1 bar
pressure. In this process, 10.0% of the liquid is converted to the gaseous form at a
pressure of 1 bar. A current of 2.00 A flows through the heater from a 12.0 V battery
55
2.8 State Functions and Path Functions
for 1.011 * 103 s to effect the transformation. The densities of liquid and gaseous
water under these conditions are 997 and 0.590 kg m-3, respectively.
a. It is commonly useful to replace a real process by a model that exhibits the
important features of the process. Design a model system and its surroundings,
like those shown in Figures 2.1 and 2.2, that would allow you to measure the heat
and work associated with this transformation. For the model system, define the
system and surroundings as well as the boundary between them.
b. Calculate q and w for the process.
c. How can you define the system for the open insulated beaker on the laboratory
bench such that the work is properly described?
Solution
a. The following figure shows the system that consists only of liquid initially and
gas and liquid in the final state. The cylinder walls and the piston form adiabatic
walls. The external pressure is held constant by a suitable weight.
b. In the system shown, the heat input to the liquid water can be equated with the
work done on the heating coil. Therefore,
q = Ift = 2.00 A * 12.0 V * 1.011 * 103 s = 24.3 kJ
w = -Pext1Vƒ - Vi2 = -Pext a
-105 Pa * a
rgas
10.0 * 10-3 kg
0.590 kg m-3
+
+
mliq,ƒ
rliq
-
mliq,i
rliq
90.0 * 10-3 kg
997 kg m-3
b
-
Mass
Heating coil
As the liquid is vaporized, the volume of the system increases at a constant external pressure. Therefore, the work done by the system on the surroundings is
mgas,ƒ
Pext 5 1.00 atm
100.0 * 10-3 kg
997 kg m-3
= -1.70 kJ
H2O(g)
A
12 V
H2O(l)
b
Note that the electrical work done on the heating coil is much larger than the P–V
work done in the expansion.
c. Define the system as the liquid in the beaker and the volume containing only
molecules of H2O in the gas phase. This volume will consist of disconnected
volume elements dispersed in the air above the laboratory bench.
2.8 STATE FUNCTIONS AND PATH FUNCTIONS
An alternate statement of the first law is that ∆U depends only on the initial and
final states and is independent of the path connecting the initial and final states. In
the following, this assertion is verified for kinetic energy, and the argument can be
extended to other forms of energy, such as potential energy and internal energy.
Consider a single molecule in the system. Imagine that the molecule of mass m
initially has the speed v1. We now change its speed incrementally following the
sequence v1 S v2 S v3 S v4. The change in the kinetic energy along this sequence
is given by
1
1
1
1
∆Ekinetic = a m1v222 - m1v122 b + a m 1v322 - m1v222 b
2
2
2
2
1
1
+ a m1v422 - m1v322 b
2
2
1
1
= a m1v422 - m1v122 b
2
2
(2.13)
Concept
The change in value of a state
function depends only on the initial
and final states and not on the path
between them.
56
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Mass
Mechanical
stops
Piston
P1,V1,T1
Even though v2 and v3 can take on any arbitrary values, they still do not influence the
result. We conclude that the change in the kinetic energy depends only on the initial
and final speed and that it is independent of the path between these values. Because this
conclusion holds for all molecules in the system, it also holds for ∆U.
This example supports the assertion that ∆U depends only on the final and initial
states and not on the path connecting these states. Any function that satisfies this condition is called a state function because it depends only on the state of the system and
not on the path taken to reach the state. This property can be expressed in a mathematical form. Any state function, for example, U, must satisfy the equation
ƒ
∆U =
T3
L
dU = Uƒ - Ui
(2.14)
i
Initial state
Mass
Piston
P2,V2,T2
where i and ƒ denote the initial and final states. This equation states that in order for
∆U to depend only on the initial and final states characterized here by i and ƒ, the value
of the integral must be independent of the path. If this is the case, U can be expressed
as an infinitesimal quantity, dU, that, when integrated, depends only on the initial and
final states. The quantity dU is called an exact differential. We defer a more detailed
discussion of exact differentials to Chapter 3.
It is useful to define a cyclic integral, denoted by the symbol A , as applying to a
cyclic path such that the initial and final states are identical. For U or any other state
function,
C
T3
Intermediate state
Mass
Piston
P3,V2,T3
T3
Final state
Figure 2.6
A system consisting of an ideal gas
contained in a piston and cylinder
assembly. The gas in the initial state
V1,T1 is compressed to an intermediate
state, whereby the temperature increases
to the value T2. It is then brought into
contact with a thermal reservoir at T3,
leading to a further rise in temperature.
The final state is V2,T3. The mechanical
stops allow the system volume to be only
V1 or V2. The color of the gas signifies
temperature, with the redder colors
indicating higher temperatures.
dU = Uƒ - Uƒ = 0
(2.15)
because the initial and final states are the same in a cyclic process.
Next, we will show that q and w are not state functions. The state of a single-phase
system at fixed composition is characterized by any two of the three variables P, T, and
V. The same is true of U. Therefore, for a system of fixed mass, U can be written in
any of the three forms U1V,T2, U1P,T2, or U1P,V2. Imagine that a gas of fixed mass
characterized by V1 and T1 is confined in a piston and cylinder system that is isolated
from the surroundings. There is a thermal reservoir in the surroundings at a temperature
T3 7 T1. As shown in Figure 2.6, we do compression work on the system starting from
an initial state in which the volume is V1 to an intermediate state in which the volume is
V2 where V2 6 V1 using a constant external pressure Pext . The work is given by
V2
w = -
L
V1
Pext dV = -Pext1V2 - V12 = -Pext ∆V
(2.16)
Because work is done on the system in the compression, w is positive and U increases.
Because the system consists of a uniform single phase, U is a monotonic function of T,
and T also increases. The change in volume ∆V has been chosen such that the temperature of the system T2 in the intermediate state after the compression satisfies the
inequality T1 6 T2 6 T3 .
We next lock the piston in place and let an amount of heat q flow between the system and surroundings at constant by bringing the system into contact with the reservoir
at temperature T3. The final state values of T and V after these two steps are T3 and V2.
This two-step process is repeated for different values of the mass. In each case, the
system is in the same final state characterized by the variables V2 and T3. The sequence
of steps that takes the system from the initial state V1,T1 to the final state V2,T3 is referred
to as a path. By changing the mass, a set of different paths is generated, all of which
originate from the state V1,T1, and end in the state V2,T3. According to the first law, ∆U
for this two-step process is
∆U = U1T3,V22 - U1T1,V12 = q + w
(2.17)
Because ∆U is a state function, its value for the two-step process just described is the
same for each of the different values of the mass.
2.9 Comparing Work for Reversible and Irreversible Processes
57
Are q and w also state functions? For this two-step process,
w = -Pext ∆V
(2.18)
and Pext is different for each value of the mass, or for each path, whereas ∆V is
constant. Therefore, w is also different for each path; we can choose one path from
V1,T1 to V2,T3 and a different path from V2,T3 back to V1,T1. Because the work is different along these paths, the cyclic integral of work is not equal to zero. Therefore, w is
not a state function.
Using the first law to calculate q for each of the paths, we obtain the result
q = ∆U - w = ∆U + Pext ∆V
(2.19)
Because ∆U is the same for each path and w is different for each path, we conclude that
q is also different for each path. Just as for work, the cyclic integral of heat is not equal
to zero. Therefore, neither q nor w is a state function, and they are called path functions.
Because both q and w are path functions, there are no exact differentials for work
and heat. Incremental amounts of these quantities are denoted by dq and dw, rather
than dq and dw, to emphasize the fact that incremental amounts of work and heat are
not exact differentials. Because dq and dw are not exact differentials, there are no such
quantities as ∆q, qƒ, and qi nor ∆w, wƒ, and wi. One cannot refer to the work or heat
possessed by a system or to the change in work or heat associated with a process. After
a transfer of heat and/or work between the system and surroundings is completed, the
system and surroundings possess internal energy but not heat or work.
The preceding discussion emphasizes that it is important to use the terms work and
heat in a way that reflects the fact that they are not state functions. Examples of systems
of interest to us are batteries, fuel cells, refrigerators, and internal combustion engines.
In each case, the utility of these systems is that work and/or heat flows between the
system and surroundings. For example, in a refrigerator, electrical energy is used to
extract heat from the inside of the device and to release it in the surroundings. One can
speak of the refrigerator as having the capacity to extract heat, but it would be wrong
to speak of it as having heat. In the internal combustion engine, chemical energy contained in the bonds of the hydrocarbon fuel molecules and in O2 is released in forming
CO2 and H2O. This change in internal energy can be used to rotate the
wheels of the vehicle, thereby inducing a flow of work between the vehicle and the surroundings. One can refer to the capability of the engine to
do work, but it would be incorrect to refer to the vehicle or the engine as
P1
containing or having work.
We next compare the work associated with the reversible and irreversible
isothermal expansion and compression of an ideal gas (see Figure 2.7).
Consider the following process. A quantity of an ideal gas is confined
in a cylinder with a weightless movable piston. The walls of the system are
diathermal, allowing heat to flow between the system and surroundings.
Therefore, the process is isothermal at the temperature of the surroundings, T. The system is initially defined by the variables T, P1, and V1. The
position of the piston is determined by Pext = P1, which can be changed
by adding or removing weights from the piston. Because the weights are
moved horizontally, no work is done in adding or removing them. The gas
is first expanded at constant temperature by decreasing Pext to the value P2
(weights are removed), where P2 6 P1. A sufficient amount of heat flows
into the system through the diathermal walls to keep the temperature at
the constant value T. The system is now in the state defined by T, P2, and
V2, where V2 7 V1. The system is then returned to its original state in an
Heat and work are path functions;
they are not state functions.
Total w for cycle
2.9 COMPARING WORK FOR
P2
w for
compression
step
Pext
REVERSIBLE AND IRREVERSIBLE
PROCESSES
Concept
w for expansion step
0
V1
V
V2
Figure 2.7
A two-step process involving compression followed
by expansion. The process is shown on a P–V diagram.
This type of figure is also known as an indicator diagram.
The work for each step and the total work can be obtained
from the diagram. The arrows indicate the direction
of change in V in the two steps. The sign of w is opposite
for these two processes.
58
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
isothermal process by increasing Pext to its original value P1 (weights are added). Heat
flows out of the system into the surroundings in this step. The system has been restored
to its original state and, because this is a cyclic process, ∆U = 0. The total work associated with this cyclic process is given by the sum of the work for each individual step:
wtot = a - Pext,i ∆Vi = wexpansion + wcompression = -P21V2 - V12 - P11V1 - V22
i
= -1P2 - P12 * 1V2 - V12 7 0 because P2 6 P1 and V2 7 V1
(2.20)
The relationship between P and V for the process under consideration is shown graphically in Figure 2.7, in what is called an indicator diagram. An indicator diagram is
useful because the work done in the expansion and contraction steps can be evaluated
from the appropriate area in the figure, which is equivalent to evaluating the integral
w = - 1 Pext dV. Note that the work done in the expansion is negative because ∆V 7 0,
and that done in the compression is positive because ∆V 6 0. Because P2 6 P1, the
magnitude of the work done in the compression process is greater than that done in the
expansion process and wtot 7 0. What can one say about qtot? The first law states that
because ∆U = qtot + wtot = 0, qtot 6 0.
The same cyclical process is carried out in a reversible cycle. A necessary condition
for reversibility is that P = Pext at every step of the cycle. This means that P changes
during the expansion and compression steps. The work associated with the expansion is
25
wisothermal expansion = -
Pi
L
V2
dV
= -nRT ln
V
V1
L
wisothermal compression = -nRT ln
15
P (bar)
L
P dV = -nRT
(2.21)
This work is shown schematically as the red area in the indicator diagram of Figure 2.8.
If this process is reversed and the compression work is calculated, the following
result is obtained:
20
V1
V2
(2.22)
It is seen that the magnitudes of the work in the forward and reverse processes are
equal. The total work done in this cyclical process is given by
10
5
Pext dV = -
w = wexpansion + wcompression = -nRT ln
Pf
= -nRT ln
0
5
10
15
V (L)
20
25
Figure 2.8
An indicator diagram for a reversible
process. Unlike Figure 2.7, the areas
under the P–V curves are the same in the
forward and reverse directions.
V2
V2
+ nRT ln
= 0
V1
V1
V2
V1
- nRT ln
V1
V2
(2.23)
Therefore, the work done in a reversible isothermal cycle is zero. Because ∆U = q + w
is a state function, q = -w = 0 for this reversible isothermal process. Looking at the
heights of the weights in the surroundings at the end of the process, we find that they
are the same as at the beginning of the process. To compare the work for reversible and
irreversible processes, the state variables need to be given specific values, as is done in
Example Problem 2.5.
EXAMPLE PROBLEM 2.5
In this example, 2.00 mol of an ideal gas undergoes isothermal expansion along
three different paths: (1) reversible expansion from Pi = 25.0 bar and Vi = 4.50 L
to Pƒ = 4.50 bar; (2) a single-step irreversible expansion against a constant external
pressure of 4.50 bar, and (3) a two-step irreversible expansion consisting initially of an
expansion against a constant external pressure of 11.0 bar until P = Pext, followed by
an expansion against a constant external pressure of 4.50 bar until P = Pext.
Calculate the work for each of these processes. For which of the irreversible
processes is the magnitude of the work greater?
2.9 Comparing Work for Reversible and Irreversible Processes
Solution
The processes are depicted in the following indicator diagram:
25
Pi
P (bar)
20
15
10
5
Pf
0
5
10
15
20
25
V (L)
We first calculate the constant temperature at which the process is carried out, the final
volume, and the intermediate volume in the two-step expansion:
T =
PiVi
25.0 bar * 4.50 L
=
= 677 K
nR
8.314 * 10-2 L bar mol-1 K-1 * 2.00 mol
Vƒ =
nRT
8.314 * 10-2 L bar mol-1 K-1 * 2.00 mol * 677 K
=
= 25.0 L
Pƒ
4.50 bar
Vint =
nRT
8.314 * 10-2 L bar mol-1 K-1 * 2.00 mol * 677 K
=
= 10.2 L
Pint
11.0 bar
The work of the reversible process is given by
w = -nRT1 ln
Vƒ
Vi
= -2.00 mol * 8.314 J mol-1 K-1 * 677 K * ln
25.0 L
= -19.3 * 103 J
4.50 L
We next calculate the work of the single-step and two-step irreversible processes:
wsingle = -Pext ∆V = -4.50 bar *
= -9.23 * 103 J
wtwo-step = -Pext ∆V = -11.0 bar *
-4.50 bar *
105 Pa
10-3 m3
* 125.00 L - 4.50 L2 *
bar
L
105 Pa
10-3 m3
* 110.2 L - 4.50 L2 *
bar
L
105 Pa
10-3 m3
* 125.00 L - 10.2 L2 *
bar
L
= -12.9 * 103 J
The magnitude of the work is greater for the two-step process than for the single-step
process, but less than that for the reversible process.
59
60
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Pressure
Pressure
Comparison of work performed under
irreversible and reversible conditions.
The graphs show (a) irreversible expansion (green areas) and (b) reversible compression (orange and yellow) work. Note
that the total work done in the irreversible
expansion and compression processes
approaches that of the reversible process
as the number of steps becomes large.
Pressure
Figure 2.9
Pressure
The amount of work that can be
extracted by a process is a maximum
if the process is reversible.
Pressure
Concept
Pressure
Volume
Volume
Volume
(a) Work done in reversible expansion (red + yellow area) compared with work done in a
multistep series of irreversible expansion processes at constant pressure (yellow area)
Volume
Volume
Volume
(b) Area under black curve is reversible compression work. Green area is work done in a
multistep series of irreversible compression processes at constant pressure
Example Problem 2.5 shows that the magnitude of w for the irreversible expansion
is less than that for the reversible expansion, but it also suggests that the magnitude
of w for a multistep expansion process increases with the number of steps. This is indeed the case, as shown in Figure 2.9. Imagine that the number of steps n increases
indefinitely. As n increases, the pressure difference Pext - P for each individual step
decreases. In the limit that n S ∞, the pressure difference Pext - P S 0, and the total
area of the rectangles in the indicator diagram approaches the area under the reversible
curve. In this limit, the irreversible process becomes reversible, and the value of the
work equals that of the reversible process.
By contrast, the magnitude of the irreversible compression work exceeds that of
the reversible process for finite values of n and becomes equal to that of the reversible
process as n S ∞. The difference between the expansion and compression processes
results from the requirement that Pext 6 P at the beginning of each expansion step,
whereas Pext 7 P at the beginning of each compression step.
On the basis of these calculations for the reversible and irreversible cycles, we introduce another criterion to distinguish between reversible and irreversible processes.
Suppose that a system undergoes a change through one or more individual steps, and
that the system is restored to its initial state by following the same steps in reverse
order. The system is restored to its initial state because the process is cyclical. If the
surroundings are also returned to their original state (all masses at the same height and
all reservoirs at their original temperatures), the process is reversible. If the surroundings are not restored to their original state, the process is irreversible.
2.10 Changing the System Energy from a Molecular-Level Perspective
We are often interested in extracting work from a system. For example, it is the
expansion of the fuel–air mixture in an automobile engine upon ignition that provides
the torque that eventually drives the wheels. Is the capacity to do work similar for
reversible and irreversible processes? This question is answered using the indicator
diagrams of Figures 2.7–2.9 for the specific case of isothermal expansion work, noting
that the work can be calculated from the area under the P–V curve. We compare the
work for expansion from V1 to V2 in a single stage at constant pressure to that for the
reversible case. For the single-stage expansion, the constant external pressure is given
by Pext = nRT>V2. However, if the expansion is carried out reversibly, the system pressure is always greater than this value. By comparing the areas in the indicator diagrams
of Figure 2.9, we see that
0 wrev 0 Ú 0 wirr 0
(2.24)
By contrast, for the compression step,
0 wrev 0 … 0 wirr 0
(2.25)
The reversible work is the lower bound for the compression work and the upper bound
for the expansion work. This result for the expansion work can be generalized to an
important statement that holds for all forms of work: The maximum work that can be
extracted from a process between the same initial and final states is that obtained
under reversible conditions.
Although the preceding statement is true, it suggests that it would be optimal to
operate an automobile engine under conditions in which the pressure inside the cylinders differs only infinitesimally from the external atmospheric pressure. This is clearly
not possible. A practical engine must generate torque on the drive shaft, and this can
only occur if the cylinder pressure is appreciably greater than the external pressure.
Similarly, a battery is only useful if one can extract a sizable rather than an infinitesimal
current. To operate such devices under useful irreversible conditions, the work output
is less than the theoretically possible limit set by the reversible process.
2.10 CHANGING THE SYSTEM ENERGY FROM
A MOLECULAR-LEVEL PERSPECTIVE
Our discussion so far has involved changes in energy for macroscopic systems. But
what happens at the molecular level if energy is added to the system? In shifting to
a molecular perspective, we move from a classical to a quantum-mechanical description of matter. First, in general the energy levels of quantum-mechanical systems are
discrete, and molecules can only possess amounts of energy that correspond to these
values. By contrast, in classical mechanics, energy is a continuous variable. Second,
we use a result from statistical mechanics showing that the relative probability of a
molecule being in a state corresponding to the allowed energy values e1 and e2 at temperature T is given by e-13e2 - e14>kBT2. This result will be discussed in Chapter 13.
To keep the mathematics simple, consider an idealized model system: a gas consisting of a single He atom confined in a one-dimensional container with a length of 8.0 nm.
The principles of quantum mechanics inform us that the translational energy of the He
atom confined in this box can only have the discrete values shown in Figure 2.10b. We
lower the temperature of this one-dimensional He gas to 0.20 K. The calculated probability of the atom being in a given energy level is shown in Figure 2.10b. If we now
do work on the surroundings by expanding the box at constant temperature, the energy
levels will change to those shown in Figure 2.10a. Note that all the energy levels are
shifted to lower values as the container is made larger. If we keep the temperature constant at 0.20 K during this expansion, the distribution of the atoms among the energy
levels changes to that shown in Figure 2.10a. This redistribution occurs because the
total energy of the He atom remains constant in the expansion because the temperature
is kept constant.
61
62
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Figure 2.10
Arrows show how
energy of each level
decreases as length of
box increases
1.4 3 10223
1.2 3 10223
Energy (joules)
Energy levels for a He atom confined
to a one-dimensional box. Boxes are
of length (a) 10.0 nm and (b) 8.00 nm.
Circles indicate the probability that the
He atom has an energy corresponding to
each of the energy levels at 0.20 K. Each
circle indicates a probability of 0.010. For
example, the probability that the energy
of the He atom corresponds to the lowest
energy level in the 10.0 nm box is 0.21.
1 3 10223
8 3 10224
6 3 10224
4 3 10224
2 3 10224
(a) a 5 10.0 nm
Concept
At the molecular level, work is
associated with changes in the
allowed energy levels, whereas heat
is associated with changes in the
population of the energy levels.
(b) a 5 8.00 nm
What happens if we keep the container at the longer length and increase the system
energy by heat flow so that T increases to 0.30 K? In this case, the energy levels are
unchanged because they depend on the container length but not on the temperature of
the gas. However, the energy of the system increases, and, as shown in Figure 2.11,
this increase occurs by a redistribution of the probability of finding the He atom in the
energy levels. The increase in system energy comes from an increase in the probability
of the He atom populating higher-energy levels and a corresponding decrease in the
probability of populating lower-energy levels.
Just as for a macroscopic system, the quantum-mechanical system energy increases
through heat or work flow. However, in doing work, the energy levels change, whereas
in transferring heat, the energy level remains constant and their population changes.
Quantum-mechanical effects become clearer in the next section when we consider how
molecules can take up energy through rotation and vibration.
1.4 3 10223
Energy (joules)
1.2 3 10223
Figure 2.11
Energy levels for the 10.0 nm box
described in Figure 2.10 caption.
Circles indicate the probability that a He
atom has an energy corresponding to each
of the energy levels at (a) 0.200 K and
(b) 0.300 K. Each circle indicates a
probability of 0.010.
1 3 10223
8 3 10224
6 3 10224
4 3 10224
2 3 10224
(a) 0.200 K
(b) 0.300 K
2.11 Heat Capacity
63
2.11 HEAT CAPACITY
We can quantify heat flow in terms of the easily measured electrical work done on a
heating coil immersed in the system, with the equation w = Ift. The response of a
single-phase system of constant composition to heat input is an increase in T, as long as
the system does not undergo a phase change such as the vaporization of a liquid.
The thermal response of a system to heat flow is described by a very important thermodynamic property called the heat capacity, which is a measure of energy needed to
change the temperature of a substance by a given amount. The symbol for heat capacity
is C. The name heat capacity is unfortunate because it implies that a substance has the
capacity to absorb heat. A much better name would be energy capacity.
Heat capacity is a material- and temperature-dependent property, as will be discussed later. Mathematically, heat capacity is defined by the relation
C1T2 = lim
∆T S 0 Tƒ
q
dq
=
- Ti
dT
(2.26)
where C is in the SI unit of J K-1. It is an extensive quantity that doubles as the number
of atoms or molecules in the system is doubled. Commonly, the molar heat capacity Cm
is used in calculations. It is an intensive quantity with the units of J K-1 mol-1. Experimentally, the heat capacity of fluids is measured by immersing a heating coil in the gas
or liquid and equating the electrical work done on the coil with the heat flow into the
sample. For solids, the heating coil is wrapped around the solid.
The value of the heat capacity depends on the experimental conditions under which
it is determined. The most common conditions are constant V or P, for which the heat
capacity is denoted CV and CP, respectively. Values of CP,m at 298.15 K for pure substances are tabulated in Table 2.3 (see Appendix A, Data Tables), and formulas for
calculating CP,m at other temperatures for selected gases, liquids, and solids are listed
in Tables 2.4–2.6.
Next, let us consider heat capacity using a molecular-level model, beginning with
gases. Figure 2.12 illustrates the energy-level structure for a molecular gas. Molecules
can absorb energy by moving faster, by rotating in three-dimensional space, by periodic oscillations (known as vibrations) of the atoms around their equilibrium structure,
and by electronic excitations. These energetic degrees of freedom are referred to as
translation, rotation, vibration, and electronic excitation. Each of the degrees of freedom has its own set of energy levels, and the probability of an individual molecule
occupying a higher-energy level increases as it gains energy. The energy levels for
atoms and molecules associated with rotation, vibration, and electronic excitation are
independent of the container size. In contrast, the energy levels for translation are
dependent on container size.
Concept
The heat capacity of a substance
depends on the values of its rotational
and vibrational energy levels, the
temperature, and whether it is a gas,
liquid, or solid.
DEelectronic
DEvibration
DEtranslation
DErotation
Figure 2.12
Vibrational
energy
levels
Electronic
energy
levels
Rotational
energy
levels
Translational
energy
levels
Schematic depiction of energy levels for
each degree of freedom of a molecule.
The dashed line between electronic energy
levels on the left indicates what appear
to be a continuous range of energies.
However, as the energy scale is magnified stepwise, discrete energy levels for
vibration, rotation, and translation can be
resolved.
64
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
TABLE 2.2 Energy-Level
Spacings for Degrees
of Freedom
Degree of
Freedom
Typical Energy-Level
Spacing
Electronic
∼ 5 * 10-19 J
Vibration
∼ 2 * 10-20 J
Rotation
∼ 2 * 10-23 J
Translation
∼ 2 * 10-41 J
The amount of energy needed to move up the ladder of energy levels is very different for the different degrees of freedom: ∆Eelectronic W ∆Evibration W ∆Erotation W
∆Etranslation. Values for these ∆E are molecule dependent, but order-of-magnitude
numbers are shown in Table 2.2.
A molecule gains or loses energy through collisions with other molecules. An
order-of-magnitude estimate of the energy that can be gained or lost by a molecule in
a collision is kBT, where kB = R>NA is the Boltzmann constant and T is the absolute
temperature. A given degree of freedom in a molecule can only take up energy through
molecular collisions if the spacing between adjacent energy levels and the temperature
satisfies the relationship ∆E ≈ kBT, which has the value 4.1 * 10-21 J at 300. K. At
300. K, ∆E ≈ kBT is always satisfied for translation and rotation, but not for vibration
and electronic excitation. We formulate the following general rule that relates the heat
capacity CV,m and the degrees of freedom in a molecule, which will be discussed in
more detail in Chapter 15.
The heat capacity CV,m for a gas at temperature T not much lower than 300. K
is R>2 for each translation and rotational degree of freedom, where R is the
ideal gas constant. Each vibrational degree of freedom for which the relation
∆E>kBT 6 0.1 is obeyed contributes R to CV,m. If ∆E>kBT 7 10, the degree of
freedom does not contribute to CV,m. For 10 7 ∆E>kBT 7 0.1, the degree of
freedom contributes partially to CV,m.
80
C2H4
CV, m ( J K21mol21 )
70
60
50
40
CO2
30
20
CO
10
He
0
200 400 600 800 1000
Temperature (K)
Figure 2.13
Molar heat capacity CV,m for selected
gases as a function of temperature.
Compared with molecules, atoms have
only translational degrees of freedom and,
therefore, have comparatively low values
for CV,m that are independent of temperature. Molecules with vibrational degrees
of freedom have higher values of CV,m at
temperatures sufficiently high to activate
the vibrations.
Figure 2.13 shows the variation of CV,m for a monatomic gas and several molecular gases. Atoms only have 3 translational degrees of freedom. Linear molecules have
an additional 2 rotational degrees of freedom and 3n@5 vibrational degrees of freedom
where n is the number of atoms in the molecule. Nonlinear molecules have 3 translational degrees of freedom, 3 rotational degrees of freedom, and 3n@6 vibrational
degrees of freedom. A He atom has only 3 translational degrees of freedom, and all
electronic transitions are of high energy compared to kT. Therefore, CV,m = 3R>2
over the entire temperature range, as shown in the figure. CO is a linear diatomic molecule that has two rotational degrees of freedom for which ∆E>kBT 6 0.1 at 200. K.
Therefore, CV,m = 5R>2 at 200. K. The single vibrational degree of freedom begins
to contribute to CV,m above 200. K but does not contribute fully for T 6 1000. K because 10 7 ∆E>kBT below 1000. K. CO2 has 4 vibrational degrees of freedom, some
of which contribute to CV,m near 200. K. However, CV,m does not reach its maximum
value of 13R>2 below 1000. K. Similarly, CV,m for C2H4, which has 12 vibrational
degrees of freedom, does not reach its maximum value of 15 R below 1000. K because
10 7 ∆E>kBT for some vibrational degrees of freedom. Electronic energy levels are
too far apart for any of the molecular gases to contribute to CV,m.
To this point, we have only discussed CV,m for gases. It is easier to measure CP,m
than CV,m for liquids and solids because liquids and solids generally expand with
increasing temperature and exert enormous pressure on a container at constant volume
(see Example Problem 3.2). An example of how CP,m depends on T for solids and
liquids is presented in Figure 2.14 for Cl2.
To make the functional form of CP,m1T2 understandable, we briefly discuss the
relative magnitudes of CP,m in the solid, liquid, and gaseous phases using an energylevel model. A solid can be thought of as a set of interconnected harmonic oscillators,
and heat uptake leads to the excitations of the collective vibrations of the solid, which
have much closer spaced energy levels than those for individual molecules. At very
low temperatures, these vibrations cannot be activated because the spacing of the
vibrational energy levels is large compared to kBT. Therefore, CP,m approaches zero as
T approaches zero.
For the solid, CP,m rises rapidly with T because the thermal energy available as T
increases is sufficient to activate the vibrations of the solid. The heat capacity increases
discontinuously as the solid melts to form a liquid. This is the case because the liquid
65
2.11 Heat Capacity
Tsys,ƒ
qP =
L
Tsys,i
Tsur,ƒ
CP1system,T2dT = -
L
Tsur,i
CP1surroundings,T2dT
(2.27)
By measuring the temperature change of a thermal reservoir in the surroundings at
constant pressure, qP can be determined as illustrated in Example Problem 2.6. In
Equation (2.27), the heat flow at constant pressure has been expressed from the perspective of both the system and the surroundings. A similar equation can be written
for a constant volume process. Water is a convenient choice of material for a heat
bath in experiments because CP is nearly constant at the value 4.18 J g -1 K-1 or
75.3 J mol-1 K-1 over the range from 0°C to 100.°C.
EXAMPLE PROBLEM 2.6
The volume of a system consisting of an ideal gas decreases at constant pressure. As a
result, the temperature of a 1.50 kg water bath in the surroundings increases by 14.2°C.
Calculate qP for the system.
Solution
Tsys,ƒ
qP =
L
Tsys,i
Tsur,ƒ
CP1system,T2dT = -
L
Tsur,i
CP1surroundings,T2dT
= -1.50 kg * 4.18 J g -1 K-1 * 14.2 K = -89.1 kJ
How are CP and CV related for a gas? Consider the processes shown in Figure 2.15
in which a fixed amount of heat flows from the surroundings into a gas. In the constant pressure process, the gas expands as its temperature increases. Therefore, the
system does work on the surroundings. As a consequence, not all the heat flow into
the system can be used to increase ∆U. No such work occurs for the corresponding constant volume process, and all the heat flow into the system can be used to
increase ∆U. Therefore, dTP 6 dTV for the same heat flow dq. For this reason,
CP 7 CV for gases.
The same argument applies to liquids and solids as long as V increases with T.
Nearly all substances follow this behavior, although notable exceptions occur, such
as liquid water between 0°C and 4°C, for which the volume increases as T decreases.
However, because ∆Vm upon heating is much smaller than for a gas, the difference
between CP,m and CV,m for a liquid or solid is much smaller than for a gas.
The preceding remarks about the difference between CP and CV have been qualitative in nature. However, the following quantitative relationship, which will be proved
in Chapter 3, holds for an ideal gas:
CP - CV = nR or CP,m - CV,m = R
(2.28)
60
50
Cp,m (J K21 mol21)
retains all the local vibrational modes of the solid, and more low-energy modes become
available upon melting. Therefore, the heat capacity of the liquid is greater than that
of the solid. As the liquid vaporizes, the local vibrational modes present in the liquid
are converted to translations that cannot take up as much energy as vibrations. Thus,
CP,m decreases discontinuously at the vaporization temperature. The heat capacity in
the gaseous state increases slowly with temperature as the vibrational modes of the
individual molecules are activated, as discussed previously. These changes in CP,m can
be calculated for a specific substance using a microscopic model and statistical thermodynamics, as will be discussed in detail in Chapter 15.
Once the heat capacity of a variety of different substances has been determined,
we have a convenient way to quantify heat flow. For example, at constant pressure, the
heat flow between the system and surroundings can be written as
40
30
20
Solid
Liquid
Gas
10
100
200
300
Temperature (K)
Figure 2.14
Variation of CP,m with temperature
for phases of Cl2. The colored fields
correspond to the solid, liquid, and gas
phases.
400
66
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
w
Figure 2.15
Comparison of constant pressure
and constant volume heating of a gas
in a piston and cylinder assembly.
(a) Constant pressure heating. Note that
not all the heat flow into the system can
be used to increase ∆U in a constant
pressure process, because the system does
work on the surroundings as it expands.
(b) Constant volume heating. Note that no
work is done for constant volume heating.
A redder color of the gas corresponds to a
higher temperature.
Mass
Piston
Mass
Piston
Pi,Ti
Pi,Tf
Initial state
Final state
q
(a) Constant pressure heating
Concept
For a given substance, for the
gaseous phase, CP 7 CV. For the
liquid or solid state, CP ≈ CV .
q
Vi,Ti
Vi,Tf
Initial state
Final state
(b) Constant volume heating
2.12 DETERMINING 𝚫U AND INTRODUCING
THE STATE FUNCTION ENTHALPY
Concept
∆U can be measured by carrying out
a process at constant volume and
determining q, ∆U = qV .
Measuring the energy taken up or released in a chemical reaction is of particular interest to chemists. How can ∆U for a thermodynamic process be measured? Although
these measurements will be the topic of Chapter 4, we briefly discuss such measurements here in order to enable you to carry out calculations on ideal gas systems in the
end-of-chapter problems. The first law states that ∆U = q + w. Imagine that the process is carried out under constant volume conditions and that nonexpansion work is not
possible. Because under these conditions w = - 1 Pext dV = 0,
∆U = qV
(2.29)
Equation (2.29) states that ∆U can be experimentally determined by measuring the
heat flow between the system and surroundings in a constant volume process.
Chemical reactions are generally carried out under constant pressure rather than
constant volume conditions. It would be useful to have an energy state function that has
a relationship analogous to Equation (2.29), but at constant pressure conditions. Under
constant pressure conditions, we can write
dU = dqP - Pext dV = dqP - P dV
(2.30)
Integrating this expression between the initial and final states:
ƒ
L
i
dU = Uƒ - Ui =
L
dqP -
L
P dV = qP - P1Vƒ - Vi2
= qP - 1PƒVƒ - PiVi2
(2.31)
2.13 CALCULATING q, w, ∆U, AND ∆H FOR PROCESSES INVOLVING IDEAL GASES
67
Note that in order to evaluate the integral involving P, we must know the functional
relationship P1V2, which in this case is Pi = Pƒ = P where P is constant. Rearranging
the last equation, we obtain
1Uƒ + PƒVƒ2 - 1Ui + PiVi2 = qP
(2.32)
H K U + PV
(2.33)
Because P, V, and U are all state functions, U + PV is a state function. This new state
function is called enthalpy and is given the symbol H.
As is the case for U, H has the units of energy, and it is an extensive property. As
shown in Equation (2.34), ∆H for a process involving only P–V work can be determined by measuring the heat flow between the system and surroundings at constant
pressure:
∆H = qP
(2.34)
This equation is the constant pressure analogue of Equation (2.29). Because chemical
reactions are conducted at constant P much more commonly than at constant V, the
energy change measured experimentally by monitoring the heat flow is ∆H rather than
∆U. We classify a chemical reaction as being exothermic if heat is liberated to the
surroundings 1∆H 6 02 or endothermic if heat flows from the surroundings into the
system 1∆H 7 02. Combustion is an example of an exothermic reaction, and melting
of a solid is an example of an endothermic process.
2.13 CALCULATING q, w, 𝚫U, AND 𝚫H FOR
PROCESSES INVOLVING IDEAL GASES
In this section, we discuss how ∆U and ∆H, as well as q and w, can be calculated
from the initial and final state variables if the path between the initial and final state
is known. The problems at the end of this chapter ask you to calculate q, w, ∆U, and
∆H for simple and multistep processes. Because an equation of state is often needed
to complete such calculations, the system will generally be an ideal gas. Using an ideal
gas as a surrogate for more complex systems has the significant advantage that the
mathematics is simplified, allowing one to concentrate on the process rather than the
manipulation of equations and the evaluation of integrals.
What does one need to know to calculate ∆U? The following discussion is restricted
to processes that do not involve chemical reactions or changes in phase. Because U is
a state function, ∆U is independent of the path between the initial and final states. To
describe a fixed amount of an ideal gas (i.e., n is constant), the values of two of the
variables P, V, and T must be known. Is this also true for ∆U for processes involving
ideal gases? To answer this question, consider the expansion of an ideal gas from an
initial state V1,T1 to a final state V2,T2. We first assume that U is a function of both V
and T. Is this assumption correct? Because ideal gas atoms or molecules do not interact
with one another, U will not depend on the distance between the atoms or molecules.
Therefore, U is not a function of V, and we conclude that ∆U must be a function of T
only for an ideal gas, ∆U = ∆U1T2.
We also know that for a temperature range over which CV is constant,
∆U = qV = CV 1Tƒ - Ti2
(2.35)
∆H = ∆U1T2 + ∆1PV2 = ∆U1T2 + ∆1nRT2 = ∆H1T2
(2.36)
Is this equation only valid for constant V? Because U is a function of only T for an ideal
gas, Equation (2.35) is also valid for processes involving ideal gases in which V is not
constant. Therefore, if one knows CV, T1, and T2, ∆U can be calculated, regardless of
the path between the initial and final states.
How many variables are required to define ∆H for an ideal gas? We write
Concept
∆H can be measured by carrying out
a process at constant pressure and
determining q, ∆H = qP .
68
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
We see that ∆H is also a function of only T for an ideal gas. In analogy to Equation (2.35),
∆H = qP = CP1Tƒ - Ti2
(2.37)
Because ∆H is a function of only T for an ideal gas, Equation (2.37) holds for all
processes involving ideal gases, whether or not P is constant, as long as it is reasonable to assume that CP is constant. Therefore, if the initial and final temperatures are
known or can be calculated, and if CV and CP are known, ∆U and ∆H can be calculated regardless of the path for processes involving ideal gases using Equations (2.35)
and (2.37), as long as no chemical reactions or phase changes occur. Because U and H
are state functions, the previous statement is true for both reversible and irreversible
processes. Recall that for an ideal gas CP - CV = nR, so that if one of CV and CP is
known, the other can be readily determined.
We next note that the first law links q, w, and ∆U. If any two of these quantities are
known, the first law can be used to calculate the third. For example, if only expansion
work takes place, one always proceeds from the equation
w = -
L
Pext dV
(2.38)
This integral can only be evaluated if the functional relationship between Pext and V is
known. A commonly encountered case is Pext = constant, such that
w = -Pext1Vƒ - Vi2
(2.39)
Because Pext ≠ P, the work considered in Equation (2.39) is for an irreversible process.
A second commonly encountered case is that the system and external pressure differ only by an infinitesimal amount. In this case, it is sufficiently accurate to write
Pext = P, and the process is reversible:
w = -
nRT
dV
L V
(2.40)
This integral can only be evaluated if T is known as a function of V. The most commonly encountered case is an isothermal process, in which T is constant. As was seen
in Section 2.3, for this case
w = -nRT
W2.1 Reversible Cyclic Processes
Vf
dV
= -nRT ln
Vi
L V
(2.41)
In solving thermodynamic problems, it is very helpful to understand the process
thoroughly before starting the calculation because it may be possible to obtain the
value of one or more of q, w, ∆U, and ∆H without a calculation for an ideal gas.
For example, ∆U = ∆H = 0 for an isothermal process because ∆U and ∆H depend
only on T. For an adiabatic process, q = 0 by definition. If only P–V work is possible, w = 0 for a constant volume process. These guidelines are illustrated in Example
Problems 2.7 and 2.8.
EXAMPLE PROBLEM 2.7
1
2
16.6 bar,
25 L
P (bar)
16.6 bar,
1.00 L
3
V (L)
Consider a system that contains 2.50 mol of an ideal gas for which CV,m =
20.79 J mol-1 K-1. The system is taken through the cycle shown in the following
diagram in the direction indicated by the arrows. The curved path corresponds to
PV = nRT, where T = T1 = T3.
a. Calculate q, w, ∆U, and ∆H for each segment and for the cycle, assuming that
the heat capacity is independent of temperature.
b. Calculate q, w, ∆U, and ∆H for each segment and for the cycle in which the
direction of each process is reversed.
2.13 CALCULATING q, w, ∆U, AND ∆H FOR PROCESSES INVOLVING IDEAL GASES
Solution
We begin by asking whether we can evaluate q, w, ∆U, or ∆H for any of the segments
without performing calculations. Because the path between states 1 and 3 is isothermal,
∆U and ∆H are zero for this segment. Therefore, from the first law, q3 S 1 = -w3 S 1.
For this reason, we only need to calculate one of these two quantities. Because ∆V = 0
along the path between states 2 and 3, w2 S 3 = 0. Therefore, ∆U2 S 3 = q2 S 3. Again,
we only need to calculate one of these two quantities. Furthermore, because the total
process is cyclic, the change in any state function is zero. Therefore, ∆U = ∆H = 0
for the cycle, no matter which direction is chosen. We now deal with each segment
individually.
Segment 1 S 2
The values of n, P1 and V1, and P2 and V2 are known. Therefore, T1 and T2 can be calculated using the ideal gas law. We use these temperatures to calculate ∆U, as follows:
∆U1 S 2 = nCV,m1T2 - T12 =
=
nCV,m
nR
20.79 J mol-1 K-1
0.08314 L bar K-1 mol-1
1P2V2 - P1V12
* 116.6 bar * 25.0 L - 16.6 bar * 1.00 L2
= 99.6 kJ
The process takes place at constant pressure, so
w = -Pext1V2 - V12 = -16.6 bar *
105 N m-2
bar
* 125.0 * 10-3 m3 - 1.00 * 10-3 m32
= -39.8 kJ
Using the first law,
q = ∆U - w = 99.6 kJ + 39.8 kJ = 139.4 kJ
We next calculate T2:
T2 =
P2V2
16.6 bar * 25.0 L
=
= 2.00 * 103 K
nR
2.50 mol * 0.08314 L bar K-1 mol-1
We next calculate T3 = T1 and then ∆H1 S 2:
T1 =
P1V1
16.6 bar * 1.00 L
=
= 79.9 K
nR
2.50 mol * 0.08314 L bar K-1 mol-1
∆H1 S 2 = ∆U1 S 2 + ∆1PV2 = ∆U1 S 2 + nR1T2 - T12
= 99.6 * 103 J + 2.5 mol * 8.314 J mol-1 K-1
*12.00 * 103 K - 79.9 K2 = 139.4 kJ
Segment 2 S 3
As previously noted, w = 0, and
∆U2 S 3 = q2 S 3 = CV1T3 - T22
= 2.50 mol * 20.79 J mol-1 K-1179.9 K - 2000 K2
= -99.6 kJ
The numerical result is equal in magnitude, but opposite in sign to ∆U1 S 2, because
T3 = T1. For the same reason, ∆H2 S 3 = - ∆H1 S 2.
69
Concept
For a process involving ideal gases,
some of q, w, ∆U, and ∆H can usually
be determined without calculation,
based on the path between initial
and final states.
70
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Segment 3 S 1
For this segment, ∆U3 S 1 = 0 and ∆H3 S 1 = 0 as noted earlier, and w3 S 1 = -q3 S 1.
Because this is a reversible isothermal compression,
w3 S 1 = -nRT ln
* ln
V1
= -2.50 mol * 8.314 J mol-1 K-1 * 79.9 K
V3
1.00 * 10-3 m3
25.0 * 10-3 m3
= 5.35 kJ
The results for the individual segments and for the cycle in the indicated direction are
given in the following table. If the cycle is traversed in the reverse fashion, the magnitudes of all quantities in the table remain the same, but all signs change.
q 1 kJ2
Path
1S2
2S3
3S1
w 1 kJ2
-39.8
0
5.35
-34.5
139.4
-99.6
-5.35
34.5
Cycle
∆U 1 kJ2
99.6
-99.6
0
0
∆H 1 kJ2
139.4
-139.4
0
0
EXAMPLE PROBLEM 2.8
In this example, 2.50 mol of an ideal gas with CV,m = 12.47 J mol-1 K-1 is expanded
adiabatically against a constant external pressure of 1.00 bar. The initial temperature
and pressure of the gas are 325 K and 2.50 bar, respectively. The final pressure is
1.25 bar. Calculate the final temperature, q, w, ∆U, and ∆H.
Solution
Because the process is adiabatic, q = 0, and ∆U = w. Therefore,
∆U = nCV,m1Tƒ - Ti2 = -Pext1Vƒ - Vi2
Using the ideal gas law,
nCV,m1Tƒ - Ti2 = -nRPext a
Tƒ anCV,m +
Tƒ
Pƒ
-
Ti
b
Pi
nRPext
nRPext
b = Ti anCV,m +
b
Pƒ
Pi
RPext
Pi
≤
RPext
+
Pƒ
CV,m +
Tƒ = Ti ±
CV,m
8.314 J mol-1 K-1 * 1.00 bar
2.50 bar
≤ = 268 K
8.314 J mol-1 K-1 * 1.00 bar
+
1.25 bar
12.47 J mol-1 K-1 +
= 325 K * ±
12.47 J mol-1 K-1
We calculate ∆U = w from
∆U = nCV,m 1Tƒ - Ti2 = 2.5 mol * 12.47 J mol-1 K-1 * 1268 K - 325 K2
= -1.78 kJ
71
2.14 Reversible Adiabatic Expansion and Compression of an Ideal Gas
Because the temperature falls in the expansion, the internal energy and enthalpy
decreases:
∆H = ∆U + ∆1PV2 = ∆U + nR1T2 - T12
= -1.78 * 103 J + 2.5 mol * 8.314 J mol-1 K-1
* 1268 K - 325 K2 = -2.96 kJ
2.14 REVERSIBLE ADIABATIC EXPANSION
AND COMPRESSION OF AN IDEAL GAS
The adiabatic expansion and compression of gases is an important meteorological process. For example, the cooling of a cloud as it moves upward in the atmosphere can be
modeled as an adiabatic process because the heat transfer between the cloud and the
rest of the atmosphere is slow on the timescale of its upward motion.
Consider the adiabatic expansion of an ideal gas. Because q = 0, the first law takes
the form
∆U = w
or CV dT = -Pext dV
Connection
The adiabatic expansion and
compression of gases is an important
meteorological process.
(2.42)
For a reversible adiabatic process, P = Pext, and
CV dT = -nRT
dV
V
or, equivalently, CV
dT
dV
= -nR
T
V
(2.43)
Integrating both sides of this equation between the initial and final states,
Tƒ
Vƒ
dT
dV
CV
= -nR
T
L
L V
Ti
(2.44)
Vi
If CV is constant over the temperature interval Tƒ - Ti, then
CV ln
Tƒ
Ti
= -nR ln
Vƒ
(2.45)
Vi
3
Adiabatic
ln a
Tƒ
Ti
b = -1g - 12 ln a
Vƒ
Vi
b
or, equivalently,
Tƒ
Ti
= a
Vƒ
Vi
b
1-g
(2.46)
where g = CP,m >CV,m. Substituting Tƒ >Ti = PƒVƒ >PiVi in the previous equation, we
obtain
PiV gi = PƒV gƒ
P (atm)
Because CP - CV = nR for an ideal gas, Equation (2.45) can be written in the form
2
1
(2.47)
for the adiabatic reversible expansion or compression of an ideal gas. Note that our
derivation is only applicable to a reversible process because we have assumed that
P = Pext.
Reversible adiabatic compression of a gas leads to heating, and reversible adiabatic expansion leads to cooling as illustrated in Example Problem 2.9. Adiabatic and
isothermal expansion and compression are compared in Figure 2.16, in which two
systems containing 1 mol of an ideal gas have the same volume at P = 1 atm. One
system undergoes adiabatic compression or expansion, and the other undergoes isothermal compression or expansion. Under isothermal conditions, heat flows out of the
system as it is compressed to P 7 1 atm, and heat flows into the system as it is expanded to P 6 1 atm to keep T constant. Because no heat flows into or out of the
system under adiabatic conditions, its temperature increases in compression and decreases in expansion. Because Tadiabatic 7 Tisothermal for a compression starting at 1 atm,
Padiabatic 7 Pisothermal for a given volume of the gas. Similarly, in a reversible adiabatic
expansion originating at 1 atm, Padiabatic 6 Pisothermal for a given volume of the gas.
Isothermal
0
10
20
30
40
V (L)
50
60
Figure 2.16
Dependence of P on V for reversible
adiabatic and isothermal compression
or expansion of an ideal gas. Two systems
containing 1 mol of N2 have the same P
and V values at 1 atm. The red curve
corresponds to reversible expansion and
compression under adiabatic conditions.
The blue curve corresponds to reversible
expansion and compression under
isothermal conditions.
72
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
EXAMPLE PROBLEM 2.9
A cloud mass moving across the ocean at an altitude of 2000. m encounters a coastal
mountain range. As it rises to a height of 3500. m to pass over the mountains, it
undergoes an adiabatic expansion. The pressure at 2000. m and 3500. m is 0.802 and
0.602 atm, respectively. If the initial temperature of the cloud mass is 288 K, what
is the cloud temperature as it passes over the mountains? Assume that CP,m for air is
28.86 J K-1 mol-1 and that air obeys the ideal gas law. If you are on the mountain,
should you expect rain or snow?
Solution
Tƒ
Vƒ
ln a b = -1g - 12 ln a b
Ti
Vi
= -1g - 12 ln a
Tƒ Pi
Tƒ
Pi
b = -1g - 12 ln a b - 1g - 12 ln a b
Ti Pƒ
Ti
Pƒ
a
CP,m
Energy
- 1b
CP,m - R
1g - 12
Pi
Pi
= ln a b = ln a b
CP,m
g
Pƒ
Pƒ
CP,m - R
28.86 J K-1 mol-1
- 1b
0.802 atm
28.86 J K mol-1 - 8.314 J K-1 mol-1
= * ln a
b
-1
-1
0.602 atm
28.86 J K mol
28.86 J K-1 mol-1 - 8.314 J K-1 mol-1
a
(a)
(b)
(c)
-1
= -0.0826
Figure 2.17
Effects of adiabatic compression on
occupation of energy levels. Blue dots
signify relative populations of different
energy levels. (a) Translational energy
levels for a system before undergoing
adiabatic compression. (b) After compression, energy levels are shifted upward,
but occupation probability of the levels
on the horizontal axis is unchanged.
(c) Subsequent cooling to the original
temperature at constant V restores energy
to its original value by decreasing the
probability of occupying higher-energy
states. Each circle corresponds to a probability of 0.10.
Tƒ = 0.9207 Ti = 265 K
You can expect snow.
It is instructive to consider an adiabatic compression or expansion from a microscopic point of view. In an adiabatic compression, the energy levels are all raised, but
the probability that a given level is accessed is unchanged, as shown in Figure 2.17.
This behavior, in contrast to that shown in Figure 2.11, is observed because in an adiabatic compression T and, therefore, U increase. For more details, see the reference
to Dickerson (1969) in Further Reading. If the gas is subsequently cooled at constant V,
the energy levels remain unchanged, but the probability that higher-energy states are
populated decreases, as shown in Figure 2.17c.
VOCABULARY
cyclic path
degrees of freedom
endothermic
enthalpy
exact differential
exothermic
first law of thermodynamics
heat
heat capacity
indicator diagram
internal energy
irreversible
isobaric
isochoric
isothermal
path
path function
quasi-static process
reversible
state function
work
Conceptual Problems
73
KEY EQUATIONS
Equation
Significance of Equation
∆U = q + w
First law of thermodynamics
2.3
w = -mg∆h
Surroundings-based work
2.4
System-based work
2.5
Work of isothemal reversible compression of ideal gas
2.7
xƒ
w =
Lxi
F # dx
wisothermal rev compression = - nRT ln
Vƒ
Vi
Equation Number
Vƒ
w = -wsur = -
LVi
Pext dV
General equation for reversible or irreversible P–V work
Following 2.9
Equation for electrical work
2.11
Definition of heat capacity
2.26
CP - CV = nR or CP,m - CV,m = R
Relation between CP and CV for ideal gas
2.28
∆U = qV , ∆H = qP
Experimental determination of ∆H, ∆U by measuring heat flow
H K U + PV
Definition of enthalpy
welectrical = Ift
C1T2 = lim
∆T S 0 Tƒ
Tƒ
Ti
= a
Vƒ
Vi
b
1-g
q
dq
=
- Ti
dT
, PiV gi = PƒV gƒ
2.29, 2.34
2.33
Relationship between T and V or P and V for a reversible
adiabatic expansion or contraction
2.46, 2.47
CONCEPTUAL PROBLEMS
Q2.1 Electrical current is passed through a resistor immersed
in a liquid set in an adiabatic container. The temperature of
the liquid is varied by 1°C. The system consists solely of the
liquid. Does heat or work flow across the boundary between
the system and surroundings? Justify your answer.
Q2.2 Two ideal gas systems undergo reversible expansion
under different conditions starting from the same P and V.
At the end of the expansion, the two systems have the same
volume. The pressure in the system that has undergone
adiabatic expansion is lower than the pressure in the system
that has undergone isothermal expansion. Explain this result
without using equations.
Q2.3 You have a liquid and its gaseous form in equilibrium
in a piston and cylinder assembly in a constant temperature
bath. Give an example of a reversible process.
Q2.4 Describe how reversible and irreversible expansions
differ by discussing the degree to which equilibrium is maintained between the system and the surroundings.
Q2.5 For a constant pressure process, ∆H = qP. Does it
follow that q is a state function? Explain.
Q2.6 A cup filled with water (the system) at 278K is placed
in a microwave oven, and the oven is turned on for 1 minute
during which it begins to boil. Which of q, w, and ∆U are
positive, negative, or zero?
Q2.7 What is wrong with the following statement? Burns
caused by steam at 100°C can be more severe than those
caused by water at 100°C because steam contains more heat
than water. Rewrite the sentence to convey the same information in a correct way.
Q2.8 Why is it incorrect to speak of the heat or work associated with a system?
Q2.9 You have a liquid and its gaseous form in equilibrium
in a piston and cylinder assembly in a constant temperature
bath. Give an example of an irreversible process.
Q2.10 What is wrong with the following statement?
Because the well-insulated house stored a lot of heat, the
temperature did not decrease much when the furnace failed.
Rewrite the sentence to convey the same information in a
correct way.
Q2.11 Explain how a mass of water in the surroundings can
be used to determine q for a process. Calculate q if the temperature of a 1.00-kg water bath in the surroundings increases
by 1.25°C. Assume that the surroundings are at a constant
pressure.
74
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
Q2.12 A chemical reaction occurs in a constant volume
enclosure separated from the surroundings by diathermal
walls. Can you say whether the temperature of the surroundings increases, decreases, or remains the same in this process?
Explain.
Q2.13 Explain the relationship between the terms exact
differential and state function.
Q2.14 Imagine a horizontal cylinder with a frictionless
piston midway between the ends held by a mechanical stop.
To the left of the piston is an ideal gas at 1 bar pressure. To
the right of the piston is a vacuum. The system consists only
of the gas. Assume that the expansion is adiabatic. Discuss
what happened when the stop is released. Assign a sign to w,
q, and ∆U after all kinetic energy has dispersed.
Q2.15 Discuss the following statement: If the temperature of
the system increased, heat must have been added to it.
Q2.16 Discuss the following statement: Letting heat flow
into an object causes its temperature to increase.
Q2.17 An ideal gas is expanded reversibly and adiabatically.
Decide which of q, w, ∆U, and ∆H are positive, negative,
or zero.
Q2.18 An ideal gas is expanded reversibly and isothermally.
Decide which of q, w, ∆U, and ∆H are positive, negative,
or zero.
Q2.19 An ideal gas is expanded adiabatically into a vacuum.
Decide which of q, w, ∆U, and ∆H are positive, negative,
or zero.
Q2.20 A bowling ball (a) rolls across a horizontal table and
(b) falls on the floor. Is the work associated with each part of
this process positive, negative, or zero? Assume that friction
can be neglected.
Q2.21 A perfectly insulating box is partially filled with water
in which an electrical resistor is immersed. An external electrical generator converts the change in potential energy of a
mass m that falls by a vertical distance h into electrical energy
that in turn is dissipated in the resistor. Assign positive, negative,
or zero values to q, w, and ∆U if the system is defined as the
resistor and the water. Everything else is in the surroundings.
Q2.22 Explain why ethene has a higher value for CV,m at
800. K than CO.
Q2.23 Explain why CP,m is a function of temperature for
ethane but not for argon in a temperature range in which
electronic excitations do not occur.
Q2.24 What is the difference between a quasi-static process
and a reversible process?
Q2.25 A dishwasher is opened immediately before the drying
cycle starts. Both ceramic mugs and plastic storage containers
are beaded with water. After five minutes, the water beads on
the mugs are gone and those on the plastic container are still
there. Explain.
Q2.26 Arrange the following molecules or atoms in order of
increasing heat capacity at 100. K.
(a) He (b) HCl (c) C2H4 (d) C6H6
Q2.27 Arrange the following molecules or atoms in order of
increasing heat capacity at 1000. K.
(a) He (b) HCl (c) C2H4 (d) C6H6
Q2.28 Liquid water one degree above its freezing point is
put into a glass bottle, and and a rubber stopper is put into the
neck. The bottle is put into a freezer, and after equilibrium is
reached, ice has formed and the bottle has broken. Determine
the signs of q, w, and ∆U for the process if the water is the
system and everything else is in the surroundings.
NUMERICAL PROBLEMS
Section 2.3
P2.1 A muscle fiber contracts by 2.5 cm and in doing so lifts
a weight. Calculate the work performed by the fiber. Assume
the muscle fiber obeys Hooke’s law F = -kx with a force
constant k of 770 N m-1.
P2.2 The formalism of Young’s modulus is sometimes used
to calculate the reversible work involved in extending or compressing an elastic material. Assume a force F is applied to
an elastic rod of cross-sectional area A0 and length L 0. As a
result of this force, the rod changes in length by ∆L. Young’s
modulus E is defined as
E =
tensile stress
=
tensile strain
F
A0
∆L
L0
=
FL 0
A0 ∆L
a. Relate k in Hooke’s law to Young’s modulus expression
given above.
b. Using your result in part (a), show that the magnitude of
the reversible work involved in changing the length L 0
of an elastic cylinder of cross-sectional area A0 by ∆L is
1 ∆L 2
a
b EA0 L 0.
2 L0
P2.3 Young’s modulus (see Problem P2.2) of muscle fiber
is approximately 2.80 * 107 Pa. A muscle fiber 3.20 cm in
length and 0.250 cm in diameter is suspended with an object
of mass M hanging at its end. Calculate the mass required to
extend the length of the fiber by 5.0%.
P2.4 DNA can be modeled as an elastic rod that can be
twisted or bent. Suppose a DNA molecule of length L is
bent so that it lies on the arc of a circle of radius Rc. The
reversible work involved in bending DNA without twisting
BL
is wbend =
, where B is the bending force constant.
2R2c
The DNA in a nucleosome particle is about 680. Å in length.
w =
Numerical Problems
P2.8 As the volume of an ideal gas confined in a piston and
cylinder assembly increases, 78.3 J of heat is absorbed from
the surroundings. The external pressure is 1.00 bar, and the
final volume of the gas is 65.6 L. If the ∆U for the process is
-215 J, calculate the initial volume of the gas. Assume ideal
Sections 2.5 and 2.6
gas behavior.
P2.5 Assume the following simplified dependence of the
P2.9 ∆U for a van der Waals gas increases by 400.0 J in an
pressure in a ventricle of the human heart as a function of the
expansion process and the magnitude of w is 115.0 J. Calvolume of blood pumped.
culate the sign of w and the magnitude and sign of q for the
process.
P2.10 2.25 mol of an ideal gas at 25.0°C expands from an
Ps
initial volume of 10.0 dm3 to a final volume of 14.0 dm3. Calculate w for this process (a) for expansion against a constant
Pd
external pressure of 1 atm and (b) for a reversible isothermal
expansion.
0
75
150
P2.11 Consider the isothermal expansion of 1.75 mol of an
V (cm3)
ideal gas at 375 K from an initial pressure of 15.0 bar to a
final pressure of 1.50 bar. Describe the process that will result
The systolic pressure, Ps, is 130. mm Hg, corresponding to
in the greatest amount of work being done by the system with
1.73 * 104 Pa. The diastolic pressure, Pd, is 90.0 mm Hg,
Pext Ú 1.50 bar and calculate w. Describe the process that will
corresponding to 1.20 * 104 Pa atm. If the volume of blood
result in the least amount of work being done by the system
pumped in one heartbeat is 75.0 cm3, calculate the work done
with Pext Ú 1.50 bar and calculate w. What is the least amount
in a heartbeat.
of
work done without restrictions on the external pressure?
P2.6 A piston and cylinder assembly with a piston radius of
P2.12 An ideal gas undergoes a single-stage expansion against
0.250 m contains 12.5 mol of an ideal gas with CV,m = 3R>2
a constant external pressure Pext at constant temperature from
at 298 K. The assembly is surrounded by air at a pressure of
T, Pi, Vi, to T, Pƒ, Vƒ.
1.00 bar. The cylinder axis is vertical, and the piston is oriented upward as in Figure 2.1. The system is at equilibrium
a. What is the largest mass m that can be lifted through the
when the length of the cylinder filled with gas is 0.750 m.
height h in this expansion?
(a) Calculate the mass of the piston. (b) Calculate the distance
b. The system is restored to its initial state in a single-state
the piston moves if the temperature of the gas is increased to
compression. What is the smallest mass m′ that must fall
750. K. (c) Calculate the work done by the piston and cylinder
through the height h to restore the system to its initial state?
assembly on the surroundings.
c. If h = 10.0 cm, Pi = 1.00 * 106 Pa, Pƒ = 1.25 * 105 Pa,
P2.7 By evaluating the integral w = - 1 Pext dV, calculate the
T = 280. K, and n = 1.75 mol, calculate the values of the
work done in each of the cycles shown below in units of Joules.
masses in parts (a) and (b).
P2.13 An ideal gas undergoes an expansion from the initial
state described by Pi, Vi, T to a final state described by Pƒ, Vƒ, T
4
4
4 in (a) a process at the constant external pressure Pƒ and (b) in
3
3
3 a reversible process. Derive expressions for the largest mass
that can be lifted through a height h in the surroundings in
2
2
2 these processes.
1
1
1 P2.14 1.500 mol of N in a state defined by T = 298. K and
2
i
Vi = 1.000 L undergoes an isothermal reversible expansion
until 1Vƒ =2 15.00
L. Calculate w assuming (a) that the gas is
1
4
1
4
4
2
3
2
3
3
describedVby
V (L)
V (L)
(L)the ideal gas law and (b) that the gas is described
(a)
(b)
(c) by the van der Waals equation of state. What is the percent
error in using the ideal gas law instead of the van der Waals
equation? The van der Waals parameters for N2 are listed in
Table 7.4 (see Appendix A, Data Tables). Carry out your cal4
culations of w to four significant figures.
3
P2.15 2.50 mol of an ideal gas with CV,m = 3R>2 initially at
2
a temperature Ti = 298 K and Pi = 1.00 bar is enclosed in an
adiabatic piston and cylinder assembly. The gas is compressed
1
by placing a 650. kg mass on the piston of diameter 19.8 cm.
Calculate the work done in this process and the distance that the
4
1
4
2
3
2
3
piston travels. Assume that the mass of the piston is negligible.
V (L)
V (L)
First calculate the external pressure and the initial volume.
P (bar)
P (bar)
3
2
1
1
(b)
(c)
P (bar)
P (bar)
P (bar)
Pressure
Nucleosomal DNA is bent around a protein complex called
the histone octamer into a circle of radius 55 Å. Calculate the
reversible work involved in bending the DNA around the histone octamer if the force constant B = 1.65 * 10-28 J m-1.
4
75
76
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.16 An ideal gas described by Ti = 275 K, Pi = 1.10 bar,
and Vi = 10.0 L is heated at constant volume until
P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.10 bar. It is then restored to its original
state by the extraction of heat at constant pressure. Depict this
closed-cycle process in a P–V diagram. Calculate w for each
step and for the total process. What values for w would you
calculate if the cycle were traversed in the opposite direction?
P2.17 A hot-air balloon is expanded by using a propane
burner to heat the gas inside the balloon. The amount of
heat released in the process is 1.25 * 108 J, and the balloon
expands from an initial volume to 2.25 * 106 L to a final
volume of 3.75 * 106 L. Assume that the external pressure is
constant at 1.00 bar. Calculate w and ∆U for the process
assuming ideal gas behavior.
P2.18 10.0 mol of an ideal gas (the system) is confined in
a piston and cylinder assembly similar to that shown in
Figure 2.2. The radius of the piston is 5.00 cm. The initial temperature is 300. K, and the initial pressure is 2.00 bar. The gas
is expanded to twice its initial volume in an isothermal reversible process. Assume that the surroundings are a vacuum.
a. Calculate w for this process considering only the movement of masses in the surroundings.
b. Make a plot of the mass on the piston as a function of x,
the height of the piston. Is m a linear function of x?
Section 2.11
P2.19 A hiker caught in a thunderstorm loses heat when her
clothing becomes wet. She is packing emergency rations that,
if completely metabolized, will release 35 kJ of heat per gram
of rations consumed. How many grams of ration must the
hiker consume to avoid a reduction in body temperature of
3.0 K as a result of heat loss? Assume the heat capacity of the
body equals that of water and that the mass of the hiker is
62 kg. State any additional assumptions.
P2.20 Count Rumford used a horse to turn cannon boring
machinery. He showed that a single horse could heat 11.6 kg of
ice water 1T = 273 K2 to T = 355 K in 2.5 hours. Assuming
the same rate of work, how high could a horse raise a 180 kg
weight in 2.0 minutes? Assume the heat capacity of water is
4.18 J K-1 g -1.
P2.21 At 298 K and 1 bar pressure, the density of water
is 0.9970 g cm-3, the coefficient of thermal expansion is
2.07 * 10-4 K-1, and CP,m = 75.30 J K-1 mol-1. If the temperature of 150 g of water is increased by 25.0 K, calculate
w, q, ∆H, and ∆U.
P2.22 A system consisting of 75 g of liquid water at 298. K
is heated using an immersion heater at a constant pressure
of 1.00 bar. If a current of 2.50 A passes through the 25.0 Ω
­resistor for 100. s, what is the final temperature of the water?
P2.23 The heat capacity of solid lead oxide is given by
CP,m = 44.35 + 1.47 * 10-3
T
in units of J K-1 mol-1
K
Calculate the change in enthalpy of 2.5 mol of PbO1s2 if it is
cooled from 650. K to 298. K at constant pressure.
P2.24 A major league pitcher throws a baseball with a speed
of 150. kilometers per hour. If the baseball weighs 195 grams
and its heat capacity is 1.7 J g -1 K-1, calculate the temperature
increase of the ball when it is stopped by the catcher’s mitt.
Assume that no heat is transferred to the catcher’s mitt and that
the catcher’s arm does not recoil when he catches the ball.
P2.25 Suppose that an adult is encased in a thermally insulating barrier and that the body retains all the heat evolved
by metabolism of foodstuffs. Calculate the increase in body
temperature after 2.5 hours. Assume that the heat capacity of
the body is 4.18 J g -1 K-1 and that the heat produced by metabolism is 9.4 kJ kg -1 hr -1.
P2.26 A vessel containing 1.50 mol of an ideal gas with
Pi = 1.00 bar and CP,m = 5R>2 is in thermal contact with
a water bath. Treat the vessel, gas, and water bath as being
in thermal equilibrium, initially at 300. K, and as separated
by adiabatic walls from the rest of the universe. The vessel, gas, and water bath have an average heat capacity of
CP = 2800. J K-1. The gas is compressed reversibly to
Pƒ = 10.0 bar. What is the temperature of the system after
thermal equilibrium has been established?
P2.27 An airflow rate of 250. cfm at a pressure of 1.00 bar is
used to heat a house. Calculate the power in kW required to
heat the air from the outside temperature of 41.0 °F to a final
temperature of 68.0 °F. Assume that the heat capacity of air is
constant at a value of 1.30 * 10-3 J cm-3 K-1.
P2.28 River water at a temperature of 290. K is used for
cooling in an industrial process. The water returned to the
river is at a temperature of 330. K. Assuming that the river
water is a closed system, what percentage of the river water
can be used for cooling if the river temperature is required by
law to increase by no more than 8.0 K?
Section 2.13
P2.29 1.75 mol of an ideal gas, for which CV,m = 3R>2, is
subjected to two successive changes in state: (1) From 15.0°C
and 125. * 103 Pa, the gas is expanded isothermally against a
constant pressure of 15.2 * 103 Pa to twice the initial volume.
(2) At the end of the previous process, the gas is cooled at
constant volume from 15.0°C to -35.0°C. Calculate q, w,
∆U, and ∆H for each of the stages. Also calculate q, w, ∆U,
and ∆H for the complete process.
P2.30 A 1.75 mol sample of an ideal gas for which
CV,m = 3R>2 undergoes the following two-step process:
(1) From an initial state of the gas described by T = 15.0°C
and P = 5.00 * 104 Pa, the gas undergoes an isothermal expansion against a constant external pressure of 2.50 * 104 Pa
until the volume has doubled. (2) Subsequently, the gas
is cooled at constant volume. The temperature falls to
-19.0°C. Calculate q, w, ∆U, and ∆H for each step and for
the overall process.
P2.31 1.50 mol of an ideal gas, for which CV,m = 3R>2,
initially at 15.0°C and 1.20 * 106 Pa undergoes a two-stage
transformation. For each of the stages described in the following list, calculate the final pressure, as well as q, w, ∆U,
and ∆H. Also calculate q, w, ∆U, and ∆H for the complete
process.
Numerical Problems
a. The gas is expanded isothermally and reversibly until the
volume triples.
b. Beginning at the end of the first stage, the temperature is
raised to 130°C at constant volume.
P2.32 The temperature of 2.15 mol of an ideal gas is changed
from 85.0°C to 15.0°C. CV,m = 3R>2. Calculate q, w, ∆U,
and ∆H.
P2.33 2.25 mol of an ideal gas is subjected to the changes
below. Calculate the change in temperature for each case if
CV,m = 3R>2.
a. q = -375 J, w = 225 J
b. q = 275. J, w = -275 J
c. q = 0, w = 375 J
P2.34 Calculate ∆H and ∆U for the transformation of
1.75 mol of an ideal gas from 15.0°C and 1.00 atm to 475.°C and
T
25.0 atm if CP,m = 20.9 + 0.042 in units of J K-1 mol-1.
K
P2.35 2.50 mol of an ideal gas for which CV,m =
20.8 J K-1 mol-1 is heated from an initial temperature of
15.0°C to a final temperature of 450.°C at constant volume.
Calculate q, w, ∆U, and ∆H for this process.
P2.36 2.50 mol of an ideal gas is compressed isothermally
from 84.0 to 22.0 L using a constant external pressure of
2.80 atm. Calculate q, w, ∆U, and ∆H.
P2.37 A pellet of Zn of mass 35.0 g is dropped into a flask
containing dilute H2SO4 at a pressure of P = 1.00 atm and
temperature of T = 298. K. What is the reaction that occurs?
Calculate w for the process.
Section 2.14
P2.38 Calculate q, w, ∆U, and ∆H if 1.50 mol of an ideal
gas with CV,m = 3R>2 undergoes a reversible adiabatic
expansion from an initial volume Vi = 19.0 dm3 to a final
volume Vƒ = 60.8 dm3. The initial temperature is 32.0°C.
P2.39 Calculate w for the adiabatic expansion of 1.25 mol of
an ideal gas at an initial pressure of 2.00 bar from an initial
temperature of 380. K to a final temperature of 300. K. Write
an expression for the work done in the isothermal reversible expansion of the gas at 300. K from an initial pressure
of 2.00 bar. What value of the final pressure would give the
same value of w as the first part of this problem? Assume
that CP,m = 5R>2.
P2.40 In the adiabatic expansion of 2.50 mol of an ideal gas
from an initial temperature of 15.0°C, the work done on the
surroundings is 950. J. If CV,m = 3R>2, calculate q, w, ∆U,
and ∆H.
P2.41 A bottle at 300. K contains an ideal gas at a pressure of
145 * 103 Pa. The rubber stopper closing the bottle is removed.
The gas expands adiabatically against Pext = 1.00 * 105 Pa,
and some gas is expelled from the bottle in the process. When
P = Pext, the stopper is quickly replaced. The gas remaining in
the bottle slowly warms up to 300. K. What is the final pressure
in the bottle for a monatomic gas, for which CV,m = 3R>2,
and a diatomic gas, for which CV,m = 5R>2?
77
P2.42 A 1.50 mol sample of an ideal gas with CV,m = 3R>2
initially at 300. K and 1.00 * 105 Pa undergoes a reversible
adiabatic compression. At the end of the process, the pressure
is 1.45 * 106 Pa. Calculate the final temperature of the gas.
Calculate q, w, ∆U, and ∆H for this process.
P2.43 In an adiabatic compression of 1.75 mol of an ideal
gas with CV,m = 5R>2, the temperature rises from 300. K to
500. K. Calculate q, w, ∆H, and ∆U.
P2.44 A 2.50 mol sample of an ideal gas for which P =
3.25 bar and T = 298 K is expanded adiabatically against
an external pressure of 0.225 bar until the final pressure is
0.225 bar. Calculate the final temperature, q, w, ∆H, and ∆U
for (a) CV,m = 3R>2 and (b) CV,m = 5R>2.
P2.45 2.25 mol of an ideal gas with CV,m = 3R>2 is
expanded adiabatically against a constant external pressure of
1.00 bar. The initial temperature and pressure are Ti = 310. K
and Pi = 15.2 bar. The final pressure is Pƒ = 1.00 bar.
Calculate q, w, ∆U, and ∆H for the process.
P2.46 A nearly flat bicycle tire becomes noticeably warmer
after it has been pumped up. Approximate this process as a
reversible adiabatic compression. Assume the initial pressure and temperature of the air before it is put in the tire
to be Pi = 1.00 bar and Ti = 300. K. The final pressure is
Pƒ = 3.00 bar. Calculate the final temperature of the air in
the tire. Assume that CV,m = 5R>2. Does your answer seem
plausible? If not, explain why.
P2.47 An automobile tire contains air at 2.00 * 105 Pa at
15.0°C. The stem valve is removed, and the air is allowed to
expand adiabatically against the constant external pressure of
one bar until P = Pexternal. For air, CV,m = 5R>2. Calculate
the final temperature. Assume ideal gas behavior. The stem
is reinserted, and the tire is allowed to warm up to 15.0°C.
What is the final pressure?
P2.48 Consider the adiabatic expansion of an ideal monoatomic gas with CV,m = 3R>2. The initial state is described by
P = 7.50 bar and T = 298 K.
a. Calculate the final temperature if the gas undergoes
a reversible adiabatic expansion to a final pressure of
P = 2.25 bar.
b. Calculate the final temperature if the same gas undergoes
an adiabatic expansion against an external pressure of
P = 2.25 bar to a final pressure P = 2.25 bar.
Explain the difference in your results for parts (a) and (b).
P2.49 The temperature of 2.25 mol of an ideal gas increases
from 15.1°C to 39.5°C as the gas is compressed adiabatically.
Calculate q, w, ∆U, and ∆H for this process assuming that
CV,m = 3R>2.
P2.50 1.50 mol of an ideal gas with CV,m = 3R>2 is
compressed adiabatically and reversibly from the initial
conditions P = 1.00 bar and 300. K to a final pressure of
95.0 bar. Calculate the final temperature and the work done
in the process.
78
CHAPTER 2 Heat, Work, Internal Energy, Enthalpy, and the First Law of Thermodynamics
P2.51 A cylindrical vessel with rigid adiabatic walls is
separated into two parts by a frictionless adiabatic piston.
Each part contains 55.0 L of an ideal monatomic gas with
CV,m = 3R>2. Initially, Ti = 300. K and Pi = 2.00 * 105 bar
in each part. Heat is slowly introduced into the left part using
an electrical heater until the piston has moved sufficiently to
the right to produce a final pressure Pƒ = 3.75 bar in the right
part. Consider the compression of the gas in the right part to
be a reversible process.
a. Calculate the work done on the right part in this process
and the final temperature in the right part.
b. Calculate the final temperature in the left part and the
amount of heat that flowed into this part.
P2.52 2.75 mol of an ideal gas is expanded from 375 K and
an initial pressure of 4.50 bar to a final pressure of 1.00 bar,
and CP,m = 5R>2. Calculate w for the following two cases:
a. The expansion is isothermal and reversible.
b. The expansion is adiabatic and reversible.
Without resorting to equations, explain why the result for part
(b) is greater than or less than the result for part (a).
P2.53 A 2.25 mol sample of carbon dioxide, for which
CP,m = 37.11 J K-1 mol-1 at 298 K, is expanded reversibly
and adiabatically from a volume of 6.00 L and a temperature
of 298 K to a final volume of 45.0 L. Calculate the final temperature, q, w, ∆H, and ∆U. Assume that CP,m is constant
over the temperature interval.
WEB-BASED SIMULATIONS, ANIMATIONS, AND PROBLEMS
Simulations, animations, and homework problem worksheets can be accessed on Mastering™ Chemistry.
W2.1 Reversible cyclic processes are simulated in which the
cycle is either rectangular or triangular on a P–V plot. For
each segment and for the cycle, q, and w are determined. For
a given cycle type, the ratio of work done on the surroundings
to the heat absorbed from the surroundings is determined for
different P and V values.
FURTHER READING
Dickerson, R. E. Molecular Thermodynamics. Menlo Park,
CA: W.A. Benjamin, 1969, Chapter 4.
Gislason, E. A., and Craig, N. C. “Cementing the Foundations
of Thermodynamics: Comparison of System-Based and
Surroundings-Based Definitions of Work and Heat.”
Journal of Chemical Thermodynamics 37 (2005): 954.
Gislason, E. A., and Craig, N. C. “Pressure-Volume Integral
Expressions for Work in Irreversible Processes.” Journal
of Chemical Education. 84 (2007): 499.
Romer, R. “Heat Is Not a Noun.” American Journal of
Physics 69 (2001): 107–109.
Schmidt-Rohr, K. “Expansion Work without the External
Pressure and Thermodynamics in Terms of Quasistatic
Irreversible Processes.” Journal of Chemical Education
91 (2014): 402.
MATH ESSENTIAL 3:
Partial Derivatives
Many quantities that we will encounter in physical chemistry are functions of several
variables. In that case, we have to reformulate differential calculus to take several variables into account. We define the partial derivative with respect to a specific variable
just as we did in Section ME2.2 by treating all other variables indicated by subscripts
as constants. For example, consider 1 mol of an ideal gas for which
P = ƒ1V, T2 =
RT
V
(ME3.1)
Note that P can be written as a function of the two variables V and T. The change in P
resulting from a change in V or T is proportional to the following partial derivatives:
a
P1V + ∆V, T2 - P1V, T2
0P
R
1
1
b = lim ∆V S 0
= lim ∆V S 0
a
- b
0V T
∆V
∆V V + ∆V
V
= lim ∆V S 0
a
R
- ∆V
RT
b = - 2
a 2
∆V V + V∆V
V
P1V, T + ∆T2 - P1V, T2
0P
1 R1T + ∆T2
RT
b = lim ∆T S 0
= lim ∆T S 0
c
d
0T V
∆T
∆T
V
V
=
R
V
(ME3.2)
The subscript y in 10ƒ>0x2y indicates that y is being held constant in the differentiation of the function ƒ with respect to x. The partial derivatives in Equation
(ME3.2) allow one to determine how a function changes when all of the variables change. For example, what is the change in P if the values of T and V both
change? In this case, P changes to P + dP where
0P
0P
R
RT
dP = a b dT + a b dV = dT - 2 dV
0T V
0V T
V
V
y
10
(ME3.3)
z
0z
0z
b dx + a b dy
0x y
0y x
(ME3.4)
The first term is the slope of the hill in the x direction, and the second term is the
slope in the y direction. These changes in the height z as you move first along
the x direction and then along the y direction are illustrated in Figure ME3.1.
Because the slope of the hill is a function of x and y, this expression for dz
is only valid for small changes dx and dy. Otherwise, higher-order derivatives
need to be considered.
30
Able Hill
40
( )y dx 1 (−z
−y )x dy
dx
( )y
−z
z 1 −x
−z
z 1 −x
x
Consider the following practical illustration of Equation (ME3.3). You are
on a hill and have determined your altitude above sea level. How much will the
altitude (denoted z) change if you move a small distance east (denoted by x)
and north (denoted by y)? The change in z as you move east is the slope of the
hill in that direction, 10z>0x2y, multiplied by the distance dx that you move. A
similar expression can be written for the change in altitude as you move north.
Therefore, the total change in altitude is the sum of these two changes or
dz = a
20
Able Hill
50 m
40 m
30 m
20 m
10 m
Sea level
50
Figure ME3.1
Able Hill contour plot and cross section. The
cross section (bottom) is constructed from the
contour map (top). Starting at the point labeled z
on the hill, you first move in the positive x direction and then along the y direction. If dx and dy
are sufficiently small, the change in height dz is
0z
0z
given by dz = a b dx + a b dy.
0x y
0y x
79
80
Math Essential 3 Partial Derivatives
Second or higher derivatives with respect to either variable can also be taken. The
mixed second partial derivatives are of particular interest. Consider the two mixed second partial derivatives of P:
RT
0c
d
2
V ¢ Æ ¢
0 0P
0P
° °
RT
R
a a b b =
= 0
0T
= a0 c - 2 d n 0T b = - 2
0T 0V T V
0T0V
0V
V
V
T
V
V
a
2
0 0P
0P
° °
a b b =
= 0
0V 0T V T
0T0V
0c
RT
d
V ¢ Æ ¢
R
R
0V
= a0 c d n 0Vb = - 2
0T
V
V
V
T
T
(ME3.5)
For all state functions ƒ and for the specific case of P, the order in which the function is
doubly differentiated does not affect the outcome, and we conclude that
a
0 0ƒ1V, T2
0 0ƒ1V, T2
a
b b = a a
b b
0T
0V
0V
0T
T V
V T
(ME3.6)
Because Equation (ME3.5) is only satisfied by state functions ƒ, it can be used
to determine if a function ƒ is a state function. If ƒ is a state function, one can write
ƒ
∆ƒ = 1i dƒ = ƒfinal - ƒinitial. This equation states that ƒ can be expressed as an infinitesimal quantity, dƒ, that, when integrated, depends only on the initial and final
states; dƒ is called an exact differential.
We can illustrate these concepts with the following calculation.
PROBLEM
a. Calculate
2
a
2
0ƒ
0ƒ
0ƒ
0ƒ
°
b ,a b ,a
b ,a
b ,
0x y 0y x 0x 2 y 0y 2 x
0a
x
0ƒ
b
0x y ¢
0y
, and
x
°
0a
0ƒ
b
0y x ¢
0x
y
for the function ƒ1x, y2 = ye + xy + x ln y.
b. Determine if ƒ1x, y2 is a state function of the variables x and y.
c. If ƒ1x, y2 is a state function of the variables x and y, what is the total differential dƒ?
Solution
0ƒ
a. a b = yex + y + ln y, 0x y
a
a
02ƒ
0x
°
b = yex,
2
0a
a
y
0ƒ
b
0x y ¢
0y
x
= ex + 1 +
b. Because we have shown that
°
02ƒ
0y
°
1
,
y
0a
0ƒ
x
b = ex + x +
y
0y x
0ƒ
b
0x y ¢
0y
x
=
2
0a
°
b = x
x
y2
0ƒ
b
0y x ¢
1
= ex + 1 +
y
0x
y
0a
0ƒ
b
0y x ¢
0x
y
ƒ1x, y2 is a state function of the variables x and y. Generalizing this result, any
well-behaved function that can be expressed in analytical form is a state function.
c. The total differential is given by
dƒ = a
0ƒ
0ƒ
b dx + a b dy
0x y
0y x
= 1yex + y + ln y2dx + aex + x +
x
b dy
y
C H A P T E R
3
The Importance of State
Functions: Internal Energy
and Enthalpy
WHY is this material important?
Expressing state functions in terms of their variables using partial derivatives is a
powerful tool. It allows us to write equations linking state functions such as U and H
with quantities that can be determined experimentally.
WHAT are the most important concepts and results?
3.1
Mathematical Properties
of State Functions
3.2
Dependence of U on V and T
3.3
Does the Internal Energy
Depend More Strongly
on V or T ?
3.4
Variation of Enthalpy with
Temperature at Constant
Pressure
3.5
How Are CP and CV Related?
3.6
Variation of Enthalpy
with Pressure at Constant
Temperature
3.7
The Joule–Thomson Experiment
3.8
Liquefying Gases Using
an Isenthalpic Expansion
The natural variables for U are T and V, but for most cases U can be regarded as a
function of T alone. The natural variables for H are T and P, but for most cases H can
be regarded as a function of T alone. The expansion of real gases can lead to cooling
or heating, depending on the value of the Joule–Thomson coefficient of the gas at the
given temperature and pressure.
WHAT would be helpful for you to review for this chapter?
Because partial derivatives are used extensively in this chapter, it would be helpful to
review the material in Math Essential 3.
3.1 MATHEMATICAL PROPERTIES
OF STATE FUNCTIONS
In Chapter 2 we demonstrated that U and H are state functions and that w and q are path
functions. We also discussed how to calculate changes in these quantities for an ideal gas.
In this chapter, the path independence of state functions is exploited to derive relationships with which ∆U and ∆H can be calculated as functions of P, V, and T for real gases,
liquids, and solids. In doing so, we develop the formal aspects of thermodynamics. We
will show that the formal structure of thermodynamics provides a powerful aid in linking
theory and experiment. The thermodynamic state functions of interest here are defined by
two variables from the set P, V, and T. In formulating changes in state functions, we will
make extensive use of partial derivatives, which are reviewed in Math Essential 3.
In the following discussion, two important results from differential calculus will
be used frequently. First consider a function z = ƒ1x, y2 that can be rearranged to
x = g1y, z2 or y = h1x, z2. For example, if P = nRT>V, then V = nRT>P and
T = PV>nR. In this case
a
0P
b =
0V T
1
0V
a b
0P T
(3.1)
81
82
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Second, the cyclic rule will also be used:
a
0P
0V
0T
b a b a b = -1
0V T 0T P 0P V
(3.2)
It is called the cyclic rule because the variables x, y, and z in the three terms follow the
orders x, y, z; y, z, x; and z, x, y. As we show next, Equations (3.1) and (3.2) can be used
to reformulate Equation (3.3):
dP = a
0P
0P
b dT + a b dV
0T V
0V T
(3.3)
Suppose this expression needs to be evaluated for a specific substance, such as N2 gas.
What quantities must be measured in the laboratory in order to obtain numerical values
for 10P>0T2V and 10P>0V2T ? Using Equations (3.1) and (3.2),
a
0P
0V
0T
b a b a b = -1
0V T 0T P 0P V
0V
a b
0T P
0P
0P
0V
a
a b = -a b a b = =
kT
0T V
0V T 0T P
0V
a b
0P T
a
Concept
The easily measured thermal
expansion and compressibility
coefficients are directly related to
partial derivatives.
and
0P
1
b = 0V T
kTV
(3.4)
where a and kT are the readily measured isobaric volumetric thermal expansion
coefficient and the isothermal compressibility, respectively, defined by
a =
1 0V
1 0V
a b
and kT = - a b
V 0T P
V 0P T
(3.5)
Both 10V>0T2P and 10V>0P2T can be measured by determining the change in volume of
the system when the pressure or temperature is varied, while keeping the second variable constant.
The minus sign in the equation for kT is chosen so that values of the isothermal compressibility are positive. For small changes in T and P, Equations (3.5) can be written
in the form with differences rather than derivatives: V1T22 = V1T1211 + a3 T2 - T1 4 2
and V1P22 = V1P1211 - kT 3 P2 - P1 4 2. Values for a and kT for selected solids and
liquids are shown in Tables 3.1 and 3.2, respectively.
Equation (3.5) is an example of how seemingly abstract partial derivatives can be
directly linked to experimentally determined quantities using the mathematical properties of state functions. Using the definitions of a and kT , we can write Equation (3.3)
in the form
dP =
a
1
dT dV
kT
kTV
(3.6)
which can be integrated to give
Tf
Vf
Vf
a
1
a
1
∆P =
dT dV ≈
1Tf - Ti2 ln
kT
kT Vi
L kT
L kTV
Ti
(3.7)
Vi
The second expression in Equation (3.7) holds if ∆T and ∆V are small enough that a
and kT are constant over the range of integration. Example Problem 3.1 shows a useful
application of this equation.
3.1 Mathematical Properties of State Functions
TABLE 3.1 Selected Isobaric Volumetric Thermal Expansion Coefficients
at 298 K
Ag(s)
106 a>1K-12
57.6
Hg1l2
Al(s)
69.3
11.4
Au(s)
42.6
CCl41l2
Cu(s)
Element
Element or Compound
104 a>1K-12
1.81
14.6
49.5
CH3COCH31l2
CH3OH1l2
14.9
Fe(s)
36.9
C2H5OH1l2
11.2
Mg(s)
78.3
10.5
Si(s)
7.5
C6H5CH31l2
W(s)
Zn(s)
13.8
C6H61l2
H2O1l2
2.04
90.6
H2O1s2
1.66
11.4
Sources: Based on Benenson, W., Harris, J. W., Stocker, H., and Lutz, H. Handbook of Physics. New York:
Springer, 2002; Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC
Press, 2002; Blachnik, R., ed. D’Ans Lax Taschenbuch für Chemiker und Physiker. 4th ed. Berlin:
Springer, 1998.
TABLE 3.2 Selected Isothermal Compressibility Coefficients at 298 K
Substance
106 kT >bar -1
Substance
106 kT >bar -1
SiO2(s)
2.57
Br21l2
C2H5OH1l2
110
Ni(s)
0.513
C6H5OH1l2
61
TiO2(s)
0.56
C6H61l2
94
Al(s)
1.33
Cu(s)
0.702
CCl41l2
C(graphite)
0.156
CH3OH1l2
Mn(s)
0.716
Co(s)
0.525
CS21l2
Au(s)
0.563
Hg1l2
Pb(s)
2.37
Fe(s)
0.56
SiCl41l2
Ge(s)
1.38
Na(s)
13.4
CH3COCH31l2
H2O1l2
TiCl41l2
64
103
120
125
92.7
45.9
3.91
165
89
Sources: Based on Benenson, W., Harris, J. W., Stocker, H., and Lutz, H. Handbook of Physics. New York:
Springer, 2002; Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton FL: CRC
Press, 2002; Blachnik, R., ed. D’Ans Lax Taschenbuch für Chemiker und Physiker. 4th ed. Berlin:
Springer, 1998.
EXAMPLE PROBLEM 3.1
You have accidentally arrived at the end of the range of an ethanol-in-glass
thermometer so that the entire volume of the glass capillary is filled. By how much
will the pressure in the capillary increase if the temperature is increased by
another 10.0°C? aglass = 2.00 * 10-51°C2-1, aethanol = 11.2 * 10-41°C2-1,
and kT, ethanol = 11.0 * 10-51bar2-1. Will the thermometer survive your
experiment?
83
84
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Solution
Using Equation (3.7),
∆P =
=
Vf
aethanol
aethanol
1
1
dT dV ≈
∆T ln
kT, ethanol
kT, ethanol Vi
L kT, ethanol
L kT, ethanolV
Vi11 + aglass ∆T2
aethanol
1
∆T ln
kT, ethanol
kT, ethanol
Vi
≈
Viaglass ∆T
aethanol
1
∆T kT, ethanol
kT, ethanol
Vi
=
1aethanol - aglass2
=
111.2 - 0.2002 * 10-41°C2-1
kT, ethanol
∆T
11.0 * 10-51bar2-1
* 10.0°C = 100. bar
In this calculation, we have used the relations V1T22 = V1T1211 + a 3 T2 - T1 4 2 and
ln11 + x2 ≈ x if x V 1.
The glass is unlikely to withstand such a large increase in pressure.
3.2 DEPENDENCE OF U ON V AND T
In this section, we will investigate how the state function U varies with T and V. To do
so, we will use an important property of state functions. If ƒ is a state function, one can
ƒ
write ∆ƒ = 1i dƒ = ƒfinal - ƒinitial. This equation states that ƒ can be expressed as
an infinitesimal quantity dƒ that when integrated depends only on the initial and final
states; df is called an exact differential. An example of a state function and its exact
differential is U and dU = dq - Pext dV.
For a given amount of a pure substance or a mixture of fixed composition, U is
determined by any two of the three variables P, V, and T. The following discussion will
demonstrate that it is particularly convenient to choose the variables T and V. Because
U is a state function, an infinitesimal change in U can be written as
dU = a
0U
0U
b dT + a b dV
0T V
0V T
(3.8)
This expression indicates that if the state variables change from T, V to T + dT,
V + dV, the change in U, dU, can be determined as follows: First, determine the slopes
of U1T,V2 with respect to T and V and evaluate them at T, V. Next, multiply the values
of these slopes by the increments dT and dV, respectively, Finally, add the two terms.
As long as dT and dV are infinitesimal quantities, higher-order derivatives can be
neglected.
How can numerical values for 10U>0T2V and 10U>0V2T be obtained? In the following, we only consider P-V work. Combining Equation (3.8) and the differential
expression of the first law,
dq - Pext dV = a
0U
0U
b dT + a b dV
0T V
0V T
(3.9)
The symbol dq is used for an infinitesimal amount of heat as a reminder that heat is not
a state function. We first consider processes at constant volume for which dV = 0, so
that Equation (3.9) becomes
dqV = a
0U
b dT
0T V
(3.10)
3.2 DEPENDENCE OF U ON V AND T
Note that in the previous equation, dqV is the product of a state function and an exact
differential. Therefore, dqV behaves like an exact differential, but only because the path
(constant V ) is specified. The quantity dq is not an exact differential.
Although the quantity 10U>0T2V looks very abstract, it can be readily measured.
For example, imagine immersing a container with rigid diathermal walls in a water
bath, where the contents of the container are the system. A process such as a chemical
reaction is carried out in the container, and the heat flow to the surroundings is measured. If heat flow dqV occurs, a temperature increase or decrease dT is observed in
the system and the water bath surroundings. Both of these quantities can be measured.
Their ratio, dqV >dT, is a special form of the heat capacity discussed in Section 2.5:
dqV
0U
= a b = CV
dT
0T V
(3.11)
where dqV >dT corresponds to a constant volume path and is called the heat capacity
at constant volume.
The quantity CV is extensive (that is, it depends on the size of the system), whereas
CV, m is an intensive quantity (that is, it does not depend on the size of the system).
As discussed in Section 2.5, CV, m is different for different substances under the same
conditions. Observations show that CV, m is always positive for a single-phase, pure
substance or for a mixture of fixed composition, as long as no chemical reactions or
phase changes take place in the system. For processes subject to these constraints, U
increases monotonically with T.
Now that we have an expression for CV , we have a way to experimentally determine changes in U with T at constant V for systems of pure substances or for mixtures
of constant composition in the absence of chemical reactions or phase changes. After
CV has been determined as a function of T as discussed in Section 2.11, the following
integral is numerically evaluated:
T2
∆UV =
L
T1
T2
CV dT = n CV, m dT
L
(3.12)
T1
Over a limited temperature range, CV, m can often be regarded as a constant. If this is
the case, Equation (3.12) simplifies to
T2
∆UV =
L
CV dT = CV ∆T = nCV, m ∆T
(3.13)
T1
Equation 3.11 can be written in a different form to explicitly relate qV and ∆U:
ƒ
L
ƒ
dqV =
i
0U
b dT or qV = ∆U
L 0T V
i
a
(3.14)
Although dq is not an exact differential, the integral has a unique value if the path is
defined, as it is in this case (constant volume). Equation (3.14) shows that ∆U for an
arbitrary process in a closed system in which only P-V work is carried out can be
determined by measuring q under constant volume conditions. As will be discussed in
Chapter 4, the technique of bomb calorimetry uses this approach to determine ∆U for
chemical reactions.
Next consider the dependence of U on V at constant T, or 10U>0V2T . This quantity
has the units of J>m3 = 1J>m2>m2 = kg ms-2 >m2 = force>area = pressure and is
called the internal pressure. To explicitly evaluate the internal pressure for different
substances, a result will be used that is derived in the discussion of the second law of
thermodynamics in Section 5.12:
a
0U
0P
b = Ta b - P
0V T
0T V
(3.15)
85
Concept
The heat capacity at constant volume
is a direct measure of the partial
derivative of U with respect to T at
constant V.
86
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Using this equation, we can write the total differential of the internal energy as
Vf , Tf
Vi , Tf
A
T
B
C
Vi , Ti
Vf , Ti
V
dU = dUV + dUT = CV dT + c T a
(3.16)
In this equation, the symbols dUV and dUT have been used, where the subscript indicates which variable is constant. Equation (3.16) is an important result that applies to
systems containing gases, liquids, or solids in a single phase (or mixed phases at a
constant composition) if no chemical reactions or phase changes occur. The advantage
of writing dU in the form given by Equation (3.16) over that in Equation (3.8) is that
10U>0V2T can be evaluated in terms of the system variables P, V, and T and their
derivatives, all of which are experimentally accessible.
Once 10U>0V2T and 10U>0T2V are known, these quantities can be used to determine
dU. Because U is a state function, the path taken between the initial and final states
is unimportant. Three different paths are shown in Figure 3.1, and dU is the same for
these and any other paths connecting Vi, Ti and Vƒ, Tƒ. To simplify the calculation, the
path chosen consists of two segments in which only one of the variables changes in a
given path segment. An example of such a path is Vi, Ti S Vf, Ti S Vf, Tf . Because T is
constant in the first segment,
Figure 3.1
Volume–temperature diagram with
three paths connecting initial and final
states. Paths A, B, and C connect Vi, Ti
and Vƒ, Tƒ. Because U is a state function,
all paths are equally valid in calculating
∆U. Therefore, a specification of the path
is irrelevant.
0P
b - P d dV
0T V
dU = dUT = c T a
0P
b - P d dV
0T V
Because V is constant in the second segment, dU = dUV = CV dT. Finally, the total
change in U is the sum of the changes in the two segments, dUtotal = dUV + dUT .
3.3 DOES THE INTERNAL ENERGY DEPEND
MORE STRONGLY ON V OR T?
Chapter 2 demonstrated that U is a function of T alone for an ideal gas. However, this
statement is not true for real gases, liquids, and solids for which the change in U with V
must be considered. In this section, we investigate if the temperature or the volume dependence of U is most important in determining ∆U for a process of interest. To evaluate
the relative importance of temperature and volume, we will consider separately systems
that consist of an ideal gas, a real gas, a liquid, and a solid. Example Problem 3.2 shows
that Equation (3.15) leads to a simple result for a system consisting of an ideal gas.
EXAMPLE PROBLEM 3.2
Evaluate 10U>0V2T for an ideal gas and modify Equation (3.16) accordingly for the
specific case of an ideal gas.
Solution
a
03 nRT>V4
0U
0P
nRT
b = Ta b - P = Ta
b - P =
- P = 0
0V T
0T V
0T
V
V
Therefore, dU = CVdT, which shows that for an ideal gas, U is a function of T only.
Example Problem 3.2 shows that U is only a function of T for an ideal gas. Specifically,
U is not a function of V. This result is understandable in terms of the potential function
of Figure 1.9. Because ideal gas molecules do not attract or repel one another, no energy
is required to change their average distance of separation (increase or decrease V ):
Tƒ
∆U =
L
Ti
CV1T2 dT
(3.17)
3.3 DOES THE INTERNAL ENERGY DEPEND MORE STRONGLY ON V OR T?
Recall that because U is only a function of T, Equation (3.17) holds for an ideal gas
even if V is not constant.
Next consider the variation of U with T and V for a real gas. The experimental
determination of 10U>0V2T was carried out in the mid-19th century by James Joule
using an apparatus consisting of two glass flasks separated by a stopcock, all of
which were immersed in a water bath. An idealized view of the experiment is shown
in Figure 3.2. As a valve between the volumes is opened, a gas initially in volume
A expands to completely fill the volume A + B. In interpreting the results of this
experiment, it is important to understand where the boundary between the system and
surroundings lies. Here, the decision was made to place the system boundary so that
it includes all the gas. Initially, the boundary lies totally within VA, but it moves during the expansion so that it continues to include all gas molecules. With this choice,
the volume of the system changes from VA before the expansion to VA + VB after the
expansion has taken place.
The first law of thermodynamics [Equation (3.9)] states that
dq - Pext dV = a
0U
0U
b dT + a b dV
0T V
0V T
However, all the gas is contained in the system; therefore, Pext = 0 because a vacuum
cannot exert a pressure. Therefore, Equation (3.9) becomes
dq = a
0U
0U
b dT + a b dV
0T V
0V T
(3.18)
To within experimental accuracy, Joule found that dTsur = 0. Because the water
bath and the system are in thermal equilibrium, dT = dTsur = 0. With this observation,
Joule concluded that dq = 0. Therefore, Equation (3.18) becomes
a
0U
b dV = 0
0V T
(3.19)
Because dV ≠ 0, Joule concluded that 10U>0V2T = 0. Joule’s experiment was not definitive because the experimental sensitivity was limited, as shown in Example Problem 3.3.
EXAMPLE PROBLEM 3.3
In Joule’s experiment to determine 10U>0V2T , the heat capacities of the gas and the
water bath surroundings were related by Csur >Csys ≈ 1000. If the precision with
which the temperature of the surroundings could be measured is {0.006°C, what is
the minimum detectable change in the temperature of the gas?
Solution
View the experimental apparatus as two interacting systems in a rigid adiabatic enclosure.
The first is the volume within vessels A and B, and the second is the water bath and the
vessels. Because the two interacting systems are isolated from the rest of the universe,
q = Cwater ∆Twater + Cgas ∆Tgas = 0
∆Tgas = -
Cwater
∆Twater = -1000 * 1{0.006°C2 = |6°C
Cgas
In this calculation, ∆Tgas is the temperature change that the expanded gas undergoes to
reach thermal equilibrium with the water bath, which is the negative of the temperature change during the expansion.
Because the minimum detectable value of ∆Tgas is rather large, this apparatus
is clearly not suited for measuring small changes in the temperature of the gas in an
expansion.
87
Thermometer
A
P5Pi
B
P50
Water bath
Figure 3.2
Schematic depiction of the Joule
experiment. In this experiment, which
is used to determine 10U>0V2T , two
spherical vessels, A and B, are separated
by a valve. Both vessels are immersed in
a water bath, the temperature of which is
monitored. The initial pressure in each
vessel is indicated.
88
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Concept
U does not depend on V for an
ideal gas.
Joule later carried out more sensitive experiments in collaboration with William
Thomson (Lord Kelvin). These experiments, which are discussed in Section 3.8, demonstrate that 10U>0V2T is small but nonzero for real gases.
Example Problem 3.2 has shown that 10U>0V2T = 0 for an ideal gas. We next calV
culate 10U>0V2T and ∆UT = 1Vm,m,i f 10U>0V2T dVm for a real gas, in which the van der
Waals equation of state is used to describe the gas, as illustrated in Example Problem 3.4.
EXAMPLE PROBLEM 3.4
For a gas described by the van der Waals equation of state, P = nRT>1V - nb2 an2 >V 2. Use this equation to complete these tasks:
a. Calculate 10U>0V2T using 10U>0V2T = T10P>0T2V - P.
V
b. Derive an expression for the change in internal energy, ∆UT = 1Vi ƒ 10U>0V2T dV,
in compressing a van der Waals gas from an initial molar volume Vi to a final
molar volume Vƒ at constant temperature.
Solution
a. T a
0P
°
b - P = T
0T V
=
0c
nRT
n2a
- 2d
V - nb
nRT
V ¢
- P =
- P
0T
V
- nb
V
nRT
nRT
n2a
n2a
+ 2 = 2
V - nb
V - nb
V
V
Vƒ
Vƒ
Vi
Vi
0U
n2a
1
1
b. ∆UT =
a b dV =
dV = n2a a b
2
Vi
Vƒ
L 0V T
LV
Note that ∆UT is zero if the attractive part of the intermolecular potential is zero.
Example Problem 3.4 demonstrates that, in general, 10U>0V2T ≠ 0 and that ∆UT
can be calculated if the equation of state of the real gas is known. This allows the
T
relative importance and value of ∆UV = 1Ti ƒ CV dT to be determined in a process in
which both T and V change, as shown in Example Problem 3.5.
EXAMPLE PROBLEM 3.5
One mole of N2 gas undergoes a change from an initial state described by T = 200. K
and Pi = 5.00 bar to a final state described by T = 400. K and Pƒ = 20.0 bar.
Treat N2 as a van der Waals gas with the parameters a = 0.137 Pa m6 mol-2 and
b = 3.87 * 10-5 m3 mol-1. We use the path N21g, T = 200. K, P = 5.00 bar2 S
N21g, T = 200. K, P = 20.0 bar2 S N21g, T = 400. K, P = 20.0 bar2, keeping in
mind that all paths will give the same answer for ∆U of the overall process.
V
a. Calculate ∆UT = 1Vi ƒ 10U>0V2T dV using the result of Example Problem 3.4.
Note that Vi = 3.28 * 10-3 m3 and Vƒ = 7.88 * 10-4 m3 at 200. K, as calculated using the van der Waals equation of state.
T
b. Calculate ∆UV = n 1Ti ƒ CV, m dT using the following relationship for CV, m in this
temperature range:
CV, m
J K-1 mol-1
= 22.50 - 1.187 * 10-2
T
T2
T3
+ 2.3968 * 10-5 2 - 1.0176 * 10-8 3
K
K
K
The ratios T n >Kn ensure that CV, m has the correct units.
c. Compare the two contributions to ∆U. Can ∆UT be neglected relative to ∆UV?
3.3 DOES THE INTERNAL ENERGY DEPEND MORE STRONGLY ON V OR T?
Solution
a. Using the result of Example Problem 3.4,
∆UT = n2a a
* a
1
1
b = 0.137 Pa m6
Vm, i
Vm, ƒ
1
1
b = -132 J
-3 3
3.28 * 10 m
7.88 * 10-4 m3
Tƒ
b. ∆UV = n
L
CV, m dT
Ti
22.50 - 1.187 * 10-2
400. K
=
L
200. K
±
-1.0176 * 10
-8
T
T2
+ 2.3968 * 10-5 2
K
K
T3
K3
≤ dT J
= 14.50 - 0.712 + 0.447 - 0.06102kJ = 4.17 kJ
c. ∆UT is 3.2% of ∆UV for this case. In this example, and for most processes, ∆UT
can be neglected relative to ∆UV for real gases.
The calculations in Example Problems 3.4 and 3.5 show that to a good approximaV
tion ∆UT = 1Vi ƒ 10U>0V2T dV ≈ 0 for real gases under most conditions. Therefore,
it is sufficiently accurate to consider U as a function of T only 3 U = U1T24 for real
gases in processes that do not involve unusually high gas densities.
Having discussed ideal and real gases, what can be said about the relative magV
T
nitude of ∆UT = 1Vi ƒ 10U>0V2T dV and ∆UV = 1Ti ƒ CV dT for processes involving
liquids and solids? From experiments, it is known that the density of liquids and solids
varies only slightly with the external pressure over the range in which these two forms
of matter are stable. This conclusion is not valid for extremely high-pressure conditions
such as those in the interior of planets and stars. However, it is safe to say that dV for a
solid or liquid is very small in most processes. Therefore,
V2
liq
∆U solid,
T
=
0U
0U
b dV ≈ a b ∆V ≈ 0
0V
0V T
L
T
V1
a
(3.20)
because ∆V ≈ 0. This result is valid even if 10U>0V2T is large.
The conclusion that can be drawn from this section is as follows: Under most
conditions encountered by chemists in the laboratory, U can be regarded as a function
of T alone for all substances. The following equations give a good approximation even
if V is not constant in the process under consideration:
Tƒ
U1Tƒ, Vƒ2 - U1Ti, Vi2 = ∆U =
L
Ti
Tƒ
CV dT = n CV, m dT
L
(3.21)
Ti
Note that Equation (3.21) is only applicable to a process in which there is no change
in the phase of the system, such as vaporization or fusion, and in which there are no
chemical reactions. Changes in U that arise from these processes will be discussed in
Chapters 4 and 8.
Concept
For real gases, liquids, and solids,
the dependence of U on T is much
stronger than on V.
89
90
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
3.4 VARIATION OF ENTHALPY WITH
TEMPERATURE AT CONSTANT PRESSURE
Pext 5 P
Mass
Mass
Piston
Piston
P, Vi ,Ti
Initial state
As for U, H can be defined as a function of any two of the three variables P, V, and T.
It was convenient to choose U to be a function of T and V because this choice led to the
identity ∆U = qV . Using a similar reasoning, we choose H to be a function of T and P.
How does H vary with P and T? The variation of H with T at constant P is discussed next,
and a discussion of the variation of H with P at constant T is deferred to Section 3.6.
Consider the constant pressure process shown schematically in Figure 3.3. For this
process defined by P = Pext,
P, Vf ,Tf
dU = dqP - P dV
Final state
Although the integral of dq is in general path dependent, it has a unique value in this
case because the path is specified, namely, P = Pext = constant. Integrating both sides
of Equation (3.22),
Figure 3.3
Constant pressure process in a piston
and cylinder assembly. The initial and
final states are shown for an undefined
process that takes place at constant
pressure.
ƒ
L
i
ƒ
dU =
L
i
(3.22)
ƒ
dqP -
L
i
P dV or Uƒ - Ui = qP - P1Vƒ - Vi2
(3.23)
Because P = Pƒ = Pi, this equation can be rewritten as
1Uƒ + PƒVƒ2 - 1Ui + PiVi2 = qP or
∆H = qP
(3.24)
The preceding equation shows that the value of ∆H can be determined for an arbitrary
process at constant P in a closed system in which only P-V work occurs by simply
measuring qP, the heat transferred between the system and surroundings in a constant
pressure process. Note the similarity between Equation (3.24) and Equation (3.14).
For an arbitrary process in a closed system in which there is no work other than P-V
work, ∆U = qV if the process takes place at constant V and ∆H = qP if the process
takes place at constant P. These two equations are the basis for the fundamental experimental techniques of bomb calorimetry and constant pressure calorimetry discussed
in Chapter 4.
A useful application of Equation (3.24) is in experimentally determining ∆H and ∆U
of fusion and vaporization for a given substance. Fusion 1solid S liquid2 and vaporization 1liquid S gas2 occur at a constant temperature if the system is held at a constant
pressure and heat flows across the system–surroundings boundary. In both of these
phase transitions, attractive interactions between the molecules of the system must be
overcome. Therefore, q 7 0 in both cases and CP S ∞ . Because ∆H = qP, ∆ fus H
and ∆ vap H can be determined by measuring the heat needed to affect the transition at
constant pressure. Because ∆H = ∆U + ∆1PV2, at constant P,
∆ vap H - ∆ vap U = P∆ vap V 7 0
(3.25)
The change in volume upon vaporization is ∆ vap V = Vgas - Vliq W 0; therefore,
∆ vap U 6 ∆ vap H. An analogous expression to Equation (3.25) can be written relating ∆ fus U and ∆ fus H. Note that ∆Vfus is much smaller than ∆Vvap and can be either
positive or negative. Therefore, ∆ fus U ≈ ∆ fus H. The thermodynamics of fusion and
vaporization will be discussed in more detail in Chapter 8.
Because H is a state function, dH is an exact differential, allowing us to link 10H>0T2P
to a measurable quantity. In analogy to the preceding discussion for dU, dH is written in
the form
dH = a
0H
0H
b dT + a b dP
0T P
0P T
(3.26)
Because dP = 0 at constant P, and dH = dqP from Equation (3.24), Equation (3.26)
becomes
dqP = a
0H
b dT
0T P
(3.27)
3.4 Variation of Enthalpy with Temperature at Constant Pressure
Equation (3.27) allows the heat capacity at constant pressure CP to be defined in
a fashion analogous to CV in Equation (3.11):
dqP
0H
CP =
= a b
dT
0T P
(3.28)
Although this equation looks abstract, CP is a readily measurable quantity. To measure
it, one need only measure the heat flow to or from the surroundings for a constant pressure process, together with the resulting temperature change in the limit in which dT
and dq approach zero and form the ratio lim 1dq>dT2P .
dT S 0
As was the case for CV , CP is an extensive property of the system and varies from
substance to substance. The temperature dependence of CP must be known in order to
calculate the change in H with T. For a constant pressure process in which there is no
change in the phase of the system and no chemical reactions,
Tƒ
∆HP =
L
Ti
Tƒ
CP1T2dT = n CP, m1T2dT
L
(3.29)
Ti
If the temperature interval is small enough, it can usually be assumed that CP is constant. In that case,
∆HP = CP ∆T = nCP,m ∆T
(3.30)
Example Problem 3.6 shows a case in which the temperature dependence of CP cannot
be neglected.
The calculation of ∆H for chemical reactions and changes in phase will be discussed in Chapters 4 and 8.
EXAMPLE PROBLEM 3.6
A 143.0 g sample of C1s2 in the form of graphite is heated from 300. to 600. K at a
constant pressure. Over this temperature range, CP, m has been determined to be
CP, m
-1
J K mol
-1
= -12.19 + 0.1126
-7.800 * 10-11
T
T2
T3
- 1.947 * 10-4 2 + 1.919 * 10-7 3
K
K
K
T4
K4
Calculate ∆H and qP. How large is the relative error in ∆H if we neglect the
temperature-dependent terms in CP, m and assume that CP, m maintains its value at
300. K throughout the temperature interval?
Solution
Tƒ
m
∆H =
C 1T2dT
M L P, m
Ti
=
T
T2
- 1.947 * 10-4 2 + 1.919
K
J
K
±
≤ dT
3
4
T
T
mol
L
-7
- 7.800 * 10-11 4
300. K * 10
K
K3
600. K
143.0 g
12.00 g mol-1
-12.19 + 0.1126
600. K
T
T2
T3
+ 0.0563 2 - 6.49 * 10-5 3 + 4.798
143.0
K
K
K
=
* ≥
¥
J = 46.9 kJ
4
5
12.00
T
T
* 10-8 4 - 1.56 * 10-11 5
K
K
300. K
-12.19
From Equation (3.24), ∆H = qP.
Concept
The heat capacity at constant
pressure is a direct measure of the
partial derivative of H with respect
to T at constant P.
91
92
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
If we had assumed CP, m = 8.617 J mol-1 K-1, which is the calculated value at
300. K, ∆H = 143.0 g>12.00 g mol-1 * 8.617 J K-1 mol-1 * 3 600. K - 300. K4 =
30.8 kJ. The relative error is 100 * 130.8 kJ - 46.9 kJ2>46.9 kJ = -34.3%. In this
case, it is not reasonable to assume that CP, m is independent of temperature.
3.5 HOW ARE CP AND CV RELATED?
To this point, two distinct heat capacities, CP and CV, have been defined. How are these
quantities related? To answer this question, the differential form of the first law is written as
dq = CV dT + a
0U
b dV + Pext dV
0V T
(3.31)
Consider a process that proceeds at constant pressure for which P = Pext . In this case,
Equation (3.31) becomes
Because dqP = CPdT,
CP = CV + a
dqP = CV dT + a
0U
b dV + P dV
0V T
(3.32)
0U
0V
0V
0U
0V
b a b + P a b = CV + c a b + P d a b
0V T 0T P
0T P
0V T
0T P
= CV + T a
0P
0V
b a b
0T V 0T P
(3.33)
To obtain Equation (3.33), both sides of Equation (3.32) have been divided by dT, and
the ratio dV>dT has been converted to a partial derivative at constant P. Equation (3.15)
has been used in the last step. Using Equation (3.5) and the cyclic rule, one can simplify
Equation (3.33) to
0V 2
a b
0T P
0P
0V
CP = CV + T a b a b = CV - T
0T V 0T P
0V
a b
0P T
CP = CV + TV
a2
a2
or CP, m = CV, m + TVm
kT
kT
(3.34)
Equation (3.34) provides another example of the usefulness of the formal theory of
thermodynamics in linking seemingly abstract partial derivatives with experimentally
available data. The difference between CP, m and CV, m can be determined at a given
temperature knowing only the molar volume, the isobaric volumetric thermal expansion coefficient, and the isothermal compressibility.
Equation (3.34) is next applied to ideal and real gases, as well as liquids and solids, in the absence of phase changes and chemical reactions. Because a and kT are
always positive for real and ideal gases, CP - CV 7 0 for these substances. For an
0P
0V
ideal gas, 10U>0V2T = 0 as shown in Example Problem 3.2, and T a b a b =
0T V 0T P
nR nR
T a b a b = nR, so that Equation (3.33) becomes
V
P
CP - CV = nR
(3.35)
This result was stated without derivation in Section 2.4. Turning to liquids and solids,
we note that the partial derivative 10V>0T2P ≈ Va is much smaller for liquids and
solids than for gases. Therefore, generally
CV W c a
0U
0V
b + Pd a b
0V T
0T P
(3.36)
3.6 Variation of Enthalpy with Pressure at Constant Temperature
so that CP ≈ CV for a liquid or a solid. As shown earlier in Example Problem 3.1, it is
not feasible to carry out heating experiments for liquids and solids at constant volume
because of the large pressure increase that occurs. Therefore, tabulated heat capacities
for liquids and solids list CP, m rather than CV, m.
3.6 VARIATION OF ENTHALPY WITH PRESSURE
AT CONSTANT TEMPERATURE
In the previous section, we learned how H changes with T at constant P. To calculate how H changes as both P and T change, 10H>0P2T must be calculated. The partial derivative 10H>0P2T is less straightforward to determine in an experiment than
10H>0T2P. As we will show, for many processes involving changes in both P and T,
10H>0T2P dT W 10H>0P2T dP, and the pressure dependence of H can be neglected
relative to its temperature dependence. However, the knowledge that 10H>0P2T is not
zero is essential for understanding the operation of a refrigerator and the liquefaction of
gases. The following discussion is applicable to gases, liquids, and solids.
Given the definition H = U + PV, we begin by writing dH as
dH = dU + P dV + V dP
(3.37)
Substituting the differential forms of dU and dH,
CPdT + a
0H
0U
b dP = CV dT + a b dV + P dV + V dP
0P T
0V T
= CV dT + c a
0U
b + P d dV + V dP
0V T
(3.38)
For isothermal processes, dT = 0, and Equation (3.38) can be rearranged to
a
0H
0U
0V
b = ca b + Pd a b + V
0P T
0V T
0P T
(3.39)
Using Equation (3.15) for 10U>0V2T ,
a
0H
0P
0V
b = Ta b a b + V
0P T
0T V 0P T
= V - Ta
0V
b ≈ V11 - Ta2
0T P
(3.40)
The second formulation of Equation (3.40) is obtained through application of the cyclic
rule [Equation (3.3)]. This equation is applicable to all systems containing pure substances or mixtures at a fixed composition, provided that no phase changes or chemical
reactions take place. The quantity 10H>0P2T is evaluated for an ideal gas in Example
Problem 3.7.
EXAMPLE PROBLEM 3.7
Evaluate 10H>0P2T for an ideal gas.
Solution
10P>0T2V = 103 nRT>V4 >0T2V = nR>V and 10V>0P2T = 1d3 nRT>P4 >dP2T =
-nRT>P 2 for an ideal gas. Therefore,
a
0H
0P
0V
nR
nRT
nRT nRT
b = T a b a b + V = T a- 2 b + V = + V = 0
0P T
0T V 0P T
V
P nRT
P
This result could have also been derived directly from the definition H = U + PV.
Thus, for an ideal gas U = U1T2 only and PV = nRT. Therefore, H = H1T2 for
an ideal gas and 10H>0P2T = 0.
93
94
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Concept
Because Example Problem 3.7 shows that H is only a function of T for an ideal gas,
Tƒ
H does not depend on P for an
ideal gas.
∆H =
L
Ti
Tƒ
CP1T2dT = n CP,m1T2dT
L
(3.41)
Ti
for an ideal gas. Because H is only a function of T, Equation (3.41) holds for an ideal
gas even if P is not constant. This result is also understandable in terms of the potential
function of Figure 1.9. Because ideal gas molecules do not attract or repel one another, no
energy is required to change their average distance of separation (increase or decrease P).
Equation (3.40) is next applied to several types of systems. We have seen that
10H>0P2T = 0 for an ideal gas. For liquids and solids, 1 W Ta for T 6 1000 K, as
can be seen in Table 3.1. Therefore, for liquids and solids, 10H>0P2T ≈ V to a good
approximation, and dH can be written as
dH ≈ CP dT + V dP
(3.42)
for systems that consist only of liquids or solids.
EXAMPLE PROBLEM 3.8
Calculate the change in enthalpy when 124 g of liquid methanol initially at 1.00 bar
and 298 K undergoes a change of state to 2.50 bar and 425 K. The density of liquid
methanol under these conditions is 0.791 g cm-3, and CP, m for liquid methanol is
81.1 J K-1 mol-1.
Solution
Because H is a state function, any path between the initial and final states will give the
same ∆H. We choose the path methanol (l, 1.00 bar, 298 K) S methanol (l, 1.00 bar,
425 K) S methanol (l, 2.50 bar, 425 K). The first step is isothermal, and the second
step is isobaric. The total change in H is
Tƒ
∆H = n
L
Pƒ
CP,mdT +
Ti
L
Pi
V dP ≈ nCP,m 1Tƒ - Ti2 + V1Pƒ - Pi2
= 81.1 J K-1 mol-1 *
+
124 g
0.791 g cm-3
*
124 g
32.04 g mol-1
* 1425 K - 298 K2
10-6 m3
105 Pa
*
12.50
bar
1.00
bar2
*
bar
cm3
= 39.9 * 103 J + 23.5 J ≈ 39.9 kJ
Note that the contribution to ∆H from the change in T is far greater than that from
the change in P.
Concept
Except for large pressure changes,
H does not depend on P for liquids
and solids.
Example Problem 3.8 shows that because molar volumes of liquids and solids are
small, H changes much more rapidly with T than with P. Under most conditions, H can
be assumed to be a function of T only for solids and liquids. Exceptions to this rule are
encountered in geophysical or astrophysical applications, for which extremely large
pressure changes can occur.
The following conclusion can be drawn from this section: under most conditions
encountered by chemists in the laboratory, H can be regarded as a function of T alone
for liquids and solids. It is a good approximation to write
T2
H1Tƒ, Pƒ2 - H1Ti, Pi2 = ∆H =
L
T1
T2
CP dT = n CP, m dT
L
T1
(3.43)
3.7 The Joule–Thomson Experiment
95
even if P is not constant in the process under consideration. The dependence of H on
P for real gases is discussed in Section 3.8 and Section 3.9 in the context of the Joule–
Thomson experiment.
Note that Equation (3.43) is only applicable to a process in which there is no change
in the phase of the system, such as vaporization or fusion, and in which there are no
chemical reactions. Changes in H that arise from chemical reactions and changes in
phase will be discussed in Chapters 4 and 8.
Having dealt with solids, liquids, and ideal gases, we are left with real gases. For
real gases, 10H>0P2T and 10U>0V2T are small, but the values of these partial derivatives
still have a considerable effect on the properties of the gases upon expansion or compression. Conventional technology for the liquefaction of gases and for the operation
of refrigerators is based on the fact that 10H>0P2T and 10U>0V2T are not zero for real
gases. To derive a useful formula for calculating 10H>0P2T for a real gas, the Joule–
Thomson experiment is discussed in the next section.
3.7 THE JOULE–THOMSON EXPERIMENT
If the valve on a cylinder of compressed N2 at 298 K is opened fully, it will become
covered with frost, demonstrating that the temperature of the valve is lowered below the
freezing point of H2O. A similar experiment, using a cylinder that contains H2, leads
to a considerable increase in temperature and, potentially, an explosion. How can these
effects be understood? To explain them, we discuss the Joule–Thomson experiment.
The Joule–Thomson experiment, shown in Figure 3.4, can be viewed as an improved version of the Joule experiment because it allows 10U>0V2T to be measured
with a much higher sensitivity than in the Joule experiment. In this experiment, gas
flows from the high-pressure cylinder on the left to the low-pressure cylinder on the
right through a porous plug in an insulated pipe. The pistons move to keep the pressure
unchanged in each region until all the gas has been transferred to the region to the
right of the porous plug. If N2 is used in the expansion process 1P1 7 P22, it is found
that T2 6 T1; in other words, the gas is cooled as it expands. What is the origin of this
effect? Consider an amount of gas equal to the initial volume V1 as it passes through the
apparatus from left to right. The total work in this expansion process is the sum of the
work performed on each side of the plug separately by the moving pistons:
0
w = wleft + wright = -
L
V1
V2
P1 dV -
L
P2 dV = P1V1 - P2V2
(3.44)
0
Pressure gauges
Figure 3.4
P1V1T1
Initial state
Porous plug
P2V2T2
Final state
The Joule–Thomson experiment. In
the experiment, a gas is forced through a
porous plug using a piston and cylinder
mechanism. The pistons move to maintain
a constant pressure in each region. There
is an appreciable pressure drop across the
plug, and the temperature change of the
gas is measured. The upper and lower
figures show the initial and final states,
respectively. As shown in the text, if the
piston and cylinder assembly forms an
adiabatic wall between the system (the
gases on both sides of the plug) and the
surroundings, the expansion is isenthalpic.
96
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
Because the pipe is insulated, q = 0, and
∆U = U2 - U1 = w = P1V1 - P2V2
(3.45)
This equation can be rearranged to
U2 + P2V2 = U1 + P1V1 or H2 = H1
Concept
The Joule Thomson experiment
provides a way to measure the partial
derivative of U with respect to V at
constant T and the partial derivative
of H with respect to P at constant T
for real gases.
TABLE 3.3 Joule–Thomson
Coefficients for Selected
Substances at 273 K and 1 atm
Gas
Ar
CH4
mJ - T 1K>MPa2
mJ - T = lim a
∆P S 0
10.9
-0.34
He
-0.62
N2
2.15
Ne
-0.30
NH3
28.2
2.69
Source: Data from Linstrom, P. J., and Mallard,
W. G., eds. NIST Chemistry Webbook: NIST
Standard Reference Database Number 69.
Gaithersburg, MD: National Institute of
Standards and Technology. Retrieved from
http://webbook.nist.gov.
∆T
0T
b = a b
∆P H
0P H
(3.47)
If mJ - T is positive, the conditions are such that the attractive part of the potential
dominates, and if mJ - T is negative, the repulsive part of the potential dominates.
Using experimentally determined values of mJ - T , 10H>0P2T can be calculated. For an
isenthalpic process,
0H
dH = CP dT + a b dP = 0
(3.48)
0P T
Dividing through by dP and making the condition dH = 0 explicit,
CP a
4.38
H2
O2
Note that the enthalpy is constant in the expansion; the expansion is isenthalpic.
For the conditions of the experiment that uses N2, both dT and dP are negative, so
10T>0P2H 7 0. The experimentally determined limiting ratio of ∆T to ∆P at constant
enthalpy is known as the Joule–Thomson coefficient:
3.66
CO2
(3.46)
0T
0H
b + a b = 0
0P H
0P T
giving a
0H
b = -CPmJ - T
0P T
(3.49)
Equation (3.49) states that 10H>0P2T can be calculated using the measurement of
material-dependent properties CP and mJ - T . Because mJ - T is not zero for a real gas,
the pressure dependence of H for an expansion or compression process for which the
pressure change is large cannot be neglected. Note that 10H>0P2T can be positive or
negative, depending on the sign of mJ - T at the P and T of interest.
If mJ - T is known from experiment, 10U>0V2T can be calculated as shown in
Example Problem 3.9. This has the advantage that a calculation of 10U>0V2T based
on measurements of CP, mJ - T and the isothermal compressibility kT is much more
accurate than a measurement based on the Joule experiment. Values of mJ - T are shown
for selected gases in Table 3.3. Keep in mind that mJ - T is a function of T, P and ∆P,
so the values listed in the table are only valid for a small pressure decrease originating
at 1 atm pressure.
EXAMPLE PROBLEM 3.9
Using Equation (3.39), 10H>0P2T = 3 10U>0V2T + P4 10V>0P2T + V, derive an
expression giving 10U>0V2T entirely in terms of measurable quantities for a gas.
Solution
a
0H
0U
0V
b = ca b + Pda b + V
0P T
0V T
0P T
0U
a b =
0V T
=
a
0H
b - V
0P T
- P
0V
a b
0P T
CPmJ - T + V
- P
kTV
In this equation, kT is the isothermal compressibility defined in Equation (3.5).
3.8 Liquefying Gases Using an Isenthalpic Expansion
700
EXAMPLE PROBLEM 3.10
Using Equation (3.39),
600
Temperature (K)
0H
0U
0V
a b = ca b + Pd a b + V
0P T
0V T
0P T
show that mJ - T = 0 for an ideal gas.
Solution
mJ - T = -
1 0H
1
0U
0V
0V
a b = ca b a b + P a b + Vd
CP 0P T
CP 0V T 0P T
0P T
N2
500
400
300
200
H2
100
1
0V
= c0 + P a b + Vd
CP
0P T
100
03 nRT>P4
1
1
nRT
= cP a
b + Vd = c+ Vd = 0
CP
0P
CP
P
T
Example Problem 3.10 shows that for an ideal gas, mJ - T is zero. It can be shown
that for a van der Waals gas in the limit of zero pressure
1
2a
a
- bb
CP,m RT
200 300 400
Pressure (atm)
500
Figure 3.5
In this calculation, we have used the result that 10U>0V2T = 0 for an ideal gas.
mJ - T =
97
(3.50)
3.8 LIQUEFYING GASES USING
AN ISENTHALPIC EXPANSION
For real gases, the Joule–Thomson coefficient mJ - T can take on either negative or positive values in different regions of P-T space. If mJ - T is positive, a decrease in pressure leads to a cooling of the gas; if it is negative, the expansion of the gas leads to a
heating. Figure 3.5 shows the variation of mJ - T with T and P for N2 and H2. All along
the solid curve, mJ - T = 0. To the left of each curve, mJ - T is positive, and to the right,
it is negative. The temperature for which mJ - T = 0 is referred to as the inversion temperature. If the expansion conditions are kept in the region in which mJ - T is positive,
∆T can be made sufficiently large as ∆P decreases in the expansion to liquefy the gas.
Note that Equation (3.50) predicts that the inversion temperature for a van der Waals
gas is independent of P, which is not in agreement with experiment.
The Joule–Thomson coefficients for N2 and H2 are shown as a function of temperature and pressure in Figure 3.5. The information in Figure 3.5 is in accord with
the observation that a high-pressure 1100 6 P 6 500 atm2 expansion of N2 at 300 K
leads to cooling, whereas similar conditions for H2 lead to heating. To cool H2 in an
expansion, it must first be precooled below 200 K, and the pressure must be less than
160 atm. He and H2 are heated in an isenthalpic expansion at 300 K for P 6 200 atm.
The Joule–Thomson effect can be used to liquefy gases such as N2. As shown in
Figure 3.6, the gas at atmospheric pressure is first compressed to a value of 50 atm
to 200 atm, which leads to a substantial increase in its temperature. It is cooled and
subsequently passed through a heat exchanger in which the gas temperature decreases
to a value within ∼50 K of the boiling point. At the exit of the heat exchanger, the gas
expands through a nozzle to a final pressure of 1 atm in an isenthalpic expansion. The
cooling that occurs because mJ - T 7 0 results in liquefaction. The gas that boils away
passes back through the heat exchanger in the opposite direction in which the gas to
be liquefied is passing. The two gas streams are separated, but they are in good thermal contact. In this process, the gas to be liquefied is effectively precooled, enabling a
single-stage expansion to achieve liquefaction.
Plot of Joule–Thomson coefficients for
N2 and H2. Along the curves in the figure
mJ - T = 0. The value of mJ - T is positive
to the left of the curves and negative to the
right. To experience cooling upon expansion at 100. atm, T must lie between 50. K
and 150. K for H2. The corresponding
temperatures for N2 are 100. K and 650. K.
Connection
The dependence of H on P at
constant T for real gases allows
gases to be liquified.
Cooler
Gas feed
Compressor
Liquid out
Figure 3.6
Schematic depiction of the liquefaction
of a gas using an isenthalpic Joule–
Thomson expansion. Heat is extracted
from the gas exiting from the compressor.
It is further cooled in the countercurrent
heat exchanger before expanding through
a nozzle. Because its temperature is sufficiently low at the exit to the countercurrent heat exchanger, liquefaction occurs.
98
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
VOCABULARY
internal pressure
isenthalpic
isobaric volumetric thermal expansion coefficient
isothermal compressibility
cyclic rule
exact differential
heat capacity at constant pressure
heat capacity at constant volume
Joule–Thomson coefficient
Joule–Thomson experiment
partial derivatives
KEY EQUATIONS
Equation
a
Significance of Equation
0P
0V
0T
b a b a b = -1
0V T 0T P 0P V
a =
1 0V
1 0V
a b
and kT = - a b
V 0T P
V 0P T
dqV
0U
= a b = CV
dT
0T V
ƒ
L
ƒ
dqV =
i
0U
b dT or qV = ∆U
L 0T V
i
a
Tƒ
U1Tƒ, Vƒ2 - U1Ti, Vi2 = ∆U =
L
Ti
LTi
CV, m dT
∆H = qP
dqP
0H
= a b
dT
0T P
CP = CV + TV
a2
a2
or CP, m = CV, m + TVm
kT
kT
T2
H1Tƒ, Pƒ2 - H1Ti, Pi2 = ∆H =
mJ - T = lim a
∆P S 0
L
Cyclic rule
3.2
Definition of volumetric thermal expansion
coefficient and the isothermal compressibility
3.5
Definition of heat capacity at constant volume
3.11
∆U can be measured by determining heat
flow at constant volume
3.14
Under most conditions, U = U1T2 alone
3.21
Changes in enthalpy can be measured by
determining heat flow at constant pressure.
3.24
Definition of heat capacity at constant
pressure
3.28
Relation between CP and CV for any substance
3.34
Under most conditions, H = H1T2 alone
3.43
Definition of Joule–Thomson coefficient
3.47
Tƒ
CV dT = n
1Uƒ + PƒVƒ2 - 1Ui + PiVi2 = qP or
CP =
Equation Number
T2
CP dT = n
T1
L
CP, m dT
T1
∆T
0T
b = a b
∆P H
0P H
CONCEPTUAL PROBLEMS
Q3.1 The heat capacity CP,m is less than CV,m for H2O1l2
near 4°C. Explain this result.
Q3.2 What is the physical basis for the experimental result
that U is a function of V at constant T for a real gas? Under
what conditions will U decrease as V increases?
Q3.3 Explain why ethene has a higher value for CV,m at
800. K than CO.
Q3.4 Why does the relation CP 7 CV always hold for a gas?
Can CP 6 CV be valid for a liquid?
Q3.5 Why is CP,m a function of temperature at 300. K for
ethane but not for argon?
Q3.6 Explain, without using equations, why 10H>0P2T is
generally small for a real gas.
Numerical Problems
Q3.7 Why is it reasonable to write dH ≈ CPdT + VdP for a
liquid or solid sample?
Q3.8 Refer to Figure 1.9 and explain why 10U>0V2T is generally small for a real gas.
Q3.9 Can a gas be liquefied through an isenthalpic expansion
if mJ - T = 0?
Q3.10 Why is qv = ∆U only for a constant volume
process? Is this formula valid if work other than P–V work
is possible?
Q3.11 Classify the following variables and functions as
intensive or extensive: T, P, V, q, w, U, H.
Q3.12 Why are q and w not state functions?
T
Q3.13 Why is the equation ∆H = 1Ti ƒCP1T2 dT =
Tƒ
n 1Ti CP,m1T2 dT valid for an ideal gas even if P is not constant in the process? Is this equation also valid for a real gas?
Why or why not?
Q3.14 What is the relationship between a state function and
an exact differential?
99
Q3.15 Is the following statement always, never, or sometimes valid? Explain your reasoning: ∆H is only defined for a
constant pressure process.
Q3.16 Is the following statement always, never, or sometimes
valid? Explain your reasoning: A thermodynamic process is
completely defined by the initial and final states of the system.
Q3.17 Is the following statement always, never, or sometimes
valid? Explain your reasoning: q = 0 for a cyclic process.
Q3.18 The molar volume of H2O1l2 decreases with increasing
temperature near 4°C. Can you explain this behavior using a
molecular-level model?
Q3.19 Why was the following qualification made in
Section 3.6? Note that Equation (3.43) is only applicable to
a process in which there is no change in the phase of the system, such as vaporization or fusion, and in which there are no
chemical reactions.
T
T
Q3.20 Is the expression ∆UV = 1T12CV dT = n 1T12CV,m dT
only valid for an ideal gas if V is constant?
NUMERICAL PROBLEMS
Section 3.1
P3.1 One consequence of climate change is the rise in sea
level as the oceans warm and as land based ice melts and
increases the volume of the oceans. The volume of the Earth’s
oceans is 1.35 * 109 km3, and the average depth is 3668 m.
The volume of the Greenland ice sheet is 2.85 * 106 km3.
Using the data tables, calculate (a) the rise in sea level expected if the ocean temperature increases by 2.0°C, the current maximum increase targeted by the UN intergovernmental
panel on climate change and (b) the rise in sea level expected
if the Greenland ice sheet melts. State any assumptions made.
P3.2 Obtain an expression for the isothermal compressibility
kT = -1>V10V>0P2T for a van der Waals gas.
P3.3 A vessel is filled completely with liquid water and
sealed at 20.0°C and a pressure of 1.00 bar. What is the pressure if the temperature of the system is raised to 65.0°C?
Under these conditions, awater = 2.04 * 10-4 K-1,
avessel = 1.42 * 10-4 K-1, and kT = 4.59 * 10-5 bar -1.
P3.4 Integrate the expression a = 1>V10V>0T2P assuming
that a is independent of pressure. By doing so, obtain an
expression for V as a function of T and a at constant P.
P3.5 The function ƒ1x, y2 is given by ƒ1x, y2 = xy sin 5x +
2
x 2 2y ln y + 3e-2x cos y. Determine
a
0ƒ
0ƒ
02ƒ
02ƒ
0 0ƒ
0 0ƒ
b , a b , a 2 b , a 2 b , a a b b and a a b b
0x y 0y x 0x y 0y x 0y 0x y x
0x 0y x y
Obtain an expression for the total differential dƒ.
P3.6 Using the result of Equation (3.4), 10P>0T2V = a>kT ,
express a as a function of kT and Vm for an ideal gas, and a as
a function of b, kT , and Vm for a van der Waals gas.
P3.7 Starting with the van der Waals equation of state,
find an expression for the total differential dP in terms of
dV and dT. By calculating the mixed partial derivatives
1010P>0V2T >0T2V and 1010P>0T2V >0V2T , determine if dP is
an exact differential.
P3.8 Use 10U>0V2T = 1aT - kTP2>kT to calculate
10U>0V2T for an ideal gas.
P3.9 A differential dz = ƒ1x, y2dx + g1x, y2dy is exact if the
integral 1 ƒ1x, y2dx + 1 g1x, y2dy is independent of the path.
Demonstrate that the differential dz = 2xydx + x 2dy is exact
by integrating dz along the paths (1,1) S (1, 8) S (6, 8) and
(1, 1) S (1, 3) S (4, 3) S (4, 8) S (6, 8). The first number
in each set of parentheses is the x coordinate, and the second
number is the y coordinate.
P3.10 Show that dr>r = -adT + kTdP where r is the density r = m>V. Assume that the mass, m, is constant.
P3.11 Because V is a state function, 1010V>0T2P >0P2T =
1010V>0P2T >0T2P. Using this relationship, show that the
isothermal compressibility and isobaric expansion coefficient
are related by 10a>0P2T = -10kT >0T2P.
P3.12 Starting with a = 11>V210V>0T2P, show that
a = -11>r210r>0T2P, where r is density.
P3.13 This problem will give you practice in using the
cyclic rule. Use the ideal gas law to obtain the three functions
P = ƒ1V, T2, V = g1P, T2, and T = h1P, V2. Show that the
cyclic rule 10P>0V2T 10V>0T2P 10T>0P2V = -1 is obeyed.
P3.14 Regard the enthalpy as a function of T and P. Use the
cyclic rule to obtain the expression
CP = - a
0H
0T
b na b
0P T
0P H
100
CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy
P3.15 Derive the equation 10P>0V2T = -1>1kTV2 from
basic equations and definitions.
P3.16 Derive the equation 10H>0T2V = CV + Va>kT from
basic equations and definitions.
0CV
02P
P3.17 Show that a
b = T a 2b
0V T
0T V
0CV
P3.18 Show that a
b = 0 for both an ideal and a van der
0V T
Waals gas.
P3.19 Calculate the isobaric volumetric thermal expansion
coefficient and the isothermal compressibility, respectively,
defined by
1 0V
1 0V
a = a b
and kT = - a b
V 0T P
V 0P T
for an ideal gas at 0°C and 1.00 bar. Repeat for an ideal gas at
20°C and 1.00 atm.
Section 3.2
P3.20 Show that the expression 10U>0V2T = T10P>0T2V - P
can be written in the form
a
0U
P
P
1
b = T 2 a0 c d n 0T b = -a0 c d n 0 c d b
0V T
T
T
T
V
V
P3.21 Derive an expression for the internal pressure of a gas
RT
a
that obeys the Berthelot equation of state, P =
Vm - b
TV 2m
P3.22 Because U is a state function, 10>0V10U>0T2V2T =
10>0T10U>0V2T2V . Using this relationship, show that
10CV >0V2T = 0 for an ideal gas.
P3.23 Derive the following relation,
0U
3a
b =
0Vm T
22TVm1Vm + b2
for the internal pressure of a gas that obeys the Redlich–Kwong
equation of state,
RT
a
1
P =
.
Vm - b
V
1V
2T m m + b2
a
P3.24 A mass of 28.0 g of H2O(s) at 273 K is dropped into
250. g of H2O(l) at 298 K in an insulated container at 1 bar
of pressure. Calculate the temperature of the system once
equilibrium has been reached. Assume that CP,m for H2O is
constant at its values for 298 K throughout the temperature
range of interest.
Section 3.3
P3.25 A mass of 55.0 g of H2O(g) at 373 K is allowed to
flow into 550. g of H2O(l) at 298 K and 1 atm. Calculate the
final temperature of the system once equilibrium has been
reached. Assume that CP,m for H2O is constant at its values
for 298 K throughout the temperature range of interest.
P3.26 A 50.0 g piece of gold at 700. K is dropped into 200 g of
H2O(l) at 298 K in an insulated container at 1 bar pressure. Calculate the temperature of the system once equilibrium has been
reached. Assume that CP,m for Au and H2O are constant at their
values for 298 K throughout the temperature range of interest.
P3.27 Calculate w, q, ∆H, and ∆U for the process in which
1.30 mol of water undergoes the transition H2O(l, 373 K) S
H2O(g, 610. K) at 1 bar of pressure. The volume of liquid
water at 373 K is 1.89 * 10-5 m3 mol-1, and the volume
of steam at 373 K and 610. K is 3.03 * 10-2 m3 mol-1and
5.06 * 10-2 m3 mol-1, respectively. For steam, CP,m can be
considered constant over the temperature interval of interest
at 33.58 J mol-1 K-1.
P3.28 The Joule coefficient is defined by 10T>0V2U =
11>CV2 3 P - T10P>0T2V 4 . Calculate the Joule coefficient for
an ideal gas and for a van der Waals gas.
Section 3.4
0U
0H
P3.29 For an ideal gas, a b and a b = 0. Prove that
0V T
0P T
CV is independent of volume and CP is independent of pressure.
Section 3.5
P3.30 Derive the following expression for calculating the
isothermal change in the constant volume heat capacity:
10CV >0V2T = T102P>0T 22V .
P3.31 Equation (3.34), CP = CV + TV1a2 >kT2, links CP and
CV with a and kT . Use this equation to evaluate CP - CV for
an ideal gas.
P3.32 Use the result of Problem P3.30 to derive a formula for
10CV >0V2T for a gas that obeys the Redlich–Kwong equation
RT
a
1
of state, P =
Vm - b
2T Vm1Vm + b2
P3.33 Use the relation
CP,m - CV,m = T a
0Vm
0P
b a b
0T P 0T V
the cyclic rule, and the van der Waals equation of state to
derive an equation for CP, m - CV, m in terms of Vm, T, and
the gas constants R, a, and b.
P3.34 For the equation of state Vm = RT>P + B1T2, show that
a
0CP, m
0P
b = -T
T
d 2B1T2
dT 2
[Hint: Use Equation 3.40 and the property of state functions
with respect to the order of differentiation in mixed second
derivatives.]
0U
0V
P3.35 Prove that CV = - a b a b
0V T 0T U
Section 3.6
P3.36 The molar heat capacity CP,m of SO21g2 is described by
the following equation over the range 300 K 6 T 6 1700 K:
CP,m
R
= 3.093 + 6.967 * 10-3
+ 1.035 * 10-9
T3
K3
T
T2
- 45.81 * 10-7 2
K
K
In this equation, T is the absolute temperature in Kelvin. The
ratios T n >Kn ensure that CP,m has the correct dimension. Assuming ideal gas behavior, calculate q, w, ∆U, and ∆H if
Further Reading
1.35 mol of SO21g2 is heated from 25.0°C to 1250°C at a
constant pressure of 1 bar. Explain the sign of w.
P3.37 For a gas that obeys the equation of state
Vm =
RT
+ B1T2
P
derive the result
a
dB1T2
0H
b = B1T2 - T
0P T
dT
101
Section 3.7
P3.39 Because 10H>0P2T = -CPmJ - T , the change in
enthalpy of a gas expanded at constant temperature can be
calculated. To do so, the functional dependence of mJ - T on P
must be known. Treating Xe as a van der Waals gas, calculate
∆H when 1 mol of Xe is expanded from 25 bar to 1.75 bar
at 375 K. Assume that mJ - T is independent of pressure and is
given by mJ - T = 3 12a>RT2 - b4 >CP,m and CP,m = 5R>2
for Xe. What value would ∆H have if the gas exhibited ideal
gas behavior?
P3.38 Use the relation 10U>0V2T = T10P>0T2V - P and the
cyclic rule to obtain an expression for the internal pressure,
10U>0V2T , in terms of P, a, T, and kT .
FURTHER READING
Apfelbaum, E. M., and Vorb’ev, V. S. “The Similarity Law for
the Joule–Thomson Inversion Curve.” Journal of Physical
Chemistry B 118 (2014): 12239–12242.
Barrante, J. R. “Applied Mathematics for Physical
Chemistry.” Chapter 4. Differential Calculus, 3rd Edition.
New York: Pearson, 2004.
Goussard, J.-O., and Roulet, B. “Free Expansion for Real
Gases.” American Journal of Physics 61 (1993): 845–848.
Halpern, A. M., and Gozashti, S. “An Improved Apparatus
for the Measurement of the Joule–Thomson Coefficient
of Gases.” Journal of Chemical Education 63 (1986):
1001–1002.
Hecht, C. E., and Zimmerman, G. “Measurement of the
Joule–Thomson Coefficient.” Journal of Chemical
Education 31 (1954): 530–533.
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C H A P T E R
4
Thermochemistry
WHY is this material important?
Thermochemistry is the branch of thermodynamics that investigates the heat flow into
or out of a reaction system. As reactants are converted into products, energy can either
be absorbed by the system or released to the surroundings. By measuring this energy
transfer, chemical bond energies and enthalpies can be determined.
4.1
Energy Stored in Chemical
Bonds Is Released or Absorbed
in Chemical Reactions
4.2
Internal Energy and Enthalpy
Changes Associated with
Chemical Reactions
4.3
Hess’s Law Is Based on Enthalpy
Being a State Function
4.4
Temperature Dependence
of Reaction Enthalpies
4.5
Experimental Determination
of ∆U and ∆H for Chemical
Reactions
4.6
(Supplemental Section)
Differential Scanning Calorimetry
WHAT are the most important concepts and results?
For a reaction that occurs at constant volume, the heat that flows into or out of the
system is equal to ∆U for the reaction. For a reaction that takes place at constant
pressure, the heat that flows into or out of the system is equal to ∆H for the reaction.
Because H and U are state function, the reaction energy or enthalpy can be written as
the energies or enthalpies of formation of the products minus those of the reactants.
This property allows ∆H and ∆U for a reaction to be calculated for many reactions
without carrying out an experiment.
WHAT would be helpful for you to review for this chapter?
You should be familiar with state functions and in particular with U and H.
4.1 ENERGY STORED IN CHEMICAL BONDS
IS RELEASED OR ABSORBED IN CHEMICAL
REACTIONS
Most of the internal energy or enthalpy of a molecule is stored in the form of potential
energy in chemical bonds. As reactants are transformed to products in a chemical reaction,
energy can be released or taken up as bonds are made or broken. For example, consider
a reaction in which N21g2 and H21g2 dissociate into atoms, and the atoms recombine to
form NH31g2. The enthalpy changes associated with individual steps and with the overall reaction 1>2 N21g2 + 3>2 H21g2 ¡ NH31g2 are shown in Figure 4.1. Note that
large enthalpy changes are associated with the individual steps but the enthalpy change
in the overall reaction is much smaller.
The change in enthalpy or internal energy resulting from chemical reactions
appears in the surroundings in the form of a temperature increase or decrease resulting
from heat flow and/or in the form of expansion or nonexpansion work. For example,
nonexpansion electrical work is possible if the oxidation of a hydrocarbon or H21g2 is
carried out in an electrochemical cell. In Chapters 6 and 11, the extraction of nonexpansion work from chemical reactions will be discussed. In this chapter, the focus is
on using measurements of heat flow to determine changes in U and H due to chemical
reactions.
103
104
CHAPTER 4 Thermochemistry
Figure 4.1
Enthalpy changes for individual
steps in the overall reaction
1>2 N21g2 + 3>2 H21g2 ¡ NH31g2.
N(g) 1 3H(g)
314 3 103 J
NH(g) 1 2H(g)
1124 3 103 J
390 3 103 J
NH2(g) 1 H(g)
466 3 103 J
1
2
N2 (g) 1 32 H2(g)
245.9 3 103 J
NH3(g)
4.2 INTERNAL ENERGY AND ENTHALPY
CHANGES ASSOCIATED WITH CHEMICAL
REACTIONS
Concept
In an exothermic process, heat is
transferred from the system to the
surroundings. In an endothermic
process, heat is transferred from the
surroundings to the system.
In the previous chapters, we discussed how ∆U and ∆H are calculated from work and
heat flow between the system and the surroundings for processes that do not involve
phase changes or chemical reactions. In this section, this discussion is extended to reaction systems.
Imagine that a reaction involving a stoichiometric mixture of reactants (the system) is
carried out in a constant pressure reaction vessel with diathermal walls that is immersed
in a water bath (the surroundings). If the temperature of the water bath increases, heat
has flowed from the system (the contents of the reaction vessel) to the surroundings
(the water bath and the vessel). In this case, we say that the reaction is exothermic. If
the temperature of the water bath decreases, the heat has flowed from the surroundings
to the system, and we say that the reaction is endothermic.
Consider the following reaction:
Fe 3O41s2 + 4 H21g2 ¡ 3 Fe1s2 + 4 H2O1l2
(4.1)
Note that the phase (solid, liquid, or gas) for each reactant and product has been specified because U and H are different for each phase. This reaction will only proceed at
a measurable rate at elevated temperatures. However, as we show later, it is useful to
tabulate values for ∆H for reactions at a pressure of 1 bar and a specified temperature,
generally 298.15 K. The pressure value of 1 bar defines a standard state, and changes
in H and U at the standard pressure of 1 bar are indicated by a superscript ~ as in ∆H ~
and ∆U ~ . The standard state for gases is a hypothetical state in which the gas behaves
ideally at a pressure of 1 bar. For most gases, deviations from ideal behavior are very
small. The enthalpy of reaction, ∆ r H, at specific values of T and P is defined as the
4.2 Internal Energy and Enthalpy Changes Associated with Chemical Reactions
105
heat exchanged between the system and the surroundings as the reactants are transformed into products at conditions of constant T and P. The subscript r indicates that
the process is a chemical reaction. By convention, heat flowing into the system is given
a positive sign. Therefore, ∆ r H is a negative quantity for an exothermic reaction and
a positive quantity for an endothermic reaction. The standard enthalpy of reaction,
∆ r H ~ , refers to one mole of the specified reaction at a pressure of 1 bar and, unless
indicated otherwise, to T = 298.15 K.
How can the reaction enthalpy and internal energy be determined? We proceed in
the following way. The reaction is carried out at 1 bar pressure, and the temperature
change ∆T that occurs in a finite size water bath, initially at 298.15 K, is measured.
The water bath is large enough that ∆T is small. If ∆T is negative as a result of the
reaction, the bath is heated to return it, the reaction vessel, and the system to 298.15 K
using an electrical heater. By doing so, we ensure that the initial and final states are the
same and, therefore, the measured ∆H is equal to ∆ r H ~ . The electrical work done on
the heater that restores the temperature of the water bath and the system to 298.15 K is
equal to ∆ r H ~ . If the temperature of the water bath increases as a result of the reaction,
a separate experiment must be carried out. In this case, the electrical work done on a
heater in the water bath at 298.15 K that increases the system and water bath temperature by ∆T is measured. For this scenario, ∆ r H ~ is equal to the negative of the electrical work done on the heater.
The experimental method for determining ∆ r H ~ was described in the previous
paragraph. Because the number of reactions among all possible reactants and products
is very large, tabulation of the reaction enthalpies for all possible chemical reactions
would be a monumental undertaking. Thus, another approach, in which ∆ r H ~ is calculated from tabulated enthalpy values for individual reactants and products, is generally
preferred. The advantage of this approach is that there are far fewer reactants and products than there are reactions among them.
As an example of how to calculate ∆ r H ~ from enthalpies of individual reactants
and products, consider the determination of ∆ r H ~ for the reaction of Equation (4.1)
at T = 298.15 K and P = 1 bar. These values for P and T are chosen because, by
convention, thermodynamic values are tabulated at T = 298.15 K and P = 1 bar.
However, ∆ r H can be calculated for other values of P and T using the methods
discussed in Chapters 2 and 3. In principle, we could express ∆ r H ~ in terms of the
individual enthalpies of reactants and products:
~
~
∆ r H ~ = H products
- H reactants
= 3H m~ 1Fe, s2 + 4H m~ 1H2O, l2 - H m~ 1Fe 3O4, s2 - 4H m~ 1H2, g2 (4.2)
The m subscripts refer to molar quantities. Although Equation (4.2) is correct, it
does not provide a useful way to calculate ∆ r H ~ . That is, there is no experimental
way to determine the absolute enthalpy for any element or compound because there
is no unique reference point with a value of zero against which individual enthalpies can be measured. Only ∆H and ∆U, as opposed to H and U, can be determined
experimentally.
Equation (4.2) can be transformed into a more useful form by introducing the
enthalpy of formation. The standard enthalpy of formation, ∆ f H ~ , is defined as the
enthalpy change of a reaction in which the only reaction product is 1 mol of the species of interest, and only pure elements in their most stable state of aggregation under
standard state conditions appear as reactants. We refer to these species as being in their
standard reference state. For example, the standard reference state of water and carbon at 298.15 K and 1 bar pressure are H2O1l2 and solid carbon in the form of graphite.
Note that with this definition ∆ f H ~ = 0 for an element in its standard reference state
because the reactants and products are identical.
Reaction enthalpies can be expressed in terms of formation enthalpies. The only
compounds that are produced or consumed in the reaction Fe 3O41s2 + 4 H21g2 ¡
3 Fe1s2 + 4 H2O1l2 are Fe 3O41s2 and H2O1l2. All elements that appear in the reaction
Concept
The standard reference state of a
substance is the most stable state at
1 bar pressure and 298.15 K.
106
CHAPTER 4 Thermochemistry
are in their standard reference states. The formation reactions for the compounds at
298.15 K and 1 bar are
∆r H ~
∆r H
~
1
O 1g2 ¡ H2O1l2
2 2
= ∆ f H ~ 1H2O, l2 - 0 - 0 = ∆ f H ~ 1H2O, l2
H21g2 +
3 Fe1s2 + 2 O21g2 ¡ Fe 3O41s2
= ∆ f H ~ 1Fe 3O4, s2 - 0 - 0 = ∆ f H ~ 1Fe 3O4, s2
(4.3)
(4.4)
If Equation (4.2) is rewritten in terms of the enthalpies of formation, a simple equation
for the reaction enthalpy is obtained:
∆ r H ~ = 4∆ f H ~ 1H2O, l2 - ∆ f H ~ 1Fe 3O4, s2
(4.5)
nAA + nBB + c ¡ nXX + nYY + c
(4.6)
Note that elements in their standard reference state do not appear in this equation
because ∆ f H ~ = 0 for these species. This result can be generalized to any chemical
transformation
which we write in the form
Concept
The enthalpy change for a reaction
can be expressed in terms of the
standard formation enthalpies of
reactants and products.
0 = a ni Xi
i
The Xi refer to all species that appear in the overall equation. The unitless stoichiometric coefficients ni are positive for products and negative for reactants. The enthalpy
change associated with this reaction is
∆ r H ~ = a ni ∆ f H i~
i
nAA 1 nBB
[
DrH
2nA DfH A 2nB DfH
[
nCC 1 nDD
[
B
(4.7)
(4.8)
The rationale behind Equation (4.8) is depicted in Figure 4.2. Two paths are considered between the reactants A and B and the products C and D in the reaction vAA +
vBB ¡ vCC + vDD. The first of these is a direct path for which ∆H ~ = ∆ r H ~ . In
the second path, A and B are first broken down into their elements, each in its standard
reference state. Subsequently, the elements are combined to form C and D. The enthalpy
~
~
change for the second route is ∆ r H ~ = g i ni ∆ f H i,products
- g i 0 ni 0 ∆ f H i,reactants
=
~
g ni ∆ f H i . Because H is a state function, the enthalpy change is the same for both paths.
i
[
nC DfH C 1nD DfH
[
D
Elements in standard reference state
Figure 4.2
Two different pathways between the
same reactants and products. The value
of ∆H for both paths is the same because
they both connect the same initial and
final states. Equation (4.8) is a consequence of this conclusion.
This is stated in mathematical form in Equation (4.8).
Writing ∆ r H ~ in terms of formation enthalpies is a great simplification over
compiling measured values of reaction enthalpies. Standard formation enthalpies for
atoms and inorganic compounds at 298.15 K are listed in Table 4.1, and standard
formation enthalpies for organic compounds are listed in Table 4.2 (see Appendix
A, Data Tables).
Another thermochemical convention involves the calculation of enthalpy changes
of electrolyte solutions. The solution reaction that occurs when a salt such as NaCl is
dissolved in water is
NaCl1s2 ¡ Na+ 1aq2 + Cl- 1aq2
Because it is not possible to form only positive or negative ions in solution, the measured enthalpy of solution of an electrolyte is the sum of the enthalpies of all anions
and cations formed. To be able to tabulate values for enthalpies of formation of individual ions, the enthalpy for the following reaction is set equal to zero at P = 1 bar for
all temperatures:
1>2 H21g2 ¡ H + 1aq2 + e- 1metal electrode2
In other words, solution enthalpies of formation of ions are measured relative to that
for H + 1aq2. The thermodynamics of electrolyte solutions will be discussed in detail in
Chapter 10.
4.3 Hess’s Law Is Based on Enthalpy Being a State Function
107
As the previous discussion shows, only the ∆ f H ~ of each reactant and product is
needed to calculate ∆ r H ~ . Each ∆ f H ~ is a difference in enthalpy between the compound and its constituent elements, rather than an absolute enthalpy. However, there
is a convention that allows absolute enthalpies to be specified using the experimentally
determined values of the ∆ f H ~ of compounds. In this convention, the absolute enthalpy
of each pure element in its standard reference state is set equal to zero. With this convention, the absolute molar enthalpy of any chemical species in its standard reference
state H m~ is equal to ∆ f H ~ for that species. To demonstrate this convention, recall the
reaction in Equation (4.4):
∆ r H ~ = ∆ f H ~ 1Fe 3O4, s2 = H m~ 1Fe 3O4, s2 - 3H m~ 1Fe, s2 - 2H m~ 1O2, g2 (4.4)
Setting H m~ = 0 for each element in its standard reference state,
∆ f H ~ 1Fe 3O4, s2 = H m~ 1Fe 3O4, s2 - 3 * 0 - 2 * 0 = H m~ 1Fe 3O4, s2
(4.9)
The value of ∆ r H ~ for any reaction involving compounds and elements is unchanged
by this convention. In fact, one could choose a different number for the absolute enthalpy of each pure element in its standard reference state, and it would still not change
the value of ∆ r H ~ . However, it is much more convenient (and easier to remember) if
one sets H m~ = 0 for all elements in their standard reference state. This convention
will be used again in Chapter 6 when chemical potential is discussed.
4.3 HESS’S LAW IS BASED ON ENTHALPY
BEING A STATE FUNCTION
As discussed in the previous section, it is extremely useful to tabulate values of ∆ f H ~
for chemical compounds at one fixed combination of P and T. Tables 4.1 and 4.2 list
this data for 1 bar and 298.15 K. With access to these values of ∆ f H ~ , ∆ r H ~ can be
calculated for all reactions among these elements and compounds at 1 bar and 298.15 K.
But how is ∆ f H ~ determined? Consider the formation reaction for C2H61g2:
2 C1graphite2 + 3 H21g2 ¡ C2H61g2
(4.10)
Graphite is the standard reference state for carbon at 298.15 K and 1 bar because it is
slightly more stable than diamond under these conditions. However, it is unlikely that
one would obtain only ethane if the reaction were carried out as written. Given this
experimental hindrance, how can ∆ f H ~ for ethane be determined? To determine ∆ f H ~
for ethane, we take advantage of the fact that ∆H is path independent. In this context,
path independence means that the enthalpy change for any sequence of reactions that
sum to the same overall reaction is identical. This statement is known as Hess’s law.
Therefore, one is free to choose any sequence of reactions that leads to the desired
outcome. Combustion reactions are well suited for these purposes because in general
they proceed rapidly, go to completion, and produce only a few products. To determine
∆ f H ~ for ethane, one can carry out the following combustion reactions:
C2H61g2 + 7>2 O21g2 ¡ 2 CO21g2 + 3 H2O1l2
∆ r H I~
(4.11)
∆ r H II~
(4.12)
H21g2 + 1>2 O21g2 ¡ H2O1l2
~
∆ r H III
(4.13)
2 * 3 C1graphite2 + O21g2 ¡ CO21g2 4
2∆ r H II~
(4.14)
2 CO21g2 + 3 H2O1l2 ¡ C2H61g2 + 7>2 O21g2
- ∆ r H I~
(4.15)
~
3∆ r H III
(4.16)
2 C1graphite2 + 3 H21g2 ¡ C2H61g2
~
2∆ r H II~ - ∆ r H I~ + 3∆ r H III
C1graphite2 + O21g2 ¡ CO21g2
These reactions are combined in the following way to obtain the desired reaction:
3 * 3 H21g2 + 1>2 O21g2 ¡ H2O1l2 4
Concept
Hess’s law states that the enthalpy
change for any sequence of reactions
that sum to the same overall reaction
has the same value.
108
CHAPTER 4 Thermochemistry
We emphasize again that it is not necessary for these reactions to be carried out at
298.15 K. The reaction vessel is immersed in a water bath at 298.15 K, and the combustion reaction is initiated. If the temperature in the vessel rises during the course of
the reaction, the heat flow that restores the system and surroundings to 298.15 K after
completion of the reaction is measured, allowing ∆ r H ~ to be determined at 298.15 K.
Several points should be made about enthalpy changes in relation to balanced overall equations describing chemical reactions. First, because H is an extensive function,
multiplying all stoichiometric coefficients with any number changes ∆ r H ~ by the same
factor. Therefore, it is important to know which set of stoichiometric coefficients has
been assumed if a numerical value of ∆ r H ~ is given. Second, because the units of
∆ f H ~ for all compounds in the reaction are kJ mol-1, the units of the reaction enthalpy
∆ r H ~ are also kJ mol-1. One might pose the question “per mole of what?” given that
all the stoichiometric coefficients may differ from each other and from the value of
one. The answer to this question is per mole of the reaction as written. Doubling all
stoichiometric coefficients doubles ∆ r H ~ .
EXAMPLE PROBLEM 4.1
The average bond enthalpy of the O —H bond in water is defined as one-half of
the enthalpy change for the reaction H2O1g2 ¡ 2 H1g2 + O1g2. The formation
enthalpies, ∆ f H ~ , for H1g2 and O1g2 are 218.0 and 249.2 kJ mol-1, respectively, at
298.15 K, and ∆ f H ~ for H2O1g2 is -241.8 kJ mol-1 at the same temperature.
a. Use this information to determine the average bond enthalpy of the O —H bond in
water at 298.15 K.
b. Determine the average bond energy ∆U of the O —H bond in water at 298.15 K.
Assume ideal gas behavior.
Solution
a. We consider the sequence
H2O1g2 ¡ H21g2+1>2 O21g2
∆ r H ~ = 241.8 kJ mol-1
1>2 O21g2 ¡ O1g2
∆ r H ~ = 249.2 kJ mol-1
H21g2 ¡ 2 H1g2
∆ r H ~ = 2 * 218.0 kJ mol-1
H2O1g2 ¡ 2 H1g2 + O1g2
∆ r H ~ = 927.0 kJ mol-1
This is the enthalpy change associated with breaking both O —H bonds under standard
conditions. We conclude that the average bond enthalpy of the O —H bond in water is
1
* 927.0 kJ mol-1 = 463.5 kJ mol-1. We emphasize that this is the average value
2
because the values of ∆ r H ~ for the reactions H2O1g2 ¡ H1g2 + OH1g2 and
OH1g2 ¡ O1g2 + H1g2 differ.
b. ∆U ~ = ∆H ~ - ∆1PV2 = ∆H ~ - ∆nRT
= 927.0 kJ mol-1 - 2 * 8.314 J mol-1K-1 * 298.15 K
= 922.0 kJ mol-1
1
* 922.0 kJ mol-1 =
2
461.0 kJ mol-1. The bond energy and the bond enthalpy are nearly identical.
The average value for ∆U ~ for the O —H bond in water is
Example Problem 4.1 shows how bond energies can be calculated from reaction
enthalpies. The value of a bond energy is of particular importance for chemists in
estimating the thermal stability of a compound as well as its stability with respect to
reactions with other molecules. Values of bond energies tabulated in the format of
the periodic table together with the electronegativities are shown in Table 4.3. (For
more details, see the article by N. K. Kildahl cited in Further Reading.) The value of the
4.4 Temperature Dependence of Reaction Enthalpies
TABLE 4.3 Mean Bond Energies
1
2
H,2.20
432
--432
459
565
Selected Bond Energies
(kJ/mol)
14
15
16
17
18
He
B,2.04
293
--389
536,636
613
C,2.55
346
602,835
411
358,799
485
N,3.04
167
418,942
386
201,607
283
O,3.44
142
494
459
142,494
190
F,3.98
Ne
Na,0.93 Mg,1.31
72
--197
--477
129
-------,377
513
Al,1.61
---272
--583
Si,1.90
222
318
318
452,640
565
P,2.19
S,2.58
Cl,3.16
240
--428
218
249
Ar
K,0.82
Sr,1.00
105
-------,460
550
Ga,1.81 Ge,2.01 As,2.18 Se,2.55 Br,2.96 Kr
Li,0.98
105
--243
--573
49
--180
--490
Be,1.57
13
208
-------,444
632
<220
---,481
322
335,544
490
113
------<469
188
272
----<470
Rb,0.82 Sr,0.95
In,1.78
100
------<523
Sn,1.80 Sb,2.05 Te,2.10 I,2.66
Cs,0.79 Ba,0.89
Tl,2.04
--------439
Pb,2.33 Bi,2.02 Po,2.00 At,2.20 Rn
45
--163
--490
44
--176
--502
84
-------,347
553
44
----467,561
578
146
------<450
--------<360
146
---,380
247
301,389
<440
240
425
363
---,523
284
121
---,295
-----<420
172
272
276
--<351
155
--565
--155
126
218
238
---<393
-----,192
--<350
190
--362
201
250
149
-295
201
278
50
Xe
84
<131
116
KEY
Element symbol
C C
C C, C C
H C
O C, O C
C F
C,2.55
346
602,835
411
358,799
485
Electronegativity
Single bond with self
Double, triple bond with self
Bond with H
Single, double bond with O
Bond with F
single-bond energy, ∆UA9B, for a combination A–B not listed in the table can be estimated using the empirical relationship proposed by Linus Pauling in Equation (4.17):
∆UA - B = 2∆UA - A * ∆UB - B + 96.481xA - xB22
(4.17)
where xA and xB are the electronegativities of atoms A and B.
4.4 TEMPERATURE DEPENDENCE
OF REACTION ENTHALPIES
Suppose that we plan to carry out a reaction that is mildly exothermic at 298.15 K at
another temperature. Is the reaction endothermic or exothermic at the second temperature? To answer this question, it is necessary to determine ∆ r H ~ at the second
temperature. We assume that no phase changes occur in the temperature interval of
interest. The enthalpy for each reactant and product at temperature T is related to the
109
110
CHAPTER 4 Thermochemistry
value at 298.15 K by Equation (4.18), which accounts for the energy supplied in order
to heat the substance to the new temperature at constant pressure:
T
H 1T2 = H 1298.15 K2 +
~
Concept
The dependence of reaction enthalpy
on temperature is determined by
the heat capacities of reactants and
products.
~
L
CP1T′2dT′
(4.18)
298.15 K
The prime in the integral indicates a “dummy variable” that is otherwise identical to the
temperature. This notation is needed because T appears in the upper limit of the integral. In Equation (4.18), H ~ 1298.15 K2 is the absolute enthalpy at 1 bar and 298.15 K.
However, because there are no unique values for absolute enthalpies, it is useful to combine similar equations for all reactants and products with the appropriate stoichiometic
coefficients to obtain the following equation for the reaction enthalpy at temperature T:
T
∆ r H ~ 1T2 = ∆ r H ~ 1298.15 K2 +
where
L
∆CP1T′2dT′
(4.19)
298.15 K
∆CP1T′2 = a ni CP,i1T′2
(4.20)
i
Recall that in our notation, ∆ r H ~ or ∆ f H ~ without an explicit temperature value implies
that T = 298.15 K. In Equation (4.20), the sum is over all reactants and products,
including both elements and compounds. A calculation of ∆ r H ~ at an elevated temperature is shown in Example Problem 4.2.
EXAMPLE PROBLEM 4.2
Calculate ∆ r H ~ 11450 K2 for the reaction 1>2 H21g2 + 1>2 Cl21g2 ¡ HCl1g2
and 1 bar pressure given that ∆ f H ~ 1HCl, g2 = -92.3 kJ mol-1 at 298.15 K and that
CP,m1H2, g2 = a29.064 - 0.8363 * 10-3
CP,m1Cl2, g2 = a31.695 + 10.143 * 10-3
CP,m1HCl, g2 = a28.165 + 1.809 * 10-3
T
T2
+ 20.111 * 10-7 2 b J K-1mol-1
K
K
T
T2
- 40.373 * 10-7 2 b J K-1mol-1
K
K
T
T2
+ 15.464 * 10-7 2 b J K-1mol-1
K
K
over this temperature range. The ratios T>K and T 2 >K2 appear in these equations in
order to have the right units for the heat capacity.
Solution
1450 K
∆ r H 11450 K2 = ∆ r H 1298.15 K2 +
~
~
∆CP1T2 = c 28.165 + 1.809 * 10-3
-
L
∆CP1T2dT
298.15 K
T
T2
+ 15.464 * 10-7 2
K
K
1
T
T2
a29.064 - 0.8363 * 10-3 + 20.111 * 10-7 2 b
2
K
K
1
T
T2
a31.695 + 10.143 * 10-3 - 40.373 * 10-7 2 b d J K-1mol-1
2
K
K
= a -2.215 - 2.844 * 10-3
T
T2
+ 25.595 * 10-7 2 b J K-1mol-1
K
K
4.5 EXPERIMENTAL DETERMINATION OF ∆U AND ∆H FOR CHEMICAL REACTIONS
∆ r H ~ 11450 K2 = -92.3 kJ mol-1
1450 K
+
L
298.15 K
a -2.215 - 2.844 * 10-3
111
T
T2
+ 25.595 * 10-7 2 b * dT J mol-1
K
K
= -92.3 kJ mol-1 - 2.836 kJ mol-1 = -95.1 kJ mol-1
In this particular case, the change in the reaction enthalpy with T is not large because
∆CP1T2 is small and not because an individual CP,i1T2 is small.
4.5 EXPERIMENTAL DETERMINATION OF
∆U AND ∆H FOR CHEMICAL REACTIONS
For chemical reactions, ∆U and ∆H are generally determined through experiment.
In this section, we discuss how these experiments are carried out. If some or all of the
reactants or products are volatile, it is necessary to contain the reaction mixture. Such
an experiment can be carried out in a bomb calorimeter, shown schematically in
Figure 4.3. In a bomb calorimeter, the reaction is carried out at constant volume. The
motivation for doing so is that if dV = 0, ∆U = qV. Therefore, a measurement of the
Concept
heat flow normalized to 1 mole of the specified reaction provides a direct measurement
of ∆ rU ~ . Bomb calorimetry is restricted to reaction mixtures that contain gases because Because ∆U = qV , the change in
it is impractical to carry out chemical reactions at constant volume for systems consist- internal energy for a reaction can be
ing solely of liquids and solids, as shown in Example Problem 3.2. In the following, determined in a bomb calorimeter.
we describe how ∆ rU ~ and ∆ r H ~ are determined for an experiment in which a single
liquid or solid reactant undergoes combustion in an excess of O21g2.
The bomb calorimeter is a good illustration of how one can define the sysThermometer
tem and surroundings to simplify the analysis of an experiment. The system is Stirrer
Diathermal Steel bomb
defined as the contents of a stainless steel thick-walled pressure vessel, the presIgnition
container
filled with
sure vessel itself, and the inner water bath. Given this definition of the system,
wires
sample and
the surroundings consist of the container holding the inner water bath, the outer
excess of O2
gas
water bath, and the rest of the universe. The outer water bath encloses the inner
bath, and, through a heating coil, its temperature is always held at the temperature of the inner bath. Therefore, no heat flow will occur between the system
and surroundings, and q = 0. Because the combustion experiment takes place
at constant volume, w = 0. Therefore, ∆U = 0. These conditions describe an
isolated system of finite size that is not coupled to the rest of the universe. We
are only interested in one part of this system, namely, the reaction mixture.
What are the individual components that determine ∆U? Consider the system
as consisting of three subsystems: the reactants in the calorimeter, the calorimeter vessel, and the inner water bath. These three subsystems are separated by
rigid diathermal walls and are in thermal equilibrium. Energy is redistributed
among the subsystems as reactants are converted to products, the temperature
of the inner water bath changes, and the temperature of the calorimeter changes.
∆U =
m H 2O
ms
∆ cU +
* CP,m1H2O, l2 * ∆T + Ccal * ∆T = 0
Ms
MH2O
Reactants
in sample
cup
(4.21)
In Equation (4.21), ∆T is the change in the temperature of the three subsystems.
The values for the following variables in Equation (4.21) are known: the mass
of water in the inner bath, mH2O ; its molecular weight, MH2O ; its heat capacity,
CP,m1H2O, l2; the mass of the sample, ms ; and its molecular weight, Ms . The
energy of combustion, ∆ cU, is defined per mole of the combustion reaction, but
because it is defined for 1 mole of reactant, the factor ms >Ms in Equation (4.21)
is appropriate. We wish to measure ∆ cU. However, to determine ∆ cU, the calorimeter constant, Ccal , must first be determined by carrying out a reaction for
which ∆ rU ~ is already known, as illustrated in Example Problem 4.3. To be
more specific, we consider a combustion reaction between a compound and an
excess of O2.
Inner water
bath
Figure 4.3
Schematic diagram of a bomb calorimeter. The
liquid or solid reactant is placed in a cup suspended
in the thick-walled steel bomb, which is filled
with O2 gas. The vessel is immersed in an inner
water bath, and its temperature is monitored. The
diathermal container is immersed in an outer water
bath (not shown) whose temperature is maintained
at the same value as the inner bath through a
heating coil. In this configuration, there is no heat
exchange between the inner water bath and the
rest of the universe.
112
CHAPTER 4 Thermochemistry
EXAMPLE PROBLEM 4.3
When 0.972 g of cyclohexane undergoes complete combustion in a bomb calorimeter,
∆T of the inner water bath is 2.98°C. For cyclohexane, ∆ cU is -3913 kJ mol-1. Given
this result, what is the value for ∆ cU for the combustion of benzene if ∆T is 2.36°C
when 0.857 g of benzene undergoes complete combustion in the same calorimeter?
The mass of the water in the inner bath is 1.812 * 103 g, and the CP,m of water is
75.3 J K-1 mol-1.
Solution
To calculate the calorimeter constant through the combustion of cyclohexane, we write
Equation (4.21) in the following form:
Ccal =
mH2O
ms
∆ cU C 1H O, l2∆T
Ms
MH2O P,m 2
∆T
0.972 g
=
84.16 g mol-1
* 3913 * 103 J mol-1 -
1.812 * 103 g
18.02 g mol-1
2.98°C
* 75.3 J mol-1 K-1 * 2.98°C
= 759 * 103 J 1°C2-1
In calculating ∆ cU for benzene, we use the value for Ccal :
∆ cU = -
Ms mH2O
a
C 1H O, l2∆T + Ccal ∆T b
ms MH2O P,m 2
1.812 * 103 g
78.12 g mol-1
* 75.3 J mol-1 K-1 * 2.36°C
= * ° 18.02 g mol-1
¢
0.857 g
3
-1
+ 7.59 * 10 J1°C2 * 2.36°C
= -3.26 * 106 J mol-1
Once ∆ cU has been determined, ∆ cH can be determined using the following
equation:
∆ c H = ∆ cU + ∆1PV2
(4.22)
For reactions involving only solids and liquids, ∆U W ∆1PV2 and ∆H ≈ ∆U.
If some of the reactants or products are gases, the small change in the temperature
that is measured in a calorimetric experiment can generally be ignored and ∆1PV2 =
∆1nRT2 = ∆nRT
∆ c H = ∆ cU + ∆nRT
(4.23)
where ∆n is the change in the number of moles of gas in the overall reaction. For the
combustion of cyclohexane,
C6H121l2 + 9 O21g2 ¡ 6 CO21g2 + 6 H2O1l2
(4.24)
and ∆n = -3. Note that at T = 298.15 K, the most stable form of cyclohexane and
water is a liquid.
∆ c H = ∆ cU - 3RT = -3913 * 103 kJ mol-1
-3 * 8.314 J K-1mol-1 * 298.15 K
= -3920 * 103 kJ mol-1
(4.25)
For this reaction, ∆ c H and ∆ cU differ by only 0.2%. Note that because the contents
of the bomb calorimeter are not at 1 bar pressure, ∆ cU rather than ∆ cU ~ is measured.
The difference is small, but it can be calculated.
4.6 Differential Scanning Calorimetry
If the reaction under study does not involve gases or highly volatile liquids, there
is no need to operate under constant volume conditions. It is preferable to carry out the
reaction at constant P using a constant pressure calorimeter. ∆H is directly determined because ∆H = qP. A vacuum-insulated vessel with a loosely fitting stopper as
shown in Figure 4.4 is adequate for many purposes and can be treated as an isolated
composite system. Equation (4.21) takes the following form for constant pressure calorimetry involving the solution of a salt in water:
m H 2O
msalt
∆H ~ =
∆ sol H ~ +
C 1H O, l2∆T + Ccal ∆T = 0
(4.26)
Msalt
MH2O P,m 2
Salt
113
Thermometer
Stopper
∆ sol H ~ is defined per mole of the solution reaction, but because the reaction includes
exactly 1 mole of reactant, the factor msalt >Msalt in Equation (4.21) is appropriate.
Because ∆1PV2 is negligibly small for the solution of a salt in a solvent, ∆ solU ~ =
∆ sol H ~ . The solution must be stirred to ensure that equilibrium is attained before ∆T is
measured.
H 2O
EXAMPLE PROBLEM 4.4
The enthalpy of solution for the reaction
Na 2SO41s2 ¡ 2 Na+1aq2 + SO24 1aq2
is determined in a constant pressure calorimeter. The calorimeter constant was determined to be 342.5 J K-1. When 1.423 g of Na 2SO4 is dissolved in 100.34 g of H2O1l2,
∆T = 0.031 K. Calculate ∆ sol H ~ for Na 2SO4 from these data. Compare your result
with that calculated using the standard enthalpies of formation in Table 4.1 (see Appendix A, Data Tables) and in Chapter 10 in Table 10.1.
Solution
∆ sol H ~ = -
Msalt mH2O
a
C 1H O, l2∆T + Ccal ∆T b
msalt MH2O P,m 2
100.34 g
* 75.3 J K-1 mol-1 * 0.031 K
142.04 g mol-1
= * ° 18.02 g mol-1
¢
1.423 g
+ 342.5 J K-1 * 0.031 K
= -2.4 * 103 J mol-1
We next calculate ∆ sol H ~ using the data tables.
~
∆ sol H ~ = 2∆ f H ~ 1Na+, aq2 + ∆ f H ~ 1SO24 , aq2 - ∆ f H 1Na 2SO4, s2
= 2 * 1-240.1 kJ mol-12 - 909.3 kJ mol-1 + 1387.1 kJ mol-1
= -2.4 kJ mol-1
The agreement between the calculated and experimental results is good.
S U PPL EM EN TAL
SEC TI O N
4.6 DIFFERENTIAL SCANNING CALORIMETRY
Differential scanning calorimetry (DSC) is a form of constant pressure calorimetry
that is well suited to routine laboratory tests in pharmaceutical and material sciences. It
is also used to study chemical changes such as polymer cross-linking, melting, and unfolding of protein molecules in which heat is absorbed or released in the transition. The
experimental apparatus for such an experiment is shown schematically in Figure 4.5.
The word differential appears in the name of the technique because the uptake of heat
is measured relative to that for a reference material, and the word scanning refers to the
fact that the temperature of the sample is varied, usually linearly with time.
Solution
Figure 4.4
Schematic diagram of a constant pressure calorimeter suitable for measuring
the enthalpy of solution of a salt in
water.
114
CHAPTER 4 Thermochemistry
Lid
Figure 4.5
Heat flux differential scanning
calorimeter. Shown here is a schematic
diagram of this type of calorimeter, which
consists of an insulated massive enclosure
and lid that are heated to the temperature
TE using a resistive heater. A support
disk in good thermal contact with the
enclosure supports the sample and reference materials. The temperatures of the
sample and reference are measured with a
thermocouple. In practice, the reference is
usually an empty sample pan.
Sample
Reference
q
q
Support disk
DT
Sample
thermocouple
Reference
thermocouple
Resistive heater
Power supply
The temperature of the enclosure TE is increased linearly with time using a power
supply. Heat flows from the enclosure through the disk to the sample because of the
temperature gradient generated by the heater. Because the sample and reference are
equidistant from the enclosure, the heat flow to each sample is the same. The reference
material is chosen such that its melting point is not in the range of that of the samples.
We consider a simplified one-dimensional model of the heat flow in the DSC in
Figure 4.6 following the treatment of Höhne et al. cited in Further Reading. The electrical current through the resistive heater increases the temperature of the calorimeter
enclosure TE to a value greater than that of the sample and reference, TS , and TR . The
heat flow per unit time from the enclosure to the sample and reference are designated
by ΦES and ΦER , respectively, which typically have the units of J g -1 s-1.
Assume that a process such as melting occurs in the sample and not in the reference. The heat flow per unit time associated with the process is given by Φ1t2. It is
time dependent because, using melting as an example, heat flow associated with the
phase change begins at the onset of melting and ceases when the sample is completely
in the liquid state. This additional heat flow changes the sample temperature by dTS,
A plot of 𝚫T versus time for an
exothermic process. For this example,
it is assumed that the heat capacities
of the sample and reference are constant
over the temperature range shown. The
heat associated with the process of
interest is proportional to the blue area.
The green area arises from the difference
in heat capacities of sample and reference.
The zero line is obtained without material
in the crucibles.
DT
Figure 4.6
2aR(CS2CR)
Zero line
t1
t2
Time
4.6 Differential Scanning Calorimetry
and consequently both TE - TS and the heat flow rate to the sample ΦES change. In the
experiment, ∆T1t2 = TS1t2 - TR1t2 is measured. The change in the heat flow to the
sample resulting from the process is
CS
dTS1t2
= ΦES1t2 - Φ1t2
dt
(4.27)
where CS is the constant pressure heat capacity of the sample. Equation (4.27) relates
the sample temperature to the heat flow generated by the process of interest. If the
process is exothermic, Φ1t2 6 0, and if it is endothermic, Φ1t2 7 0. We rewrite
Equation (4.27) to explicitly contain the experimentally accessible function ∆T1t2.
CS
dTR1t2
d∆T1t2
+ CS
= ΦES1t2 - Φ1t2
dt
dt
(4.28)
The corresponding equation for the reference in which only heating occurs is
CR
dTR1t2
= ΦER1t2
dt
(4.29)
The essence of DSC is that a difference between the heat flux to the sample and reference
is measured. We, therefore, subtract Equation (4.29) from Equation (4.28) and obtain
ΦES1t2 - ΦER1t2 = 1CS - CR2
dTR1t2
d∆T1t2
+ CS
+ Φ1t2
dt
dt
(4.30)
The quantities ΦES and ΦER are directly proportional to the temperature differences
TE - TS and TE - TR and are inversely proportional to the thermal resistance, Rthermal,
between the heated enclosure and the sample and reference. The thermal resistance is
an intrinsic property of the calorimeter and can be obtained through calibration of the
instrument.
ΦES =
TE - TS
Rthermal
and ΦER =
TE - TR
Rthermal
(4.31)
Therefore,
Φ1t2 = -
∆T1t2
dTR1t2
d∆T1t2
- 1CS - CR2
- CS
Rthermal
dt
dt
(4.32)
Equation (4.32) links the heat flow per unit mass generated by the process of interest,
Φ1t2, to the measured quantity ∆T1t2. Typically TR is increased linearly with time,
and Equation (4.32) simplifies to
Φ1t2 = -
TR1t2 = T0 + at
∆T1t2
d∆T1t2
- a1CS - CR2 - CS
Rthermal
dt
(4.33)
We see that ∆T1t2 is not simply related to Φ1t2, which is to be determined in the
experiment. The second term in Equation (4.33) arises because the heat capacities of
the sample and reference are not equal. Additionally, the heat capacity of the sample
changes as it undergoes a phase change.
The third term in Equation (4.33) is proportional to d∆T1t2>dt and has the effect of
broadening ∆T1t2 relative to Φ1t2 and shifting it to longer times. A schematic picture
of a DSC scan in this model is shown in Figure 4.6.
Because CRa = ΦER1t2 from Equation (4.29) and ∆T are proportional as shown in
Equation (4.31), a graph of CR versus time has the same shape as shown in Figure 4.6.
An analogous equation can be written for the sample, so that
ΦES - ΦER = a1CS - CR2
Therefore, a graph of 1CS - CR2 versus time also has the same shape as Figure 4.6.
115
116
CHAPTER 4 Thermochemistry
The goal of the experiment is to determine the heat absorbed or evolved in
the process per unit mass, which is given by
DH
qP = ∆H =
Ttrue
(a) DH is shown as a function of the true
temperature for a melting process.
CP
Ttrue
F or CPapparent
(b) The heat capacity becomes infinite at the
transition temperature because the
temperature remains constant as heat
flows into the sample.
Tmeasured
F or CPapparent
(c) The heat capacity curve in (b) is distorted
because the measured temperature increases,
while the sample temperature remains
constant during the melting transition.
Tmeasured
(d) Further distortions to the hypothetical
scan (c) and a shift of the peak to higher
temperature occur because of the thermal
inertia of the calorimeter.
Figure 4.7
Differential scanning calorimetry (DSC) curves.
(a) ∆H is shown as a function of the true temperature for a melting process. (b) The heat capacity
becomes infinite at the transition temperature
because the temperature remains constant as heat
flows into the sample. (c) The heat capacity
curve in (b) is distorted because the measured
temperature increases, while the sample temperature remains constant during the melting transition.
(d) Further distortions to the hypothetical scan (c)
and a shift of the peak to higher temperature occur
because of the thermal inertia of the calorimeter.
= -
t2
Lt1
Φ1t2dt
t2
Rthermal Lt1
1
∆T1t2dt -
t2
Rthermal Lt1
1
1-a1CS - CR22dt
(4.34)
In Equation (4.34), the baseline contribution to the integral has been subtracted
because it has no relevance for the process of interest.
Interpreting DSC curves in terms of heat capacities must be done with caution
as illustrated for a melting transition in Figure 4.7. At the melting temperature,
the enthalpy rises abruptly as shown in Figure 4.7a (and discussed in Chapter 8).
The heat capacity is the derivative of the enthalpy with respect to temperature
and has the form shown in Figure 4.7b. As discussed in Section 2.5, the heat
capacity of the liquid is higher than that of the solid. However, a measured DSC
curve has the shape shown in Figure 4.7d rather than that in Figure 4.7b.
There are two reasons for this discrepancy. Heat is taken up by the sample
during the melting transition, but the sample temperature does not change. However, the sample crucible temperature continues to increase as heat flows to the
sample. Therefore, the apparent heat capacity would have the shape shown in
Figure 4.7c if there were no resistances to heat flow in the calorimeter. The thermal resistances present in the calorimeter spread out the curve shown in Figure
4.7c and shift it to higher temperatures, as shown in Figure 4.7d. Clearly, the
temperature dependence of the apparent heat capacity obtained directly from a
DSC scan is significantly different from the temperature dependence of the true
heat capacity. Deconvolution procedures must be undertaken to obtain accurate
heat capacities. Because the heat absorbed or evolved in the process per unit
mass is proportional to the area under the DSC scan, it is less affected by the
instrument distortions than the heat capacity.
DSC is the most direct method for determining the energetics of biological
macromolecules undergoing conformational transitions, which is important in
understanding their biological activity. In particular, DSC has been used to
determine the temperature range over which proteins undergo the conformational changes associated with reversible denaturation, a process in which
a protein unfolds. The considerations for a melting process discussed earlier
apply as well as for denaturation, although melting enthalpies are generally
much larger than denaturation enthalpies. From the preceding discussion, when
a protein solution is heated at a constant rate and at constant pressure, DSC
reports the apparent heat capacity of the protein solution. Protein structures are
stabilized by the cooperation of numerous weak forces. Assume that a protein
solution is heated from a temperature T1 to a temperature T2 . If the protein
denatures within this temperature interval reversibly and cooperatively, the
CP1T2 versus T curve has the form shown in Figure 4.8.
The heat capacity of denaturation is defined as
D
N
∆C den
P = C P 1T2 - C P 1T2
(4.35)
which is the difference between the heat capacity of the denatured protein, C D
P 1T2,
and the heat capacity of the native (i.e., structured) protein, C N
1T2.
The
value
of
P
den
∆C den
can
be
determined
as
shown
in
Figure
4.8.
The
value
of
∆C
is
a
posiP
P
tive quantity and for many globular proteins varies from 0.3 to 0.7 J K-1 g -1.
The higher heat capacity of the protein in the denatured state can be understood
in the following way. In the structured state, internal motions of the protein are
characterized by coupled bending and rotations of several bonds that occur at
frequencies on the order of kBT>h, where kB is Boltzmann’s constant and h is
Planck’s constant. When the protein denatures, these “soft” vibrations are shifted
to lower frequencies, higher-frequency bond vibrations are excited, and the heat
4.6 Differential Scanning Calorimetry
117
Figure 4.8
CP (T)
Apparent heat capacity of a protein
undergoing reversible, thermal denaturation. Denaturation occurs within the
temperature interval T1 -T2. Heat capacity
is shown as a function of temperature.
The temperature at the heat capacity
maximum is, to a good approximation,
the melting temperature Tm , defined as
the temperature at which half the total
protein has been denatured. The excess
heat capacity ∆CP = CP1T2 - C N
P 1T2
is composed of two parts: the intrinsic
excess heat capacity dC int
P and the transition excess heat capacity dC trs
P.
dCPtrs
DCP
CPD
DCPden
CPN
T1
dCPint
Tm
T2
capacity increases. In addition to increased contributions from vibrational motions,
when a protein structure is disrupted, nonpolar amino acid residues formerly isolated
from solvent within the interior of the protein are exposed to water. Water molecules
now order about these nonpolar amino acids, further increasing the heat capacity.
The excess heat capacity is defined as
∆CP = CP1T2 - C N
P 1T2
(4.36)
where ∆CP is obtained at a given temperature from the difference between the point on
the heat capacity curve CP1T2 and the linearly extrapolated value for the heat capacity
of the native state C N
P 1T2 (see Figure 4.8). The excess heat capacity is in turn composed
of two parts:
trs
∆CP1T2 = dC int
(4.37)
P + dC P
Within the interval T1 9T2 the protein structure does not unfold suddenly. There occurs
instead a gradual unfolding of the protein, such that at any given temperature a fraction
of the structured protein remains and an ensemble of unfolded species is produced. The
heat capacity and enthalpy change observed arise from this ensemble of physical forms
of the partially unfolded protein, and as such these properties are averages over possible configurations of the protein. The component of the heat capacity that accounts
for the sum of molecular species produced in the course of making a transition from the
folded to the unfolded state is the intrinsic excess heat capacity or dC int
P . The second
component of the heat capacity is called the transition excess heat capacity dC trs
P . The
transition excess heat capacity results from fluctuations of the system as the protein
changes from different states in the course of denaturation.
The intrinsic and transition excess heat capacities are determined by first extrapD
olating the functions C N
P 1T2 and C P 1T2 into the transition zone between T1 and T2.
This is indicated in Figure 4.8 by the solid line connecting the heat capacity baselines
above and below Tm . Once this baseline extrapolation is accomplished, dC int
P is the
difference at a given temperature between the extrapolated baseline curve and C N
P 1T2.
The value of ∆C trs
is
obtained
from
the
difference
between
C
1T2
and
the
extrapoP
P
lated baseline curve. The excess enthalpy of thermal denaturation is finally given by
∆Hden = 1 dC trs
P dT. In other words, the excess enthalpy of thermal denaturation is the
T2
T1
area under the peak in Figure 4.8 above the extrapolated baseline curve.
118
CHAPTER 4 Thermochemistry
VOCABULARY
bomb calorimeter
bond energy
bond enthalpy
constant pressure calorimeter
denaturation
differential scanning calorimetry (DSC)
endothermic
enthalpy of reaction
excess heat capacity
exothermic
Hess’s law
intrinsic excess heat capacity
standard enthalpy of formation
standard enthalpy of reaction
standard reference state
standard state
transition excess heat capacity
KEY EQUATIONS
Equation
Significance of Equation
∆ r H ~ = a ni ∆ f H i~
Reaction enthalpy can be written as a sum
over formation enthalpies of reactants and
products.
i
Hess’s law
∆U =
4.8
The enthalpy change for any sequence
of reactions that sum to the same overall
reaction is identical. This statement is
known as Hess’s law.
T
∆ r H ~ 1T2 = ∆ r H ~ 1298.15 K2 +
Equation Number
L
298.15 K
∆CP1T′2dT′
mH2O
ms
∆ cU +
* CP,m1H2O, l2 * ∆T + Ccal * ∆T = 0
Ms
MH2O
Temperature dependence of reaction
enthalpies. Sum is over both compounds
and elements.
4.19
Energy change in bomb calorimeter
4.21
CONCEPTUAL PROBLEMS
Q4.1 In calculating ∆ r H ~ at 285.15 K, only the ∆ f H ~ of the
compounds that take part in the reactions (listed in Tables 4.1
and 4.2 of Appendix A, Data Tables) are needed. Is this statement also true if you want to calculate ∆ r H ~ at 500 K?
Q4.2 What is the point of having an outer water bath in a
bomb calorimeter (see Figure 4.3), especially if its temperature is always equal to that of the inner water bath?
Q4.3 If the ∆ f H ~ for the chemical compounds involved in a
reaction are available at a given temperature, how can the
reaction enthalpy be calculated at another temperature?
Q4.4 Does the enthalpy of formation of compounds containing a certain element change if the enthalpy of formation of
the element under standard state conditions is set equal to
100 kJ mol-1 rather than to zero? If it changes, how will it
change for the compound AnBm if the formation enthalpy of
element A is set equal to 100 kJ mol-1?
Q4.5 Why are elements included in the sum in Equation 4.20
when they are not included in calculating ∆ r H ~ at 298 K?
Q4.6 Why are heat capacities of reactants and products
required for calculations of ∆ r H ~ at elevated temperatures?
Q4.7 Under what conditions are ∆H and ∆U for a reaction
involving gases and/or liquids or solids equal?
Q4.8 Why is it valid to add the enthalpies of any sequence of
reactions to obtain the enthalpy of the reaction that is the sum
of the individual reactions?
Q4.9 In a calorimetric study, the temperature of the system
rises to 325 K before returning to its initial temperature of
298 K. Why doesn’t this temperature rise affect your measurement of ∆ r H ~ ?
Q4.10 Which is greater in a bond breaking reaction, ∆H or
∆U?
Q4.11 The reactants in the reaction 2NO1g2 + O21g2 S
2NO21g2 are initially at 298 K. Why is the reaction enthalpy
the same (a) if the reaction is constantly kept at 298 K or
(b) if the reaction temperature is not controlled and the heat
flow to the surroundings is measured only after the temperature of the products is returned to 298 K?
Q4.12 What is the advantage of a differential scanning calorimeter over a bomb calorimeter in determining the enthalpy
of fusion of a series of samples?
Numerical Problems
Q4.13 You wish to measure the heat of solution of NaCl in
water. Would the calorimetric technique of choice be at
constant pressure or constant volume? Why?
Q4.14 Is the following statement correct? If it is not, rewrite
it so that it is correct. Because the enthalpy of formations of
elements is zero, ∆ f H ~ 1O1g22 = 0.
Q4.15 If the ∆ f H ~ for the chemical compounds involved in
a reaction are known at a given temperature, how can ∆ r H ~
be calculated at another temperature?
Q4.16 Is the following statement correct? If not, rewrite it
so that it is correct. If ∆ r H ~ for a chemical reaction does not
change appreciably with temperature, the heat capacities for
reactants and products must be small.
Q4.17 Under what conditions are ∆H and ∆U for a reaction
involving gases and/or liquids or solids identical?
119
Q4.18 Dogs cool off in hot weather by panting. Write a
chemical equation to describe this process and calculate
∆r H ~ .
Q4.19 Is ∆H for breaking the first C—H bond in methane
equal to the average C—H bond enthalpy in this molecule?
Explain your answer.
Q4.20 Humans cool off through perspiration. How does the
effectiveness of this process depend on the relative humidity?
Q4.21 Vaporization refers to the transition between a liquid
and a gas; fusion refers to the transition between a solid and a
liquid; and sublimation refers to the direct transition between
a solid and a gas. Invoking the properties of state functions,
show that ∆ sub H = ∆ fus H + ∆ vap H.
NUMERICAL PROBLEMS
Section 4.2
P4.1 Given the data in Table 4.1 (see Appendix A, Data
Tables) and the following information, calculate the bond enthalpies and energies for C–Cl, P–H, H–Br, N=O, C=O, C=O:
Substance
CCl4(g) PH3(g) HBr(g) NO2(g) CO2(g) CO(g)
-1
∆ f H 1kJ mol 2 -96.0
~
5.4
-36.3
33.2
-393.5 -110.5
P4.2 The total surface area of the Earth covered by ocean
is 3.35 * 108 km2. Carbon is fixed in the oceans via photosynthesis carried out by marine plants according to the
reaction 6CO21g2 + 6H2O1l2 S C6H12O61s2 + 6O21g2.
An estimate of the mass of carbon fixed in the oceans is
39.8 * 103 kg km-2 per year. Calculate the annual enthalpy
change resulting from photosynthetic carbon fixation in the
ocean per mole of C. Assume P = 1 bar and T = 298 K.
P4.3 Given the data in Table 4.2 of the Data Tables, calculate
the bond enthalpy and energy of the following:
a. The C–H bond in CH4
b. The C–C single bond in C2H6
c. The C=C double bond in C2H4
P4.4 The total arable surface area of the Earth is approximately
15.8 * 106 km2. Every year, the mass of carbon fixed by photosynthesis by vegetation covering this land surface according to
the reaction 6CO21g2 + 6H2O1l2 S C6H12O61s2 + 6O21g2
is about 480. metric tons km-2. Calculate the annual enthalpy
change resulting from photosynthetic carbon fixation over
the land surface per mole of carbon. Assume P = 1 bar and
T = 298 K.
P4.5 Calculate ∆ r H ~ and ∆ rU ~ at 298.15 K for the following reactions:
a. N21g2 + 2O21g2 S 2NO21g2
b. C2H61g2 + 2H2O1l2 S 2CO1g2 + 5H21g2
c. 4NH31g2 + 5O21g2 S 4NO1g2 + 6H2O1l2
d. KOH1aq2 + HCl1aq2 S KCl1aq2 + H2O1l2
Assume complete dissociation of KOH, HCl, and KCl.
e. CO1g2 + H2O1g2 S CO21g2 + H21g2
f. 2Al1s2 + Fe 2O31s2 S Al2O31s2 + 2Fe1s2
P4.6 Nitrogen is a vital component of proteins and ­nucleic
acids and thus is necessary for life. The atmosphere is
­composed of roughly 80% N2, but most organisms cannot
directly utilize N2 for biosynthesis. Bacteria capable of “fixing”
­nitrogen (i.e., converting N2 to a chemical form, such as NH3,
that can be utilized in the biosynthesis of proteins and nucleic
acids) are called diazotrophs. The ability of some plants, such
as legumes, to fix nitrogen is due to a symbiotic relationship
between the plant and nitrogen fixing bacteria that live in the
plant’s roots. Assume the hypothetical reaction for fixing
­nitrogen biologically is
N21g2 + 3H2O1l2 S 2NH31aq2 + 32 O21g2
a. Calculate the standard enthalpy change for the biosynthetic
fixation of nitrogen at T = 298 K. For NH31aq2, ammonia
dissolved in aqueous solution, ∆ f H ~ = -80.3 kJ mol-1.
b. In some bacteria, glycine is produced from ammonia by
the reaction
NH31aq2 + 2CH41g2 + 52 O21g2
S NH2CH2COOH1s2 + 3H2O1l2
Calculate the standard enthalpy change for this reaction.
For glycine, ∆ f H ~ = -537.2 kJ mol-1. Assume
T = 298.15 K.
c. Calculate the standard enthalpy change for the production
of 1 mole of glycine and water from nitrogen, oxygen, and
methane.
120
CHAPTER 4 Thermochemistry
P4.7 Calculate the average C—H bond enthalpy in methane
using the data tables. Calculate the percent error in equating
the average C—H bond energy in Table 4.3 with the bond
enthalpy.
P4.8 A camper stranded in snowy weather loses heat by wind
convection. The camper is packing emergency rations consisting of 54% sucrose, 33% fat, and 13% protein by weight.
Assume that the enthalpies for combustion of fat and protein
are -39.13 kJ g -1 and -22.0 kJ g -1, respectively. How much
emergency rations must the camper consume in order to compensate for a reduction in body temperature of 4.8 K? Assume
the heat capacity of the body equals that of water, and assume
the camper weighs 75 kg. State any additional assumptions.
P4.9 In order to get in shape for mountain climbing, an avid
hiker with a mass of 58 kg ascends the stairs in the world’s
tallest structure, the 828-m-tall Burj Khalifa in Dubai, United
Arab Emirates. Assume that she eats energy bars on the way
up and that her body is 25% efficient in converting the energy
content of the bars into the work of climbing. How many
energy bars does she have to eat if a single bar produces
1.25 * 103 kJ of energy upon metabolizing?
P4.10 We return to the 58 kg hiker of Problem P4.9, who
is climbing the 828-m-tall Burj Khalifa in Dubai. If the
­efficiency of converting the energy content of the bars into the
work of climbing is 25%, the remaining 75% of the energy
released through metabolism is heat released to her body. She
eats two energy bars, one of which produces 1.25 * 103 kJ of
energy upon metabolizing. Assume that the heat capacity of
her body is equal to that for water. Calculate her temperature
at the top of the structure. Is your result reasonable? Can you
think of a mechanism by which her body might release e­ nergy
to avoid a temperature increase? Repeat the calculation for
her companion hiker who weighs 75 kg.
P4.11 In a constant pressure calorimeter, 33.3 mL of
0.100M AgNO3 is mixed with 33.3 mL of 0.100M HCl. The
reaction Ag +1aq2 + Cl-1aq2 takes place. The temperature
of the reactants was 25.60°C, and the final temperature is
26.47°C. Calculate the molar enthalpy change for this reaction. Compare your result with that calculated from the
data tables.
Section 4.3
P4.12 Use the following data at 298.15 K to complete this
problem:
∆ r H ~ 1 kJ mol −1 2
1>2 H21g2 + 1>2 O21g2 S OH1g2 38.95
H21g2 + 1>2 O21g2 S H2O1g2
-241.814
H21g2 S 2H1g2 435.994
O21g2 S 2O1g2 498.34
Assuming ideal gas behavior, calculate ∆ r H ~ and ∆ rU ~ for
the following three reactions.
a. OH1g2 S H1g2 + O1g2
b. H2O1g2 S 2H1g2 + O1g2
c. H2O1g2 S H1g2 + OH1g2
P4.13 Several reactions and their standard reaction enthalpies
at 298.15 K are given here:
∆ r H ~ 1 kJ mol −1 2
CaC21s2 + 2H2O1l2 S Ca1OH221s2 + C2H21g2 -127.9
Ca1s2 + 1>2O21g2 S CaO1s2
-635.1
CaO1s2 + H2O1l2 S Ca1OH221s2 -65.2
The standard enthalpies of combustion of graphite and C2H21g2
are -393.51 and -1299.58 kJ mol-1, respectively. Calculate
the standard enthalpy of formation of CaC21s2 at 25°C.
P4.14 From the following data at 298.15 K, as well as data
in Table 4.1, calculate the standard enthalpy of formation of
H2S1g2 and of FeS21s2:
∆ r H ~ 1 kJ mol −1 2
Fe1s2 + 2H2S1g2 S FeS21s2 + 2H21g2
H2S1g2 + 3>2O21g2 S H2O1l2 + SO21g2
-137.0
-562.0
P4.15 From the following data at 298.15 K, calculate the
standard enthalpy of formation of FeO1s2 and of Fe 2O31s2:
∆ r H ~ 1 kJ mol −1 2
Fe 2O31s2 + 3C1graphite2 S 2Fe1s2 + 3CO1g2 492.6
FeO1s2 + C1graphite2 S Fe1s2 + CO1g2 155.8
C1graphite2 + O21g2 S CO21g2
-393.51
CO1g2 + 1>2O21g2 S CO21g2
-282.98
Calculate the standard enthalpy of formation of FeO1s2 and
of Fe 2O31s2.
P4.16 Use the average bond energies in Table 4.3 to estimate ∆ rU ~ for the reaction CO21g2 + 4H21g2 S CH41g2 +
2H2O1g2. Also calculate ∆ rU ~ from the tabulated values
of ∆ f H ~ for reactant and products (see Appendix A, Data
Tables). Calculate the percent error in estimating ∆ rU ~ from
the average bond energies for this reaction.
P4.17 Use the tabulated values of the enthalpy of combustion
of benzene and the enthalpies of formation of CO21g2 and
H2O1l2 to determine ∆ f H ~ for benzene.
P4.18 Compare the heat evolved at constant pressure per
mole of oxygen in the combustion of sucrose 1C12H22O112
and palmitic acid 1C16H32O22 with the combustion of a typical protein, for which the empirical formula is C4.3H6.6NO.
Assume for the protein that the combustion yields N21g2,
CO21g2, and H2O1l2. Assume that the enthalpies for combustion of sucrose, palmitic acid, and a typical protein are
5647 kJ mol-1, 10,035 kJ mol-1, and 22.0 kJ g -1, respectively.
Based on these calculations, determine the average heat
evolved per mole of oxygen consumed, assuming combustion
of equal moles of sucrose, palmitic acid, and protein.
Section 4.4
P4.19 At 1000. K, ∆ r H ~ = -123.77 kJ mol-1 for the reaction
N21g2 + 3H21g2 S 2NH31g2, with CP,m = 3.502R, 3.466R,
Numerical Problems
and 4.217R for N21g2, H21g2, and NH31g2, respectively.
From this information, calculate ∆ f H ~ of NH31g2 at 375 K.
Repeat the calculation for room temperature (20°C). Assume
that the heat capacities are independent of temperature.
P4.20 Calculate ∆ f H ~ for CO21g2 at -50.0°C assuming that
the heat capacities of reactants and products are constant over
the temperature interval at their values at 298.15 K.
P4.21 Derive a formula for ∆ r H ~ for the reaction CO1g2 +
1>2O21g2 S CO21g2 assuming that the heat capacities of
reactants and products do not change with temperature.
P4.22 Calculate the standard enthalpy of formation of FeS21s2
at 425.°C from the data tables and the following data at
298.15 K. Assume that the heat capacities are independent of
temperature.
Substance
∆ f H ~ 1kJ mol-12
Fe1 s 2 FeS2 1s 2 Fe 2O3 1s 2 S(rhombic) SO2 1 g2
- 824.2
CP,m >R
3.02
7.48
-296.81
2.72
You are also given that for the reaction 2FeS21s2 + 111>22O21g2
S Fe 2O31s2 + 4SO21g2, ∆ r H ~ = -1655 kJ mol-1.
P4.23 At 298 K, ∆ r H ~ = 131.28 kJ mol-1 for the reaction
C1graphite2 + H2O1g2 S CO1g2 + H21g2, with CP,m =
8.53, 33.58, 29.12, and 28.82 J K-1 mol-1 for graphite, H2O1g2,
CO1g2, and H21g2, respectively. From this information,
calculate ∆ r H ~ at 145°C. Assume that the heat capacities are
independent of temperature.
P4.24 Consider the reaction TiO21s2 + 2 C1graphite2 +
2 Cl21g2 S 2 CO1g2 + TiCl41l2 for which ∆ r H ~ =
-80.0 kJ mol-1. Given the following data at 25°C,
a. Calculate ∆ r H ~ at 135.8°C, the boiling point of TiCl4.
b. Calculate ∆ f H ~ for TiCl41l2 at 25°C.
TiO2 1 s 2 Cl2 1 g2 C(graphite) CO1g2 TiCl4 1 l2
Substance
-1
∆ f H 1kJ mol 2
~
-1
CP, m 1J K
- 944
-1
mol 2 55.06
-110.5
33.91
8.53
29.12
145.2
Assume that the heat capacities are independent of temperature.
P4.25 From the following data, calculate ∆ r H ~ 1391.4 K2 for
the reaction CH3COOH1g2 + 2O21g2 S 2H2O1g2 +
2CO21g2:
∆ r H ~ 1 kJ mol −1 2
CH3COOH1l2 + 2O21g2 S 2H2O1l2 + 2CO21g2 -871.5
H2O1l2 S H2O1g2 40.656
CH3COOH1l2 S CH3COOH1g2 24.4
Values for ∆ r H ~ for the first two reactions are at 298.15 K,
and for the third reaction at 391.4 K.
Substance
CP, m >R
CH3COOH1 l2 O2 1g2 CO2 1g2 H2O1l2
14.9
3.53
4.46
9.055
H2O1g2
4.038
P4.26 Calculate ∆ r H ~ at 520. K for the reaction 4 NH31g2 +
6 NO1g2 S 5 N21g2 + 6 H2O1g2 using the temperature
dependence of the heat capacities from the data tables.
121
Compare your result with ∆ r H ~ at 298.15 K. Is the difference large or small? Why?
P4.27 Given the following heat capacity data, calculate ∆ f H ~
of CO21g2 at 500°C. Compare your result to the result of
Problem P4.20.
Substance
CP,m >J mol-1 K-1
C(graphite)
8.52
O2 1 g2
29.4
CO2 1g2
37.1
P4.28 Calculate ∆H for the process in which Cl21g2 initially
at 298.15 K at 1 bar is heated to 585 K at 1 bar. Use the temperature-dependent heat capacities in the data tables. How large
is the relative error if the molar heat capacity is assumed to be
constant at its value of 298.15 K over the temperature interval?
Section 4.5
P4.29 A sample of K1s2 of mass 4.330 g undergoes combustion in a constant volume calorimeter. The calorimeter
constant is 1849 J K-1, and the measured temperature rise in
the inner water bath containing 1450. g of water is 2.529 K.
Calculate ∆ f U ~ and ∆ f H ~ for K2O.
P4.30 Calculate ∆ c H ~ and ∆ cU ~ for the oxidation of
­toluene (g). Also calculate
∆ c H ~ - ∆ cU ~
∆c H ~
P4.31 Benzoic acid of mass 1.87 g reacts with oxygen in a
constant volume calorimeter to form H2O1l2 and CO21g2 at
298 K. The mass of the water in the inner bath is 1.44 * 103 g.
The temperature of the calorimeter and its contents rises 3.48 K
as a result of this reaction. Calculate the calorimeter constant.
P4.32 A sample of Na 2SO41s2 is dissolved in 225 g of water
at 298 K such that the solution is 0.288 molar in Na 2SO4.
A temperature rise of 0.129°C is observed. The calorimeter
constant is 330. J K-1. Calculate the enthalpy of solution of
Na 2SO4 in water at this concentration. Compare your result
with that calculated using the data in Table 4.1 (Appendix A,
Data Tables).
P4.33 If 7.300 g of methanol, CH3OH(l) is burned completely in a bomb calorimeter at 298.15 K, the heat produced
is 165.13 kJ.
a. Calculate ∆ c H ~ for methanol at 298.15 K.
b. Calculate ∆ f H ~ of methanol at 298.15K.
P4.34 A 0.1850 g sample of sucrose, C12H22O11, is burned
in a bomb calorimeter at 298 K. In order to produce the same
temperature rise in the calorimeter as the reaction, 3048.3 J
must be expended.
a. Calculate ∆ cU ~ and ∆ c H ~ for the combustion of 1 mol of
sucrose.
b. Using the data tables and your answer to (a), calculate
∆ f H ~ for sucrose.
c. The rise in temperature of the calorimeter and its contents
as a result of the reaction is 2.058 K. Calculate the heat
capacity of the calorimeter and its contents.
P4.35 Calcium chloride is soluble in water according to
the reaction CaCl21s2 S Ca2+1aq2 + 2Cl-1aq2. Assuming
122
CHAPTER 4 Thermochemistry
that no heat is transferred to the surroundings, calculate the
change in temperature when 12.5 g of CaCl2 are dissolved in
150. g of water.
CP(T)
Section 4.6
P4.36 The following data are a DSC scan of a solution of a
T4 lysozyme mutant. From the data, determine Tm. Determine
also the excess heat capacity ∆CP at T = 308 K. In addition,
trs
determine the intrinsic dC int
P and transition ∆C P excess heat
capacities at T = 308 K. In your calculations use the extrapolated curves, shown as dotted lines in the DSC scan, where
the y axis shows CP.
P4.37 Using the protein DSC data in Problem P4.36, calculate the enthalpy change between T = 288 K and T = 318 K.
Give your answer in units of kJ per mole. Assume the molar
mass of the protein is 14000. grams per mole. Hint: You can
perform the integration of the heat capacity by estimating
the area under the DSC curve and above the baseline in
Problem P4.36. This can be done by dividing the area into
small rectangles and summing the areas of the rectangles.
Comment on the accuracy of this method.
0.418 J K21g21
288
298
308
Temperature (K)
318
FURTHER READING
Höhne, G. W., Hemminger, W. F., and Flammersheim, H.-J.
Differential Scanning Calorimetry, 2nd Edition. Berlin:
Springer, 2003.
Jacobson, N. “Use of Tabulated Thermochemical Data for
Pure Compounds.” Journal of Chemical Education
78 (2001): 814–819.
Kildahl, N. K. “Bond Energy Data Summarized.” Journal of
Chemical Education 72 (1995): 423– 424.
C H A P T E R
5
Entropy and the
Second and Third Laws
of Thermodynamics
5.1
What Determines the Direction
of Spontaneous Change
in a Process?
5.2
The Second Law of
Thermodynamics, Spontaneity,
and the Sign of ∆S
5.3
Calculating Changes in Entropy
as T, P, or V Change
5.4
Understanding Changes in
Entropy at the Molecular Level
WHY is this material important?
5.5
The Clausius Inequality
Mixtures of chemically reactive species can evolve toward reactants or toward products, defining the direction of spontaneous change. How do we know which of these
reactions is spontaneous? Entropy, designated by S, is the state function that predicts
the direction of natural, or spontaneous, change. In this chapter, you will learn how to
carry out entropy calculations.
5.6
The Change of Entropy
in the Surroundings and
∆Stot = ∆S + ∆Ssur
5.7
Absolute Entropies and the
Third Law of Thermodynamics
5.8
Standard States in Entropy
Calculations
5.9
Entropy Changes in Chemical
Reactions
WHAT are the most important concepts and results?
Heat flows from hotter bodies to colder bodies, and gases mix rather than separate.
Entropy always increases for a spontaneous change in an isolated system. For a spontaneous change in a system interacting with its environment, the sum of the entropy
of the system and that of the surroundings always increases. Thermodynamics can be
used to devise ways to reduce our energy use and to transition away from fossil fuels.
WHAT would be helpful for you to review for this chapter?
Because differential and integral calculus and partial derivatives are used extensively in
this chapter, it would be helpful to review the material on calculus in Math Essential 2
and 3.
5.10 Heat Engines and the
Carnot Cycle
5.11 (Supplemental Section) How
Does S Depend on V and T ?
5.12 (Supplemental Section)
Dependence of S on T and P
5.13 (Supplemental Section) Energy
Efficiency, Heat Pumps,
Refrigerators, and Real Engines
5.1 WHAT DETERMINES THE DIRECTION OF
SPONTANEOUS CHANGE IN A PROCESS?
The first law of thermodynamics constrains the range of possible processes to those
that conserve energy and rules out machines that work indefinitely without an energy
source (perpetual motion machines) that have been the dream of inventors over centuries. Any system that is not at equilibrium will evolve with time. We refer to the
direction of evolution as the spontaneous direction. Spontaneous does not mean that
the process occurs immediately, but rather that it will occur with high probability if any
barrier to the change is overcome. For example, the transformation of a piece of wood
to CO2 and H2O in the presence of oxygen is spontaneous, but it only occurs at elevated
temperatures because an activation energy barrier must be overcome for the reaction to
proceed. In this chapter, we discuss the second law of thermodynamics, which allows
us to predict if a process is spontaneous. Before stating the second law, we discuss two
processes that satisfy the first law because they conserve energy and attempt to identify
criteria that can predict the spontaneous direction of evolution.
Concept
A process is defined to be
spontaneous even if it occurs only
for some conditions because of an
energy barrier.
123
124
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Energy
uptake
by floor
Figure 5.1
Path of a bouncing rubber ball. The
bouncing ball rebounds each time with a
smaller amplitude. The green semicircles
in the floor indicate energy uptake by the
floor.
qP
T1
T2
Figure 5.2
Adjacent metal rods of differing temperatures. The rods, for which T1 7 T2,
differ in temperature by ∆T, are placed in
thermal contact at constant pressure. This
composite system is contained in a rigid
adiabatic enclosure (not shown) and is,
therefore, an isolated system.
Consider the rubber ball shown in Figure 5.1 that falls in a gravitational field and
bounces upward after colliding with the floor.
We assume for simplicity that the ball and the floor consist of the same material.
After the initial drop, the ball falls again and bounces back a number of times. We know
from experience that each successive bounce will be lower than the previous one. How
do we explain this observation? Because of its kinetic energy, the falling ball has excess
energy relative to the floor. In an inelastic collision with the floor, the net energy transfer
is to the lower-energy floor and not to the higher-energy ball. Because the ball has lost
energy, each successive bounce does not attain the height of the previous bounce. What
would it take for each successive bounce to be higher? This would require that there be
a decrease of floor energy in the region where the ball hits and a corresponding increase
in the energy of the ball in such a way that the first law is not violated. In this case, the
collision partner with higher-energy gains rather than loses energy. Experience tells us
that this process is not spontaneous. We generalize this result and conclude that in a collision, excess energy is redistributed to the lower-energy collision partner, and not to the
higher-energy collision partner. This conclusion leads us to the following insight: As a
system evolves toward equilibrium in the spontaneous direction, energy becomes more
evenly distributed in space rather than being concentrated in a small region.
In Figure 5.2, we show a second process in which a hot metal rod is put in contact
with a cold metal rod, giving rise to a temperature difference between the two ends.
What happens to the temperature difference as the system evolves toward equilibrium?
Our experience is that the hot end becomes colder and the cold end becomes warmer
until the temperature difference vanishes. We never observe that the hot end continues
to become hotter while the cold end becomes colder, even though such a process could
still obey the first law. As for the previous example, in the spontaneous direction, energy
flows from the hot to the cold rod until the energy is distributed uniformly in space.
These two examples suggest that the driving force in a spontaneous process is
spreading out energy that occurs through heat flow rather than through work. We would
like to identify a new thermodynamic function that allows us to predict the direction of
spontaneous change. It should be a state function, so that the prediction depends only
on the initial and final states associated with the process and not on the path between
them. Because heat flow seems to be a key factor, we write the first law in the following form for an ideal gas undergoing changes in V and T in a reversible process.
dqrev = dU - dwrev
= CV1T2dT + PdV
nRT
= CV1T2dT +
dV
V
(5.1)
Recall from Math Essential 3 that the criterion for an exact differential is that the mixed
second partial derivatives are equal. Because
a
0CV1T2
0V
b ≠ a
T
01nRT>V2
0T
b
V
(5.2)
dqrev is not an exact differential. However, if we divide both sides of Equation (5.1) by T,
Concept
dqrev
is the differential form of a state
T
function that we call entropy.
CV1T2
dqrev
nR
=
dT +
dV
T
T
V
(5.3)
dqrev
is an exact differential1 because 103 CV1T2>T 4 >0V2T =
T
101nR>V2>0T2V = 0. This means that there is a state function that we call entropy,
dqrev
which we denote by S for which the differential form is
. For our example, the
T
change in S is given by
we see that
∆S1V, T2 =
1
L
dS =
Tƒ
Vƒ
CV1T2
dqrev
nR
=
dT +
dV
T
L T
LTi
LVi V
(5.4)
Although we have assumed ideal gas behavior, the conclusion that dqrev >T is an exact differential is more
general. See Lee et al. (2015) in Further Reading for details.
5.2 THE SECOND LAW OF THERMODYNAMICS, SPONTANEITY, AND THE SIGN OF ∆S
If Tƒ - Ti is not large, we can assume that CV is constant over the temperature interval
and can remove it from the integral to obtain
∆S1V, T2 = CV ln
Tƒ
Ti
+ nR ln
Vƒ
Vi
(5.5)
for the change in entropy of an ideal gas undergoing a change in volume and temperature.
5.2 THE SECOND LAW OF THERMODYNAMICS,
SPONTANEITY, AND THE SIGN OF 𝚫S
We have shown that dqrev >T is an exact differential, and we have defined a state function S related to the exact differential. However, we have not yet made a link between
∆S and spontaneity. We now do so by considering two processes—heat flow in the
metal rod shown in Figure 5.2 and the collapse of an ideal gas to one-half its volume,
in a quantitative manner.
The first process concerns the spontaneous direction of change in a metal rod with a
temperature gradient that we considered in the previous section. To model this process,
consider the isolated composite system considered in Section 5.1. Two systems, in the
form of identical metal rods with different temperatures T1 7 T2, are brought into thermal contact.
In the following discussion, we assume a very small amount of heat dqP is withdrawn from one rod and deposited in the other rod under constant pressure conditions.
Because dqP is very small, we can assume that T1 and T2 remain constant. To calculate
dS for this irreversible process, we must imagine a reversible process in which the initial and final states are the same as for the irreversible process because dS = dqrev >T.
For example, in an imaginary reversible process, the hotter rod is coupled to a reservoir
whose temperature is lowered very slowly. The temperatures of the rod and the reservoir differ only infinitesimally throughout the process in which an amount of heat,
dqP, is withdrawn from the rod. The decrease in temperature of the rod, dT, is related
to qP by
dqP = CPdT
(5.6)
It has been assumed that dT = T1 - T2 V T2 and T1.
Because the path is defined (constant pressure), dqP is independent of how rapidly
the heat is withdrawn (the path). More formally, because dqP = dH and because H is
a state function, dqP is independent of the path between the initial and final states.
Therefore, dqP = dqrev if the temperature increment dT is identical for the reversible
and irreversible processes. We next couple the second rod at the lower temperature T2
to a heat reservoir and raise the reservoir temperature slowly by dT. We deposit the
same amount of heat in the second rod that was withdrawn from the first rod.
Next, we calculate the entropy change for this process in which heat flows from one
rod to the other. Keep in mind that although we are considering a reversible process,
S is a state function. Therefore, if the initial and final states are the same, ∆S has the
same value for any process, whether reversible or irreversible. Because the composite
system is isolated, dq1 + dq2 = 0, and dq1 = -dq2 = dqP . The entropy change of
the composite system is the sum of the entropy changes in each rod:
dS =
dqrev, 1
T1
+
dqrev, 2
T2
=
dq1
dq2
T2 - T1
1
1
+
= dqP a - b = dqP a
b (5.7)
T1
T2
T1
T2
T2
Because T1 7 T2, the quantity in the parentheses is negative. We calculate ∆S for the
two possible directions:
• If heat flows from the hotter to the colder rod, the temperature difference becomes
smaller. dqP 6 0 and dS 7 0.
• If heat flows from the colder to the hotter rod, the temperature difference becomes
larger. dqP 7 0 and dS 6 0.
Concept
System entropy increases for the
direction of spontaneous change
in an isolated system.
125
126
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Ti,Vi
Ti,1/2 Vi
Initial state
Final state
(a) Irreversible process
Piston
Ti,Vi
Piston
Ti,1/2 Vi
Initial state
Final state
(b) Reversible isothermal compression
Figure 5.3
Two processes in which a confined
gas is reduced to one-half of its initial
volume. (a) An irreversible process is
shown in which an ideal gas confined
in a container with rigid adiabatic walls
is spontaneously reduced to one-half its
initial volume. (b) A reversible isothermal
compression in a piston and cylinder
assembly is shown for the same initial
and final states as in the irreversible process in part (a). Reversibility is achieved
by adjusting the rate at which the beaker
on top of the piston is filled with water
relative to the evaporation rate.
Note that dS and dqP have the same magnitude, but different signs, for the two
directions of change. This shows that S is a useful function for measuring the direction
of spontaneous change in an isolated system because it gives us different values for
the two possible directions. All experimental evidence tells us that heat flows from the
hotter to the colder body and the temperature gradient becomes smaller with time. In
the absence of constraints, systems evolve to distribute energy uniformly throughout
space. It can be concluded from this example that the process in which ∆S 7 0 is the
direction of spontaneous change in an isolated system.
Next, imagine a second process in which an ideal gas collapses to one-half its initial
volume without a force acting on it. This process and its reversible analog are shown
in Figure 5.3. Recall that U is independent of V for an ideal gas. Because U does not
change as V decreases, and U is a function of T only for an ideal gas, the temperature
remains constant in the irreversible process. Therefore, the irreversible process shown
in Figure 5.3a is both adiabatic and isothermal and is described by Vi, Ti S 11>2Vi2, Ti.
From the ideal gas law, we know that the final pressure is twice the initial pressure. The
imaginary reversible process that we use to carry out the calculation of ∆S is shown in
Figure 5.3b. In this process, which must have the same initial and final states as the irreversible process, water is slowly and continuously added to the beaker on the piston to
ensure that P = Pext throughout the process. The ideal gas undergoes a reversible isothermal transformation described by Vi, Ti S 11>2Vi2, Ti. Because ∆U = 0, q = -w.
We calculate ∆S for this process:
∆S =
1
qrev
dqrev
wrev
2 Vi
=
= = nR ln
= -nR ln 2 6 0
Ti
Ti
Vi
L T
(5.8)
Experience tells us that the opposite process, in which the gas spontaneously expands so
that it occupies twice the volume, is the spontaneous direction of change. For this case
∆S = nR ln
2Vi
= nR ln 2 7 0
Vi
(5.9)
Again, we find that the process with ∆S 7 0 is the direction of spontaneous change
in this isolated system. We can also understand this process in terms of a redistribution
of energy; the kinetic energy of the gas molecules is spatially spread out by the gas
uniformly filling the volume.
The results obtained for the two isolated systems that we have discussed are generalized in the following statement of the second law of thermodynamics:
For any process that proceeds in an isolated system, there is a unique direction
of spontaneous change and ∆S 7 0 for the spontaneous process.
∆S 6 0 for the opposite or nonspontaneous direction of change, and ∆S = 0 only
for a reversible process. In a reversible process, there is no direction of spontaneous
change because the system is always in equilibrium. We cannot emphasize too strongly
that ∆S 7 0 is a criterion for spontaneous change only if the system does not exchange
energy in the form of heat or work with its surroundings. Note that if any process
occurs in the isolated system, it is by definition spontaneous and the entropy increases.
Whereas U can neither be created nor destroyed, S for a process occurring in an isolated system is created 1∆S 7 02 but never destroyed 1∆S 6 02.
5.3 CALCULATING CHANGES IN ENTROPY
AS T, P, OR V CHANGE
The most important thing to remember in carrying out entropy calculations is that ∆S
must be calculated along a reversible path because it is defined by 1 1dqrev >T2. For
an irreversible process, ∆S must be calculated for a reversible process that proceeds
5.3 CALCULATING CHANGES IN ENTROPY AS T, P, OR V CHANGE
127
between the same initial and final states as the irreversible process. Because S is a state
function, ∆S is path independent, provided that the transformation is between the same
initial and final states in both processes. Therefore, your calculation for the reversible
path also holds for the irreversible process.
We first consider two processes that require no calculation. First, for any reversible
adiabatic process, qrev = 0, so that ∆S = 1 1dqrev >T2 = 0. Second, for any cyclic
process, ∆S = A 1dqrev >T2 = 0, because the change in any state function for a cyclic
process is zero. Next, consider ∆S for the reversible isothermal expansion or compression of an ideal gas, described by Vi, Ti S Vƒ, Ti. Because ∆U = 0 for this process, as
was discussed in Section 2.5,
qrev = -wrev = nRT ln
∆S =
Vƒ
Vi
and
Vƒ
dqrev
1
= qrev = nR ln
T
Vi
L T
(5.10)
(5.11)
Note that ∆S 7 0 for an expansion 1Vƒ 7 Vi2 and ∆S 6 0 for a compression
1Vƒ 6 Vi2.
Next, consider ∆S for an ideal gas that undergoes a reversible change in T at constant V or P. For a reversible process described by Vi, Ti S Vi, Tƒ, it also holds that
dqrev = CV dT, and furthermore that
∆S =
Tƒ
nCV,m dT
dqrev
=
≈ nCV,m ln
T
Ti
L T
L
(5.12)
For a constant pressure process described by Pi, Ti S Pi, Tƒ, it also holds that
dqrev = CP dT, and furthermore that
∆S =
Tƒ
nCP,m dT
dqrev
=
≈ nCP,m ln
T
Ti
L T
L
(5.13)
The last expressions in Equations (5.12) and (5.13) are valid if the temperature interval
is small enough that the temperature dependence of CV,m and CP,m can be neglected.
Again, although expressions for ∆S have been derived for a reversible process, Equations (5.12) and (5.13) hold for any reversible or irreversible process between the same
initial and final states for an ideal gas.
The results of the last three derived expressions can be combined in the following
way. Because the macroscopic variables V, T or P, T completely define the state of an
ideal gas, any change Vi, Ti S Vƒ, Tƒ can be separated into two segments, Vi, Ti S Vƒ, Ti
and Vƒ, Ti S Vƒ, Tƒ. A similar statement about P and T is also valid. Because ∆S is independent of the path, any reversible or irreversible process for an ideal gas described
by Vi, Ti S Vƒ, Tƒ can be treated as consisting of two segments, one of which occurs at
constant volume and the other at constant temperature. For this two-step process, ∆S
is given by
∆S = nR ln
Vƒ
Vi
+ nCV,m ln
Tƒ
Ti
(5.14)
This result was previously obtained in Equation (5.5). Similarly, for any reversible or
irreversible process for an ideal gas described by Pi, Ti S Pƒ, Tƒ as shown in Example
Problem 5.1.
∆S = -nR ln
Pƒ
Pi
+ nCP,m ln
Tƒ
Ti
(5.15)
In writing Equations (5.14) and (5.15), it has been assumed that the temperature dependence of CV,m and CP,m can be neglected over the temperature range of interest.
Concept
Because entropy is a state function,
∆S is the same for a reversible and
irreversible process between the
same initial and final states.
128
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
EXAMPLE PROBLEM 5.1
Using the equation of state and the relationship between CP,m and CV,m for an ideal
gas, show that Equation (5.14) can be transformed into Equation (5.15).
Solution
∆S = nR ln
Vƒ
Vi
= -nR ln
Tƒ
+ nCV,m ln
Pƒ
Pi
Ti
= nR ln
+ n1CV,m + R2ln
Tƒ
Ti
Tƒ Pi
Ti Pƒ
+ nCV,m ln
= -nR ln
Pƒ
Pi
Tƒ
Ti
+ nCP,m ln
Tƒ
Ti
Next consider ∆S for phase changes. Experience shows that a liquid is converted to
a gas at a constant boiling temperature through heat input if the process is carried out at
constant pressure. Because qP = ∆H, ∆S for the reversible process is given by
∆ vap S =
∆ vap H
dqrev
qrev
=
=
Tvap
Tvap
L T
(5.16)
Similarly, for the phase change solid S liquid,
∆ fus S =
Concept
dqrev
qrev
∆ fus H
=
=
Tfus
Tfus
L T
(5.17)
Finally, consider ∆S for an arbitrary process involving real gases, solids, and liquids
for which the volumetric thermal expansion coefficient a and the isothermal compressibility kT are known, but the equation of state is not known. The calculation of ∆S
for such processes is described in Supplemental Section 5.11 and 5.12, in which the
properties of S as a state function are fully exploited. The results are stated here. For the
system undergoing the change Vi, Ti S Vƒ, Tƒ,
Even if the equation of state is not
known, ∆S can be calculated if a and
kT are known.
Tƒ
Vƒ
Ti
Vi
Tƒ
CV
a
a
∆S =
dT +
dV = CV ln +
1Vƒ - Vi2
k
k
T
T
T
L
L T
i
(5.18)
The last step in Equation (5.18) is only valid if CV, kT , and a are constant over
the temperature and volume intervals of interest, which is only the case for a liquid or
solid. For the system undergoing a change Pi, Ti S Pƒ, Tƒ,
Tƒ
Pƒ
Ti
Pi
CP
∆S =
dT Va dP
L T
L
(5.19)
For a solid or liquid, the last equation can be simplified to
∆S = CP ln
Tƒ
Ti
- Va1Pƒ - Pi2
(5.20)
if CP, V, and a are assumed constant over the temperature and pressure intervals of
interest. The integral forms of Equations (5.18) and (5.20) are valid for ideal and real
gases, liquids, and solids. Calculations using these equations are presented in Example
Problems 5.2 through 5.4.
EXAMPLE PROBLEM 5.2
One mole of CO gas is transformed from an initial state characterized by Ti = 320. K
and Vi = 80.0 L to a final state characterized by Tƒ = 650. K and Vƒ = 120.0 L.
Assuming ideal gas behavior, calculate ∆S for this process. For CO,
CV,m
J mol
-1
-1
K
= 31.08 - 0.01452
T
T2
T3
+ 3.1415 * 10-5 2 - 1.4973 * 10-8 3
K
K
K
5.3 CALCULATING CHANGES IN ENTROPY AS T, P, OR V CHANGE
Solution
Consider the following reversible process in order to calculate ∆S. The gas is first heated
reversibly from 320. to 650. K at a constant volume of 80.0 L. Subsequently, the gas is
reversibly expanded at a constant temperature of 650. K from a volume of 80.0 L to a
volume of 120.0 L. The entropy change for this process is obtained using Equation (5.14).
Tƒ
∆S =
Vƒ
CV
dT + nR ln
Vi
L T
Ti
-1
∆S1J K 2 =
650. K
1.00 mol *
L
320. K
a31.08 - 0.01452
T
T2
T3
+ 3.1415 * 10-5 2 - 1.4973 * 10-8 3 b
K
K
K
dT
T
+ 1.00 mol * 8.314 J K-1 mol-1 * ln
120.0 L
80.0 L
= 22.024 J K-1 - 4.792 J K-1 + 5.027 J K-1 - 1.207 J K-1
+ 3.371 J K-1
= 24.4 J K-1
EXAMPLE PROBLEM 5.3
In this problem, 2.50 mol of CO2 gas is transformed from an initial state characterized
by Ti = 450. K and Pi = 1.35 bar to a final state characterized by Tƒ = 800. K and
Pƒ = 3.45 bar. Calculate ∆S for this process using Equation (5.19). Assume ideal gas
behavior and use the ideal gas value for a. For CO2,
CP,m
J mol-1 K-1
= 18.86 + 7.937 * 10-2
T
T2
T3
- 6.7834 * 10-5 2 + 2.4426 * 10-8 3
K
K
K
Solution
Consider the following reversible process in order to calculate ∆S. The gas is first
heated reversibly from 450. to 800. K at a constant pressure of 1.35 bar. Subsequently,
the gas is reversibly compressed at a constant temperature of 800. K from a pressure
of 1.35 bar to a pressure of 3.45 bar. We first derive an expression for a.
a =
Tƒ
1 0V
1 03 nRT>P4
1
a b = a
b =
V 0T P
V
0T
T
P
Pƒ
Tƒ
Pƒ
Tƒ
Pi
Ti
Pi
Ti
Pƒ
CP,m
CP,m
CP
dP
∆S1J K 2 =
dT Va dP = n
dT - nR
= n
dT - nR ln
Pi
L T
L
L T
L P
L T
-1
Ti
= 2.50 mol
*
800. K a18.86
L
+ 7.937 * 10-2
450. K
T
T2
T3
- 6.7834 * 10-5 2 + 2.4426 * 10-8 3 b
K
K
K
dT
T
-2.50 mol * 8.314 * ln
3.45 bar
J K-1 mol-1
1.35 bar
= 27.13 J K-1 + 69.45 J K-1 - 37.10 J K-1 + 8.57 J K-1 -19.50 J K-1
= 48.6 J K-1
129
130
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
EXAMPLE PROBLEM 5.4
In this problem, 3.00 mol of liquid mercury is transformed from an initial state characterized by Ti = 300. K and Pi = 1.00 bar to a final state characterized by Tƒ = 600. K
and Pƒ = 3.00 bar.
a. Calculate ∆S for this process; a = 1.81 * 10-4 K-1, r = 13.54 g cm-3, and
CP,m for Hg1l2 = 27.98 J mol-1 K-1.
b. What is the ratio of the pressure-dependent term to the temperature-dependent
term in ∆S? Explain your result.
Solution
a. Because the volume changes only slightly with temperature and pressure over the
range indicated,
Tƒ
Pƒ
Ti
Pi
Tƒ
CP
∆S =
dT Va dP ≈ nCP,m ln
- nVm,ia1Pƒ - Pi2
Ti
L T
L
= 3.00 mol * 27.98 J mol-1 K-1 * ln
-3.00 mol *
600. K
300. K
200.59 g mol-1
13.54 g cm
-3
106 cm3
*
m3
* 1.81 * 10-4 K-1
* 2.00 bar * 105 Pa bar -1
= 58.2 J K-1 - 1.61 * 10-3 J K-1 = 58.2 J K-1
b. The ratio of the pressure-dependent to the temperature-dependent term is
-3 * 10-5. Because the volume change with pressure is very small, the contribution of the pressure-dependent term is negligible in comparison with the
temperature-dependent term.
As Example Problem 5.4 shows, ∆S for a liquid or solid for a process in which both
P and T change is dominated by the temperature dependence of S. Unless the change in
pressure is very large, ∆S for liquids and solids can be considered to be a function of
temperature only.
5.4 UNDERSTANDING CHANGES IN ENTROPY
AT THE MOLECULAR LEVEL
In the first part of this chapter, we used the behavior of a bouncing ball, two rods placed
in thermal contact, and an ideal gas expanding to fill its accessible volume to show
that there is a natural tendency to spread out energy in spontaneous processes. At the
molecular level, what is the basis for this tendency? This tendency can be explained in
terms of probability as will be discussed in Chapter 12.
At the molecular level, systems have discrete energy levels rather than a continuous
energy spectrum. Consider the example of a He atom confined to a one-dimensional
box. The allowed energy levels for He atoms in boxes of length 8.00 and 10.00 nm
are shown in Figure 5.4. Think of the probabilities depicted in Figure 5.4 by circles
as individual He atoms. The total energy of the system is the sum of the individual
energies. Assigning the number of He atoms to the various energy levels specifies a
configuration, and the weight of a configuration is the number of ways (microstates)
5.4 Understanding Changes in Entropy at the Molecular Level
Figure 5.4
Arrows show how
energy levels change as
box length increases
from 8.00 nm to 10.0 nm.
Allowed energy levels for a He atom
in a one-dimensional box. Box length
is 10.0 nm in (a) and 8.00 nm in (b). The
length of the chain of circles on each
energy level gives the probability that
each energy level is occupied. One circle
corresponds to a probability of 0.01. The
connecting lines between the boxes show
how the energy levels change as the box
length increases from 8.00 nm to 10.0 nm.
1.4 3 10223
1.2 3 10223
Energy (joules)
131
1.0 3 10223
8.0 3 10224
6.0 3 10224
4.0 3 10224
2.0 3 10224
(a) 10.0 nm
(b) 8.00 nm
that a configuration can be generated by placing He atoms in the energy levels. The
configurations depicted in Figure 5.4 for the two boxes are associated with the greatest
number of microstates and are, therefore, the most probable configurations.
Ludwig Boltzmann (1844–1906) derived the following relationship between the
entropy and the number of microstates consistent with a given total energy.
S = k ln W
(5.21)
In this equation, W is the number of microstates and k is the Boltzmann constant, which
is equal to the ideal gas constant divided by Avogadro’s number. The important conclusion to be drawn from this equation is that entropy increases as the number of
microstates available to the system increases. The increase can occur either through
lowering the energy levels or increasing the temperature, which allows more energy
levels to be occupied.
We use the length of the box in analogy to the volume of a three-dimensional system to explain why S increases as V increases. For a given total energy, there are more
ways to distribute the atoms in energy levels as the box length increases because the
energy levels become more closely spaced and lower in energy. Therefore, more energy
levels and more microstates are accessible for a given total energy as the system
volume increases. This explanation is the molecular-level basis of the increase in S as
V increases.
Keeping the box length constant but increasing the temperature to increase the
system energy also leads to a redistribution of the atoms in the most probable configuration, as shown in Figure 5.5. We see that more energy levels are accessible to
the system as the temperature increases. Employing the same reasoning as above, we
see that more microstates will be available to the system as the temperature increases,
which explains why S increases as T increases.
We will revisit the molecular level definition of entropy and the dependence
of S on V and T at a molecular level in a more rigorously quantitative manner in
Chapter 15.
Concept
At the microscopic level, entropy
is associated with the number of
microstates consistent with a given
total energy.
132
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Figure 5.5
1.4 3 10223
1.2 3 10223
Energy (joules)
Probabilities for a He atom of occupying the allowed energy levels in a onedimensional box. Energy levels are for
(a) 0.200 K and (b) 0.300 K. Box lengths
for both diagrams are 10.0 nm. The length
of the chain of circles on each energy
level is proportional to the probability that
each energy level is occupied. One circle
corresponds to a probability of 0.01.
1.0 3 10223
8.0 3 10224
6.0 3 10224
4.0 3 10224
2.0 3 10224
(a) 0.200 K
(b) 0.300 K
5.5 THE CLAUSIUS INEQUALITY
In Section 5.2, it was shown, by using two examples, that ∆S 7 0 provides a criterion
to predict the natural direction of change in an isolated system. This result can also be
obtained without considering a specific process. Consider the differential form of the
first law for a process in which only P–V work is possible:
dU = dq - Pext dV
(5.22)
Equation (5.22) is valid for both reversible and irreversible processes. If the process is
reversible, we can write Equation (5.22) in the following form:
dU = dqrev - P dV = T dS - P dV
(5.23)
Because U is a state function, dU is independent of the path, and Equation (5.23) holds
for both reversible and irreversible processes, as long as there are no phase transitions
or chemical reactions, and only P–V work occurs.
To derive the Clausius inequality, we equate the expressions for dU in Equations
(5.22) and (5.23):
Concept
For any irreversible process in an
isolated system, ∆S 7 0.
dqrev - dq = 1P - Pext2dV
(5.24)
dqrev - dq = TdS - dq Ú 0 or TdS Ú dq
(5.25)
If P - Pext 7 0, the system will spontaneously expand, and dV 7 0. If P - Pext 6 0,
the system will spontaneously contract, and dV 6 0. For both possibilities,
1P - Pext2dV 7 0. Therefore, we conclude that
The equality holds only for a reversible process. We rewrite the Clausius inequality
in Equation (5.25) for an irreversible process in the form
dS 7
dq
T
(5.26)
However, for an irreversible process in an isolated system, dq = 0. Therefore, we have
proved that for any irreversible process in an isolated system, ∆S 7 0.
5.6 THE CHANGE OF ENTROPY IN THE SURROUNDINGS AND ∆Stot = ∆S + ∆Ssur
133
How can the result from Equations (5.22) and (5.23) that dU = dq - Pext dV =
TdS - PdV be reconciled with the fact that work and heat are path functions? The
answer is that dw Ú -PdV and dq … TdS, where the equalities hold only for a
reversible process. The result dq + dw = TdS - PdV states that the amount by
which the work is greater than -PdV and the amount by which the heat is less than
TdS in an irreversible process involving only PV work are exactly equal. Therefore,
the differential expression for dU in Equation (5.23) is obeyed for both reversible
and irreversible processes. In Chapter 6, the Clausius inequality is used to generate
two new state functions, the Gibbs energy and the Helmholtz energy. These functions
allow predictions to be made about the spontaneous direction of change in processes
for which the system interacts with its environment, using only information about
the system.
5.6 THE CHANGE OF ENTROPY
IN THE SURROUNDINGS AND
𝚫S tot = 𝚫S + 𝚫Ssur
As shown in Section 5.2, the entropy of an isolated system increases in a spontaneous
process. Is it always true that a process is spontaneous if ∆S for the system is positive?
In this section, a criterion for spontaneity is developed that takes into account the
entropy change in both the system and the surroundings.
In general, a system interacts only with the part of the universe that is very close.
Therefore, one can think of the system and the interacting part of the surroundings as
forming an interacting composite system that is isolated from the rest of the universe,
as shown in Figures 2.1 and 5.6. The part of the surroundings that is relevant for entropy calculations is a thermal reservoir at a fixed temperature, T. The mass of the
reservoir is sufficiently large that its temperature is only changed by an infinitesimal
amount dT when heat is transferred between the system and the surroundings. Therefore, the surroundings always remain in internal equilibrium during heat transfer.
Next consider the entropy change of the surroundings, whereby the surroundings are
at either constant V or constant P. We assume that the system and surroundings are at
the same temperature. If this were not the case, heat would flow across the boundary
until T is the same for system and surroundings, unless the system is surrounded by
adiabatic walls, in which case qsur = 0. The amount of heat absorbed by the surroundings, qsur, depends on the process occurring in the system. If the surroundings are
at constant V, qsur = ∆Usur, and if the surroundings are at constant P, qsur = ∆Hsur .
Because H and U are state functions, the amount of heat entering the surroundings
is independent of the path; q is the same whether the transfer occurs reversibly or
irreversibly. Therefore,
dSsur =
dqsur
qsur
, or for a macroscopic change, ∆Ssur =
T
T
(5.27)
Note that the heat that appears in Equation (5.27) is the actual heat transferred. By
contrast, in calculating ∆S for the system, dqrev for a reversible process that connects
the initial and final states of the system must be used, not the actual dq for the process
because dS is defined by dqrev >T and not dq>T. It is essential to understand this reasoning in order to carry out calculations for ∆S and ∆Ssur .
Using the correct value for heat so as to correctly calculate the entropy change of
the system, as opposed to that of the surroundings, is further explored in Figure 5.6. In
this figure, a gas (the system) is enclosed in a piston and cylinder assembly with diathermal walls. The gas is reversibly compressed by the external pressure generated by
water slowly accumulating on the top of the piston. The piston and cylinder assembly
is in contact with a water bath thermal reservoir that keeps the temperature of the
gas fixed at the value T. In Example Problem 5.5, ∆S and ∆Ssur is calculated for this
reversible compression.
Concept
For a system that interacts with
its surroundings, the spontaneity
condition is ∆S tot = ∆S + ∆Ssur 7 0.
w
T
Piston
P,V
q
Figure 5.6
Sample of an ideal gas (the system) confined in a piston and cylinder assembly
with diathermal walls. The assembly is
in contact with a thermal reservoir that
holds the temperature at a value of 300. K.
Water dripping into the beaker on the
piston increases the external pressure
slowly enough to ensure a reversible compression. The directions of work and heat
flow are indicated.
134
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
EXAMPLE PROBLEM 5.5
One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a
volume of 25.0 L to a volume of 10.0 L. Because the water bath is very large, the temperature of this thermal reservoir in the surroundings remains essentially constant at
300. K during the process. Calculate ∆S, ∆Ssur, and ∆Stot.
Solution
Because this is an isothermal reversible process, ∆U = 0, and q = qrev = -w. From
Section 2.5,
Vƒ
qrev = -w = nRT
Vƒ
dV
= nRT ln
Vi
L V
Vi
= 1.00 mol * 8.314 J mol-1 K-1 * 300. K * ln
10.0 L
= -2.29 * 103 J
25.0 L
The entropy change of the system is given by
∆S =
dqrev
qrev
-2.29 * 103 J
=
=
= -7.62 J K-1
T
300. K
L T
The entropy change of the surroundings is given by
∆Ssur =
qsur
q
2.29 * 103 J
= - =
= 7.62 J K-1
T
T
300. K
The total change in the entropy is given by
∆Stot = ∆S + ∆Ssur = -7.62 J K-1 + 7.62 J K-1 = 0
Because the process in Example Problem 5.5 is reversible, there is no direction of
spontaneous change and, therefore, ∆Stot = 0. In Example Problem 5.6, this calculation is repeated for an irreversible process that goes between the same initial and final
states of the system.
EXAMPLE PROBLEM 5.6
One mole of an ideal gas at 300. K is isothermally compressed by a constant
external pressure equal to the final pressure in Example Problem 5.5. Because
P ≠ Pext, this process is irreversible. The initial volume is 25.0 L, and the final
volume is 10.0 L. The temperature of the surroundings is 300. K. Calculate ∆S,
∆Ssur , and ∆Stot .
Solution
The initial and final states of the system in this problem are the same as those in
Example Problem 5.5. Because S is a state function, we know that ∆S for this
­irreversible process must be the same as for the reversible process, therefore,
∆S = -7.62 J>K. To calculate ∆Ssur, we first calculate the external pressure and the
initial pressure in the system:
Pext =
Pi =
nRT
1 mol * 8.314 J mol-1 K-1 * 300. K
=
= 2.49 * 105 Pa
V
1 m3
10.0 L * 3
10 L
nRT
1 mol * 8.314 J mol-1 K-1 * 300. K
=
= 9.98 * 104 Pa
V
1 m3
25.0 L * 3
10 L
5.7 Absolute Entropies and the Third Law of Thermodynamics
135
Because Pext 7 Pi, we know that the direction of spontaneous change will be the compression of the gas to a smaller volume. Because ∆U = 0,
q = -w = Pext1Vƒ - Vi2 = 2.49 * 105 Pa * 110.0 * 10-3 m3 - 25.0 * 10-3 m32
= -3.74 * 103 J
The entropy change of the surroundings is given by
∆Ssur =
qsur
q
3.74 * 103 J
= - =
= 12.45 J K -1
T
T
300. K
It is seen that ∆S 6 0 and ∆Ssur 7 0. The total change in the entropy is given by
∆Stot = ∆S + ∆Ssur = -7.62 J K -1 + 12.45 J K -1 = 4.83 J K -1
The entropy of the system can decrease in a spontaneous process, as it does in this
case, as long as ∆Stot 7 0.
5.7 ABSOLUTE ENTROPIES AND THE
THIRD LAW OF THERMODYNAMICS
All elements and many compounds exist in three different states of aggregation. One
or more solid phases are the most stable forms at low temperature, and when the temperature is increased to the melting point, a constant temperature transition to the liquid phase is observed. As the temperature is increased further, a constant temperature
phase transition to a gas is observed at the boiling point. At temperatures greater than
the boiling point, the gas is the stable form.
The entropy of an element or a compound is experimentally determined using heat
capacity data through the relationship dqrev,P = CP dT. Just as for the thermochemical
data discussed in Chapter 4, entropy values are generally tabulated for a standard temperature of 298.15 K and a standard pressure of 1 bar. As an example, we will describe
such a determination for the entropy of O2 at 298.15 K, first in a qualitative fashion and
then quantitatively in Example Problem 5.7.
The experimentally determined heat capacity of O2 is shown in Figure 5.7 as a function of temperature for a pressure of 1 bar. O2 has three solid phases, and transitions
between them are observed at 23.66 and 43.76 K. The solid form that is stable above
43.76 K melts to form a liquid at 54.39 K. The liquid vaporizes to form a gas at 90.20 K.
These phase transitions are indicated as dashed lines in Figure 5.7. Experimental measurements of CP,m are available above 12.97 K. Below this temperature, the data are
extrapolated to zero kelvin by assuming that in this very low temperature range CP,m
varies with temperature as T 3. This extrapolation is based on a model of the vibrational
spectrum of a crystalline solid that will be discussed in Chapter 15. The explanation for
the dependence of CP,m on T is the same as that presented for Cl2 in Section 2.11.
Concept
The entropy of the universe is always
increasing.
50
Cp,m (JK21 mol21)
The previous calculations lead to the following conclusion: If the system and the
part of the surroundings with which it interacts are viewed as an isolated composite system, the criterion for spontaneous change is ∆Stot = ∆S + ∆Ssur 7 0. We can
generalize this result by expanding the size of the composite system to infinity and
conclude that a decrease in the entropy of the universe will never be observed because
∆Stot Ú 0 for any change that actually occurs. Any process that occurs in the universe
is by definition spontaneous and leads to an increase of Stot.
Note that a spontaneous process in a system that interacts with its surroundings
is not characterized by ∆S 7 0 but by ∆Stot 7 0. The entropy of the system 1∆S2
can decrease in a spontaneous process, as long as the entropy of the surroundings
1∆Ssur2 increases by a greater amount. In Chapter 6, the spontaneity criterion ∆Stot =
∆S + ∆Ssur 7 0 will be used to generate two state functions, the Gibbs energy and the
Helmholtz energy. These functions allow one to predict the direction of change in
systems that interact with their environment using only the changes in system state
functions.
40
30
Solid I
Solid II
Solid III
Liquid
20
Gas
10
0
20
40
60
80 100 120
Temperature (K)
Figure 5.7
Experimentally determined heat capacity for O2 as a function of temperature.
Dots are data from Giauque and Johnston
[J. American Chemical Society 51 (1929),
2300], and the red solid curve below 90 K
are polynomial fits to these data. The red
line above 90 K is a fit to data from the
NIST Chemistry Webbook. The blue
curve is an extrapolation from 12.97 to 0 K
as described in the text. Vertical dashed
lines indicate constant temperature-phase
transitions, and the most stable phase at a
given temperature is indicated in the figure.
136
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Under constant pressure conditions, the molar entropy of the gas can be expressed
in terms of the molar heat capacities of the solid, liquid, and gaseous forms and the
enthalpies of fusion and vaporization as
Tƒ
Sm1T2 = Sm10 K2 +
Concept
The entropy of a pure, perfectly
crystalline substance (element or
compound) is zero at zero Kelvin.
+
∆ vap H
Tb
L
C solid
P,m dT′
0
T
L
+
Tb
T′
+
∆Hƒusion
Tƒ
Tb
+
L
Tƒ
C liquid
P,m dT′
T′
gas
C P,m dT′
(5.28)
T′
If the substance has more than one solid phase, each will give rise to a separate integral.
Note that the entropy changes associated with the phase transitions solid S liquid and
liquid S gas, discussed in Section 5.4, must be included in the calculation as well as
any solid S solid transitions. To obtain a numerical value for Sm1T2, the heat capacity
must be known down to zero Kelvin, and Sm10 K2 must also be known.
We first address the issue of the entropy of a solid at zero Kelvin. The third law of
thermodynamics can be stated in the following form, attributed to Max Planck:
The entropy of a pure, perfectly crystalline substance (element or compound) is
zero at zero Kelvin.
Different forms of the same substance, if pure and perfectly crystalline, such as graphite,
diamond, and C60 all have zero entropy at absolute zero.
Although a more detailed discussion of the third law using a microscopic model
will be presented in Chapter 15, a brief overview is presented here. Recall that in a
perfectly crystalline atomic (or molecular) solid, the position of each atom is known.
Because the individual atoms are indistinguishable, exchanging the positions of two
atoms does not lead to a new state. Therefore, a perfect crystalline solid has only one
microstate at zero kelvin and S = k ln W = k ln 1 = 0. Some crystalline solids such
as CO(s) are not perfectly ordered as they are cooled to 0 K because the energy difference between parallel and antiparallel CO molecules is very small. At the freezing
temperature, the thermal energy of CO molecules is insufficient to surmount the energy
barrier to rotation. In such a case, disorder is frozen in, and the entropy approaches a
nonzero value as T S 0 K, which is referred to as a residual entropy. The importance
of the third law is that it allows determinations of the absolute entropies of elements
and compounds to be carried out for any value of T. To calculate S at a temperature T
using Equation (5.28), the CP,m data of Figure 5.7 are graphed in the form CP,m >T as
shown in Figure 5.8.
Entropy can be obtained as a function of temperature by numerically integrating
the area under the curve in Figure 5.8 and adding the entropy changes associated with
1
CP,m
(JK22 mol21)
T
Solid I
Figure 5.8
CP,m , T as a function of temperature
for O2. The vertical dashed lines indicate
constant temperature-phase transitions,
and the most stable phase at a given
temperature is indicated in the figure.
0.8
0.6
0.4
0.2
Solid
II
Liquid
Solid
III
0
Gas
50
100
150
200
Temperature (K)
250
300
5.7 Absolute Entropies and the Third Law of Thermodynamics
Figure 5.9
The molar entropy of O2 as a function
of temperature. The vertical dashed
lines indicate constant temperature-phase
transitions, and the most stable phase at
a given temperature is indicated in the
figure.
150
Solid I
100
Gas
Solid II
[
m (JK
21 mol21)
200
S
137
50
Liquid
Solid III
0
50
100
150
Temperature (K)
200
250
300
phase changes at the transition temperatures. Calculations of Sm~ for O2 at 298.15 K are
carried out in Example Problem 5.7, and Sm~ 1T2 is shown in Figure 5.9.
From the previous discussion, several general statements, listed below as bullet
points, regarding the relative magnitudes of the entropy of different substances and
different phases of the same substance can now be presented. These statements will be
justified on the basis of a microscopic model in Chapter 15.
• Because CP >T in a single-phase region and ∆S for melting and vaporization are
•
•
•
always positive, Sm for a given substance is greatest for the gas-phase species. The
gas
molar entropies follow the order Sm
7 Sliquid
7 Ssolid
m
m .
The molar entropy increases with the size of a molecule because the number of
degrees of freedom increases with the number of atoms. A nonlinear gas-phase molecule has three translational degrees of freedom, three rotational degrees of freedom,
and 3n - 6 vibrational degrees of freedom. A linear molecule has three translational,
two rotational, and 3n - 5 vibrational degrees of freedom. For a molecule in a liquid,
the three translational degrees of freedom are converted to local vibrational modes.
A solid has only vibrational modes. It can be modeled as a three-dimensional array
of coupled harmonic oscillators, as shown in Figure 5.10. This solid has a wide spectrum of vibrational frequencies, and solids with a large binding energy have higher
frequencies than more weakly bound solids. Because modes with high frequencies
are not activated at low temperatures, Ssolid
is larger for weakly bound solids than
m
for strongly bound solids at low and moderate temperatures.
The entropy of all substances is a monotonically increasing function of temperature.
EXAMPLE PROBLEM 5.7
The heat capacity of O2 has been measured at 1 atm pressure over the interval
12.97 K 6 T 6 298.15 K. The data have been fit to the following polynomial series
in T>K:
0 6 T 6 12.97 K:
CP,m1T2
J mol-1 K-1
J mol
-1
K
-1
= -5.666 + 0.6927
J mol-1 K-1
Concept
Concept
2
3
T
T
T
- 5.191 * 10-3 2 + 9.943 * 10-4 3
K
K
K
23.66 K 6 T 6 43.76 K:
CP,m1T2
Schematic model of a solid. A useful
model of a solid is a three-dimensional
array of coupled springs. In solids with
a high binding energy, the springs are
very stiff.
Molar entropies follow the order
liquid
S gas
7 S solid
m 7 Sm
m .
T3
= 2.11 * 10-3 3
K
12.97 K 6 T 6 23.66 K:
CP,m1T2
Figure 5.10
= 31.70 - 2.038
T
T2
T3
+ 0.08384 2 - 6.685 * 10-4 3
K
K
K
The entropy of all substances is a
monotonically increasing function
of temperature.
138
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
43.76 K 6 T 6 54.39 K:
CP,m1T2
J mol-1 K -1
54.39 K 6 T 6 90.20 K:
CP,m1T2
J mol-1 K -1
= 81.268 - 1.1467
= 46.094
T
T2
T3
+ 0.01516 2 - 6.407 * 10-5 3
K
K
K
90.20 K 6 T 6 298.15 K:
CP,m1T2
J mol
-1
K
-1
= 32.71 - 0.04093
T
T2
T3
+ 1.545 * 10-4 2 - 1.819 * 10-7 3
K
K
K
The transition temperatures and the enthalpies for the transitions indicated in Figure 5.8
are as follows:
Solid III S solid II
Solid II S solid I
Solid I S liquid
Liquid S gas
93.80 J mol-1
743 J mol-1
445.0 J mol-1
6815 J mol-1
23.66 K
43.76 K
54.39 K
90.20 K
a. Using these data, calculate Sm~ for O2 at 298.15 K.
b. What are the three largest contributions to Sm~ ?
Solution
heating solid phase III
to 12.97 K
T
0K
heating solid phase II
T
23.66 K
12.97 K
743 J mol-1
+
+
43.76 K
T
54.39 K
L
C solid,I
P,m dT
43.76 K
T
µ
L
54.39 K
C liquid
P,m dT
93.80 J
23.36 K
vaporization
e
µ
445.0 J mol-1
+
54.39 K
90.20 K
+
heating solid phase I
transition II S I
heating liquid
fusion
+
T
e
L
C solid,II
P,m dT
L
C solid,III
dT
P,m
µ
+
e
43.76 K
23.66 K
•
C solid,III
dT
P,m
transition
III S II
e
L
a. Sm~ 1298.15 K2 = Sm~ 10 K2 +
e
12.97 K
heating solid
phase III
+
6815 J mol-1
90.20 K
heating gas
e
298.15 K
+
L
90.20 K
gas
C P,m dT
T
= 0 + 1.534 J mol-1 K-1
+ 3.964 J mol-1 K-1 +
+ 16.98 J mol-1 K-1 +
+ 8.181 J mol-1 K-1 +
+ 75.59 J mol-1 K-1 +
= 204.9 J mol-1 K-1
+ 6.649 J mol-1 K-1
19.61 J mol-1 K-1
10.13 J mol-1 K-1
27.06 J mol-1 K-1
35.27 J mol-1 K-1
There is an additional small correction for nonideality of the gas at 1 bar. The currently
accepted value is Sm~ 1298.15 K2 = 205.152 J mol-1 K-1 (Linstrom, P. J., and Mallard,
W. G., Eds., NIST Chemistry Webbook: NIST Standard Reference Database Number 69,
National Institute of Standards and Technology, Gaithersburg, MD, retrieved from
http://webbook.nist.gov.)
b. The three largest contributions to Sm are ∆S for the vaporization transition, ∆S for
heating of the gas from the boiling temperature to 298.15 K, and ∆S for heating
of the liquid from the melting temperature to the boiling point.
139
5.9 Entropy Changes in Chemical Reactions
The preceding discussion and Example Problem 5.7 show how numerical values of
the entropy for a specific substance can be determined at a standard pressure of 1 bar
for different values of the temperature. These numerical values can then be used to
calculate entropy changes in chemical reactions, as will be shown in Section 5.9.
Concept
Tabulated values of entropy refer to a
standard pressure of 1 bar.
5.8 STANDARD STATES IN ENTROPY
[
Sm – S
As discussed in Chapter 4, changes in U and H are calculated using the result that ∆Hƒ
values for pure elements in their standard state at a pressure of 1 bar and a temperature
of 298.15 K are zero. For S, the third law provides a natural definition of zero, namely,
the crystalline state at zero kelvin. Therefore, the absolute entropy of a compound as
a given temperature can be experimentally determined from heat capacity measurements, as described in the previous section. Entropy is also a function of pressure, and
tabulated values of entropy refer to a standard pressure of 1 bar. The value of S varies
most strongly with P for a gas. From Equations (5.14) and (5.15), for an ideal gas at
constant T,
Vƒ
Pƒ
∆Sm = R ln
= -R ln
(5.29)
Vi
Pi
m
CALCULATIONS
0
1
Choosing Pi = P ~ = 1 bar,
Sm1P2 = Sm~ - R ln
P1bar2
P~
2
3
(5.30)
Figure 5.11 shows a plot of the molar entropy of an ideal gas as a function of pressure.
It is seen that as P S 0, Sm S ∞. This is a consequence of the fact that as P S 0,
V S ∞. As Equation (5.30) shows, entropy becomes infinite in this limit.
Equation (5.30) provides a way to calculate the entropy of a gas at any pressure.
For solids and liquids, S varies so slowly with variations in P, as shown in Section 5.3
and Example Problem 5.4, that the pressure dependence of S can usually be neglected.
Figure 5.11
Molar entropy of an ideal gas as a function of gas pressure. By definition, at
1 bar, Sm = Sm~ , the standard state molar
entropy. As discussed in the text, as
P S 0, entropy becomes infinite.
5.9 ENTROPY CHANGES IN CHEMICAL
REACTIONS
The entropy change in a chemical reaction is a major factor in determining the equilibrium concentration in a reaction mixture. In an analogous fashion to calculating ∆H R~
and ∆U R~ for chemical reactions, ∆SR~ is equal to the difference in the entropies of
products and reactions, which can be written as
∆ rS ~ = a vi Si~
i
(5.31)
In Equation (5.31), the stoichiometric coefficients vi are positive for products and
negative for reactants. For example, in the reaction
Fe 3O41s2 + 4H21g2 S 3Fe1s2 + 4H2O1l2
4
P (bar)
(5.32)
the entropy change under standard state conditions of 1 bar and 298.15 K is given by
~
~
~
~
~
∆ r S298.15
= 3S298.15
1Fe, s2 + 4S298.15
K1H2O, l2 - S298.151Fe 3O4, s2 - 4S298.151H2, g2
= 3 * 27.28 J K-1 mol-1 + 4 * 69.61 J K-1 mol-1 - 146.4 J K-1 mol-1
- 4 * 130.684 J K-1 mol-1
= -308.9 J K-1 mol-1
Note that unlike formation enthalpies, S ~ for elements in their standard reference state
are not zero. For this reaction, ∆SR is large and negative, primarily because gaseous
species are consumed in the reaction, and none are generated. For reactions involving
gases, ∆SR~ is generally positive for ∆n 7 0, and negative for ∆n 6 0 where ∆n is the
change in the number of moles of gas in the overall reaction.
Concept
The entropy change for a chemical
reaction is equal to the difference
in the entropies of products and
reactions.
140
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Tabulated values of S ~ are generally available at the standard temperature of 298.15 K,
and values for selected elements and compounds are listed in Tables 4.1 and 4.2 (see
Appendix A, Data Tables). However, it is often necessary to calculate ∆S ~ at other
temperatures. Such calculations are carried out using the temperature dependence of S
discussed in Section 5.4 and shown in Example Problem 5.8:
T
∆ r S ~ 1T2 = ∆ r S ~ 1298.15 K2 +
∆CP
dT′
L T′
(5.33)
298.15
This equation is valid only if no phase changes occur in the temperature interval
between 298.15 K and T. If phase changes occur, the associated entropy changes must
be included as they were in Equation (5.28).
EXAMPLE PROBLEM 5.8
The standard entropies of CO, CO2, and O2 at 298.15 K are
~
S298.15
1CO, g2 = 197.67 J K-1 mol-1
~
S298.15
1CO2, g2 = 213.74 J K-1 mol-1
~
S298.15
1O2, g2 = 205.138 J K-1 mol-1
The temperature dependence of constant pressure heat capacity for CO, CO2, and O2
is given by
CP,m1CO, g2
= 31.08 - 1.452 * 10-2
T
T2
T3
+ 3.1415 * 10-5 2 - 1.4973 * 10-8 3
K
K
K
CP,m1CO2, g2
= 18.86 + 7.937 * 10-2
T
T2
T3
- 6.7834 * 10-5 2 + 2.4426 * 10-8 3
K
K
K
CP,m1O2, g2
= 30.81 - 1.187 * 10-2
-1
JK
mol
-1
J K-1 mol-1
J K-1 mol-1
T
T2
+ 2.3968 * 10-5 2
K
K
Calculate ∆ r S ~ for the reaction CO1g2 + 1>2 O21g2 S CO21g2 at 475 K.
Solution
∆CP,m
-1
JK
mol
-1
= a18.86 - 31.08 -
1
* 30.81b
2
+ a7.937 + 1.452 +
1
T
* 1.187b * 10-2
2
K
- a6.7834 + 3.1415 +
1
T2
* 2.3968b * 10-5 2
2
K
+ 12.4426 + 1.49732 * 10-8
= -27.63 + 9.983 * 10-2
T3
K3
T
T2
T3
- 1.112 * 10-4 2 + 3.940 * 10-8 3
K
K
K
∆ r S ~ = S ~ 1CO2, g, 298.15 K2 - S ~ 1CO, g, 298.15 K2 = 213.74 J K-1 mol-1 - 197.67 J K-1 mol-1 = -86.50 J K-1 mol-1
1
* S ~ 1O2, g, 298.15 K2
2
1
* 205.138 J K-1 mol-1
2
5.10 Heat Engines and the Carnot Cycle
T
∆ r S ~ 1T2 = ∆ r S ~ 1298.15 K2 +
L
298.15 K
∆Cp
T′
141
dT′
= -86.50 J K-1 mol-1
475 K
+
L
a- 27.63 + 9.983 * 10-2
298.15 K
T
T2
T3
- 1.112 * 10-4 2 + 3.940 * 10-8 3 b
K
K
K
dT J K-1 mol-1
T
= -86.50 J K-1 mol-1 + 1-12.866 + 17.654 - 7.605 + 1.05942J K-1 mol-1
= -86.50 J K-1 mol-1 - 1.757 J K-1 mol-1 = -88.3 J K-1 mol-1
The value of ∆ r S ~ is negative at both temperatures because the number of moles of
gaseous species is reduced in the reaction.
5.10 HEAT ENGINES AND THE CARNOT CYCLE
The concept of entropy arose as 19th-century scientists attempted to maximize the work
output of engines. In this section, we obtain an alternate formulation of the second law
of thermodynamics based on a theoretical study of idealized engines. A real engine,
such as an automobile engine, operates in a cyclical process of fuel intake, compression, ignition and expansion, and exhaust, which occurs several thousand times per
minute and is used to perform work on the surroundings. Because the work produced
by such an engine is a result of the heat released in a combustion process, it is referred
to as a heat engine. An idealized version of a heat engine is depicted in Figure 5.12.
The system consists of a working substance (in this case an ideal gas) confined in a
piston and cylinder assembly with diathermal walls. This assembly can be brought into
contact with a hot reservoir at Thot or a cold reservoir at Tcold. The expansion or contraction of the gas caused by changes in its temperature drives the piston in or out of the
cylinder. This linear motion is converted to circular motion, and the rotary motion is
used to do work in the surroundings.
The efficiency of a heat engine is of particular interest in practical applications.
Experience shows that work can be converted to heat with 100% efficiency. Consider
an example from calorimetry, discussed in Chapter 4, in which electrical work is
performed on a resistive heater immersed in a water bath. We observe that all of the
electrical work done on the heater has been converted to heat, resulting in an increase
in the temperature of the water and the heater. What is the maximum theoretical
efficiency of the reverse process, the conversion of heat to work? As shown below, it
is less than 100%. There is a natural asymmetry in the efficiency of converting work
to heat and converting heat to work. Thermodynamics provides an explanation for
this asymmetry.
Concept
An analysis of the Carnot cycle shows
that there is a natural asymmetry in
the efficiency of converting work to
heat and converting heat to work.
Figure 5.12
Thot
Tcold
A schematic depiction of a heat engine.
The working substance (in this example
an ideal gas) is confined by a piston and
cylinder assembly. As the assembly is
brought into contact with hot or cold
reservoirs, temperature changes cause the
piston to move, which generates a linear
motion of the connecting rod, that is in
turn mechanically converted to a rotary
motion, which is used to perform work.
142
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Figure 5.13
a
Pa
Isothermal expansion
Thot
Pressure
A reversible Carnot cycle for a sample
of an ideal gas working substance projected onto an indicator diagram. The
cycle consists of two adiabatic and two
isothermal segments. The arrows indicate
the direction in which the cycle is traversed. The insets show the volume of gas
and the coupling to the reservoirs at the
beginning of each successive segment of
the cycle. The coloring of the contents of
the cylinder indicates the presence of the
gas and not its temperature. The volume
of the cylinder shown is that at the beginning of the appropriate segment.
Adiabatic
compression
Tcold
Thot
Pb
b
Thot
Tcold
Pd
d
Pc
Isothermal
compression
Va
Tcold
Thot
c
Tcold
0
Adiabatic expansion
Vd
Vb
Vc
Volume
Concept
It is not possible to construct a cyclic
process in an indicator diagram that
results in nonzero net work using
only adiabatic or only isothermal
segments.
As discussed in Section 2.7, the maximum work output in an isothermal expansion
occurs in a reversible process. For this reason, we will next calculate the efficiency of
a reversible heat engine because the efficiency of a reversible engine is an upper bound
to the efficiency of a real engine. This reversible engine converts heat into work by
exploiting the spontaneous tendency of heat to flow from a hot reservoir to a cold reservoir. It performs work on the surroundings by operating in a cycle of reversible expansions and compressions of an ideal gas in a piston and cylinder assembly. (We discuss
automotive engines in Section 5.13.)
This reversible cycle is shown in Figure 5.13 in a P–V diagram. The expansion
and compression steps are designed so that the engine returns to its initial state after
four steps. Recall that the area within the cycle equals the work done by the engine.
Beginning at point a, the first segment is a reversible isothermal expansion in which the
gas absorbs heat from the reservoir at Thot, and does work on the surroundings. In the
second segment, the gas expands further, this time adiabatically. Work is also done on
the surroundings in this step. At the end of the second segment, the gas has cooled to
the temperature Tcold. The third segment is an isothermal compression in which the surroundings do work on the system and heat is absorbed by the cold reservoir. In the final
segment, the gas is compressed to its initial volume, this time adiabatically. Work is
done on the system in this segment, and the temperature returns to its initial value, Thot.
In summary, heat is taken up by the engine in the first segment at Thot and is released to
the surroundings in the third segment at Tcold. Work is done on the surroundings in the
first two segments and on the system in the last two segments. An engine is only useful
if net work is done on the surroundings—that is, if the magnitude of the work done in
the first two steps is greater than the magnitude of the work done in the last two steps.
The efficiency of the engine can be calculated by comparing the net work per cycle
with the heat taken up by the engine from the hot reservoir.
Before carrying out this calculation, we will discuss the rationale for the design of
this reversible cycle in more detail. To avoid losing heat to the surroundings at temperatures between Thot and Tcold, the adiabatic segments 2 1b S c2 and 4 1d S a2
are used to move the gas between these temperatures. To absorb heat only at Thot and
release heat only at Tcold, segments 1 1a S b2 and 3 1c S d2 must be isothermal. The
reason for using alternating isothermal and adiabatic segments is that no two isotherms
at different temperatures intersect and no two adiabats starting from two different temperatures intersect. Therefore, it is impossible to create a closed cycle of nonzero area
in an indicator diagram out of isothermal or adiabatic segments alone. However, net
work can be done using alternating adiabatic and isothermal segments. The reversible
cycle depicted in Figure 5.13 is called a Carnot cycle, after the French engineer, Nicolas
Carnot, who first studied such cycles in the early 19th century.
5.10 Heat Engines and the Carnot Cycle
143
TABLE 5.1 Heat, Work, and 𝚫U for the Reversible Carnot Cycle
Segment Initial State
Final State
aSb
Pa, Va, Thot
Pb,Vb, Thot
bSc
q
∆U
w
Pc,Vc, Tcold
wab1- 2
∆Uab = 0
Pb, Vb, Thot
qab1+2
cSd
Pc, Vc, Tcold
Pd,Vd, Tcold
Pa,Va, Thot
0
wcd1+2
∆Ucd = 0
dSa
Pd, Vd, Tcold qcd1-2
wbc1-2
Cycle
Pa, Va, Thot
Pa,Va, Thot
wda1+ 2
∆Uda = wda1+2
qab + qcd1+ 2 wab + wbc + wcd + wda1-2 ∆Ucycle = 0
0
∆Ubc = wbc1-2
The efficiency of the Carnot cycle can be determined by calculating q, w, and ∆U
for each segment of the cycle, assuming that the working substance is an ideal gas. The
results are shown in a qualitative fashion in Table 5.1. The appropriate signs for q and
w are indicated. If ∆V 7 0, w 6 0 for the segment, and the corresponding entry for
work has a negative sign. For the isothermal segments, q and w have opposite signs
because ∆U = q + w = 0. From Table 5.1, it is seen that
wcycle = wcd + wda + wab + wbc and qcycle = qab + qcd
(5.34)
Because ∆Ucycle = 0,
wcycle = -1qcd + qab2
(5.35)
wcycle 6 0, therefore 0 qab 0 7 0 qcd 0
(5.36)
By comparing the areas under the two expansion segments with those under the
two compression segments in the indicator diagram in Figure 5.13, one can see that
the total work as seen from the system is negative, meaning that work is done on the
surroundings in each cycle. Using this result, we arrive at an important conclusion that
relates the heat flow in the two isothermal segments:
More heat is withdrawn from the hot reservoir than is deposited in the cold reservoir. It
is useful to make a model of this heat engine that indicates the relative magnitude and
direction of the heat and work flow; such a model is shown in Figure 5.14a. The figure
makes it clear that not all of the heat withdrawn from the higher temperature reservoir
is converted to work done by the system (the engine) on the surroundings.
Hot reservoir
Hot reservoir
Work
Figure 5.14
Work
Cold reservoir
(a) Schematic model of the heat engine
operating in a reversible Carnot cycle.
(b) According to the second law of
thermodynamics, the heat engine
shown cannot be constructed.
Carnot heat engine cycle. (a) A schematic model of a heat engine operating in
a reversible Carnot cycle. (b) The second
law of thermodynamics asserts that it is
impossible to construct a heat engine as
shown that operates using a single heat
reservoir and converts the heat withdrawn
from the reservoir into work with 100%
efficiency as shown.
144
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
The efficiency, e, of the reversible Carnot engine is defined as the ratio of the work
output to the heat withdrawn from the hot reservoir. Referring to Table 5.1,
Concept
No heat engine can convert heat
entirely into work.
e = -
wcycle
qab
= 1 -
=
qab + qcd
qab
0 qcd 0
6 1 because 0 qab 0 7 0 qcd 0 , qab 7 0, and qcd 6 0
0 qab 0
(5.37)
Equation (5.37) shows that the efficiency of a heat engine operating in a reversible
Carnot cycle is always less than one. Equivalently, not all of the heat withdrawn from
the hot reservoir can be converted to work. This conclusion is valid for all engines and
illustrates the asymmetry in converting heat to work and work to heat.
These considerations on the efficiency of reversible heat engines led to the Kelvin–
Planck formulation of the second law of thermodynamics:
It is impossible for a system to undergo a cyclic process whose sole effects are
the flow of heat into the system from a heat reservoir and the performance of an
equivalent amount of work by the system on the surroundings.
An alternative, but equivalent, statement of the second law by Clausius is as follows:
Pressure
It is impossible for a system to undergo a cyclic process whose sole effects are
the flow of heat into the system from a hot reservoir and the flow of an equivalent amount of heat out of the system into a cold reservoir.
Volume
Figure 5.15
Approximation of a reversible cycle. A
reversible cycle of arbitrary shape, indicated by the ellipse, can be approximated
to any desired accuracy by a large number
of Carnot cycles, each having a sequence
of alternating adiabatic and isothermal
segments.
The second law asserts that the heat engine depicted in Figure 5.14b cannot be
constructed. The second law has been put to the test many times by inventors who have
claimed that they have invented an engine that has an efficiency of 100%. No such
claim has ever been validated. To test the assertion made in this statement of the second
law, imagine that such an engine has been invented. We mount it on a boat in Seattle
and set off on a journey across the Pacific Ocean. Heat is extracted from the ocean,
which is the single heat reservoir, and is converted entirely to work in the form of a
rapidly rotating propeller. Because the ocean is huge, the decrease in its temperature as
a result of withdrawing heat is negligible. By the time we arrive in Japan, not a gram
of diesel fuel has been used because all the heat needed to power the boat has been
extracted from the ocean. The money that was saved on fuel is used to set up an office
and begin marketing this wonder engine. Does this scenario sound too good to be true?
It is. Such an impossible engine is called a perpetual motion machine of the second
kind because it violates the second law of thermodynamics. A perpetual motion
machine of the first kind violates the first law.
The first statement of the second law can be understood using an indicator diagram.
For an engine to produce work, the area of the cycle in a P–V diagram must be greater
than zero. However, this is impossible in a simple cycle using a single heat reservoir.
If Thot = Tcold in Figure 5.13, the cycle a S b S c S d S a collapses to a line, and
the area of the cycle is zero. An arbitrary reversible cycle can be constructed that does
not consist of individual adiabatic and isothermal segments. However, as shown in
Figure 5.15, any reversible cycle can be approximated by a succession of adiabatic and
isothermal segments, an approximation that becomes exact as the length of each segment approaches zero. It can be shown that the efficiency of such a cycle is also given
by Equation (5.37), so that the efficiency of all heat engines operating in any reversible
cycle between the same two temperatures, Thot and Tcold, is identical.
A more useful form than Equation (5.37) for the efficiency of a reversible heat engine can be derived if the working substance in the engine is an ideal gas. Calculating
5.10 Heat Engines and the Carnot Cycle
145
the work flow in each of the four segments of the Carnot cycle using the results of
Sections 2.7 and 2.9,
wab = -nRThot ln
Vb
Va
wab 6 0 because Vb 7 Va
wbc = nCV,m1Tcold - Thot2 wbc 6 0 because Tcold 6 Thot
wcd
Vd
= -nRTcold ln
Vc
(5.38)
wcd 7 0 because Vd 6 Vc
wda = nCV,m1Thot - Tcold2 wda 7 0 because Thot 7 Tcold
As derived in Section 2.14, the volume and temperature in the reversible adiabatic
segments are related by
g-1
ThotV b
g-1
= TcoldV c
g-1
and TcoldV d
g-1
= ThotV a
(5.39)
You will show in the end-of-chapter problems that Vc and Vd can be eliminated from
the set of Equations (5.38) to yield
wcycle = -nR1Thot - Tcold2 ln
Vb
6 0
Va
(5.40)
Because ∆Ua S b = 0, the heat withdrawn from the hot reservoir is
qab = -wab = nRThot ln
Vb
Va
(5.41)
and the efficiency of the reversible Carnot heat engine with an ideal gas as the working
substance can be expressed solely in terms of the reservoir temperatures
e =
0 wcycle 0
qab
=
Thot - Tcold
Tcold
= 1 6 1
Thot
Thot
(5.42)
The efficiency of this reversible heat engine can approach one only as Thot S ∞ or
Tcold S 0, neither of which can be accomplished in practice. Therefore, heat can never
be totally converted to work in a reversible cyclic process. Because wcycle for an engine
operating in an irreversible cycle is less than the work attainable in a reversible cycle,
eirreversible 6 ereversible 6 1.
EXAMPLE PROBLEM 5.9
Calculate the maximum work that can be done by a reversible heat engine operating
between 500. and 200. K if 1000. J is absorbed at 500. K.
Solution
The fraction of the heat that can be converted to work is the same as the fractional fall
in the absolute temperature. This is a convenient way to link the efficiency of an engine
with the properties of the absolute temperature.
e = 1 -
Tcold
200. K
= 1 = 0.600
Thot
500. K
w = eqab = 0.600 * 1000. J = 600. J
In this section, only the most important features of heat engines have been discussed.
It can also be shown that the efficiency of a reversible heat engine is independent of the
working substance. Therefore, the result in Equation (5.42) is not restricted to an ideal
gas working substance; rather, it is valid for any working substance and depends only
on Thot and Tcold. For a more in-depth discussion of heat engines, the interested reader
is referred to Heat and Thermodynamics, seventh edition, by M. W. Zemansky and
R. H. Dittman (McGraw-Hill, 1997). We will return to a discussion of the efficiency
of engines when we discuss refrigerators, heat pumps, and real engines in Section 5.13.
Concept
The efficiency of a reversible Carnot
engine can be calculated from the
temperatures of the hot and cold
reservoirs.
146
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
S U P P L E M E N TA L
S E C T I O N
5.11 HOW DOES S DEPEND ON V AND T?
Section 5.4 showed how entropy varies with P, V, and T for an ideal gas. In this section,
we derive general equations for the dependence of S on V and T that can be applied to
solids, liquids, and real gases. We do so by using the property that dS is an exact differential. A similar analysis of S as a function of P and T is carried out in Section 5.12.
Consider Equation (5.23), rewritten in the form
Concept
Using the properties of partial
derivatives and state functions, the
dependence of S on V and T can be
expressed in therms of measurable
quantities.
dS =
1
P
dU + dV
T
T
(5.43)
Because 1>T and P>T are greater than zero, the entropy of a system increases with the
internal energy at constant volume and increases with the volume at constant internal
energy. However, because internal energy is not generally a variable under experimental
control, it is more useful to obtain equations for the dependence of dS on V and T.
We first write the total differential dS in terms of the partial derivatives with respect
to V and T:
0S
0S
dS = a b dT + a b dV
(5.44)
0T V
0V T
To evaluate 10S>0T2V and 10S>0V2T , we start with dS = dU>T + 1P>T2dV, allowing
Equation (5.44) for dS to be rewritten in the form
dS =
CV
1
0U
P
1
0U
c CV dT + a b dV d + dV =
dT + c P + a b d dV
T
0V T
T
T
T
0V T
(5.45)
Equating the coefficients of dT and dV in Equations (5.44) and (5.45),
a
CV
0S
b =
0T V
T
and a
0S
1
0U
b = cP + a b d
0V T
T
0V T
(5.46)
The temperature dependence of entropy at constant volume can be calculated straightforwardly using the first equality in Equation (5.46):
dS =
CV
dT, constant V
T
(5.47)
The expression for 10S>0V2T in Equation (5.46) is not in a form that allows for a direct
comparison with experiment. The property that dS is an exact differential (see Section 3.1)
allows a more useful relation to be derived. Because
a
0 0S
0 0S
a b b = a a b b
0T 0V T V
0V 0T V T
(5.48)
one can take the mixed second derivatives of the expressions in Equation (5.46),
a
0 0S
1 0 0U
a b b = a a b b
0V 0T V T
T 0V 0T V T
0 0S
1 0P
0 0U
1
0U
a a b b = ca b + a a b b d - 2 cP + a b d
0T 0V T V
T 0T V
0T 0V T V
0V T
T
(5.49)
Substituting the expressions for the mixed second derivatives in Equation (5.49) into
Equation (5.48), canceling the double mixed derivative of U that appears on both sides
of the equation, and simplifying the result, we obtain the following equation:
P + a
0U
0P
b = Ta b
0V T
0T V
(5.50)
This equation provides the expression for the internal pressure 10U>0V2T that was used
without a derivation in Section 3.2. It provides a way to calculate the internal pressure
of the system if the equation of state for the substance is known.
5.12 DEPENDENCE OF S ON T AND P
147
Comparing the result in Equation (5.50) with the second equality in Equation (5.46),
we obtain a practical equation for the dependence of entropy on volume under constant
temperature conditions:
a
10V>0T2P
0S
0P
a
b = a b = =
kT
0V T
0T V
10V>0P2T
(5.51)
where a is the coefficient for thermal expansion at constant pressure and kT is the isothermal compressibility coefficient. Both of these quantities are readily obtained from
experiments. In simplifying this expression, the cyclic rule for partial derivatives has
been used.
The result of these mathematical manipulations is that dS can be expressed in terms
of dT and dV as
dS =
CV
a
dT +
dV
kT
T
(5.52)
Integrating both sides of this equation,
Tƒ
Vƒ
Ti
Vi
CV
a
∆S =
dT +
dV
k
T
L
L T
(5.53)
This result applies to a single-phase system of a liquid, solid, or gas that undergoes a
transformation from Ti, Vi to Tƒ, Vƒ, provided that no phase changes or chemical reactions occur in the system.
S U PPL EM EN TAL
SEC TI O N
5.12 DEPENDENCE OF S ON T AND P
Because chemical transformations are normally carried out at constant pressure rather
than constant volume, we need to know how S varies with T and P. The total differential dS is written in the form
dS = a
0S
0S
b dT + a b dP
0T P
0P T
(5.54)
Concept
Starting from the relation U = H - PV, we write the total differential dU as
dU = T dS - P dV = dH - P dV - V dP
(5.55)
This equation can be rearranged to give an expression for dS:
dS =
1
V
dH - dP
T
T
(5.56)
The previous equation is analogous to Equation (5.43) but contains the variable P
rather than V.
dH = a
0H
0H
0H
b dT + a b dP = CP dT + a b dP
0T P
0P T
0P T
(5.57)
Substituting this expression for dH into Equation (5.56),
dS =
CP
1 0H
0S
0S
dT + c a b - V d dP = a b dT + a b dP
T
T 0P T
0T P
0P T
(5.58)
Because the coefficients of dT and dP must be the same on both sides of Equation (5.58),
a
CP
0S
b =
0T P
T
and a
0S
1 0H
b = ca b - Vd
0P T
T 0P T
(5.59)
Using the properties of partial
derivatives and state functions, the
dependence of S on P and T can
be expressed in terms of measurable
quantities.
148
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
The ratio CP >T is positive for all substances, allowing one to conclude that S is a monotonically increasing function of the temperature.
Just as for 10S>0V2T in Section 5.12, the expression for 10S>0P2T is not in a form
that allows a direct comparison with experimental measurements. Just as in our evaluation of 10S>0V2T , we equate the mixed second partial derivatives of 10S>0T2P and
10S>0P2T :
a
0 0S
0 0S
a b b = a a b b
0T 0P T P
0P 0T P T
(5.60)
These mixed partial derivatives can be evaluated using Equation (5.59):
a
a
0 0S
1 0CP
1 0 0H
a b b = a
b = a a b b
0P 0T P T
T 0P T
T 0P 0T P T
0 0S
1
0 0H
0V
1
0H
a b b = ca a b b - a b d - 2 ca b - Vd
0T 0P T P
T 0T 0P T P
0T P
0P T
T
(5.61)
(5.62)
Equating Equations (5.61) and (5.62),
1 0 0H
1
0 0H
0V
1
0H
a a b b = ca a b b - a b d - 2 ca b - Vd
T 0T 0P T P
T 0T 0P T P
0T P
0P T
T
(5.63)
Simplifying this equation results in
a
0H
0V
b - V = -T a b
0P T
0T P
(5.64)
Using this result and Equation (5.59), we can write the pressure dependence of the
entropy at constant temperature in a form that easily allows an experimental determination of this quantity to be made:
a
0S
0V
b = - a b = -Va
0P T
0T P
(5.65)
Using these results, we can write the total differential dS in terms of experimentally
accessible parameters as
dS =
CP
dT - Va dP
T
(5.66)
Integrating both sides of this equation,
Tƒ
Pƒ
CP
∆S =
dT Va dP
L T
L
Ti
(5.67)
Pi
This result applies to a single-phase system of a pure liquid, solid, or gas that undergoes a transformation from Ti Pi to Tƒ, Pƒ, provided that no phase changes or chemical
reactions occur in the system.
SUP P LE ME NTAL
SE CTION
5.13 ENERGY EFFICIENCY, HEAT PUMPS,
REFRIGERATORS, AND REAL ENGINES
Thermodynamics provides the tools to study the release of energy through chemical
reactions or physical processes such as harvesting light to perform work or generate
heat. As the Earth’s population continues to increase in the coming decades, the need
for energy in various forms will rise rapidly. Electricity is a major component of this
6.0
High-emissions scenario
Moderate-emissions scenario
5.0
Low-emissions scenario
4.0
2020–2029
Year 2000 constant
concentrations
20th century
3.0
2090–2099
Moderateemissions
scenario
1.0
0
–1.0
2000
Year
2100
(a) Global surface warming for years 1900 to 2000
(black curve) and projected to 2100 (colored
curves, various scenarios).
Moderate-emissions scenario
High-emissions scenario
2.0
1900
149
Highemissions
scenario
Low-emissions scenario
Global surface warming (°C)
5.13 Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines
Lowemissions
scenario
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5
(°C)
(b) Uncertainty in predicted
temperature increase for
a given scenario is
represented by the height
of bars between panels.
(c) Increase in temperature on a regional scale
for different scenarios of greenhouse gas
emissions.
Figure 5.16
energy demand, and fossil fuels are expected to be the major source for electricity
production for the foreseeable future. The increased combustion of fossil fuels will
continue to contribute to the rapid increase in the CO2 concentration in the atmosphere
that began with the industrial revolution.
This increased CO2 concentration is expected to lead to a significant increase in
the global average surface temperature, as shown for different scenarios of the rate of
atmospheric CO2 increase in Figure 5.16. For details on the scenarios and the possible
consequences of this temperature increase, which include flooding of land areas near
sea level, acidification of the oceans, and increasingly severe droughts and storms, see
the Intergovernmental Panel on Climate Change website at www.ipcc.ch/.
In order to slow or to reverse the buildup of greenhouse gases in the atmosphere,
we must capture significant portions of the greenhouses gases produced in the combustion of fossil fuels and move to energy sources, such as wind and solar, that do not generate greenhouse gases. We must also find ways to produce more industrial output and
work with less energy. Table 5.2 shows that per capita energy use in the United States
is substantially higher than that in other regions of the world as well as Europe which
has a similar climate and living standard.
The energy flow in the United States for 2013 through various sectors of the
economy is shown in Figure 5.17. The parallel pathway for CO2 emissions into
the atmosphere is shown in Figure 5.18. By comparing the useful and lost energy,
TABLE 5.2 Per Capita Energy Use, 2009
Nation or Region
Per Capita Energy Use (mega BTU)
1 J = 1054 BTU
Ratio to U.S. per Capita Use
United States
308
1
Eurasia
137
0.444
Europe
133
0.431
Africa
16
0.0519
World
71
0.231
Source: U.S. Energy Information Administration
Historical and projected global surface
temperatures.
Source: UN Intergovernmental Panel on
Climate Change
Figure 5.17
U.S. energy flow trends for 2013. The energy contributions from different sources to each economic sector, such as electricity generation, are
shown, together with net useful and rejected (wasted) energy for each sector. Note that 59% of the total consumed energy in all sectors is wasted.
Source: Lawrence Livermore National Laboratory, U.S. Department of Energy
Figure 5.18
2013 U.S. carbon dioxide emissions from energy consumption. The pathways of CO2 generation in economic sectors are shown, together
with a breakdown in the three fossil fuel sources of natural gas, coal, and petroleum.
Source: Lawrence Livermore National Laboratory, U.S. Department of Energy
5.13 Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines
it is seen that 59% of the energy is lost in the form of heat, including friction in
engines and turbines, resistive losses in the distribution of electricity, and heat loss
from poorly insulated buildings. Note in particular the inefficiency of the transportation sector, where 79% of expended energy is wasted. As discussed in Section 5.10,
the conversion of heat to work cannot be achieved with an efficiency of 100%, even
if dissipative processes such as friction are neglected. Therefore, significant losses
are inevitable, but substantial increases in energy efficiency are possible, as will be
discussed below. Electricity generation by wind turbines, photovoltaic panels, hydroelectric plants, and fuel cells (see Section 11.13) is of particular importance because
it involves the conversion of one form of work to another. The efficiency for these
methods is not subject to the limitations imposed on the conversion of heat to work
by the second law.
What can we learn from thermodynamics that will help us use less energy and produce more work with less energy? To address this question, consider how energy is
used in the residential sector, as shown in Figure 5.19. Heating buildings and water
accounts for 47% of residential energy usage, and electrical energy is most commonly
used for these purposes. Both building and water heating can be provided with significantly less energy input using an electrically powered heat pump, as explained below.
In an idealized reversible heat pump, that can be used either to heat or refrigerate,
the Carnot cycle in Figure 5.13 is traversed in the opposite direction. Thus, the signs of
w and q in the individual segments and the signs of the overall w and q are reversed.
Heat is now withdrawn from the cold reservoir (the refrigerator) and deposited in the
home, which is the hot reservoir (Figure 5.20a). Because this is not a spontaneous process, work must be done on the system to effect this direction of heat flow. The heat and
work flow for room heating is shown in Figure 5.20b.
A heat pump is used to heat a building by extracting heat from a colder thermal reservoir, such as a lake, the ground, or the ambient air. The coefficient of performance
Hot reservoir
151
Concept
Adoption of electrical vehicles, heat
pumps, and electricity generation
using wind and sunlight can greatly
reduce the rate of climate change.
Building
heating
34%
Water
heating
13%
Refrigerator
8%
Appliances
and lighting
34%
Electric
A/C
11%
Figure 5.19
U.S. household energy use.
Source: U.S. Energy Information Agency
Hot reservoir
Work
Work
Figure 5.20
Cold reservoir
Cold reservoir
(a)
(b)
Application of the principle of a reverse
Carnot heat engine to refrigeration and
household heating. The reverse Carnot
heat engine can be used to induce heat flow
from a cold reservoir to a hot reservoir
with the input of work. The hot reservoir
in both (a) and (b) is a room. The cold
reservoir can be configured as the inside
of a refrigerator or outside ambient air,
earth, or water for a heat pump.
152
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
of a heat pump, hhp, is defined as the ratio of the heat pumped into the hot reservoir to
the work input to the heat pump:
hhp =
qhot
qhot
Thot
=
=
w
qhot + qcold
Thot - Tcold
(5.68)
Assume that Thot = 294 K and Tcold = 278 K, typical for a mild winter day. The maximum hhp value is calculated to be 18. Such high values cannot be attained for real heat
pumps operating in an irreversible cycle with dissipative losses. Typical values for commercially available heat pumps lie in the range of 2 to 3. This means that for every joule
of work done on the system, 2 to 3 J of heat is made available for building heating.
Heat pumps become less effective as Tcold decreases, as shown by Equation (5.68).
Therefore, geothermal heat pumps that use ∼55°F soil 6 to 10 feet below the Earth’s
surface, rather than ambient air, as the cold reservoir are much more efficient in very
cold climates than air source heat pumps. Note that a coefficient of performance of
3 for a heat pump means that a house can be heated using 33% of the electrical power
consumption that would be required to heat the same house using electrical baseboard
heaters. This is a significant argument for using heat pumps for residential heating.
Heat pumps can also be used to heat water. In addition to improving the efficiency
of heating space and water, houses can also be designed to significantly reduce total
energy use. A combination of a well-insulated house with rooftop photovoltaic panels
can result in a house that generates as much energy as it consumes in a year in most
regions of the United States.
A refrigerator is also a heat engine operated in reverse (Figure 5.20a). The interior
of the refrigerator is the system, and the room in which the refrigerator is situated is
the hot reservoir. Because more heat is deposited in the room than is withdrawn from
the interior of the refrigerator, the overall effect of such a device is to increase the temperature of the room. However, the usefulness of the device is that it provides a cold
volume for food storage. The coefficient of performance, hr, of a reversible Carnot
refrigerator is defined as the ratio of the heat withdrawn from the cold reservoir to the
work supplied to the device:
hr =
qcold
qcold
Tcold
=
=
w
qhot + qcold
Thot - Tcold
(5.69)
This formula shows that as Tcold decreases from 0.9 Thot to 0.1 Thot, hr decreases from
9 to 0.1, showing that in order to provide a lower temperature, more work is required
to extract a given amount of heat.
A household refrigerator typically operates at 255 K in the freezing compartment
and 277 K in the refrigerator section. Using the lower of these temperatures for the cold
reservoir and 294 K as the temperature of the hot reservoir (the room), we find that the
maximum hr value is 6.5. This means that for every joule of work done on the system,
6.5 J of heat can be extracted from the contents of the refrigerator. This is the maximum
coefficient of performance and is only applicable to a refrigerator operating in a reversible Carnot cycle with no dissipative losses. When we take losses into account, we find
that it is difficult to achieve hr values greater than ∼1.5 in household refrigerators. This
shows the significant loss of efficiency in an irreversible dissipative cycle.
We have seen that a heat engine cannot convert heat into electricity with 100% efficiency because some of the heat is injected into the cold reservoir. The cold reservoir in
fossil-fuel-fired electrical generation plants is the atmosphere. Generally, heat and electricity are produced separately. However, it is possible to increase the efficiency of the
overall process by collecting the waste heat from electricity production and using it to
heat buildings, as shown in Figure 5.21, in a process called cogeneration. An example
of cogeneration is burning a fuel such as coal to generate steam used to drive a turbine
that generates electricity. The steam exiting the turbine is still at a high temperature and
can be used to heat buildings. In New York City, many buildings are heated using
cogeneration. Because of losses in piping heat over long distances, cogeneration is most
suitable for urban areas. The U.S. Department of Energy has set a goal of producing
20% of electricity by cogeneration by the year 2030.
5.13 Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines
Figure 5.21
Exhaust gases
Illustration of cogeneration. Conventionally, electricity and heat for buildings
are produced separately. Waste heat is
produced in each process. Through cogeneration, most of the waste heat generated
in electricity production can be used to
heat buildings.
Air
intake
Catalytic
converter
Hot water
to building
Water
pipes
Natural
gas fuel
Electricity
to building
Engine
Cold water
in from
building
153
Generator
In addition to improvements in heating and electrical generation, possible increases
in energy efficiency in the household sector include the areas of cooking and lighting.
Traditional electric cooktops use either natural gas or electricity to create heat, which
is transferred to the cooking pot and its contents by conduction and convection. About
45% of the energy produced by the combustion of natural gas and 35% of the electrical energy are lost because the air rather than the pot is heated. The most efficient
method for cooking is induction cooking in which an induction coil generates a magnetic field that induces heating in metal cookware placed on top of it. The efficiency
of this method is 90%. Traditional lighting technology is very inefficient, with incandescent lights converting only 2% of the electrical work required to heat the tungsten
filament to visible light. The remaining 98% of the radiated energy appears as heat.
Fluorescent lights, which are much more efficient, contain a small amount of Hg vapor
that emits UV light from an excited state created by an electrical discharge. The UV
light is absorbed by a fluorescent coating on the surface of the bulb, which emits visible
light. The fraction of the emitted light in the visible spectrum is much larger than that
in incandescent lighting. For this reason, fluorescent lighting is a factor of 5 to 6 more
efficient. Illumination with light emitting diodes (LED lighting) is a rapidly growing
sector of lighting. LED lights have an efficiency similar to fluorescent lighting, but they
are more compact and can be highly directed or diffuse sources of light.
Because the transportation sector is a major user of energy, we will next discuss
real engines. We will use the Otto engine and the diesel engine as examples. The Otto
engine is the most widely used engine in automobiles. The engine cycle consists of four
strokes, as shown in Figure 5.22. In the intake stage, the intake valve opens as the piston
is moving downward, drawing a fuel–air mixture into the cylinder (see yellow arrow).
At the compression stage, the intake valve is closed, and the mixture is compressed as
the piston moves upward. Just after the piston has reached its highest point, the fuel–air
mixture is ignited by a spark plug, and the rapid heating that results from the combustion process causes the gas to expand and the pressure to increase. The combustion
process drives the piston down in a power stroke. This is shown as the power stage
in Figure 5.22. Finally, the combustion products are forced out of the cylinder by the
upward-moving piston as the exhaust valve is opened (exhaust stage). To arrive at a
154
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Spark plug
Intake valve
Exhaust
valve
Cylinder
Piston
Connecting
rod
Crankshaft
Stroke 1: Intake
Intake valve opens as
piston moves downward,
drawing in fuel-air
mixture.
Stroke 2: Compression
Intake valve closes and
mixture is compressed
as piston moves
upward.
Stroke 3: Power
Spark plug ignites
compressed mixture
and combustion moves
piston downward.
Stroke 4: Exhaust
Exhaust valve opens as
piston moves upward,
forcing combustion
products out of cylinder.
Figure 5.22
Schematic illustration of the four-stroke cycle of an Otto engine. See text for detailed
description and discussion. The left valve (see yellow arrow) is the intake valve, and the right
valve (see red arrow) is the exhaust valve.
Otto cycle
Pressure
a
Adiabatic
qhot
b
d
Adiabatic
qcold
c
e
Volume
(a) Otto engine
qhot
a
b
Diesel cycle
maximum theoretical efficiency, we will analyze the reversible Otto cycle, shown in
Figure 5.23a. We conclude this section with a discussion of the diesel engine cycle,
shown in Figure 5.23b.
The reversible Otto cycle begins with the intake stroke e S c, which is assumed
to take place at constant pressure. At this point, the intake valve is closed, and the
piston compresses the fuel–air mixture along the adiabatic path c S d in the second
step. This path can be assumed to be adiabatic because the compression occurs too
rapidly to allow much heat to be transferred out of the cylinder. Ignition of the fuel–air
mixture takes place at d. The rapid increase in pressure occurs at constant volume. In
this reversible cycle, the combustion is modeled as a quasi-static heat transfer from a
series of reservoirs at temperatures ranging from Td to Ta . The power stroke is modeled
as the adiabatic expansion a S b. At this point, the exhaust valve opens and the gas is
expelled. This step is modeled as the constant volume pressure decrease b S c. The
upward movement of the piston expels the remainder of the gas along the line c S e,
after which the cycle begins again.
The efficiency of this reversible cyclic engine can be calculated as follows. Assuming
CV to be constant along the segments d S a and b S c, we write
qhot = CV 1Ta - Td2 and qcold = CV 1Tb - Tc2
(5.70)
The efficiency is given by
Pressure
Adiabatic
c
Adiabatic
0 qcold 0
qhot + qcold
Tb - Tc
= 1 = 1 - a
b
qhot
qhot
Ta - Td
(5.71)
The temperatures and volumes along the reversible adiabatic segments are related by
qcold
d
e
e =
Volume
(b) Diesel engine
Figure 5.23
Pressure–volume diagrams for idealized
reversible cycles of engines. (a) Otto
engine and (b) diesel engine.
TcV gc - 1 = TdV gd - 1 and TbV gc - 1 = TaV gd - 1
(5.72)
because Vb = Vc and Vd = Ve . Recall that g = CP >CV . Ta and Tb can be eliminated
from Equation (5.72) to give
e = 1 -
Tc
Td
(5.73)
where Tc and Td are the temperatures at the beginning and end of the compression stroke
c S d. Temperature Tc is fixed at ∼300 K, and the efficiency can be increased only
5.13 Energy Efficiency, Heat Pumps, Refrigerators, and Real Engines
by increasing Td. This is done by increasing the compression ratio, Vc >Vd . However,
if the compression ratio is too high, Td will be sufficiently high that the fuel ignites
before the end of the compression stroke. A reasonable upper limit for Td is 600 K,
and that for Tc = 300 K, e = 0.50. This value is an upper limit because a real engine
does not operate in a reversible cycle and because heat is lost along the c S d segment.
Achievable efficiencies in Otto engines used in passenger cars lie in the range of 0.20
to 0.30. However, additional losses occur in the drive chain, tire deformation, and displacement of air as the vehicle moves. As shown in Figure 5.18, the overall efficiency
of the transportation sector is only 21%.
In the diesel engine depicted in Figure 5.23b, higher compression ratios are possible because only air is let into the cylinder in the intake stroke. The fuel is injected
into the cylinder at the end of the compression stroke, thus avoiding spontaneous ignition of the fuel–air mixture during the compression stroke d S a. Because T ∼ 950 K
for the compressed air, combustion occurs spontaneously without a spark plug along
the constant pressure segment a S b after fuel injection. Because the fuel is injected
over a time period in which the piston is moving out of the cylinder, this step can be
modeled as a constant pressure heat intake. In this segment, it is assumed that heat is
absorbed from a series of reservoirs at temperatures between Ta and Tb in a quasi-static
process. In the other segments, the same processes occur as described for the reversible
Otto cycle.
The heat intake along the segment a S b is given by
qhot = Cp1Tb - Ta2
(5.74)
and qcold is given by Equation (5.70). Therefore, the efficiency is given by
e = 1 -
1 Tc - Td
a
b
g Tb - Ta
(5.75)
Similarly to the treatment of the Otto engine, Tb and Tc can be eliminated from Equation (5.75), and the expression
Vb g
Va g
b - a b
Vd
1 Vd
e = 1 g Vb
Va
a b - a b
Vd
Vd
a
(5.76)
can be derived. For typical values Vb >Vd = 0.2, Va >Vd = 1>15, g = 1.5, and e = 0.64.
The higher efficiency achievable with the diesel cycle in comparison with the Otto
cycle is a result of the higher temperature attained in the compression cycle. Real
diesel engines used in trucks and passenger cars have efficiencies in the range of
0.30 to 0.35.
Just as for building heating, energy usage in the U.S. transportation sector can be
drastically reduced. The U.S. personal vehicle fleet currently averages approximately
23 miles per gallon of gasoline. The corresponding figure for the European Union,
where gasoline prices are more than twice as high as in the United States, is 43 miles
per gallon. Hybrid gasoline–electric vehicles, which use electrical motors to power the
vehicle at low speeds, switch automatically to an Otto engine at higher speeds, and
use regenerative braking to capture the energy used to slow the vehicle, are currently
the most efficient vehicles, averaging more than 50 miles per gallon. Purely electric
vehicles (EVs) and plug-in hybrids, which allow the batteries that power the electrical
operation mode to be recharged at the owner’s residence, are now available, and their
energy use in gasoline equivalent is more than 100 miles per gallon. An additional
benefit of EVs is that if they are charged with electricity generated without burning
fossil or biofuels, their operation does not generate greenhouse gases, Advances in
battery technology are likely to make the use of electrical vehicles more widespread in
the near future. Automobiles that use fuel cells (see Section 11.12) as a power source
are of particular interest for future development.
155
156
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
VOCABULARY
entropy
heat engine
heat pump
Otto engine
perpetual motion machine of the first kind
perpetual motion machine of the second kind
Carnot cycle
Clausius inequality
coefficient of performance
cogeneration
configuration
diesel engine
refrigerator
second law of thermodynamics
spontaneous direction
third law of thermodynamics
weight of a configuration
KEY EQUATIONS
Equation
∆S =
Significance of Equation
dqrev
L T
∆S = nR ln
Vƒ
Vi
∆S = -nR ln
∆ vap S =
Definition of entropy
+ nCV,m ln
Pƒ
Tƒ
Ti
+ nCP,m ln
Pi
Tƒ
Ti
∆ vap H
qrev
dqrev
=
=
Tvap
Tvap
L T
Equation Number
5.4 and 5.8
Entropy change for ideal gas in process Vi, Ti S Vƒ, Tƒ
5.14
Entropy change for ideal gas in process Pi, Ti S Pƒ, Tƒ
5.15
Entropy change for vaporization and fusion
5.16 and 5.17
qrev
∆ fus H
dqrev
∆ fus S =
=
=
T
T
Tfus
L
fus
Tƒ
Vƒ
Ti
Vi
CV
a
∆S =
dT +
dV
k
T
L
L T
Entropy change for arbitrary process involving gases,
liquids, and solids
5.18
S = k ln W
Molecular-level definition of entropy
5.21
dq
T
Clausius inequality relates entropy change and dq
5.26
dqsur
or for a macroscopic change,
T
qsur
=
T
How to calculate entropy change in surroundings
dS 7
dSsur =
∆Ssur
Tƒ
Sm1T2 = Sm10 K2 +
Tb
+
L
L
C solid
P,m dT′
T′
0
liquid
C P,m dT′
T′
Tƒ
Sm1P2 = Sm~ - R ln
∆ r S ~ = a vi Si~
+
∆ vap H
Tb
+
∆Hfusion
Tƒ
T
+
5.27
L
Tb
gas
C P,m dT′
Calculating absolute entropy of gas from heat
capacities
5.28
Dependence of entropy of ideal gas on pressure
5.30
Calculating entropy change in chemical reaction
5.31
Temperature dependence of reaction entropies
5.33
T′
P1bar2
P~
i
T
∆ r S 1T2 = ∆ r S 1298.15 K2 +
~
~
∆CP
dT′
L T′
298.15
Conceptual Problems
e =
0 wcycle 0
qab
=
Thot - Tcold
Tcold
= 1 6 1
Thot
Thot
Tƒ
Vƒ
Ti
Vi
Tƒ
Pƒ
Ti
Pi
CV
a
∆S =
dT +
dV
L T
L kT
CP
∆S =
dT Va dP
L T
L
Efficiency of reversible Carnot heat engine
5.42
Entropy change for liquid, solid, or gas that undergoes
a transformation from Ti, Vi to Tƒ, Vƒ.
5.53
Entropy change for liquid, solid, or gas that undergoes
a transformation from Ti Pi to Tƒ, Pƒ .
5.67
157
CONCEPTUAL PROBLEMS
Q5.1 Under what conditions is ∆S 6 0 for a spontaneous
process?
Q5.2 Why are ∆ fus S and ∆ vap S always positive?
Q5.3 An ideal gas in thermal contact with the surroundings
is cooled in an irreversible process at constant pressure. Are
∆S, ∆Ssur, and ∆Stot positive, negative, or zero? Explain your
reasoning.
Q5.4 The amplitude of a pendulum consisting of a mass on
a long wire is initially adjusted to have a very small value.
The amplitude is found to decrease slowly with time. Is this
process reversible? Would the process be reversible if the
amplitude did not decrease with time?
Q5.5 A process involving an ideal gas is carried out in which
the temperature changes at constant volume. For a fixed value
of ∆T, the mass of the gas is doubled. The process is repeated
with the same initial mass, and ∆T is doubled. For which of
these processes is ∆S greater? Why?
Q5.6 You are told that ∆S = 0 for a process in which the
system is coupled to its surroundings. Can you conclude that
the process is reversible? Justify your answer.
Q5.7 Under what conditions does the equality ∆S = ∆H>T
hold?
Q5.8 Is the following statement true or false? If it is false,
rephrase it so that it is true. The entropy of a system cannot
increase in an adiabatic process.
Q5.9 Which of the following processes is spontaneous?
a. The reversible isothermal expansion of an ideal gas
b. The vaporization of superheated water at 102°C and 1 bar
c. The constant pressure melting of ice at its normal freezing
point by the addition of an infinitesimal quantity of heat
d. The adiabatic expansion of a gas into a vacuum
Q5.10 One joule of work is done on a system, raising its
temperature by one degree centigrade. Can this increase in
temperature be harnessed to do one joule of work? Explain.
Q5.11 Your roommate decides to cool the kitchen by
opening the refrigerator. Will this strategy work? Explain
your reasoning.
Q5.12 An ideal gas undergoes an adiabatic expansion into a
vacuum. Are ∆S, ∆Ssur, ∆Stot positive, negative, or zero?
Explain your reasoning.
Q5.13 When a saturated solution of a salt is cooled, a precipitate crystallizes out. Is the entropy of the crystalline precipitate
greater or less than the dissolved solute? Explain why this
process is spontaneous.
Q5.14 A system undergoes a change from one state to another
along two different pathways, one of which is reversible and
the other of which is irreversible. What can you say about the
relative magnitudes of qrev and qirr?
Q5.15 An ideal gas in a piston and cylinder assembly with
adiabatic walls undergoes an expansion against a constant
external pressure. Are ∆S, ∆Ssur, and ∆Stot positive, negative,
or zero? Explain your reasoning.
Q5.16 Is the equation
Tƒ
Vƒ
Ti
Vi
Tƒ
CV
a
a
∆S =
dT +
dV = CV ln
+
1V - Vi2
kT ƒ
Ti
L T
L kT
valid for an ideal gas?
Q5.17 Why is the efficiency of a Carnot heat engine the upper
bound to the efficiency of an internal combustion engine?
Q5.18 Two vessels of equal volume, pressure, and temperature both containing Ar are connected by a valve. Allowing
mixing of the two volumes, what is the change in entropy
when the valve is opened? Is ∆S the same if one of the volumes contained Ar and the other contained Ne?
Q5.19 Without using equations, explain why ∆S for a liquid
or solid is dominated by the temperature dependence of S as
both P and T change.
Q5.20 Solid methanol in thermal contact with the surroundings is reversibly melted at the normal melting point at a pressure of 1 atm. Are ∆S, ∆Ssur, and ∆Stot positive, negative, or
zero? Explain your reasoning.
Q5.21 Can incandescent lighting be regarded as an example
of cogeneration during the heating season? In a season where
air conditioning is required?
158
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
NUMERICAL PROBLEMS
Section 5.3
P5.1 One mole of H2O1l2 is compressed from a state
described by P = 1.00 bar and T = 298 K to a state
described by P = 550 bar and T = 490 K. In addition,
a = 2.07 * 10-4 K-1, and the density of liquid water can be
assumed to be constant at the value 998 kg m-3. Calculate ∆S
for this transformation, assuming that kT = 0.
P5.2 1.75 mol of an ideal gas with CV,m = 5R>2 is transformed from an initial state at T = 590 K and P = 2.20 bar
to a final state at T = 273 K and P = 5.50 bar. Calculate
∆U, ∆H, and ∆S for this process.
P5.3 2.50 mol of an ideal gas with CV,m = 3R>2 undergoes
a process in which the initial state is described by Ti = 420 K
and Pi = 1.00 bar and the final state is described by
Tƒ = 798 K and Pƒ = 80 bar. Calculate ∆S for (a) a reversible process and (b) for an irreversible process at a constant
external pressure of 80 bar.
P5.4 The diagram below shows three paths for 2.75 mol of a
monatomic ideal gas between the initial (0.500 bar, 250. K)
and final (6.50 bar, 650. K) states. Calculate ∆S for paths a
and c and describe how you could calculate ∆S for path b.
700
a
T (K)
600
500
b
400
c
300
1
2
3
4
P (bar)
5
6
P5.5 2.50 mol of N2 at 25.0°C and 6.20 bar undergoes a
transformation to the state described by 350°C and 2.00 bar.
­Calculate ∆S if
CP,m
J mol-1 K-1
= 30.81 - 11.87 * 10-3
- 1.0176 * 10-8
T
T2
+ 2.3968 * 10-5 2
K
K
T3
K3
P5.6 Calculate ∆S for the isothermal compression of 1.50 mol
of Cu(s) from 1.00 bar to 1200. bar at 298 K. a = 0.492 *
10-4 K-1, kT = 0.78 * 10-6 bar -1, and the density is
8.92 g cm-3. Repeat the calculation assuming that kT = 0.
CP
dT - VadP, calculate
T
the decrease in temperature that occurs if 2.25 mol of water at
298 K and 1325 bar is brought to a final pressure of 1.00 bar
in a reversible adiabatic process. Assume that kT = 0.
P5.7 Using the expression dS =
P5.8 2.50 mol of an ideal gas with CV,m = 3R>2 undergoes
the transformations described in the following list from an
initial state described by T = 310. K and P = 6.00 bar. Calculate q, w, ∆U, ∆H, and ∆S for each process.
a. The gas undergoes a reversible adiabatic expansion until
the final pressure is one-fourth its initial value.
b. The gas undergoes an adiabatic expansion against a constant external pressure of 1.50 bar until the final pressure
is one-fourth its initial value.
c. The gas undergoes an expansion against a constant external pressure of zero bar until the final pressure is equal to
one-fourth of its initial value.
P5.9 At the transition temperature of 95.4°C, the enthalpy of
transition from rhombic to monoclinic sulfur is 0.38 kJ mol-1.
a. Calculate the entropy of transition under these conditions.
b. At its melting point, 119°C, the enthalpy of fusion of
monoclinic sulfur is 1.23 kJ mol-1. Calculate the entropy
of fusion.
c. The values given in parts (a) and (b) are for 1 mol of sulfur;
however, in crystalline and liquid sulfur, the molecule is
present as S8. Convert the values of the enthalpy and entropy
of fusion in parts (a) and (b) to those appropriate for S8.
P5.10 One mole of a van der Waals gas at 25.0°C is expanded
isothermally and reversibly from an initial volume of 0.015 m3
to a final volume of 0.100 m3. For the van der Waals gas,
10U>0V2T = a>V 2m. Assume that a = 0.556 Pa m6 mol-2 and
that b = 64.0 * 10-6 m3 mol-1. Calculate q, w, ∆U, ∆H,
and ∆S for the process.
P5.11 Calculate ∆H and ∆S if the temperature of 2.50 mol
of Hg1l2 is increased from 0.00°C to 125.0°C at 1 bar. Over
this temperature range, CP,m = 30.093 - 4.944 *
T
10-3 J mol - 1 K-1.
K
P5.12 Calculate ∆S if the temperature of 2.15 mol of an
ideal gas with CV = 5R>2 is increased from 298 K to 680.
K under conditions of (a) constant pressure and (b) constant
volume.
P5.13 Between 0°C and 100°C, the heat capacity of Hg1l2
is given by
CP,m1Hg, l2
-1
JK
mol
-1
= 30.093 - 4.944 * 10-3
T
K
Calculate ∆H and ∆S if 2.15 mol of Hg1l2 are raised in temperature from 20.0°C to 80.0°C at constant P.
P5.14 The heat capacity of a-quartz is given by
CP,m1a@quartz, s2
-1
JK
mol
-1
= 46.94 + 34.31 * 10-3
- 11.30 * 105
T2
K2
T
K
Numerical Problems
The coefficient of thermal expansion is given by a =
0.3530 * 10-4 K-1 and Vm = 22.6 cm3 mol-1. Calculate
∆Sm for the transformation a-quartz (0.00°C, 1 atm) S
a-quartz (575.°C, 850. atm).
P5.15
a. Calculate ∆S if 2.15 mol of liquid water is heated from
0.00°C to 15.5°C under constant pressure if CP,m =
75.3 J K-1 mol-1.
b. The melting point of water at the pressure of interest is
0.00°C, and the enthalpy of fusion is 6.010 kJ mol-1. The
boiling point is 100.°C, and the enthalpy of vaporization
is 40.65 kJ mol-1. Calculate ∆S and ∆S per mol for the
transformation H2O1s, 0.00°C2 S H2O1g, 100.°C2.
P5.16 An ideal gas sample containing 2.25 mol for which
CV,m = 3R>2 undergoes the following reversible cyclical
process from an initial state characterized by T = 298 K and
P = 1.00 bar:
a. It is expanded reversibly and adiabatically until the volume
triples.
b. It is reversibly heated at constant volume until T increases
to 298 K.
c. The pressure is increased in an isothermal reversible compression until P = 1.00 bar.
Calculate q, w, ∆U, ∆H, and ∆S for each step in the cycle
and for the total cycle.
P5.17 The standard entropy of Pb(s) at 298.15 K is
64.80 J K-1 mol-1. Assume that the heat capacity of Pb(s) is
given by
CP,m1Pb, s2
= 22.13 + 0.01172
J mol-1 K-1
T
T2
+ 1.00 * 10-5 2
K
K
The melting point is 327.4°C, and the heat of fusion under
these conditions is 4770. J mol-1. Assume that the heat
capacity of Pb1l2 is given by
CP,m1Pb, l2
-1
JK
mol
-1
= 32.51 - 0.00301
T
K
a. Calculate the standard entropy of Pb1l2 at 850.°C.
b. Calculate ∆H for the transformation Pb1s, 25.0°C2 S
Pb1l, 850.°C2.
P5.18 1.75 mol of an ideal gas with CV = 13>22 R undergoes the transformations described in the following list from
an initial state described by T = 298 K and P = 1.00 bar.
Calculate q, w, ∆U, ∆H, and ∆S for each process.
a. The gas is heated to 595 K at a constant external pressure
of 1.00 bar.
b. The gas is heated to 595 K at a constant volume corresponding to the initial volume.
c. The gas undergoes a reversible isothermal expansion at
298 K until the pressure is one-third of its initial value.
159
Section 5.6
P5.19 Calculate ∆S, ∆Stot, and ∆Ssur when the volume of
180 g of CO initially at 273 K and 1.00 bar increases by a
factor of two in (a) an adiabatic reversible expansion, (b) an
adiabatic expansion against Pext = 0, and (c) an isothermal
reversible expansion. Take CP,m to be constant at the value
29.14 J mol-1 K-1 and assume ideal gas behavior. State
whether each process is spontaneous. The temperature of the
surroundings is 273 K.
P5.20 A copper block of mass 8.0 kg initially at 375. K is
immersed in 12.0 kg of water initially at 298 K. Calculate ∆S
for the water, the Cu block, and the overall entropy change.
Is the process spontaneous? Assume that the heat capacity of
Cu remains constant at its 298.15 K value over the temperature
interval of interest.
P5.21 A 34.0 g mass of ice at 273 K is added to 145 g of
H2O1l2 at 298 K at constant pressure. Is the final state of the
system ice or liquid water? Calculate ∆S for the process. Is
the process spontaneous?
P5.22 One mole of H2O1l2 is supercooled to -4.12°C at 1 bar
pressure. The equilibrium freezing temperature of water at this
pressure is 0.00°C. The transformation H2O1l2 S H2O1s2 is
suddenly observed to occur. By calculating ∆S,∆Ssur, and
∆Stot, verify that this transformation is spontaneous at
-4.12°C. The heat capacities are given by CP,m1H2O, l2 =
75.3 J K-1 mol-1 and CP,m1H2O, s2 = 37.7 J K-1 mol-1,
and ∆ fus H = 6.008 kJ mol-1 at 0.00°C. Assume that the surroundings are at -4.12°C. [Hint: Consider the two pathways
at 1 bar: (a) H2O1l, -4.12°C2 S H2O1s, -4.12°C2
and (b) H2O1l, -4.12°C2 S H2O1l, 0.00°C2 S
H2O1s, 0.00°C2 S H2O1s, -4.12°C2. Because S is a state
function, ∆S must be the same for both pathways.]
P5.23 Calculate ∆Ssur and ∆Stot for the processes described
in parts (a) and (b) of Problem P5.8. Which of the processes is
a spontaneous process? The state of the surroundings for each
part is 298 K, 1.50 bar.
P5.24 Calculate ∆Ssur and ∆Stot for part (c) of Problem P5.8.
Is the process spontaneous? The state of the surroundings is
T = 298 K, P = 0.250 bar.
P5.25 25.00 g of steam at 373 K is added to 390. g of H2O1l2
at 298 K at constant pressure of 1 bar. Is the final state of the
system steam or liquid water? Calculate ∆S for the process.
Assume that the final state is liquid water. If this is not the
case, the calculated temperature will be greater than 373 K.
P5.26 The average heat evolved by the oxidation of foodstuffs
in an average adult per hour per kilogram of body weight is
7.20 kJ kg -1 h-1. Assume the weight of an average adult
is 75.0 kg. Suppose the total heat evolved by this oxidation is
transferred into the surroundings during a period lasting one
week. Calculate the entropy change of the surroundings
associated with this heat transfer. Assume the surroundings
are at T = 310. K.
160
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
Section 5.7
P5.27 From the data shown in the accompanying table, derive the absolute entropy of crystalline glycine at T = 300. K.
You can perform the integration numerically using either a
spreadsheet program or a curve-fitting routine and a graphing
calculator (see Example Problem 5.7).
for ∆Sden given above, calculate the excess entropy of denaturation. In your calculations, use the dashed curve below the
yellow area as the heat capacity baseline, which defines dC trs
P
as shown in Figure 4.8. Assume the molecular weight of the
protein is 14000. g mol-1. You can perform the integration by
counting squares.
Heat Capacity
C P,m 1 J K−1 mol −1 2
Temperature (K)
~
0.3
2.4
7.0
13.0
25.1
35.2
43.2
50.0
56.0
61.6
67.0
72.2
77.4
82.8
88.4
94.0
99.7
CP(T)
10.
20.
30.
40.
60.
80.
100.
120.
140.
160.
180.
200.
220.
240.
260.
280.
300.
0.418 J K21g21
288
298
308
Temperature (K)
318
P5.28 The following heat capacity data have been reported
for L-alanine:
T 1K2
10.
-1
C P,m 1J K
~
20.
40.
60.
80.
100.
140.
180.
220.
260.
300.
mol 2 0.49 3.85 17.45 30.99 42.59 52.50 68.93 83.14 96.14 109.6 122.7
-1
By a graphical treatment, obtain the molar entropy of L-alanine
at T = 300. K. You can perform the integration numerically
using either a spreadsheet program or a curve-fitting routine
and a graphing calculator (see Example Problem 5.7).
P5.29 Calculate the entropy of one mole of water vapor at
175°C and 0.625 bar using the information in the data tables
in Appendix A.
P5.30 Using your result from Problem P5.28, extrapolate
the absolute entropy of L-alanine to physiological conditions,
T = 310. K. Assume the heat capacity is constant between
T = 300. K and T = 310. K.
P5.31 For protein denaturation, the excess entropy of denaT2
turation is defined as ∆ denS =
dC trs
P
dT, where dC trs
P , is
L T
T1
the transition excess heat capacity. The way in which dC trs
P
can be extracted from differential scanning calorimetry (DSC)
data is discussed in Section 4.6 and shown in Figure 4.8.
The DSC data below are for a protein mutant that denatures
between T1 = 288 K and T2 = 318 K. Using the equation
Section 5.9
P5.32 Consider the formation of glucose from carbon
dioxide and water, that is, the reaction of the photosynthetic
process: 6CO21g2 + 6H2O1l2 S C6H12O61s2 + 6O21g2.
The following table of information will be useful in working
this problem:
T = 298 K
∆ f H ~ >kJ mol-1
S >J mol
~
-1
CP,m >J mol
-1
K
-1
-1
K
CO2 1 g2 H2O 1 l2 C6H12O6 1 s 2 O2 1 g2
-393.5
-285.8
-1273.1
0.0
213.8
70.0
209.2
205.2
37.1
75.3
219.2
29.4
Calculate the entropy and enthalpy changes for this chemical
system at T = 298.15 K and T = 365 K. Also calculate the
entropy change of the surroundings and the universe at both
temperatures.
P5.33 Calculate ∆ r S ~ for the reaction 3>2 H21g2 + 1>2 N21g2
S NH31g2 at 685. K. Omit terms in the temperature-­
dependent heat capacities greater than T 2 >K2.
Numerical Problems
P5.34 Calculate ∆ r S ~ for the reaction H21g2 + Cl21g2 S
2 HCl1g2 at 675 K. Omit terms in the temperature-dependent
heat capacities greater than T 2 >K 2.
P5.35 Under anaerobic conditions, glucose is broken down
in muscle tissue to form lactic acid according to the reaction:
C6H12O61s2 S 2CH3CHOHCOOH1s2. Thermodynamic data
at T = 298.15 K for glucose and lactic acid are given below.
Glucose(s)
Lactic
Acid(s)
𝚫 f H ~ 1kJ mol−1 2 CP,m 1J K−1 mol−1 2 S ~ 1J K−1 mol−1 2
-1273.1
-673.6
219.2
127.6
209.2
192.1
Calculate ∆S for the system, the surroundings, and ∆Stot at
T = 310. K. Assume the heat capacities are constant between
T = 298 K and T = 310. K.
P5.36 The amino acid glycine dimerizes to form the dipeptide glycylglycine according to the the reaction
2 Glycine1s2 S Glycylglycine1s2 + H2O1l2
Calculate, ∆S, ∆Ssurr , and ∆Stotal at T = 298.15 K. Useful
thermodynamic data follow:
Glycine Glycylglycine Water
∆ f H ~ 1kJ mol-12
S ~ 1J K-1 mol-12
-537.2
-746.0
-285.8
103.5
190.0
70.0
Section 5.10
P5.37 The Chalk Point, Maryland, generating station supplies
electrical power to the Washington, D.C., area. Units 1 and 2
have a gross generating capacity of 710. MW (megawatt).
The steam pressure is 25 * 106 Pa, and the superheater outlet
temperature 1Th2 is 540.°C. The condensate temperature 1Tc2
is 30.0°C.
a. What is the efficiency of a reversible Carnot engine
operating under these conditions?
b. If the efficiency of the boiler is 91.2%, the overall efficiency
of the turbine, which includes the Carnot efficiency and
its mechanical efficiency, is 46.7%, and the efficiency of
the generator is 98.4%, what is the efficiency of the total
generating unit? (Another 5.0% needs to be subtracted for
other plant losses.)
c. One of the coal burning units produces 355 MW. How
many metric tons 11 metric ton = 1 * 106 g2 of coal
per hour are required to operate this unit at its peak output
if the enthalpy of combustion of coal is 29.0 * 103 kJ kg -1?
P5.38 Consider the reversible Carnot cycle shown in
Figure 5.13 with 1.25 mol of an ideal gas with CV = 5R>2
as the working substance. The initial isothermal expansion
occurs at the hot reservoir temperature of Thot = 825 K from
an initial volume of 2.25 L 1Va2 to a volume of 10.5 L 1Vb2.
The system then undergoes an adiabatic expansion until the
temperature falls to Tcold = 298 K. The system then undergoes an isothermal compression and a subsequent adiabatic
compression until the initial state described by Ta = 825 K
and Va = 2.25 L is reached.
161
a. Calculate Vc and Vd.
b. Calculate w for each step in the cycle and for the total
cycle.
c. Calculate e and the amount of heat that is extracted from
the hot reservoir to do 1.00 kJ of work in the surroundings.
P5.39 Using your results from Problem P5.38, calculate q,
∆U, and ∆H for each step in the cycle and for the total cycle
described in Figure 5.13.
P5.40 Beginning with Equation (5.38), use Equation (5.39)
to eliminate Vc and Vd to arrive at the result wcycle =
nR1Thot - Tcold2 ln Vb >Va.
P5.41 Using your results from Problems P5.38 and P5.39,
calculate ∆S, ∆Ssur, and ∆Stot for each step in the cycle and
for the total Carnot cycle described in Figure 5.13.
Section 5.13
P5.42 An electrical motor is used to operate a Carnot refrigerator with an interior temperature of 0.00°C. Liquid water
at 0.00°C is placed into the refrigerator and transformed to
ice at 0.00°C. If the room temperature is 295. K, what mass
of ice can be produced in one day by a 250 W motor that is
running continuously? Assume that the refrigerator is perfectly insulated and operates at the maximum theoretical
­efficiency.
P5.43 An air conditioner is a refrigerator with the inside of
the house acting as the cold reservoir and the outside atmosphere acting as the hot reservoir. Assume that an air conditioner consumes 1.25 * 103 W of electrical power and that
it can be idealized as a reversible Carnot refrigerator. If the
coefficient of performance of this device is 1.75, how much
heat can be extracted from the house in a day?
P5.44 The maximum theoretical efficiency of an internal
combustion engine is achieved in a reversible Carnot cycle.
Assume that the engine is operating in the Otto cycle and that
CV,m = 5R>2 for the fuel–air mixture initially at 298 K (the
temperature of the cold reservoir). The mixture is compressed
by a factor of 7.5 in the adiabatic compression step. What
is the maximum theoretical efficiency of this engine? How
much would the efficiency increase if the compression
ratio could be increased to 17? Do you see a problem in
doing so?
P5.45 Calculate ∆S, ∆Ssur, and ∆Stot per day for the
air-conditioned house described in Problem P5.43. Assume
that the interior temperature is 65.0°F and the exterior temperature is 99.0°F. The corresponding absolute temperature
values for 65.0 and 99.0 degrees are 291.48 K and 310.37 K,
respectively.
P5.46 A refrigerator is operated by a 0.25-hp 11 hp =
746 watts2 motor. If the interior is to be maintained at 2.75°C
and the room temperature is 34.5°C, what is the maximum heat
leak (in watts) that can be tolerated? Assume that the coefficient
of performance is 50.% of the maximum theoretical value.
What happens if the leak is greater than your calculated maximum value?
162
CHAPTER 5 Entropy and the Second and Third Laws of Thermodynamics
P5.47 The mean solar flux at the Earth’s surface is
∼2.00 J cm-2 min-1. In a nonfocusing solar collector, the
temperature can reach a value of 95.5°C. A heat engine is
operated using the collector as the hot reservoir and a cold
reservoir at 298 K. Calculate the area of the collector needed
to produce 1000. watts. Assume that the engine operates at
the maximum Carnot efficiency.
P5.48 The interior of a refrigerator is held at 34°F and the
interior of a freezer is held at -1.00°F. If the room temperature
is 68°F, by what factor is it more expensive to extract the same
amount of heat from the freezer than from the refrigerator?
Assume that the theoretical limit for the performance of a
reversible refrigerator is valid in this case.
FURTHER READING
Ben-Naim, A. Entropy and the Second Law. Singapore: World
Scientific Publishing Co., 2012.
Lambert, F. “Configurational Entropy Revisited.” Journal of
Chemical Education 84 (2007): 1548–1550.
Lambert, F. “Entropy Is Simple, Qualitatively.” Journal of
Chemical Education 79 (2002): 1241–1246.
Lambert, F. “Shuffled Cards, Messy Desks and Disorderly
Dorm Rooms—Examples of Entropy Increase? Nonsense!”
Journal of Chemical Education 76 (1999): 1385–1387.
Lee, S., Lee, K., and Lee, J. “An Alternative Presentation
of the Second Law of Thermodynamics.” Journal of
Chemical Education 92 (2015): 771–773.
Zemansky, M. W., and Dittman, R. H. Heat and
Thermodynamics (7th ed.). New York: McGraw-Hill,
1997, Chapter 7.
C H A P T E R
6
Chemical Equilibrium
6.1
Gibbs Energy and Helmholtz
Energy
6.2
Differential Forms of U, H, A,
and G
6.3
Dependence of Gibbs and
Helmholtz Energies on P, V, and T
6.4
Gibbs Energy of a Reaction
Mixture
6.5
Calculating the Gibbs Energy
of Mixing for Ideal Gases
6.6
Calculating the Equilibrium
Position for a Gas-Phase
Chemical Reaction
6.7
Introducing the Equilibrium
Constant for a Mixture of Ideal
Gases
6.8
Calculating the Equilibrium
Partial Pressures in a Mixture
of Ideal Gases
6.9
Variation of KP with Temperature
WHY is this material important?
In this chapter, spontaneity is discussed in the context of the approach to equilibrium
of a reactive mixture of gases. Two new state functions are introduced that express
spontaneity in terms of the properties of the system only, so that the surroundings do
not have to be considered explicitly.
WHAT are the most important concepts and results?
The Helmholtz energy provides a criterion for determining if a reaction mixture will
evolve toward reactants or products if the system is at constant V and T. The Gibbs
energy is the criterion for determining if a reaction mixture will evolve toward reactants
or products if the system is at constant P and T. Using the Gibbs energy, we derive a
thermodynamic equilibrium constant KP that predicts the equilibrium concentrations
of reactants and products in a mixture of reactive ideal gases.
WHAT would be helpful for you to review for this chapter?
6.10 Equilibria Involving Ideal Gases
It would be helpful to review the material in Chapter 5 because the current chapter is
based on the discussion of spontaneity in that chapter.
6.11 Expressing the Equilibrium
6.1 GIBBS ENERGY AND
Constant in Terms of Mole
Fraction or Molarity
6.12 Expressing U and H and Heat
Capacities Solely in Terms of
Measurable Quantities
HELMHOLTZ ENERGY
In Chapter 5, it was shown that the direction of spontaneous change for an arbitrary
process is predicted by ∆S + ∆Ssur 7 0. In this section, this spontaneity criterion is
used to derive two new state functions, the Gibbs and Helmholtz energies. These new
state functions provide the basis for all further discussions of spontaneity. Formulating
spontaneity in terms of the Gibbs and Helmholtz energies rather than ∆S + ∆Ssur 7 0
has the important advantage that spontaneity and equilibrium can be defined using only
properties of the system rather than those of both the system and surroundings. We will
also show that the Gibbs and Helmholtz energies allow us to calculate the maximum
work that can be extracted from a process such as a chemical reaction.
The fundamental expression governing spontaneity is the Clausius inequality
[Equation (5.26)], written in the form
TdS Ú dq
and Solid or Liquid Phases
6.13 (Supplemental Section)
A Case Study: The Synthesis
of Ammonia
6.14 (Supplemental Section)
Measuring ∆G for the Unfolding
of Single RNA Molecules
(6.1)
The equality is satisfied only for a reversible process. Because dq = dU - dw,
TdS Ú dU - dw or, equivalently, -dU + dw + TdS Ú 0
(6.2)
As discussed in Section 2.2, a system can do different types of work on the surroundings. Keep in mind that the work done by the system on the surroundings is equal in
magnitude and opposite in sign to the work done by the surroundings on the system.
163
164
CHAPTER 6 Chemical Equilibrium
It is particularly useful to distinguish between expansion work, in which the work arises
from a volume change in the system, and nonexpansion work (for example, electrical
work). As we will see in Example Problem 6.2, nonexpansion work is the more useful
of the two types of work. We rewrite Equation (6.2) in the form
-dU - Pext dV + dwnonexp + TdS Ú 0
(6.3)
Equation (6.3) expresses the condition of spontaneity for an arbitrary process in terms
of the changes in the state functions U, V, S, and T as well as the path-dependent functions Pext dV and dwnonexp.
To make a connection with the discussion of spontaneity in Chapter 5, consider a
special case of Equation (6.3). For an isolated system, w = 0 and dU = 0. Therefore,
Equation (6.3) reduces to the familiar result derived in Section 5.5:
dS Ú 0
(6.4)
Because chemists are interested in systems that interact with their environment, we
define equilibrium and spontaneity for such systems. As was done in Chapters 1
through 5, it is useful to consider transformations at constant temperature and either
constant volume or constant pressure. Keep in mind that constant T and P (or V) does
not imply that these variables are constant throughout the process, but rather that they
are the same for the initial and final states of the process.
We first consider isothermal processes. For an isothermal process, d1TS2 = TdS,
and Equation (6.3) can be written in the following form:
-dU + TdS Ú -dwexp - dwnonexp or, equivalently,
d1U - TS2 … dwexp + dwnonexp
(6.5)
The combination of state functions U - TS, which has the units of energy, defines a
new state function that we call the Helmholtz energy, abbreviated A. By using this
definition, the general condition of spontaneity for isothermal processes becomes
dA - dwexp - dwnonexp … 0
(6.6)
Because the equality applies for a reversible transformation, Equation (6.6) provides
a way to calculate the maximum work that a system can do on the surroundings in an
isothermal process by considering a reversible process, in which case the inequality in
Equation (6.6) becomes an equality.
dwmax = dwexp + dwnonexp = dA
(6.7)
Example Problem 6.1 illustrates the usefulness of A for calculating the maximum work
available through a chemical reaction.
EXAMPLE PROBLEM 6.1
Concept
The value of the change in Helmholtz
energy is the maximum work that can
be extracted from a process.
You have made a scientific breakthrough and can construct a fuel cell based on the
electrochemical oxidation of a hydrocarbon fuel. This fuel cell will allow you to
convert electrical work to other work such as driving the wheels of a car without the
second law losses that arise from combustion in a heat engine. Two possible choices
for a fuel are methane and octane. (a) Calculate the maximum work available through
the electrochemical oxidation of these two hydrocarbons on a per mole and a per gram
basis at 298.15 K and 1 bar pressure. (b) Another option to do work on the surroundings is to burn the hydrocarbon fuel in a heat engine, as discussed in Section 5.10.
Calculate the maximum work available in the combustion of these hydrocarbons in
a heat engine with an efficiency of 50.0%. The standard enthalpies of combustion
~
are ∆ com H ~ 1CH4, g2 = -891 kJ mol-1, ∆H com
1C8H18, l2 = -5471 kJ mol-1, and
-1 -1
~
Sm 1C8H18, l2 = 361.1 J mol K . Use tabulated values in Appendix A for other
data needed for these calculations. Assume ideal gas behavior. Are there any factors
not mentioned that should be taken into account in making a decision between these
two fuels?
6.1 Gibbs Energy and Helmholtz Energy
Solution
a. At constant T, ∆A and ∆H are related by ∆A = ∆U - T∆S = ∆H - ∆1PV2 T∆S = ∆H - ∆nRT - T∆S
The oxidation reactions are
Methane: CH41g2 + 2O21g2 S CO21g2 + 2H2O1l2
Octane: C8H181l2 + 25>2 O21g2 S 8CO21g2 + 9H2O1l2
∆ ox A~ 1CH4, g2 = ∆ oxU ~ 1CH4, g2 - T ¢
Sm~ 1CO2, g2 + 2Sm~ 1H2O, l2
≤
-Sm~ 1CH4, g2 - 2Sm~ 1O2, g2
= ∆ com H ~ 1CH4, g2 - ∆nRT
-T1Sm~ 1CO2, g2 + 2Sm~ 1H2O, l2 - Sm~ 1CH4, g2
-2Sm~ 1O2, g22
= -891 * 103 J mol-1 + 2 * 8.314 J mol-1 K-1 * 298.15 K
-298.15 K * a
∆ ox A~ 1CH4, g2 = -814 kJ mol-1
213.8 J mol-1 K-1 + 2 * 70.0 J mol-1 K-1
b
-186.3 J mol-1 K-1 - 2 * 205.2 J mol-1 K-1
∆ ox A~ 1C8H18, l2 = ∆ comU ~ 1C8H18, l2
-T a8Sm~ 1CO2, g2 + 9Sm~ 1H2O, l2 - Sm~ 1C8H18, l2 -
25 ~
S 1O , g2b
2 m 2
-T a8Sm~ 1CO2, g2 + 9Sm~ 1H2O, l2 - Sm~ 1C8H18, l2 -
25 ~
S 1O , g2b
2 m 2
= ∆ com H ~ 1C8H18, l2 - ∆nRT
= -5471 * 103 J mol-1 +
9
* 8.314 J mol-1 K-1 * 298.15 K
2
8 * 213.8 J mol-1 K-1 + 9 * 70.0 J mol-1 K-1
-298.15 K * °
¢
25
-361.1 J mol-1 K-1 * 205.2 J mol-1 K-1
2
∆ ox A~ 1C8H18, l2 = -5285 kJ mol-1
The maximum work for both reactions is negative, which is necessary to do work
by the fuel cell on the surroundings. On a per mol basis, octane (molecular weight
114.25 g mol-1) is capable of producing a factor of 6.5 more work than methane
(molecular weight 16.04 g mol-1). However, on a per gram basis, methane and
octane are nearly equal in their ability to produce work (-50.6 kJ g -1 versus
-46.3 kJ g -1). You might want to choose octane because it can be stored as a
liquid at atmospheric pressure. By contrast, a pressurized tank is needed to store
methane as a liquid at 298.15 K.
b. If these hydrocarbons are burned and used in a heat engine to generate work, the
maximum work is given by
wmax = qhot * efficiency = ∆ com H ~ * efficiency
= -5471 kJ mol-1 * 0.500 = -2736 kJ mol-1 for C8H18, l
= -891 kJ mol-1 * 0.500 = -446 kJ mol-1 for CH4, g
The efficiency enters in the calculation in part (b) because, as we showed in
Chapter 5, heat cannot be converted into work with 100% efficiency. This limitation does not apply to the fuel cell where electrical work is converted to another
type of work. Ignoring friction, which appears in both methods of generating work,
one type of work can in principle (but not practically) be converted to another type
of work with 100% efficiency. Comparing the answers to parts (a) and (b), we see
why optimizing the performance of fuel cells, which at present are not widely
used, is an active area of research.
165
166
CHAPTER 6 Chemical Equilibrium
Concept
∆A provides a criterion for spontaneity
in terms of system variables only for
a process at constant V and T.
Having demonstrated the usefulness of A in calculating the maximum amount of
work that can be generated in an arbitrary process, we return to the issue of spontaneity.
In discussing the Helmholtz energy, dT = 0 was the only constraint applied. We now
apply the additional constraint dV = 0, in which case we are considering a process at
constant T and V. Because V is constant, dwexp = 0. If nonexpansion work also is not
possible in the transformation, dwnonexp = dwexp = 0, and the condition that defines
spontaneity and equilibrium for a process at constant T and V becomes
dA … 0
(6.8)
Equation (6.8) states that a process for which T and V have the same initial and final
values is spontaneous if dA 6 0. The system is at equilibrium with its surroundings if
dA = 0, and neither direction of change is spontaneous.
Chemical reactions are more commonly studied under constant pressure than constant volume conditions. Therefore, the condition for spontaneity is considered next
for an isothermal constant pressure process. At constant P and T, d1PV2 = Pext dV and
d1TS2 = TdS. In this case, using the relation H = U + PV, Equation (6.3) can be
written in the form
d1U + PV - TS2 = d1H - TS2 … dwnonexp
(6.9)
The combination of state functions H - TS, which has the units of energy, defines a
new state function called the Gibbs energy, abbreviated G. The condition for spontaneity for an isothermal process at constant pressure becomes
dG - dwnonexp … 0
(6.10)
For a reversible process, the equality holds, and the change in the Gibbs energy is a measure of the maximum nonexpansion work that can be produced in the transformation.
Expansion work is included in the definition of Gibbs energy.
A particularly important application of Equation (6.10) is to calculate the electrical work produced by a reaction in an electrochemical cell or fuel cell, as shown in
Example Problem 6.2. This topic will be discussed in more detail in Chapter 11.
EXAMPLE PROBLEM 6.2
Concept
The value of the change in Gibbs
energy is the maximum nonexpansion
work that can be extracted from
a process.
(a) Calculate the maximum nonexpansion work that can be produced by the fuel cell
oxidation reactions in Example Problem 6.1. (b) Compare your results with those of
Example Problem 6.1. What can you conclude about the relative amounts of expansion and nonexpansion work available by carrying out these reactions?
Solution
a. ∆ oxG ~ 1CH4, g2 = ∆ com H ~ 1CH4, g2
-T ¢
Sm~ 1CO2, g2 + 2Sm~ 1H2O, l2
≤
-Sm~ 1CH4, g2 - 2Sm~ 1O2, g2
= -891 * 103 J mol-1
-298.15 K * ¢
213.8 J mol-1 K-1 + 2 * 70.0 J mol-1 K-1
≤
-186.3 J mol-1 K-1 - 2 * 205.2 J mol-1 K-1
∆ oxG ~ 1CH4, g2 = -819 kJ mol-1
∆ oxG ~ 1C8H18, l2 = ∆ com H ~ 1C8H18, l2
-T a8Sm~ 1CO2, g2 + 9Sm~ 1H2O, l2 - Sm~ 1C8H18, l2 = -5471 * 103 J mol-1
25 ~
S 1O , g2b
2 m 2
8 * 213.8 J mol-1 K-1 + 9 * 70.0 J mol-1 K-1
¢
25
-361.1 J mol-1 K-1 * 205.2 J mol-1 K-1
2
-1
~
∆ oxG 1C8H18, l2 = -5296 kJ mol
-298.15K * °
6.2 DIFFERENTIAL FORMS OF U, H, A, AND G
b. For methane, wexp + wnon = -814 kJ mol-1 and wnon = -819 kJ mol-1.
Therefore, wexp = 5 kJ mol-1. For octane, wexp + wnon = -5285 kJ mol-1 and
wnon = -5296 kJ mol-1. Therefore, wexp = 11 kJ mol-1. This comparison tells
us that the magnitude of expansion work is much smaller than that of nonexpansion work. The sign of the expansion work is positive (work done on the system
rather than by the system) because the number of moles of gaseous species
decreases in the reaction for both methane and octane. Note that we are calculating the expansion work at 298 K. In an automobile engine, the combustion
temperature is approximately 2000 K, so that expansion work is generated by
heating a fuel–air mixture using the enthalpy of combustion. The hot gas mixture
exerts a force on the pistons that is converted to a rotary motion of the wheels.
After having demonstrated the usefulness of G for determining the amount of nonexpansion work available from an arbitrary process, we consider the usefulness of G
for determining the spontaneous direction of a process. We consider a transformation at
constant P and T for which nonexpansion work is not possible. In this case, the condition for spontaneity becomes
dG … 0
(6.11)
In summary, Gibbs and Helmholtz energies are complementary state functions in
predicting spontaneity. If nonexpansion work does not occur, a process for which the
initial and final V and T values are the same is spontaneous if dA 6 0 and a process
for which the initial and final P and T values are the same is spontaneous if dG 6 0.
We emphasize that these criteria are useful because they are based on system properties; importantly, we need not be concerned about what happens in the surroundings.
However, you should keep in mind that the fundamental criterion for spontaneity is
still ∆S + ∆Ssur 7 0. By expressing ∆Ssur as -dU>T for a process at constant V and
T and as -dH>T for a process at constant P and T, we have been able to write ∆Ssur in
terms of changes in state functions of the system. We can see this clearly by dividing
∆G = ∆H - T∆S by T:
-
∆G
∆H
= + ∆S = ∆Ssurr + ∆Ssystem
T
T
(6.12)
6.2 DIFFERENTIAL FORMS OF U, H, A, AND G
To this point, we have defined four different state functions U, H, A, and G, all of
which have the units of energy. How do these functions differ from each other, and for
what purpose are they most useful? The total energy U is a measure of all of the kinetic
and potential energy in the system. The enthalpy H tells us how much heat flows out of
or into the system if a process is carried out at constant pressure. For example, ∆H for
the combustion of natural gas allows us to calculate the burn rate in a furnace used to
heat a home in winter. The Helmholtz energy A tells us the maximum total work that
the process can produce. It also provides a criterion for spontaneity for processes that
occur at constant T and V. The Gibbs energy G tells us the maximum nonexpansion
work that the process can produce. This is a critical piece of information in designing
batteries and fuel cells. The Gibbs energy also provides a criterion for spontaneity for
processes that occur at constant T and P.
In this section, we discuss how these energy state functions depend on the macroscopic system variables. To do so, the differential forms dU, dH, dA, and dG are developed. As we will see, these differential forms are essential in calculating how U, H, A,
and G vary with state variables such as P and T. Starting from the definitions
U
H
A
G
=
=
=
=
q + w
U + PV
U - TS
H - TS = U + PV - TS
(6.13)
Concept
∆G provides a criterion for
spontaneity in terms of system
variables only for a process at
constant P and T.
167
168
CHAPTER 6 Chemical Equilibrium
the following total differentials can be formed:
dU
dH
dA
dG
=
=
=
=
TdS
TdS
TdS
TdS
+
PdV
PdV + PdV + VdP = TdS + VdP
PdV - TdS - SdT = -SdT - PdV
VdP - TdS - SdT = -SdT + VdP
(6.14)
(6.15)
(6.16)
(6.17)
These differential forms express the internal energy as U(S,V ), the enthalpy as H(S,P),
the Helmholtz energy as A(T,V ), and the Gibbs energy as G(T,P). Although other
combinations of variables can be used, these natural variables are used because the
differential expressions are compact.
What information can be obtained from the differential expressions in Equations
(6.14) through (6.17)? Because U, H, A, and G are state functions, two different equivalent expressions, such as those written for dU here, can be formulated:
dU = TdS - PdV = a
0U
0U
b dS + a b dV
0S V
0V S
(6.18)
For Equation (6.17) to be valid, the coefficients of dS and dV on both sides of the
equation must be equal. Applying this reasoning to Equations (6.14) through (6.17), we
obtain the following expressions:
a
Concept
a
Equations (6.19)–(6.22) give
information on how U, H, A, and G
change with their natural variables.
0U
0U
b = T and a b = -P
0S V
0V S
0H
0H
b = T and a b = V
0S P
0P S
a
a
0A
0A
b = -S and a b = -P
0T V
0V T
0G
0G
b = -S and a b = V
0T P
0P T
(6.19)
(6.20)
(6.21)
(6.22)
These expressions state how U, H, A, and G vary with their natural variables. For example, because T and V always have positive values, Equation (6.20) states that H increases if either the entropy or the pressure of the system increases. We discuss how
to use these relations for macroscopic changes in the system variables in Section 6.3.
The differential expressions in Equations (6.14) through (6.17) can be used in a
second way. From Section 3.1, we know that because dU is an exact differential:
¢
0 0U 1S,V2
0 0U 1S,V2
a
b ≤ = ¢ a
b ≤
0V
0S
0S
0V
V S
S V
Equating the mixed second partial derivative derived from Equations (6.14) through
(6.17), we obtain the following four Maxwell relations:
Concept
The Maxwell relations connect
seemingly unrelated quantities.
a
0T
0P
b = -a b
0V S
0S V
a
0S
0P
a
b = a b =
kT
0V T
0T V
a
-a
0T
0V
b = a b
0P S
0S P
0S
0V
b = a b = Va
0P T
0T P
(6.23)
(6.24)
(6.25)
(6.26)
A derivation for one of the Maxwell relations is shown in Example Problem 6.3.
Equations (6.23) and (6.24) refer to a partial derivative at constant S. What conditions
must a transformation at constant entropy satisfy? Because dS = dqrev >T, a transformation at constant entropy refers to a reversible adiabatic process.
6.3 DEPENDENCE OF GIBBS AND HELMHOLTZ ENERGIES ON P, V, AND T
EXAMPLE PROBLEM 6.3
Derive the Maxwell relation a
0T
0P
b = -a b
0V S
0S V
Solution
Because the variables S and V appear in the equation, we look for a differential of U,
H, A, or G that includes these variables. Equation (6.18) shows that this is the case for
U. We next form the mixed partial derivatives of U as follows:
a
0 0U 1S,V2
0 0U 1S,V2
a
b b = ¢ a
b ≤ .
0V
0S
0S
0V
V S
S V
Substituting dU = TdS - PdV in the previous expression,
¢
a
0 03 TdS - PdV4
0 03 TdS - PdV4
a
b ≤ = ¢ a
b ≤
0V
0S
0S
0V
V S
S V
0T
0P
b = -a b
0V S
0S V
The Maxwell relations have been derived using only the property that U, H, A,
and G are state functions. These four relations are extremely useful in transforming
seemingly obscure partial derivatives to other partial derivatives that can be directly
measured. For example, these relations will be used to express U, H, and heat capacities solely in terms of measurable quantities such as kT, a, and the state variables P, V,
and T in Section 6.12.
6.3 DEPENDENCE OF GIBBS AND HELMHOLTZ
ENERGIES ON P, V, AND T
The state functions A and G are particularly important for chemists because of their
roles in determining the direction of spontaneous change in a reaction mixture. For this
reason, we need to know how A changes with T and V, and how G changes with T and
P to be able to predict if the direction of spontaneous change will change with changes
in P, T, and V.
We begin by asking how A changes with T and V. From Section 6.2,
a
0A
0A
b = -S and a b = -P
0T V
0V T
(6.21)
where S and P always take on positive values. Therefore, the general statement can be
made that the Helmholtz energy of a pure substance decreases as either the temperature
or the volume increases.
Because most reactions of interest to chemists are carried out under constant
pressure rather than constant volume conditions, we will devote more attention to the
properties of G than to those of A. From Section 6.2,
a
0G
0G
b = -S and a b = V
0T P
0P T
(6.22)
Thus, Gibbs energy decreases as temperature increases, but it increases as pressure
increases.
How can Equation (6.22) be used to calculate changes in G with macroscopic
changes in the variables T and P? We will consider each of these variables separately.
Because G is a state function, the total change in G as both T and P are varied is the
sum of the separate contributions. We first discuss the change in G with P.
Concept
G decreases as T increases and
increases as P increases.
169
170
CHAPTER 6 Chemical Equilibrium
For a macroscopic change in P at constant T, the second expression in Equation (6.22)
is integrated at constant T, yielding
P
L~
P
P
dG = G1T, P2 - G 1T, P 2 =
~
~
L~
(6.27)
Vd P′
P
where we have chosen the initial pressure to be the standard state pressure P ~ = 1 bar.
This equation takes on different forms for liquids and solids, on the one hand, and for
gases, on the other hand. For liquids and solids, the volume is, to a good approximation, independent of P over a limited range in P and
P
G1T, P2 = G 1T, P 2 +
~
~
L~
P
Vd P′ ≈ G ~ 1T, P ~ 2 + V1P - P ~ 2
(6.28)
By contrast, the volume of a gaseous system changes appreciably with pressure. In calculating the change of G with P at constant T, any path connecting the same initial and
final states gives the same result. Choosing the reversible path and assuming ideal gas
behavior, we find that
P
G1T, P2 = G ~ 1T2 +
L~
P
P
VdP′ = G ~ 1T2 +
nRT
P
dP′ = G ~ 1T2 + nRT ln ~
P′
P
L~
P
(6.29)
The functional dependence of the molar Gibbs energy Gm on P for an ideal gas is
shown in Figure 6.1, where Gm approaches minus infinity as the pressure approaches
zero. This is a result of the volume dependence of S that was discussed in Section 5.5,
∆S = nR ln1Vf > Vi2 at constant T. As P S 0, V S ∞ and S S ∞ . Therefore,
G = H - TS S - ∞ in this limit.
We next investigate the dependence of G on T. As we will see in the next section, the
thermodynamic equilibrium constant K is related to G>T. Therefore, it is more useful
to obtain an expression for the temperature dependence of G>T than for the temperature
dependence of G. Using the chain rule, we see that this dependence is given by
a
03 G>T 4
0T
b
P
=
=
d3 1>T 4
1 0G
a b + G
T 0T P
dT
1 0G
G
S
G
G + TS
H
a b - 2 = - - 2 = = - 2
2
T 0T P T
T T
T
T
(6.30)
Figure 6.1
Molar Gibbs energy of an ideal gas
relative to its standard state value.
The value of G1T, P2 - G ~ 1T2 is shown
as a function of pressure at 298.15 K.
G1T, P2 - G ~ 1T2 approaches minus
infinity as P approaches 0.
[Gm(298.15 K,P) 2 G
[
m(298.15
K)] (J)
In the second line of Equation (6.30), we have used Equation (6.22), 10G>0T2P = -S,
and the definition G = H - TS. Equation (6.30) is known as the Gibbs–Helmholtz
equation. Because
d11>T2
1
= - 2
dT
T
4000
2000
0
1
22000
24000
2
P/P
[
3
4
5
6.4 Gibbs Energy of a Reaction Mixture
171
the Gibbs–Helmholtz equation can also be written in the form
a
03 G>T 4
03 1>T 4
b
P
= a
03 G>T 4
0T
b a
P
dT
H
b = - 2 1-T 22 = H
d3 1>T 4
T
(6.31)
The preceding equation also applies to the change in G and H associated with a process
such as a chemical reaction. Replacing G by ∆G and integrating Equation (6.31) at
constant P,
T2
LT1
T2
da
∆G
1
b =
∆Hd a b
T
T
LT1
∆G1T22
∆G1T12
1
1
=
+ ∆H1T12 a - b
T2
T1
T2
T1
(6.32)
(6.33)
It has been assumed in the second equation that ∆H is independent of T over the
temperature interval of interest. If this is not the case, the integral must be evaluated
numerically, using tabulated values of ∆ f H ~ and temperature-dependent expressions
of CP,m for reactants and products.
EXAMPLE PROBLEM 6.4
The value of ∆ f G ~ for Fe1g2 is 370.7 kJ mol-1 at 298.15 K, and the value of ∆ f H ~
for Fe1g2 is 416.3 kJ mol-1 at the same temperature. Assuming that ∆ f H ~ is constant
in the interval 250–400 K, calculate ∆ f G ~ for Fe1g2 at 400. K.
Solution
∆ f G ~ 1T22 = T2 c
∆ f G ~ 1T12
1
1
+ ∆ f H ~ 1T12 * a - b d
T1
T2
T1
370.7 * 103 J mol-1
+ 416.3 * 103 J mol-1
298.15 K
= 400. K * ≥
¥
1
1
*a
b
400. K
298.15 K
∆ f G ~ 1400. K2 = 355.1 kJ mol-1
Analogies to Equations (6.27) and (6.32) for the dependence of A on V and T are left to
the end-of-chapter problems.
6.4 GIBBS ENERGY OF A REACTION MIXTURE
To this point, the discussion has been limited to systems at a fixed composition. The rest
of the chapter focuses on using the Gibbs energy to understand equilibrium in a reaction mixture under constant pressure conditions, which correspond to typical laboratory
experiments. Because reactants are consumed and products are generated in chemical
reactions, the expressions derived for state functions such as U, H, S, A, and G must be
revised to include changes in composition. We focus on G in the following discussion.
For a reaction mixture containing species 1, 2, 3, c, G is not only a function of
the variables T and P. Because it depends on the number of moles of each species, G is
written in the form G = G1T, P, n1, n2, n3, c2. The total differential dG is
dG = a
0G
0G
0G
b
dT + a b
dP + a
b
dn
0T P,n1,n2 c
0P T,n1,n2 c
0n1 T,P,n2 c 1
+a
0G
0G
b
dn2 + a
b
dn + c
0n2 T,P,n1 c
0n3 P,n1,n2 c 3
(6.34)
Concept
The Gibbs–Helmholtz equation is
used to calculate the variation of ∆G
with temperature.
172
CHAPTER 6 Chemical Equilibrium
where the variables dni have the units of moles. Note that if the concentrations do not
change, all of the dni = 0, and Equation (6.34) reduces to
dG = a
0G
0G
b dT + a b dP.
0T P
0P T
Equation (6.34) can be simplified by defining the chemical potential, mi, as
mi = a
0G
b
0ni P,T,nj ≠ ni
(6.35)
It is important to realize that although mi is defined mathematically in terms of an infinitesimal change in G as the amount dni of species i changes by an infinitesimal amount,
mi is the change in the Gibbs energy of the mixture per mole of substance i added
at constant concentration. These two requirements are not contradictory. To keep the
concentration constant, we add one mole of substance i to a huge vat containing many
moles of the various species. In this case, 10G>0ni2P,T,nj ≠ ni, where dni S 0, or the ratio
1∆G> ∆ni2P,T,nj ≠ ni, where ∆ni is 1 mol have the same value. Using the notation of
Equation (6.35), we can write Equation (6.34) as follows:
dG = a
Concept
0G
0G
b
dT + a b
dP + a mi dni
0T P,n1,n2 c
0P T,n1,n2 c
i
(6.36)
To ensure that you are familiar with the units for all quantities in Equation (6.36), we
recommend that you carry out a unit analysis of this equation. Now imagine integrating
Equation (6.36) at constant composition and at constant T and P from an infinitesimal
size of the system where ni S 0 and, therefore, G S 0 to a macroscopic size where
the Gibbs energy has the value G. Because T and P are constant, the first two terms in
Equation (6.36) do not contribute to the integral. Furthermore, because the composition
is constant, mi is constant, and
G
L
The chemical potential of a chemical
species is an intensive quantity
numerically equal to the Gibbs
energy of the pure substance.
0
dG′ = a mi
i
ni
L
dn′i
0
G = a nimi
(6.37)
i
Note that mi depends on the number of moles of each species present. If the system
consists of a single pure substance A, G = nAGm,A because G is an extensive quantity.1 Applying Equation (6.35),
mA = a
03 nAGm,A 4
0G
b
= a
b
= Gm,A
0nA P,T
0nA
P,T
(6.38)
showing that mA is an intensive quantity equal to the molar Gibbs energy of A, Gm,A,
for a pure substance. As we will show in the next section, mA in a mixture depends on
the concentration of species A.
Why is mi called the chemical potential of species i? This can be understood by assuming that the chemical potential for species i has the values mIi in region I, and mIIi
in region II of a given mixture with mIi 7 mIIi. If dni moles of species i are transported
from region I to region II, at constant T and P, the change in G is given by
dG = -mIi dni + mIIi dni = 1mIIi - mIi2dni 6 0
(6.39)
Because dG 6 0, this process is spontaneous. For a given species, transport will occur
spontaneously from a region of high chemical potential to one of low chemical potential.
The flow of material will continue until the chemical potential has the same value in
1
As discussed in Chapter 2, we can only measure changes in energy rather than absolute values of energy.
However, in some cases it is useful to define a reference zero and report absolute values of energy functions
such as H and G. As for formation enthalpies, we set H m~ = Gm~ = 0 for all elements in their standard reference state at 298.15 K. With this convention, Gm~ = ∆ f G ~ for all compounds in their standard reference state.
6.5 Calculating the Gibbs Energy of Mixing for Ideal Gases
all regions of the mixture. Note the analogy between this process and the flow of mass
in a gravitational potential field or the flow of charge in an electrostatic potential field.
Therefore, the term chemical potential is appropriate. In this discussion, we have
defined a new criterion for equilibrium in a multicomponent mixture: At equilibrium,
the chemical potential of each species is the same throughout a mixture.
173
Xe
6.5 CALCULATING THE GIBBS ENERGY
OF MIXING FOR IDEAL GASES
Ar
In this section, we describe how the chemical potential of a reactant or product
species changes through mixing with other species. Using these results, we show in
Section 6.7 that the partial pressures of all constituents of a gaseous reaction mixture
are related by the thermodynamic equilibrium constant, KP. Recall from Equation (6.29)
that the molar Gibbs energy of a pure ideal gas depends on its pressure as G1T, P2 =
G ~ 1T2 + nRT ln1P>P ~ 2. Therefore, the chemical potential of component i in a mixture of ideal gases can be written in the form
mi1T, P2 = mi~ 1T2 + RT ln
Pi
P~
(6.40)
where mi~ 1T2 is the chemical potential of the pure gas at a pressure of 1 bar and temperature T and Pi is the partial pressure of component i. We see that the chemical potential
of a gas in a mixture depends logarithmically on its partial pressure.
We next obtain a quantitative relationship between ∆ mixG and the mole fractions
of the individual constituents of the mixture. Consider the system shown in Figure 6.2.
The four compartments contain the gases He, Ne, Ar, and Xe at the same temperature
and pressure. The volumes of the four compartments differ. To calculate ∆ mixG, we
must compare G for the initial state shown in Figure 6.2 and the final state in which all
gases are uniformly distributed in the container. For the initial state, in which we have
four pure separated substances,
Gi = GHe + GNe + GAr + GXe = nHemHe + nNemNe + nArmAr + nXemXe
~
= nHe amHe
+ RT ln
P
P
P
~
~
b + nNe amNe
+ RT ln ~ b + nAr amAr
+ RT ln ~ b
~
P
P
P
~
+ nXe amXe
+ RT ln
P
b P~
(6.41)
For the final state in which all four components are dispersed in the mixture,
~
Gƒ = nHe amHe
+ RT ln
PNe
PHe
PAr
~
~
b + nNe amNe
+ RT ln ~ b + nAr amAr
+ RT ln ~ b
P~
P
P
~
+ nXe amXe
+ RT ln
PXe
b
P~
(6.42)
The Gibbs energy of mixing is Gƒ - Gi or
∆ mixG = Gƒ - Gi = nHe RT ln
PNe
PHe
PXe
PAr
+ nNe RT ln
+ nAr RT ln
+ nXe RT ln
P
P
P
P
(6.43)
Because the individual gas molecules in the mixture do not interact, the final pressure
and temperature of the mixture remain at their initial values. For each component i of
the mixture, we can write Pi = xi P, where xi is the mole fraction of component i and P
is the total pressure. Similarly, ni = xi n where n is the total number of moles in the gas
mixture. Using this notation, ∆ mixG becomes
∆ mixG = nRTxHe ln xHe + nRTxNe ln xNe + nRTxAr ln xAr + nRTxXe ln xXe
= nRT a xi ln xi
i
Ne
(6.44)
He
Figure 6.2
An isolated system consists of four separate subsystems that contain He, Ne,
Ar, and Xe, each at the same pressure.
The barriers separating these subsystems
can be removed, leading to mixing.
174
CHAPTER 6 Chemical Equilibrium
Note that because all the xi 6 1, ln xi 6 0, and each term in Equation (6.44) is negative. Therefore, ∆ mixG 6 0, showing that mixing is a spontaneous process.
Equation (6.44) allows us to calculate ∆ mixG for any given set of the mole fractions
xi. It is easiest to graphically visualize the results for a binary mixture of species A and
B. For this case, xB = 1 - xA and
0
D mixG (J mol21)
2250
2500
2750
∆ mixG = nRT 3 xA ln xA + 11 - xA2ln11 - xA24
A plot of ∆ mixG versus xA is shown in Figure 6.3 for a binary mixture. Note that ∆ mixG
is zero for xA = 0 and xA = 1 because only pure substances are present in these limits.
Also, ∆ mixG has a minimum for xA = 0.5, when A and B are present in equal amounts.
The entropy of mixing can be calculated from Equation (6.44):
21000
21250
21500
21750
0.2 0.4 0.6 0.8
Mole fraction xA
1
∆ mixS = - a
0∆ mixG
b = -nR a xi ln xi
0T
P
i
(6.46)
∆ mixS is positive because all of the ln xi 6 0. For a two-component mixture,
Figure 6.3
Gibbs energy of mixing of ideal gases
A and B as a function of xA, with
nA + nB = 1 mol and T = 298.15 K.
5
D mixS (J K21 mol21)
(6.45)
∆ mixS = -nR1xA ln xA + xB ln xB2
(6.47)
if both components and the mixture are at the same pressure.
As shown in Figure 6.4, the entropy of mixing is greatest for xA = 0.5. What is
the cause of the increase in entropy? Each component of the mixture expands from its
initial volume to the same larger final volume. Therefore, ∆ mixS arises from the dependence of S on V at constant T, as shown in Equation (6.24). We verify this assertion in
Example Problem 6.5.
EXAMPLE PROBLEM 6.5
4
Two gases at the same pressure and separated by a barrier have volumes ViA and ViB.
The barrier is removed, allowing each of the gases to occupy the volume Vƒ = ViA + ViB
in an isothermal process. (a) Calculate ∆S for this process. Compare your result with
Equation (6.47). (b) Use calculus to calculate which value of xA maximizes ∆S.
3
2
Solution
a. From Equation (5.14), ∆S for an isothermal expansion is ∆S = nR ln Vƒ >Vi .
Therefore, ∆S for this process is
1
0.2
0.4
0.6
0.8
Mole fraction xA
Figure 6.4
Entropy of mixing of ideal gases A
and B as a function of mole fraction
of component A at 298.15 K, with
nA + nB = 1.
1
∆S = R anA ln
Vƒ
ViA
+ nB ln
Vƒ
ViB
b
Using the relations nA = n xA and xA = ViA >Vƒ , we obtain
∆S = R anxA ln
1
1
+ nxB ln b = -nR1xA ln xA + xB ln xB2
xA
xB
This is the same result that was obtained in Equation (6.47).
b. We differentiate ∆S = -nR1xA ln xA + 11 - xA2ln11 - xA22 with respect to xA,
and set the result equal to, zero.
d5 -nR3 xA ln xA + 11 - xA2ln11 - xA24 6 >dxA =
-nR3 ln xA + 1 - ln11 - xA2 - 14 = 0
xA
xA
ln
= 0 therefore
= 1 and xA = 0.5
1 - xA
1 - xA
EXAMPLE PROBLEM 6.6
Consider the system shown in Figure 6.2. Assume that the separate compartments
contain 1.0 mol of He, 3.0 mol of Ne, 2.0 mol of Ar, and 2.5 mol of Xe at 298.15 K.
The pressure in each compartment is 1 bar.
a. Calculate ∆ mixG.
b. Calculate ∆ mixS.
6.6 Calculating the Equilibrium Position for a Gas-Phase Chemical Reaction
175
Solution
a. ∆ mixG = nRT a xi ln xi
i
= 8.5 mol * 8.314 J K-1 mol-1 * 298.15 K
*a
1.0 1.0
3.0 3.0
2.0 2.0
2.5 2.5
ln
+
ln
+
ln
+
ln
b
8.5 8.5
8.5 8.5
8.5 8.5
8.5 8.5
= -2.8 * 104 J
b. ∆ mixS = -nR a xi ln xi
i
= -8.5 mol * 8.314 J K-1 mol-1
* a
1.0 1.0
3.0 3.0
2.0 2.0
2.5 2.5
ln
+
ln
+
ln
+
ln
b
8.5 8.5
8.5 8.5
8.5 8.5
8.5 8.5
= 93 J K-1
Example Problem 6.6 illustrates how to calculate ∆ mixG and ∆ mixS. What is the
driving force for the mixing of gases? We saw in Section 6.2 that there are two contributions to ∆G: an enthalpic contribution ∆H and an entropic contribution T∆S.
By calculating ∆ mixH from ∆ mixH = ∆ mixG + T∆ mixS using Equations (6.44) and
(6.46), you will see that for the mixing of ideal gases, ∆ mixH = 0. Because the molecules in an ideal gas do not interact, there is no enthalpy change associated with
mixing. We conclude that the mixing of ideal gases is driven entirely by ∆ mixS as
shown above.
Although the mixing of gases is always spontaneous, the same is not true of liquids.
A very strong attractive potential is essential for a gas to condense to a liquid phase.
Liquids can be either miscible or immiscible. How can this fact be explained? If the
A-B attractions of the molecules of liquids A and B are greater than A-A or B-B attractions, mixing is exothermic and ∆ mixH 6 0. For gases or liquids, ∆ mixS 7 0, and
therefore, ∆ mixG = ∆ mixH - T∆ mixS 6 0. However, if the A-B attractions are less
than A-A or B-B attractions, mixing is endothermic and ∆ mixH 7 0. If ∆ mixH 7 0
and ∆ mixH 7 T∆ mixS, ∆ mixG 7 0 and the mixing of the two liquids is not spontaneous. However, mixing is still spontaneous if ∆ mixH 7 0 and ∆ mixH 6 T∆ mixS. In this
case, the attractive interaction between A and B molecules is slightly less than the A-A
and B-B interactions. The increase in the entropic component of the Gibbs energy outweighs the increase in the enthalpic component and mixing is spontaneous.
6.6 CALCULATING THE EQUILIBRIUM POSITION
2
FOR A GAS-PHASE CHEMICAL REACTION
Given a set initial number of moles of all pure separated gaseous components in a
reaction mixture, we will see that there is only one set of partial pressures, Pi, that
corresponds to equilibrium when the gases are combined. In this section, we combine
the results of the two previous sections to determine the equilibrium position in a
reaction mixture.
Consider the balanced chemical reaction
aA + bB + xC + c N dM + eN + gO + c
2
(6.48)
See the articles by L. M. Raff and by E. A. Gislason and N. C. Craig listed in Further Reading for a more
detailed discussion of conditions for spontaneity and equilibrium.
Concept
Mixing of ideal gases is driven by the
resulting increase in entropy.
176
CHAPTER 6 Chemical Equilibrium
in which Greek letters represent stoichiometric coefficients and the uppercase Roman
letters represent reactants and products. We can write an abbreviated expression for
this reaction in the form
a viXi = 0
i
(6.49)
where a, b, c, g are represented by the dimensionless stoichiometric coefficients vi
and A, B, c, O are represented by Xi. In Equation (6.49), the stoichiometric coefficients of the products are positive and those of the reactants are negative.
What determines the equilibrium partial pressures of the reactants and products?
Imagine that the reaction proceeds in the direction indicated in Equation (6.48) by an
infinitesimal amount. We describe the change in the Gibbs energy dr G for this process.
The subscript r refers to the process being a chemical reaction. Starting at the initial
value of the concentrations of all species, the reaction proceeds spontaneously to the
right if and only if the Gibbs energy decreases as the reaction proceeds. If the reaction
is spontaneous in this direction at the initial concentrations, it continues to proceed to
the right, with all concentrations changing in accord with the stoichiometric coefficients until the Gibbs energy reaches a minimum value. For this unique combination of
the concentrations of reactants and products, the system is at equilibrium. Any further
movement of the reaction to the right leads to an increase in the Gibbs energy, which is
no longer a spontaneous change.
Now, let us formulate this general picture mathematically. Assuming that P and T
are constant as the reaction advances, starting from Equation (6.36), we can write the
differential change in the Gibbs energy, dG, as
dr G = a midni
i
Concept
Extent of reaction, j, is useful in
describing the equilibrium position
in a chemical reaction.
(6.50)
where the various dni are determined by the stoichiometric equation for the reaction.
We let the reaction move slightly to the right by j moles of a particular product, which
we choose arbitrarily. We call j the extent of reaction. As the reaction advances, the
number of moles of each species changes according to
ni = n*i + ni j
(6.51)
where n*i is the number of moles of species i present before the reaction moved slightly
to the right. Writing Equation (6.51) in a differential form gives
dni = ni dj
and Equation (6.50) becomes
(6.52)
dr G = a ni mi dj or
i
a
0r G
b
= a ni mi = ∆m
0j P,T
i
(6.53)
where mi are the chemical potentials of the various reactants and products at the com0r G
position of the reaction mixture. We see that the derivative a
b , which is the
0j P,T
slope of a plot of G versus j, determines the equilibrium position for the reaction. At
0r G
the minimum in the Gibbs energy for the reaction mixture, a
b
= 0.
0j P,T
To demonstrate this analysis quantitatively, let us consider the simple reaction
A S B. As j increases from zero to one mole, one mole of A is converted to one mole
of B. From Equation (6.44), we see that the Gibbs energy for the reaction mixture has
two contributions for each reaction species. The first arises from the pure unmixed substances and the second from mixing. Gunmixed and Gmixture are defined by Equation (6.54).
Gunmixed1T, P2 = xB mB~ 1T2 + 11 - xB2mA~ 1T2
Gmixture1T, P2 = xB mB~ 1T2 + xB RT ln xB + xAmA~ 1T2 + xART ln xA
= xB mB~ 1T2 + xB RT ln xB + 11 - xB2mA~ 1T2 + 11 - xB2RT ln11 - xB2
(6.54)
6.6 Calculating the Equilibrium Position for a Gas-Phase Chemical Reaction
Figure 6.5
4000
Plot of Gmixture , Gunmixed , and 𝚫 mixG as
a function of the extent of reaction. As j
increases from zero to one mole, one mole
of A is converted to one mole of B in the
reaction A S B. Gmixture has two contributions, Gunmixed from the pure components and ∆ mixG from the mixing of A
and B, which we know from Section 6.5
has a minimum value at j = 0.5 mol.
3000
Gunmixed
Gunmixed,Gmixture,DmixG (J mol21)
177
2000
1000
Gmixture
0
0.2
0.4
0.6
0.8
Extent of
1.0 reaction
j (moles)
21000
DmixG
22000
For our simple reaction xA = 1 - xB and j is numerically equal to xB, although
~
it has the units of moles. If we assume that mA~ = G m,A
= ∆ f G A~ = 3.80 kJ>mol,
~
~
~
mB = G m,B = ∆ f G B = 1.00 kJ>mol and that T = 298 K, we can evaluate Gmixture,
Gunmixed, and ∆ mixG as a function of j. The results are shown in Figure 6.5.
0r G
We see that for j = 0.756 mol, a
b
= 0, which corresponds to the equi0j P,T
0r G
librium position. For all values j 6 0.756 mol, a
b
is negative and, therefore,
0j P,T
the reaction proceeds spontaneously in the direction of increasing j. For all values of
0r G
j 7 0.756 mol, a
b
is positive, and the reaction proceeds spontaneously in the
0j P,T
direction of decreasing j.
The position of the minimum in Gmixture is strongly dependent on the difference
G A~ - G B~ . Figure 6.6 shows that as G A~ - G B~ increases, the minimum moves closer to
the reaction species, which has the lower G ~ value. However, it is the mixing term that
gives rise to a minimum in a plot of Gmixture versus j. If the mixing did not change the
value of Gmixture, the minimum in the plot of Gmixture versus j would have a minimum
value at j = 1 mol and the reaction system would consist of pure B at equilibrium.
50800
50600
50400
Gmixture
Gunmixed
50700
50600
50500
Gmixture
(b)
50500.1
50500
Gunmixed
Gmixture
0.996 0.997 0.998 0.999 1.0
Extent of Reaction (moles)
0.97 0.98 0.99 1.00
Extent of Reaction (moles)
0.85 0.90 0.95 1.00
Extent of Reaction (moles)
(a)
50800
Gunmixed,Gmixture (J mol21)
Gunmixed
Gunmixed,Gmixture (J mol21)
Gunmixed,Gmixture (J mol21)
51200
50200
Plots of free energy versus extent of
progress for various values of Gm,A.
The plots were constructed for the reaction A S B for Gm,B = 50.5 kJ mol-1
and Gm,A values of (a) 55.0 kJ mol-1,
(b) 60.0 kJ mol-1, and (c) 85.0 kJ mol-1.
Only the portion of the graph for j values
near one is shown for (b) and (c), so that
the minimum in Gmixture is near j = 0.86
in (a), 0.98 in (b), and 0.999999 in (c) can
be seen. Note that jeq approaches one as
~
~
G m,A
- G m,B
increases.
50900
51400
51000
Figure 6.6
(c)
178
CHAPTER 6 Chemical Equilibrium
Concept
If there were no entropy of mixing,
a stoichiometric reaction mixture
would consist entirely of reactants
or products.
The most important points presented in this section on using G to predict spontaneity at constant P and T, and extending the discussion to include using A to predict
spontaneity at constant V and T, can be generalized by the following statements:
• Reaction spontaneity can only be determined for specific values of the partial
pressures of all reactants and products. In general, it is not possible to determine
spontaneity over a range of partial pressures.
For small changes of the partial pressures in a reaction mixture at constant P and T
in which no nonexpansion work is possible, the condition for spontaneity is
•
drG 6 0 or a
0r G
b
= a ni mi 6 0
0j P,T
i
(6.55)
0r G
b
= a nimi = 0, the reaction mixture is at equilibrium, and neither
0j P,T
i
direction of change is spontaneous.
• If a
For small changes of the partial pressures in a reaction mixture at constant V and T
in which no nonexpansion work is possible, A rather than G is the function that determines spontaneity, and the following points hold:
• The condition for spontaneity is
dr A 6 0 or a
0r A
b
6 0
0j V,T
(6.56)
0r A
b
= 0, the reaction mixture is at equilibrium, and neither direction of
0j V,T
change is spontaneous.
• If a
EXAMPLE PROBLEM 6.7
Is the reaction 2NOg1g2 N N2O41g2 spontaneous if the partial pressures of NO21g2
and N2O41g2 are 0.500 bar at T = 298.15 K?
Solution
As the answer is independent of the total mass, we assume one mole of reactant and
product.
a
0r G
~
b
= a ni mi = ∆ f G N~ 2O41T2 + RT ln xN2O4 - 2∆ f G NO
1T2 - 2RT ln xNO2
2
0j P,T
i
= 99.8 kJ mol-1 + 8.314 J mol-1 K-1 * 298.15 K ln
1
2
- 2 * 51.3 kJ mol-1 - 2 * 8.314 J mol-1 K-1 * 298.15 K ln
1
2
= -1.08 kJ mol-1
Because a
0r G
b is negative, the reaction is spontaneous under these conditions.
0j P,T
6.7 INTRODUCING THE EQUILIBRIUM CONSTANT
FOR A MIXTURE OF IDEAL GASES
In this section, we focus on the pressure dependence of mi in order to calculate the
equilibrium position for a typical reaction mixture. Consider the following reaction that
takes place between ideal gas species A, B, C, and D at a total pressure of 1 bar:
aA + bB N gC + dD
(6.57)
6.7 Introducing the Equilibrium Constant for a Mixture of Ideal Gases
179
In a reaction mixture with a total pressure of 1 bar, the partial pressures of each species,
Pi, is less than the standard state pressure of 1 bar. Because all four species are present
0r G
in the reaction mixture, a
b
for the arbitrary set of partial pressures PA, PB, PC,
0j P,T
and P is given by
D
a
PC
0r G
PD
b
= a ni mi = g amC~ + RT ln ~ b + d amD~ + RT ln ~ b
0j P,T
P
P
i
At equilibrium, a
PA
PB
-a amA~ + RT ln ~ b - b amB~ + RT ln ~ b
P
P
(6.58)
0r G
b
= 0 and
0j P,T
PC
PD
PA
PB
- dRT ln ~ + aRT ln ~ + bRT ln ~
~
P
P
P
P
(6.59)
We can rearrange the previous equation to give
gmC~ + dmD~ - amA~ - bmB~ = -gRT ln
PC g PD d
b a ~b
P~
P
~
∆m = -RT ln
a
PA
PB b
a ~b a ~b
P
P
a
where ∆m ~ is given by
(6.60)
∆m ~ 1T2 = gmC~ 1T2 + dmD~ 1T2 - amA~ 1T2 - bmB~ 1T2 = a vi ∆ f G i~ 1T2
i
(6.61)
Because the chemical potential m ~ 1T2 is the molar Gibbs energy for the pure component
at 1 bar pressure and temperature T, ∆m ~ can be calculated from the ∆ f G ~ 1298.15 K2
tabulated in Appendix A.
The combination of the partial pressures of reactants and products in Equation (6.60)
is called the reaction quotient of pressures, which is abbreviated QP and defined
as follows:
PC g PD d
a ~b a ~b
P
P
QP =
(6.62)
a
PA
PB b
a ~b a ~b
P
P
With these definitions of QP and ∆m ~ , Equation (6.58) becomes
a
0r G
b
= ∆ r G ~ 1T2 + RT ln QP
0j P,T
(6.63)
0r G
b can be separated into two terms. QP depends
0j P,T
only on the partial pressures of reactants and products, and ∆ r G ~ 1T2 depends only on T.
0r G
At equilibrium, a
b
= 0 and ∆ r G ~ = -RT ln QP.3 We denote this special
0j P,T
system configuration by adding a superscript eq (for equilibrium) to each partial pressure and renaming QP as KP. It is important to distinguish between QP and KP. QP is
the reaction quotient of pressures for any arbitrary state of the reaction mixture. The
reaction quotient of pressures obtained by shifting the reactants toward products or
Note this important result that a
0r G
b
and ∆ r G = ∆ r G ~ 1T2 + RT ln QP, which incorrectly
0j P,T
implies a difference in the G values of products and reactants rather than a derivative of G with respect to the
0r G
extent of reaction. A better notation is to write ∆ r m = a
b and ∆ r m = ∆ r G ~ 1T2 + RT ln QP.
0j P,T
3
Some authors write the equality ∆ r G = a
Concept
The reaction quotient of pressures
can be used to calculate the direction
of spontaneous change in a reaction
mixture of ideal gases.
180
CHAPTER 6 Chemical Equilibrium
vice versa to reach the equilibrium state is KP. The quantity KP is called the thermodynamic equilibrium constant:
g P eq d
P eq
C
D
b
a ~b
~
P
P
0 = ∆ r G ~ 1T2 + RT ln eq a eq b or, equivalently,
PA
PB
a ~b a ~b
P
P
∆ r G ~ 1T2 = -RT ln KP
or
∆ r G ~ 1T2
ln KP = (6.64)
RT
Because ∆ r G ~ 1T2 is a function of T only, KP is also a function of T only. The thermodynamic equilibrium constant KP does not depend on P. Note also that KP is a dimensionless number because each partial pressure is divided by P ~ .
Next, we will show how Equation (6.63) can be used to predict the direction of
spontaneous change for a given set of partial pressures of the reactants and products. If
the partial pressures of the products C and D are large and those of the reactants A and
B are small compared to their values at equilibrium, QP will be large. As a result, the
0r G
value of RT ln QP will be large and positive, and a
b
= ∆ r G ~ + RT ln QP 7 0.
0j P,T
a
Concept
The thermodynamic equilibrium
constant, KP , is a dimensionless
quantity that depends only on T.
In this case, the reaction as written in Equation (6.57) from left to right is not spontaneous, but the reverse of the reaction is spontaneous. Next, consider the opposite
extreme. If the partial pressures of the reactants A and B are large and those of
the products C and D are small compared to their values at equilibrium, QP will
be small. As a result, the value of RT ln QP will be large and negative. In this case,
0r G
a
b
= ∆ r G ~ + RT ln QP 6 0 and the reaction is spontaneous as written from
0j P,T
left to right, as some of the reactants combine to form products. These concepts are
illustrated in Example Problems 6.8 and 6.9.
EXAMPLE PROBLEM 6.8
0rG
b for the reaction conditions in Example Problem 6.7 at
0j P,T
525 K. Is the reaction spontaneous at this temperature?
Calculate ∆ r m = a
Solution
0rG
b
= ∆ r m at the elevated temperature, we use Equation (6.33)
0j P,T
assuming that ∆ r H is independent of T:
To calculate a
∆ r m1T22 = T2 c
∆ r m1298.15K2
1
1
+ ∆ r H1T12 a bd
298.15K
T2
298.15K
~
∆ r H1298.15K2 = ∆ f H N~ 2O41T2 - 2∆ f H NO
1T2
2
= 9.16 * 103 J mol-1 - 2 * 33.2 * 103 J mol-1
= -57.2 kJ mol-1
∆ r m1525 K2 = 525 K * c
-1.08 * 103 J mol - 1
298.15 K
- 57.2 * 103 J mol - 1 a
∆ r m1525 K2 = 45.4 kJ mol-1
1
1
bd
525 K
298.15 K
Because ∆ r m1525 K2 7 0, the reaction as written is not spontaneous, but the
reverse reaction is spontaneous.
6.7 Introducing the Equilibrium Constant for a Mixture of Ideal Gases
It is important to understand that the equilibrium condition of Equation (6.64) links
the partial pressures of all reactants and products. Imagine that an additional amount of
species D is added to the system initially at equilibrium. The system will reach a new
eq
eq
eq
equilibrium state in which P eq
A , P B , P C , and P D all have new values because the value
c P eq d
a P eq b
P eq
P eq
C
D
A
b
a
b
n
a
b
a B~ b has not changed.
P~
P~
P~
P
~
Tabulated values of ∆ f G for compounds are used to calculate ∆ rG ~ at P ~ = 1 bar
and T = 298.15 K, as shown in Example Problem 6.9. Just as for enthalpy, ∆ f G ~ for
a pure element in its standard reference state is equal to zero because the reactants and
products in the formation reaction are identical.
of KP = a
∆ rG ~ = a vi ∆ f G i~
(6.65)
i
EXAMPLE PROBLEM 6.9
Calculate ∆ rG ~ for the reaction Fe 3O41s2 + 4H21g2 S 3Fe1s2 + 4H2O1l2 at
298.15 K.
Solution
∆ rG ~ = 3∆ f G ~ 1Fe, s2 + 4∆ f G ~ 1H2O, l2 - ∆ f G ~ 1Fe 3O4, s2 - 4∆ f G ~ 1H2, g2
= 3 * 0 kJ mol-1 - 4 * 237.1 kJ mol-1 + 1015.4 kJ mol-1 - 4 * 0 kJ mol-1
= 67.00 kJ mol-1
At other temperatures, ∆ rG ~ 1T2 is calculated using Equation (6.33) as shown in
Example Problem 6.10.
EXAMPLE PROBLEM 6.10
Calculate ∆ rG ~ for the reaction in Example Problem 6.9 at 360. K.
Solution
To calculate ∆ rG ~ at the elevated temperature, we use Equation (6.32) assuming that
∆ r H ~ is independent of T:
∆ rG ~ 1T22 = T2 c
∆ rG ~ 1298.15K2
298.15K1
+ ∆ r H ~ 1T12 a
1
1
bd
T2
298.15K
∆ r H 1298.15K2 = 3∆ f H 1Fe, s2 + 4∆ f H ~ 1H2O, l2 - ∆ f H ~ 1Fe 3O4, s2
~
~
- 4∆ f H ~ 1H2, g2
= 3 * 0 J mol-1 - 4 * 285.8 * 103 J mol-1 + 1118.4
* 103 J mol-1 - 4 * 0 J mol-1
= -24.80 kJ mol-1
∆ rG ~ 1360. K2 = 360. K * c
67.00 * 103 J mol-1
- 24.80
298.15 K
* 103 J mol-1 a
∆ rG ~ 1360. K2 = 80.9 kJ mol-1
1
1
bd
360. K
298.15 K
Example Problem 6.11 shows how KP is calculated using tabulated ∆ f G ~ values.
181
182
CHAPTER 6 Chemical Equilibrium
EXAMPLE PROBLEM 6.11
a. Using data from Table 4.1 (see Appendix A, Data Tables), calculate KP at 298.15 K
for the reaction CO1g2 + H2O1l2 S CO21g2 + H21g2.
b. Equimolar amounts of reactants and products are placed in a reaction vessel at constant pressure. Based on the value that you obtained for part (a), do you expect the
mixture to consist mainly of CO21g2 and H21g2 or mainly of CO1g2 + H2O1l2
at equilibrium?
Solution
a. ln KP = = -
1
1 ∆ f G ~ 1CO2, g2 + ∆ f G ~ 1H2, g2 - ∆ f G ~ 1H2O, l2
∆ rG ~ = a
b
RT
RT - ∆ f G ~ 1CO, g2
1
-394.4 * 103 J mol-1 + 0 + 237.1
*
a
b
* 103 J mol-1 + 137.2 * 103 J mol-1
8.314 J mol-1 K-1 * 298.15 K
= 8.1087
KP = 3.32 * 103
b. Because KP W 1, the mixture will consist mainly of the products CO21g2 + H21g2
at equilibrium.
6.8 CALCULATING THE EQUILIBRIUM PARTIAL
PRESSURES IN A MIXTURE OF IDEAL GASES
Concept
Given a set of initial partial pressures,
there is a unique combination of
reactant and product partial pressures
at equilibrium for a given value of T.
As shown in the previous section, the partial pressures of the reactants and products
in a mixture of gases at equilibrium cannot take on arbitrary values because they are
related through KP. In this section, we show how the equilibrium partial pressures can
be calculated for ideal gases. Similar calculations for real gases will be discussed in the
next chapter. In Example Problem 6.12, we consider the dissociation of chlorine:
Cl21g2 N 2Cl1g2
(6.66)
It is useful to set up the calculation in a tabular form, as shown in Example Problem 6.12.
EXAMPLE PROBLEM 6.12
In this example, n0 moles of chlorine gas are placed in a reaction vessel whose
temperature can be varied over a wide range so that molecular chlorine can partially
dissociate to atomic chlorine.
a. Define the degree of dissociation as b = deq >n0, where 2deq is the number of
moles of Cl(g) present at equilibrium and n0 represents the number of moles of
Cl21g2 that would be present in the system if no dissociation occurred. Derive an
expression for KP in terms of n0, deq, and P.
b. Derive an expression for b as a function of KP and P.
Solution
a. We set up the following table:
Initial number of moles
Moles present at equilibrium
Mole fraction present at equilibrium, xi
Partial pressure at equilibrium, Pi = xi P
Cl2 1 g2
n0
n0 - deq
n0 - deq
n0 + deq
n0 - deq
a
bP
n0 + deq
N
2Cl 1 g2
0
2deq
2deq
n0 + deq
2deq
a
bP
n0 + deq
6.9 VARIATION OF KP WITH TEMPERATURE
183
We next express KP in terms of n0, deq , and P:
KP1T2 =
=
a
a
P eq
Cl
b
P~
P eq
Cl2
P~
2
=
b
ca
4d 2eq
2deq
b
P 2
d
P~
n0 + deq
n0 - deq P
a
b
n0 + deq P ~
4d 2eq
P
P
=
~
2
2
1n0 + deq21n0 - deq2 P
1n02 - d eq P ~
This expression is converted into one in terms of b :
4d 2eq
KP1T2 =
b. aKP1T2 + 4
1n022 - d 2eq
P
b b2 = KP1T2
P~
b =
R
KP1T2
KP1T2 + 4
4b2 P
P
=
~
P
1 - b2 P ~
P
P~
Because KP1T2 depends strongly on temperature, b will also be a strong function
of temperature. Note that b also depends on both KP and P for this reaction.
Whereas KP1T2 is independent of P, b as calculated in Example Problem 6.12 does
depend on P. In the particular case considered, b decreases as P increases for constant
T. As will be shown in Section 6.11, b depends on P whenever ∆v ≠ 0 for a reaction.
6.9 VARIATION OF KP WITH TEMPERATURE
In Section 6.7, we showed that KP and ∆ rG ~ are related by ln KP = - ∆ rG ~ >RT =
- ∆ r H ~ >RT + ∆ r S ~ . Here we illustrate how these individual functions vary with
temperature for two reactions of industrial importance, beginning with ammonia
synthesis. This reaction is used to produce almost all of the fertilizer used in growing food crops and will be discussed in detail in Section 6.13. The overall reaction is
3 H21g2 + N21g2 S 2NH31g2.
First, we will derive an expression for the dependence of ln KP on T. Starting with
Equation (6.64), we write
d1∆ rG ~ >RT2
d ln KP
1 d1∆ rG ~ >T2
= = dT
dT
R
dT
(6.67)
Using the Gibbs–Helmholtz equation [Equation (6.30)], the preceding equation reduces to
d ln KP
∆rH ~
1 d1∆ rG ~ >T2
= =
dT
R
dT
RT 2
(6.68)
Because tabulated values of ∆ f G ~ are available at 298.15 K, we can calculate ∆ rG ~
and KP at this temperature; KP can be calculated at the temperature Tf by integrating
Equation (6.68) between the appropriate limits:
ln KP1Tƒ2
L
ln KP1298.15 K2
Tƒ
1
d ln KP =
R L
298.15 K
∆rH ~
T2
dT
(6.69)
Figure 6.7a shows that ∆ r H is large, negative, and weakly dependent on temperature.
Because the number of moles of gaseous species decreases as the reaction proceeds, ∆ r S ~
is negative and is weakly dependent on temperature. The temperature dependence of
~
Concept
Variation of KP with T can be
calculated using the Gibbs–Helmholtz
equation.
184
CHAPTER 6 Chemical Equilibrium
[
DrS
200
[
D rH
100
300
0
[
2DrH /T
400
2100
T(K)
1000
1200
[
2DrG /T
100
2200
2DrH /T
[
0
400
500
2100
600
700
900
800
T(K)
1000
2300
DrH
[
2400
DrS
[
2200
600
800
2DrG /T
[
200
(a) Ammonia synthesis
(b) Coal gasification
Figure 6.7
Temperature dependence of thermodynamic quantities for two reactions of importance to
industry. The plots are for reactions involved in (a) ammonia synthesis and (b) coal gasification.
In both plots, the quantities - ∆ r H ~ >T, ∆ r S ~ , ∆ r H ~ , and - ∆ r G ~ >T = R ln KP are shown as a
function of temperature. The units of ∆ r H ~ are kJ mol-1, and the units of ∆ r S ~ are J mol-1 K-1.
The units of - ∆ r H ~ >T and - ∆ r G ~ >T are kJ mol-1 K-1.
ln KP = - ∆ r G ~ >RT is determined by ∆ r H ~ . Because ∆ r H ~ is negative, KP decreases
as T increases. We see that KP 7 1 for T 6 450 K, and it decreases with increasing T
throughout the temperature range shown.
Figure 6.7b shows analogous results for the coal gasification reaction
H2O1g2 + C1s2 N CO1g2 + H21g2. This reaction produces a mixture of CO1g2 and
H21g2, which is known as syngas. It can be used to produce diesel engine fuel in the
form of alkanes through the Fischer–Tropsch process in the reaction 12n + 12H21g2 +
nCO1g2 S CnH12n + 221l2 + nH2O1l2. For the coal gasification reaction, the number
of moles of gaseous species increases as the reaction proceeds so that ∆ r S ~ is positive. ∆ r H ~ is large, positive, and not strongly dependent on temperature. Again, the
temperature dependence of KP is determined by ∆ r H ~ . Because ∆ r H ~ is positive, KP
increases as T increases.
The following conclusions can be drawn from Figure 6.7 and Equations (6.67)–(6.69):
• The temperature dependence of KP is determined by ∆ r H ~ .
• If ∆ r H ~ 7 0, KP increases with increasing T; if ∆ r H ~ 6 0, KP decreases with
increasing T.
• ∆ r S ~ has little effect on the dependence of KP on T.
• If ∆ r S ~ 7 0, ln KP is increased by approximately the same amount relative to
- ∆ r H ~ >RT at all temperatures. If ∆ r S ~ 6 0, ln KP is decreased by approximately
the same amount relative to - ∆ r H ~ >RT at all temperatures.
If the temperature Tƒ is not much different from 298.15 K, it can be assumed that
∆ r H ~ is constant over the temperature interval for most reactions. As was shown in
Figure 6.7, this assumption is better than it might appear at first glance. Although H is
strongly dependent on temperature, the temperature dependence of ∆ r H ~ is governed by
the difference in heat capacities ∆CP between reactants and products (see Section 4.4).
If the heat capacities of reactants and products are nearly the same, ∆ r H ~ is nearly
independent of temperature. With the assumption that ∆ r H ~ is independent of temperature, Equation (6.69) becomes
ln KP1Tƒ2 = ln KP1298.15 K2 -
∆r H ~ 1
1
a b
R
Tƒ
298.15 K
(6.70)
6.10 Equilibria Involving Ideal Gases and Solid or Liquid Phases
EXAMPLE PROBLEM 6.13
Using the result of Example Problem 6.12 and the data tables, consider the dissociation equilibrium Cl21g2 N 2Cl1g2.
a. Calculate KP at 800. K, 1500. K, and 2000. K for P = 0.010 bar.
b. Calculate the degree of dissociation b at 800. K, 1500. K, and 2000. K.
Solution
a. ∆ r G ~ = 2∆ f G ~ 1Cl, g2 - ∆ f G ~ 1Cl2, g2 = 2 * 105.7 * 103 J mol-1 - 0
= 211.4 kJ mol-1
∆ r H ~ = 2∆ f H ~ 1Cl, g2 - ∆ f H ~ 1Cl2, g2 = 2 * 121.3 * 103 J mol-1 - 0
= 242.6 kJ mol-1
ln KP1Tƒ2 = = -
∆rG ~
∆r H ~ 1
1
a b
RT
R
Tƒ
298.15 K
211.4 * 103 J mol-1
242.6 * 103 J mol-1
-1
-1
8.314 J K mol * 298.15 K
8.314 J K-1 mol-1
* a
1
1
b
Tƒ
298.15 K
ln KP1800. K2 = -
211.4 * 103 J mol-1
242.6 * 103 J mol-1
8.314 J K-1 mol-1 * 298.15 K
8.314 J K-1 mol-1
* a
1
1
b = -23.888
800. K
298.15 K
KP1800. K2 = 4.22 * 10-11
The values for KP at 1500. and 2000. K are 1.03 * 10-3 and 0.134, respectively.
b. The value of b at 2000. K is given by
b =
KP1T2
P
KP1T2 + 4 ~
R
P
=
0.134
= 0.878
A 0.134 + 4 * 0.01
The values of b at 1500. and 800. K are 0.159 and 3.23 * 10-5, respectively.
The degree of dissociation of Cl21g2 increases with temperature as shown in
Example Problem 6.13. This is always the case for an endothermic reaction.
6.10 EQUILIBRIA INVOLVING IDEAL GASES
AND SOLID OR LIQUID PHASES
In the preceding sections, we discussed chemical equilibrium in a homogeneous system
of ideal gases. However, many chemical reactions involve a gas phase in equilibrium
with a solid or liquid phase. An example is the thermal decomposition of CaCO31s2 :
CaCO31s2 N CaO1s2 + CO21g2
(6.71)
In this case, a pure gas is in equilibrium with two solid phases at the pressure P. At
equilibrium,
∆m = a ni mi = 0
i
0 = meq1CaO, s, P2 + meq1CO2, g, P2 - meq1CaCO3, s, P2
(6.72)
185
186
CHAPTER 6 Chemical Equilibrium
Because the equilibrium pressure is P ≠ P ~ , the pressure dependence of m for each
species must be taken into account. From Section 6.3, we know that the pressure
dependence of G for a solid or liquid is very small:
Concept
KP for an equilibrium involving
gases and liquids or solids contains
contributions only from gaseous
species.
meq1CaO, s, P2 ≈ m ~ 1CaO, s2 and meq1CaCO3, s, P2 ≈ m ~ 1CaCO3, s2
(6.73)
You will verify the validity of Equation (6.73) in the end-of-chapter problems. Using
the dependence of m on P for an ideal gas, we see that Equation (6.72) becomes
0 = m ~ 1CaO, s2 + m ~ 1CO2, g2 - m ~ 1CaCO3, s2 + RT ln
∆rG ~
PCO2
or
P~
PCO2
= m ~ 1CaO, s2 + m ~ 1CO2, g2 - m ~ 1CaCO3, s2 = -RT ln ~
P
(6.74)
Rewriting this equation in terms of KP, we obtain
ln KP = ln
PCO2
P~
= -
∆rG ~
RT
(6.75)
We generalize this result as follows. KP for an equilibrium involving gases with liquids
and/or solids contains only contributions from the gaseous species. However, ∆ r G ~ is
calculated taking all species into account.
EXAMPLE PROBLEM 6.14
Given the preceding discussion and using the tabulated values of ∆ f G ~ and ∆ f H ~
in Appendix A, calculate the CO21g2 pressure in equilibrium with a mixture of
CaCO31s2 and CaO(s) at 1000., 1100., and 1200. K.
Solution
For the reaction CaCO31s2 N CaO1s2 + CO21g2, ∆ r G ~ = 131.1 kJ mol-1 and
∆ r H ~ = 178.5 kJ mol-1 at 298.15 K. We use Equation (6.75) to calculate
KP1298.15 K2 and Equation (6.70) to calculate KP at elevated temperatures.
ln
PCO2
P~
11000. K2 = ln KP11000. K2
= ln KP1298.15 K2 -
=
∆ r H ~ 1298.15 K2
1
1
a
b
R
1000. K
298.15 K
-131.1 * 103 J mol-1
178.5 * 103 J mol-1
-1
-1
8.314 J K mol * 298.15 K
8.314 J K-1 mol-1
*a
1
1
b
1000. K
298.15 K
= -2.348; PCO211000. K2 = 0.0956 bar
The values for PCO2 at 1100. K and 1200. K are 0.673 bar and 3.42 bar, respectively.
If the reaction involves only liquids or solids, the pressure dependence of the chemical potential is generally small and can be neglected. However, it cannot be neglected
if P W 1 bar, as shown in Example Problem 6.15.
EXAMPLE PROBLEM 6.15
At 298.15 K, ∆ f G ~ 1C, graphite2 = 0 and ∆ f G ~ 1C, diamond2 = 2.90 kJ mol-1.
Therefore, graphite is the more stable solid phase at this temperature at P = P ~ = 1
bar. Given that the densities of graphite and diamond are 2.25 and 3.52 kg>L, respectively, at what pressure will graphite and diamond be in equilibrium at 298.15 K?
6.11 Expressing the Equilibrium Constant in Terms of Mole Fraction or Molarity
187
Solution
At equilibrium ∆G = G1C, graphite2 - G1C, diamond2 = 0. Using the pressure
dependence of G given in Equation (6.22), 10Gm >0P2T = Vm, we establish the condition for equilibrium:
∆G = ∆ f G ~ 1C, graphite2 - ∆ f G ~ 1C, diamond2
graphite
diamond
+ 1V m
- Vm
21∆P2 = 0
graphite
diamond
0 = 0 - 2.90 * 103 kJ mol-1 + 1V m
- Vm
21P - 1 bar2
P = 1 bar +
= 1 bar +
12.00 * 10-3 kg mol-1
2.90 * 103 kJ mol-1
1
1
MC a
b
rgraphite
rdiamond
2.90 * 103 kJ mol-1
1
1
* a
b
2.25 * 103 kg m-3
3.52 * 103 kg m-3
= 105 Pa + 1.51 * 109 Pa = 1.51 * 104 bar
Fortunately for all those with diamond rings, although the conversion of
diamond to graphite at 1 bar and 298 K is spontaneous, the rate of conversion is
vanishingly small.
6.11 EXPRESSING THE EQUILIBRIUM CONSTANT
IN TERMS OF MOLE FRACTION OR
MOLARITY
Chemists often find it useful to express the concentrations of reactants and products in
units other than partial pressures. Two examples of other units that we will consider in
this section are mole fraction and molarity. Note, however, that this discussion is still
limited to a mixture of ideal gases. The extension of chemical equilibrium to include
neutral and ionic species in aqueous solutions will be made in Chapter 10, after the
concept of activity has been introduced.
We first express the equilibrium constant in terms of mole fractions. The mole fraction xi and the partial pressure Pi are related by Pi = xi P. Therefore,
g P eq d
g x eqP d
P eq
x eq
C
CP
D
b
a
b
a
b
a D~ b
g eq d
1x eq
P~
P~
P~
P
P g+d- a-b
C 2 1x D 2
KP =
=
a
b
eq a
eq b =
eq
eq
a eq b P ~
PA
PB
xA P a xB P b
1x eq
A 2 1x B 2
a ~b a ~b
a ~ b a ~ b
P
P
P
P
a
= Kx a
P
b
P~
Kx = KP a
∆v
P -∆v
b
P~
(6.76)
Recall that ∆v is the difference in the stoichiometric coefficients of products and
reactants. Note that just as for KP, Kx is a dimensionless number.
Because the molarity ci is defined as ci = ni >V = Pi >RT, we can write
Pi >P ~ = 1RT>P ~ 2ci. To work with dimensionless quantities, we introduce the ratio
ci >c ~ , which is related to Pi >P ~ by
Pi
c ~ RT ci
=
~
P~ c~
P
(6.77)
Concept
The equilibrium constant Kx depends
on T and the difference between the
stoichiometric coefficients of reactant
and products.
188
CHAPTER 6 Chemical Equilibrium
c ~ is the standard state concentration of 1 mol/L. Using this notation, we can express
Kc in terms of KP:
g P eq d
g ceq d
P eq
ceq
C
C
D
b
a
b
a
b
a D~ b
P~
P~
c~
c
KP =
eq a
eq b =
eq a eq b
PA
PB
cA
cB
a ~b a ~b
a ~b a ~b
P
P
c
c
a
Kc = KP a
a
c ~ RT g + d - a - b
c ~ RT ∆v
b
=
K
a
b
c
P~
P~
c ~ RT -∆v
b
P~
(6.78)
Equations (6.76) and (6.78) show that Kx and Kc are in general different from KP. They
are only equal in the special case that ∆v = 0.
Suppose that we have a mixture of reactive gases at equilibrium. Does the equilibrium shift toward reactants or products as T or P is changed? Because jeq increases
if KP or Kx increases and decreases if KP or Kx decreases, we need only consider the
dependence of KP on T or Kx on T and P. We first consider the dependence of KP on T.
We know from Section 6.9 that for an exothermic reaction, d ln KP >dT 6 0, and jeq will
shift toward the reactants as T increases. For an endothermic reaction, d ln KP >dT 7 0,
and jeq will shift toward the products as T increases.
The dependence of jeq on pressure can be determined from the relationship between Kx and P:
Kx = KP a
P -∆v
b
P~
(6.79)
Because KP is independent of pressure, the pressure dependence of Kx arises solely
from 1P>P ~ 2-∆v. If ∆v 7 0, the number of moles of gaseous products increases as the
reaction proceeds. In this case, Kx decreases as P increases, and jeq shifts back toward
the reactants. If the number of moles of gaseous products decreases as the reaction
proceeds, Kx increases as P increases, and jeq shifts forward toward the products. If
∆v = 0, jeq is independent of pressure.
A combined change in T and P leads to a superposition of the effects just discussed.
According to the French chemist Le Chatelier, reaction systems at chemical equilibrium respond to an outside stress, such as a change in T or P, by countering the stress.
Consider the Cl21g2 N 2Cl1g2 reaction for which ∆ r H 7 0, discussed in Example
Problems 6.12 and 6.13. There, jeq responds to an increase in T in such a way that heat
is taken up by the system; in other words, more Cl21g2 dissociates. This counters the
stress imposed on the system by an increase in T. Similarly, jeq responds to an increase
in P in such a way that the volume of the system decreases. Specifically, jeq changes in
the direction that ∆v 6 0, and for the reaction under consideration, Cl(g) is converted
to Cl21g2. This shift in the position of equilibrium counters the stress brought about by
an increase in P.
6.12 EXPRESSING U, H, AND HEAT CAPACITIES
SOLELY IN TERMS OF MEASURABLE
QUANTITIES
In this section, we use the Maxwell relations derived in Section 6.2 to express U, H,
and heat capacities solely in terms of measurable quantities such as kT, a and the state
variables P, V, and T. Often, it is not possible to determine an equation of state for a
substance over a wide range of the macroscopic variables. This is the case for most
solids and liquids. However, material constants such as heat capacities, the thermal
expansion coefficient, and the isothermal compressibility may be known over a limited
range of their variables. If this is the case, the enthalpy and internal energy can be
expressed in terms of the material constants and the variables V and T or P and T, as
shown later. The Maxwell relations play a central role in deriving these formulas.
6.12 EXPRESSING U, H, AND HEAT CAPACITIES SOLELY IN TERMS OF MEASURABLE QUANTITIES
189
We begin with the following differential expression for dU that holds for solids,
liquids, and gases, provided that no nonexpansion work, phase transitions, or chemical
reactions occur in the system:
dU = TdS - PdV
(6.80)
Invoking a transformation at constant temperature and dividing both sides of the equation by dV,
0U
0S
0P
a b = Ta b - P = Ta b - P
0V T
0V T
0T V
(6.81)
To arrive at the second expression in Equation (6.81), we have used the third Maxwell
relation (Equation (6.25)). We next express 10P>0T2V in terms of a, the isobaric volumetric thermal expansion coefficient, and kT , the isothermal compressibility:
a
10V>0T2P
0P
a
b = =
kT
0T V
10V>0P2T
(6.82)
Equation (6.82) allows us to write Equation (6.81) in the form
a
aT - kT P
0U
b =
kT
0V T
(6.83)
Through this equation, we have linked 10U>0V2T , which as discussed in Section 3.2, is
called the internal pressure, to easily measured material properties. This allows us to
calculate the internal pressure of a liquid or a gas, as shown in Example Problem 6.16.
EXAMPLE PROBLEM 6.16
At 298 K, the thermal expansion coefficient and the isothermal compressibility of liquid water are a = 2.04 * 10-4 K-1 and kT = 45.9 * 10-6 bar -1.
a. Calculate 10U>0V2T for water at 320. K and P = 1.00 bar.
b. If an external pressure equal to 10U>0V2T were applied to 1.00 m3 of liquid water at
320. K, how much would its volume change? What is the relative change in volume?
c. As shown in Example Problem 3.4, 10U>0Vm2T = a>V 2m for a van der Waals gas.
Calculate 10U>0Vm2T = a>V 2m for N21g2 at 320 K and P = 1.00 atm, given that
a = 1.35 atm L2 mol-2. Compare the value that you obtain with that of part (a)
and discuss the difference.
Solution
a. a
aT - kTP
0U
b =
kT
0V T
=
2.04 * 10-4 K-1 * 320. K - 45.9 * 10-6 bar -1 * 1.00 bar
45.9 * 10-6 bar -1
= 1.4 * 103 bar
b.
kT = -
1 0V
a b
V 0P T
Pƒ
∆VT = -
L
Pi
V1T2kT dP ≈ -V1T2kT1Pƒ - Pi2
= -1.00 m3 * 45.9 * 10-6 bar -1 * 11.4 * 103 bar - 1 bar2
= -6.4 * 10-2 m3
∆VT
6.4 * 10-2 m3
= = -6.4,
V
1 m3
Concept
It is possible to express U, H, and
heat capacities solely in terms of
measurable quantities if the equation
of state is not known.
190
CHAPTER 6 Chemical Equilibrium
c. At atmospheric pressure, we can calculate Vm using the ideal gas law and
a
1.35 atm L2 mol-2 * 11.00 atm22
0U
a
aP 2
b = 2 = 2 2 =
0Vm T
Vm
RT
18.206 * 10-2 atm L mol-1 K-1 * 320. K22
= 1.96 * 10-3 atm
The internal pressure is much smaller for N2 gas than for H2O liquid. This is the case
because H2O molecules in a liquid are relatively strongly attracted to one another,
whereas the interaction between gas phase N2 molecules is very weak. Gases, which
have a small internal pressure, are more easily compressed by an external pressure
than liquids or solids, which have a high internal pressure.
Using the result in Equation (6.83) for the internal pressure, we can write dU in
the form
aT - kT P
0U
0U
dU = a b dT + a b dV = CV dT +
dV
(6.84)
kT
0T V
0V T
Equation (6.84) is useful because it contains only the material constants CV, a, kT, and
the macroscopic variables T and V. In Example Problems 6.17 and 6.18, we show that
∆U, ∆H, and CP - CV can be calculated even if the equation of state for the material
of interest is not known, as long as the thermal expansion coefficient and the isothermal
compressibility are known.
EXAMPLE PROBLEM 6.17
A 1000. g sample of liquid Hg undergoes a reversible change from an initial state
Pi = 1.00 bar, Ti = 300. K to a final state Pƒ = 300. bar, Tƒ = 600. K. The density
of Hg is 13,534 kg m-3, a = 1.81 * 10-4 K-1, CP,m = 27.98 J mol-1 K-1, and
kT = 3.91 * 10-6 bar -1, in the range of the variables in this problem. Recall from
Section 3.5 that CP - CV = 1TV a2 >kT2, and that CP ≈ CV for liquids and solids.
a. Calculate ∆U and ∆H for the transformation. Describe the path assumed for the
calculation.
b. Compare the relative contributions to ∆U and ∆H from the change in pressure
and the change in temperature. Can the contribution from the change in pressure
be neglected?
Solution
Because U is a state function, ∆U is independent of the path. We choose the reversible
path Pi = 1.00 bar, Ti = 300. K S Pi = 1.00 bar, Tƒ = 600. K S Pƒ = 300. bar,
Tƒ = 600. K. Note that because we have assumed that all the material constants are
independent of P and T in the range of interest, the numerical values calculated will
depend slightly on the path chosen. However, if the P and T dependence of the material
constant were properly accounted for, all paths would give the same value of ∆U.
Tƒ
∆U =
L
Vƒ
CP dT +
Ti
aT - kT P
dV
kT
L
Vi
Using the definition kT = -1>V10V>0P2T , ln Vƒ >Vi = -kT1Pƒ - Pi2, and
V1P2 = Vi e-kT 1P - Pi2. We can also express P1V2 as P1V2 = Pi - 11>kT2ln1V>Vi2
Tƒ
∆U =
L
Ti
Tƒ
Vƒ
CP dT +
aT
1
V
- aPi ln b d dV
k
k
V
T
L T
i
Vi
c
Vƒ
Vƒ
Vi
Vi
aT
1
V
=
CP dT + a
- Pi b
dV +
ln dV
kT
kT L Vi
L
L
Ti
6.12 EXPRESSING U, H, AND HEAT CAPACITIES SOLELY IN TERMS OF MEASURABLE QUANTITIES
Using the standard integral 1 ln 1x>a2dx = -x + x ln 1x>a2
Tƒ
∆U =
L
Ti
Cp dT + a
aT
1
V Vƒ
- Pi b 1Vƒ - Vi2 +
c -V + V ln d
kT
kT
Vi Vi
= CP1Tƒ - Ti2 + a
Vi =
Vƒ
aT
1
- Pi b 1Vƒ - Vi2 +
aVi - Vƒ + Vƒ ln b
kT
kT
Vi
1.000 kg
m
= 7.389 * 10-5 m3
=
r
13534 kg m-3
Vƒ = Vi e-kT 1Pƒ - Pi2 = 7.389 * 10-5 m3 exp1-3.91 * 10-6 bar -1 * 299 bar2
= 7.380 * 10-5 m3
Vƒ - Vi = -9 * 10-8 m3
∆U =
1000. g
200.59 g mol-1
-a
* 27.98 J mol-1 K-1 * 1600. K - 300. K2
1.81 * 10-4 K-1 * 300. K
105 Pa
1.00
barb
*
* 9 * 10-8 m3
1 bar
3.91 * 10-6 bar -1
9 * 10-8 m3 + 7.380 * 10-5 m3
1
105 Pa
+
*
* ±
≤
7.380 * 10-5 m3
1 bar
3.91 * 10-6 bar -1
* ln
-5 3
7.389 * 10 m
= 41.8 * 103 J - 100 J + 1 J ≈ 41.7 * 103 J
As shown in Section 3.1,
V1P, 600. K2 = 3 1 + a1600. K - 300. K2 4 V1Pi, 300. K2e-kT 1P - Pi2 = V′ e-kT 1P - Pi2
Tƒ
∆H =
L
Pƒ
Cp dT +
Ti
L
Tƒ
VdP =
Pi
= CP1Tƒ - Ti2 +
=
1000. g
200.59 g mol-1
L
Pƒ
kT P i
Cp dT + V′ e
Ti
L
e-kT P dP
Pi
V′i ekT Pi -k P
1e T i - e-kT Pƒ2
kT
* 27.98 J mol-1 K-1 * 1600. K - 300. K2
+ 7.389 * 10-5 m3 * 11 + 300. K * 1.81 * 10-4 K-12
*
exp13.91 * 10-6 bar -1 * 1 bar2
3.91 * 10-6 bar -1 * 11 bar>105 Pa2
* 3 exp1-3.91 * 10-6 bar -1 * 1 bar2
- exp1-3.91 * 10-6 bar -1 * 300. bar24
∆H = 41.8 * 103 J + 2.29 * 103 J = 44.1 * 103 J
The temperature-dependent contribution to ∆U is 98% of the total change in U, and
the corresponding value for ∆H is ∼95%. The contribution from the change in
pressure to ∆U and to ∆H is very small.
191
192
CHAPTER 6 Chemical Equilibrium
EXAMPLE PROBLEM 6.18
In Example Problem 6.17, it was assumed that CP = CV .
a2
, and the experimentally determined
kT
value CP,m = 27.98 J mol-1 K-1 to obtain a value for CV,m for Hg1l2 at 300. K.
b. Did the assumption that CP,m ≈ CV,m in Example Problem 6.15 introduce an
appreciable error in your calculation of ∆U and ∆H ?
a. Use Equation (3.38), CP,m = CV,m + TVm
Solution
a. CP,m - CV,m =
TVma2
kT
300. K *
=
0.20059 kg mol-1
13534 kgm3
* 11.81 * 10-4 K-122
3.91 * 10-6 bar -1 * 11 bar>105 Pa2
= 3.73 J K-1 mol-1
CV,m = CP,m -
TVma2
kT
= 27.98 J K-1 mol-1 - 3.73 J K-1 mol-1 = 24.25 J K-1 mol-1
b. ∆U is in error by ∼15%, and ∆H is unaffected by assuming that CP,m ≈ CV,m.
It is a reasonable approximation to set CP = CV for a liquid or solid.
S U P P L E M E N TA L
S E C T I O N
6.13 A CASE STUDY: THE SYNTHESIS
OF AMMONIA
Concept
Nearly all inorganic nitrogen input
to agricultural soils is provided by
the Haber–Bosch ammonia synthesis
process.
In this section, we discuss a specific reaction and show both the power and the limitations
of thermodynamic principles and analysis in developing a strategy to maximize the rate
of a desired reaction. Ammonia synthesis is the primary route to the manufacture of fertilizers used in agriculture. The commercial process used today is essentially the same as
that invented by the German chemists Robert Bosch and Fritz Haber in 1908. In terms of
its impact on human life, the Haber–Bosch synthesis may be the most important chemical process ever invented because the increased food production possible with fertilizers
based on NH3 allowed the large increase in Earth’s population in the 20th century to
occur. In 2000, more than 2 million tons of ammonia were produced per week using the
Haber–Bosch process, and more than 98% of the inorganic nitrogen input to soils used in
agriculture worldwide as fertilizer is generated using the Haber–Bosch process.
What useful information can the ammonia synthesis reaction from thermodynamics
give us? The overall reaction is
1>2 N21g2 + 3>2 H21g2 N NH31g2
(6.85)
Because the reactants are pure elements in their standard states at 298.15 K,
∆ r H ~ = ∆ f H ~ 1NH3, g2 = -45.9 * 103 J mol-1 at 298.15 K
∆rG
3
= ∆ f G 1NH3, g2 = -16.5 * 10 J mol
-1
(6.86)
at 298.15 K
(6.87)
Assuming ideal gas behavior, we see that the equilibrium constant KP is
PNH3
a ~ b
1xNH32
P
P -1
=
a
b
KP =
3
3
1
1
~
PN2 2 PH2 2
1xN2221xH222 P
a ~b a ~b
P
P
(6.88)
~
~
6.13 A Case Study: The Synthesis of Ammonia
We will revisit this equilibrium and include deviations from ideal behavior in Chapter 7.
Because the goal is to maximize xNH3, Equation (6.88) is written in the form
1
3
xNH3 = 1xN222 1xH222 a
P
bK
P~ P
(6.89)
What conditions of temperature and pressure are most appropriate to maximizing the
yield of ammonia? Because
d ln KP
∆r H ~
=
6 0
(6.90)
dT
RT 2
KP and, therefore, xNH3 decrease as T increases. Because ∆v = -1, xNH3 increases with
P. We conclude that the best conditions to carry out the reaction are high P and moderate T. In fact, industrial production is carried out for P 7 100 atm and T ∼ 700 K.
Our discussion has focused on reaching equilibrium because this is the topic of inquiry
in thermodynamics. However, note that many synthetic processes are deliberately carried
out far from equilibrium in order to selectively generate a desired product. For example,
ethylene oxide, which is produced by the partial oxidation of ethylene, is an industrially
important chemical used to manufacture antifreeze, surfactants, and fungicides. Although
the complete oxidation of ethylene to CO2 and H2O has a more negative value of ∆ r G ~
than the partial oxidation, it is undesirable because it does not produce useful products.
These predictions provide information about the equilibrium state of the system but
lack one important component. How long will it take to reach equilibrium? Unfortunately, thermodynamics gives us no information about the rate at which equilibrium is
attained. Consider the following example. Using values of ∆ f G ~ from the data tables,
convince yourself that the oxidation of potassium and coal (essentially carbon) are both
spontaneous processes. For this reason, potassium has to be stored in an environment free
of oxygen. However, coal is stable in air indefinitely. Thermodynamics tells us that both
processes are spontaneous at 298.15 K, but experience shows us that only one proceeds
at a measurable rate. However, by heating coal to an elevated temperature, the oxidation
reaction becomes rapid. This result implies that there is an energetic activation barrier
to the reaction that can be overcome with the aid of thermal energy. As we will see, the
same is true of the ammonia synthesis reaction. Although Equations (6.86) and (6.87)
predict that xNH3 is maximized as T approaches 0 K, the reaction rate is vanishingly small
unless T 7 500 K.
To this point, we have only considered the reaction system using state functions.
However, a chemical reaction consists of a number of individual steps, called a reaction
mechanism. How might the ammonia synthesis reaction proceed? One possibility is
the gas-phase reaction sequence:
N21g2 S 2N1g2
(6.91)
H21g2 S 2H1g2
(6.92)
N1g2 + H1g2 S NH1g2
(6.93)
NH1g2 + H1g2 S NH21g2
(6.94)
NH21g2 + H1g2 S NH31g2
(6.95)
In each of the recombination reactions, other species that are not involved in the
reaction can facilitate energy transfer. The enthalpy changes associated with each of
the individual steps are known and are indicated in Figure 6.8. This diagram is useful because it gives much more information than the single-value ∆ r H ~ . Although
the overall process is slightly exothermic, the initial step, which is the dissociation of
N21g2 and H21g2, is highly endothermic. Heating a mixture of N21g2 and H21g2 to high
enough temperatures to dissociate a sizable fraction of the N21g2 would require such a
high temperature that all the NH31g2 formed would dissociate into reactants. This can
be shown by carrying out a calculation such as that in Example Problem 6.11. The conclusion is that the gas-phase reaction between N21g2 and H21g2 is not a practical route
to ammonia synthesis.
The way around this difficulty is to find another reaction sequence for which the
enthalpy changes of the individual steps are not prohibitively large. For the ammonia
193
194
CHAPTER 6 Chemical Equilibrium
Figure 6.8
N(g) 1 3H(g)
Enthalpy diagram for the reaction
mechanism in the synthesis of ammonia.
See Equations (6.91) through (6.95). The
successive steps in the reaction proceed
from left to right in the diagram.
314 3 103 J
NH(g) 1 2H(g)
390 3 103 J
1124 3 103 J
NH2(g) 1 H(g)
466 3 103 J
1
2
N2(g) 1
3
2
245.9 3 103 J
H2(g)
NH3(g)
Progress of reaction
synthesis reaction, such a route is a heterogeneous catalytic reaction, using iron as a
catalyst. The mechanism for this path between reactants and products is
N21g2 + □ S N21a2
(6.96)
N21a2 + □ S 2N1a2
H21g2 + 2□ S 2H1a2
(6.98)
N1a2 + H1a2 S NH1a2 + □
(6.99)
NH1a2 + H1a2 S NH21a2 + □
(6.100)
(6.97)
NH21a2 + H1a2 S NH31a2 + □
(6.101)
NH31a2 S NH31g2 + □
(6.102)
The symbol □ denotes an ensemble of neighboring Fe atoms, also called surface sites, which
are capable of forming a chemical bond with the indicated entities. The designation (a)
indicates that the chemical species is adsorbed (chemically bonded) to a surface site.
The enthalpy change for the overall reaction N21g2 + 3>2 H21g2 S NH31g2 is
the same for the mechanisms in Equations (6.91) through (6.95) and (6.96) through
(6.102) because H is a state function. This is a characteristic of a catalytic reaction. A
catalyst can affect the rate of the forward and backward reaction but not the position
of equilibrium in a reaction system. The enthalpy diagram in Figure 6.9 shows that the
Homogeneous gas-phase reactions
N(g) 1 3H(g)
314 3 103 J
NH(g) 1 2H(g)
Figure 6.9
1124 3 103 J
Enthalpy diagram for the homogeneous
gas-phase and heterogeneous catalytic
1
3
2 N2(g) 1 2 H2(g)
reactions for the ammonia synthesis
reaction. The activation barriers for the
0
individual steps in the surface reaction
are shown. The successive steps in the
reaction proceed from left to right in the
diagram. See the reference to G. Ertl in
1
3
Further Reading for more details.
2 N2(a) 1 2 H2(a)
Source: Adapted from G. Ertl, Catalysis
Reviews—Science and Engineering
21 (1980): 201–223.
Rate-limiting step
390 3 103 J
NH2(g) 1 H(g)
466 3 103 J
NH3(g)
245.9 3 103 J
Progress of reaction
Heterogeneous catalytic reactions
NH2(a) 1 H(a)
NH(a) 1 2H(a)
N(a) 1 3H(a)
NH3(a)
NH3(g)
245.9 3 103 J
6.13 A Case Study: The Synthesis of Ammonia
enthalpies of the individual steps in the gas phase and surface catalyzed reactions are
very different. These enthalpy changes have been determined in experiments but are not
generally listed in thermodynamic tables.
In the catalytic mechanism, the enthalpy for the dissociation of the gas-phase N2
and H2 molecules is greatly reduced by stabilization of the atoms through their bond to
the surface sites. However, a small activation barrier to the dissociation of adsorbed N2
remains. This barrier is the rate-limiting step in the overall reaction. As a result of this
barrier, only one in a million gas-phase N2 molecules incident on the catalyst surface per
unit time dissociates and is converted into ammonia. Such a small reaction probability
is not uncommon for industrial processes based on heterogeneous catalytic reactions.
In industrial reactors, on the order of ∼107 collisions of the reactants with the catalyst
occur in the residence time of the reactants in the reactor. Therefore, even reactions with
a reaction probability per collision of 10-6 can be carried out with a high yield.
How was Fe chosen as the catalyst for ammonia synthesis? Even today, catalysts
are usually optimized using a trial-and-error approach. Haber and Bosch tested hundreds
of catalyst materials, and more than 2500 substances have been tried since. On the basis
of this extensive screening, it was concluded that a successful catalyst must fulfill two
different requirements: (1) it must bind N2 strongly enough that the molecule can dissociate into N atoms, but (2) it must not bind N atoms too strongly. If that were the case,
the N atoms could not react with H atoms to produce gas-phase NH3. As Haber and
Bosch knew, both osmium and ruthenium are better catalysts than iron for NH3 synthesis, but these elements are far too expensive for commercial production. Why are these
catalysts better than Fe? Quantum-mechanical calculations by Jacobsen et al. (2001)
[See Further Reading for the reference] explained the “volcano plot” of the activity
versus the N2 binding energy shown in Figure 6.10.
Jacobsen et al. (2001) concluded that molybdenum is a poor catalyst because N is
bound too strongly, whereas N2 is bound too weakly. Nickel is a poor catalyst for the
opposite reason: N is bound too weakly and N2 is bound too strongly. Osmium and
ruthenium are the best elemental catalysts because they fulfill both requirements well.
Jacobsen et al. suggested that a catalyst that combined the favorable properties of Mo
and Co to create Co-Mo surface sites would be a good catalyst, and they concluded
that the ternary compound Co3Mo3N best satisfied these requirements. On the basis of
these calculations, the catalyst was synthesized with the results shown in Figure 6.11.
The experiments show that Co3Mo3N is far more active than Fe and more active than
Ru over most of the concentration range. These experiments also show that the turnover
frequency for Co3Mo3N lies at the point shown on the volcano plot of Figure 6.10. This
catalyst is one of the few that have been developed on the basis of theoretical calculations as opposed to the trial-and-error method.
Turnover frequency (s21)
101
“CoMo”
Concept
The ideal ammonia catalyst binds N 2
molecules relatively strongly but does
not bind N atoms strongly.
Ru
Os
Figure 6.10
100
Fe
1021
Co
1022
1023
1024
195
Mo
Ni
1025
2100.0
275.0
250.0
225.0
20.0
25.0
[DE2DE(Ru)] (kJ/mol N2)
50.0
75.0
Calculated turnover frequency for
NH3 production as a function of the
calculated binding energy of N2. NH3
production is the rate of NH3 molecules
per surface site per second. Values are
relative to those on ruthenium. Reaction
conditions are 400°C, 50 bar total pressure,
and the gas composition of H2 : N2 = 3 :1
contains 5% NH3. “CoMo” is the compound Co3Mo3N, which best satisfied the
requirements of the catalyst. For more
details, see the reference to Jacobsen in
Further Reading.
Source: Adapted from Jacobsen et al, Journal
of the American Chemical Society 123 (2001):
8404–8405.
196
CHAPTER 6 Chemical Equilibrium
Figure 6.11
Source: Adapted from Jacobsen et al, Journal
of the American Chemical Society 123 (2001):
8404–8405.
8
Ru
7
NH3 concentration (%)
Experimental data for turnover
frequency for NH3 production as a
function of the NH3 concentration in
the mixture. The Co3Mo3N catalyst is
superior to both Fe and Ru as predicted by
the calculations. The reaction conditions
are those for Figure 6.10. The number
of active surface sites is assumed to be
1% of the total number of surface sites.
See the reference to Jacobsen in Further
Reading for more details.
Co3Mo3N
6
5
Fe
4
3
2
0
2.5
5.0
7.5 10.0 12.5 15.0
Turnover frequency [s21]
17.5
20.0
An unresolved issue is the nature of the surface sites indicated by □ in Equations
(6.96) through (6.102). Are all adsorption sites on an NH3 synthesis catalyst surface
equally reactive? To answer this question, it is important to know that metal surfaces
at equilibrium consist of large, flat terraces separated by monatomic steps. It was long
suspected that the steps rather than the terraces play an important role in ammonia synthesis. This was convincingly shown (see reference to Dahl in Further Reading) in an
elegant experiment. Dahl et al. evaporated a small amount of gold on a Ru crystal. As
had been demonstrated (see reference to Hwang in Further Reading) the inert Au atoms
migrate to the step sites on Ru and thereby block these sites, rendering them inactive in
NH3 synthesis.
Dahl et al. (2001) showed that the rate of ammonia synthesis was lowered by a
factor of 109 after the step sites—which make up only 1% to 2% of all surface sites—
were blocked by Au atoms! This result clearly shows that in this reaction only step
sites are active. They are more active because they have fewer nearest neighbors than
terrace sites. This enables the step sites to bind N2 more strongly, promoting dissociation into N atoms that react rapidly with H atoms to form NH3.
This discussion of catalysis of ammonia synthesis has shown the predictive power of
thermodynamic principles in analyzing chemical reactions. Calculating ∆ r G ~ allowed
us to predict that the reaction of N2 and H2 to form NH3 is a spontaneous process at
298.15 K. Calculating ∆ r H ~ allowed us to predict that the yield of ammonia increases
with increasing pressure and decreases with increasing temperature. However, thermodynamics cannot be used to determine how rapidly equilibrium is reached in a reaction
mixture. If thermodynamic tables of species such as N21a2, N(a), NH(a), NH21a2, and
NH31a2 were available, it would be easier for chemists to develop synthetic strategies
that are likely to proceed at an appreciable rate. However, whether an activation barrier
exists in the conversion of one species to another—and how large the barrier is—are
not predictable by thermodynamic analysis. Therefore, synthetic strategies must rely
on chemical intuition, experiments, and quantum-mechanical calculations to develop
strategies that reach thermodynamic equilibrium at a reasonable rate.
S U P P L E M E N TA L
S E C T I O N
6.14 MEASURING 𝚫G FOR THE UNFOLDING
OF SINGLE RNA MOLECULES
Concept
∆G for unfolding biomolecules can
be measured by mechanically pulling
the molecule apart.
In general, ∆G, ∆H, and ∆U are measured using thermochemical methods as discussed
in Chapter 4. However, many biochemical systems are irreversibly changed upon heating, so that calorimetric methods are unsuitable for such systems. With the advent of
atomic-scale manipulation of molecules, which began with the invention of scanning
6.14 MEASURING ∆G FOR THE UNFOLDING OF SINGLE RNA MOLECULES
220
Extension (nm)
240
260
Streptavidin bead
(a)
(b)
P5abcDA
C A5’
AC
A GA
G C
G C
C GU
P5a A
U
G C
U A
C G
G U
U A
3-helix
G C
junction
G C
AAAGGGUAU
C
GUU C C
G A
G A
P5c
U G
A U
Hairpin G C
A U
loop
G C
P5b U A
U G
U G
C G
A G
AA
RNA/DNA
Handle A
Biotin
Antidigoxigenin bead
probe microscopies (see Chapter 16), it has become possible to measure ∆G for a
change using an individual molecule. An interesting illustration is the measurement of
∆G for the unfolding of single RNA molecules. RNA molecules must fold into very
specific shapes in order to carry out their function as catalysts for biochemical reactions. However, the mechanisms for how the molecule achieves its equilibrium shape—
which is important because the energetics of folding is a key parameter in modeling the
folding process—are not well understood.
Liphardt et al. (see Further Reading) obtained ∆G for this process by unzipping a
single RNA molecule. The measurement relies on Equation (6.10), which states that
Gibbs energy is a measure of the maximum nonexpansion work that can be produced
in a transformation. Figure 6.12 shows how the force associated with the unfolding of
RNA into two strands can be measured. Each of the two strands that make up the RNA
is attached to a chemical “handle” that allows the individual strands to be firmly linked
to 2-mm-diameter polystyrene beads. By using techniques based on scanning probe
microscopy, the distance between the beads is increased by distances in the nanometer
range, and the force needed to do so is measured. The plot of force versus distance for
the P5abc∆A RNA is shown in Figure 6.13 for a rate of 1 pN s-1. The red and black
curves are nearly superimposed, showing that the process is essentially reversible. (No
process that occurs at a finite rate is truly reversible.) Finally, we will discuss the form
of the force versus distance curve.
The initial increase in distance between ∼210 and ∼225 nm is due to the stretching of the handles. However, a marked change in the shape of the curve is observed for
F ≈ 13 pN, for which the distance increases abruptly. This behavior is the signature of the
unfolding of the two strands. The negative slope of the curve between 230 and 260 nm
is determined by the experimental technique used to stretch the RNA. In a “perfect”
experiment, this portion of the curve would be a horizontal line, corresponding to an
increase in length at constant force. The observed length increase is identical to the
length of the molecule, showing that the molecule unfolds all at once in a process in
which all bonds between the complementary base pairs are broken. The analysis of the
results to obtain ∆G from the force curves is complex and will not be discussed here.
The experimentally obtained value for ∆G is 193 kJ>mol, showing that a change in
the thermodynamic state function G can be determined by a measurement of work
on a single molecule. The unfolding can also be carried out by increasing the temperature of the solution containing the RNA to ∼80°C, in which case the process
is referred to as melting. In the biochemical environment, unfolding is achieved
through molecular forces exerted by enzymes called DNA polymerases, which act as
molecular motors.
197
RNA/DNA
Handle B
Digoxigenin
Figure 6.12
RNA molecule and measurement of
the force needed to unfold strands.
(a) Schematic diagram of one of the three
types of RNA studied by Liphardt et al.
are shown. The letters G, C, A, and
U refer to the bases guanine, cytosine,
adenine, and uracil. The strands are bound
together by hydrogen bonding between the
complementary base pairs A and U, and G
and C. The pronounced bulge at the center
of P5abc∆A is a hairpin loop. Such loops
are believed to play a special role in the
folding of RNA. (b) An individual RNA
molecule is linked by handles attached to
the end of each strand to functionalized
polystyrene spheres that are moved relative to one another to induce unfolding of
the RNA mechanically. See the reference
to J. Liphardt in Further Reading for more
details.
Source: Adapted from Liphardt et al, Science,
292 (2001): 733–737.
280
16
Force/pN
15
14
Figure 6.13
Measured force versus distance curve
for the P5abc𝚫A form of RNA. The
black trace represents the stretching of the
molecule, and the red curve represents the
refolding of the molecule as the beads are
brought back to their original separation.
The rate of increase in force is 1 pN s-1.
13
12
1 pN s21
Source: Adapted from J. Liphardt et al, Science,
292 (2001): 733–737.
198
CHAPTER 6 Chemical Equilibrium
VOCABULARY
activation barrier
chemical potential
extent of reaction
Gibbs energy
Gibbs–Helmholtz equation
Helmholtz energy
maximum nonexpansion work
Maxwell relations
molar Gibbs energy
natural variables
reaction mechanism
reaction quotient of pressures
thermodynamic equilibrium constant
KEY EQUATIONS
Equation
Significance of Equation
d1U - TS2 … dwexp + dwnonexp
Definition of Helmholtz energy
A = U - TS
Equation Number
6.5
and identification with maximum total work
dA … 0
Spontaneity criterion for process at constant T and V
d1U + PV - TS2 = d1H - TS2 … dwnonexp
Definition of Gibbs energy
G = H - TS
6.8
6.9
and identification with maximum nonexpansion work
dG … 0
Spontaneity criterion for process at constant T and
P, no nonexpansion work
6.11
dU = TdS - PdV
dH = TdS - PdV + PdV + VdP = TdS + VdP
dA = TdS - PdV - TdS - SdT = - SdT - PdV
Differential forms of important state functions
6.14–6.17
Maxwell relations provide relationships between
different functions that are useful in linking seemingly unconnected state functions.
6.23–6.26
dG = TdS + VdP - TdS - SdT = - SdT + VdP
a
a
a
0T
0P
b = -a b
0V S
0S V
0T
0V
b = a b
0P S
0S P
0P
a
0S
b = -a b =
kT
0V T
0T V
-a
0S
0V
b = - a b = Va
0P T
0T P
P
G1T, P2 = G 1T2 +
~
a
03G>T4
0T
∆G1T22
T2
mi = a
b
=
P
= -
L~
P
H
T
∆G1T12
T1
0G
b
0ni P,T,nj ≠ ni
2
+ ∆H1T12 a
G = a ni mi
i
∆ mixG = nRT a xi ln xi
i
VdP′ = G ~ 1T2 + nRT ln
1
1
- b
T2
T1
P
P~
Pressure dependence of the Gibbs energy for an
ideal gas
6.29
Gibbs–Helmholtz equation allows KP to be calculated at different temperatures
6.30
Calculation of Gibbs energy at elevated temperatures
assuming ∆H is constant over temperature range
6.33
Definition of chemical potential
6.35
Gibbs energy of gas mixture
6.37
Gibbs energy of mixing ideal gases
6.44
Conceptual Problems
∆ mixS = - a
a
0∆ mixG
b = -nR a xi ln xi
0T
P
i
0r G
b
= a ni mi = ∆m
0j P,T
i
drG 6 0 or a
0r G
b
= a ni mi 6 0
0j P,T
i
PC g PD d
b a ~b
P~
P
∆m ~ = a vi ∆ f G i~ 1T2 = -RT ln
PA a PB b
i
a ~b a ~b
P
P
a
0 = ∆ r G ~ 1T2 + RT ln
a
a
P eq
C
P
P eq
A
~
P
~
c
b a
a
b a
P eq
D
P
P eq
B
~
P
~
b
b
6.46
Slope of plot of G versus extent of reaction equals
change of chemical potential of mixture
6.53
Criterion for spontaneity of reaction at constant T
and P
6.55
Relates reaction quotient of pressures to ∆m ~ and
Gibbs energies of formation of reactants and
products
6.60 and 6.61
Defines thermodynamic equilibrium constant KP
6.64
Relates ∆ r G ~ to Gibbs energies of formation of reactants and products
6.65
Calculation of KP at elevated temperatures assuming
∆H is constant over temperature range
6.70
Defines equilibrium constant in terms of mole fractions
6.76
Defines equilibrium constant in terms of molarity
6.78
d
b
or, equivalently, ∆ r G 1T2 = -RT ln KP
~
Entropy change of mixing ideal gases
199
or
ln KP = -
∆ r G ~ 1T2
RT
∆ r G ~ = a vi ∆ f G i~
i
ln KP1Tƒ2 = ln KP1298.15 K2 Kx = KP a
Kc = KP a
P -∆v
b
P~
∆r H ~ 1
1
a b
R
Tƒ
298.15 K
c ~ RT -∆v
b
P~
CONCEPTUAL PROBLEMS
Q6.1 It is found that KP is independent of T for a particular
chemical reaction. What does this tell you about the reaction?
Q6.2 The reaction A + B S C + D is at equilibrium for
j = 0.1. What does this tell you about the ∆ f G ~ for reactants
and products?
Q6.3 Under what condition is KP = Kx?
Q6.4 What is the relationship between the KP for the
two reactions 3>2H21g2 + 1>2N21g2 S NH31g2 and
3H21g2 + N21g2 S 2NH31g2?
Q6.5 Under what conditions is dA … 0 a condition that
defines the spontaneity of a process?
Q6.6 By invoking the pressure dependence of the chemical
potential, show that if a valve separating a vessel of pure A
from a vessel containing a mixture of A and B is opened, mixing will occur. Both A and B are ideal gases, and the initial
pressure in both vessels is 1 bar.
Q6.7 Under what conditions is dG … 0 a condition that
defines the spontaneity of a process?
Q6.8 Is the equation 10U>0V2T = 1aT - kTP2>kT valid for
liquids, solids, and gases?
Q6.9 Why is it reasonable to assume that the chemical potential of a pure liquid or solid substance is independent of the
pressure in considering chemical equilibrium?
Q6.10 The reaction A + B S C + D is at equilibrium for
j = 0.5. What does this tell you about the ∆ f G ~ for reactants
and products?
Q6.11 Which thermodynamic state function gives a measure
of the maximum electrical work that can be carried out in a
fuel cell?
Questions Q6.12–Q6.17 refer to the reaction system
CO 1 g2 + 1 , 2O2 1 g2 u CO2 1 g2 at equilibrium for which
𝚫 r H ~ = −283.0 kJ mol −1.
Q6.12 Predict the change in the partial pressure of CO2 as
temperature is increased at constant total pressure.
Q6.13 Predict the change in the partial pressure of CO2 as
pressure is increased at constant temperature.
200
CHAPTER 6 Chemical Equilibrium
Q6.14 Predict the change in the partial pressure of CO2 as
Xe gas is introduced into the reaction vessel at constant
pressure and temperature.
Q6.15 Predict the change in the partial pressure of CO2 as
Xe gas is introduced into the reaction vessel at constant
volume and temperature.
Q6.16 Predict the change in the partial pressure of CO2 as a
platinum catalyst is introduced into the reaction vessel at
constant volume and temperature.
Q6.17 Predict the change in the partial pressure of CO2 as
O2 is removed from the reaction vessel at constant pressure
and temperature.
Q6.18 Is the equation ∆G = ∆H - T∆S applicable to all
processes?
Q6.19 Is the equation ∆A = ∆U - T∆S applicable to all
processes?
Q6.20 Under what conditions is Kx 7 KP?
Q6.21 Under what conditions is the distribution of products
in an ideal gas reactions system at equilibrium unaffected by
an increase in the pressure?
Q6.22 If KP is independent of pressure, why does the degree
of dissociation in the reaction Cl21g2 N 2Cl1g2 depend
on pressure?
Questions Q6.23–Q6.26 refer to the reaction system
H2 1 g2 + Cl2 1 g2 N 2HCl1 g2 at equilibrium. Assume ideal
gas behavior.
Q6.23 How does the total number of moles in the reaction
system change as T increases?
Q6.24 If the reaction is carried out at constant V, how does
the total pressure change if T increases?
Q6.25 Is the partial pressure of H21g2 dependent on T? If so,
how will it change as T decreases?
Q6.26 If the total pressure is increased at constant T, how
will the relative amounts of H21g2 and HCl1g2 change?
Q6.27 If additional Cl21g2 is added to the reaction system
at constant V and T, how will the degree of dissociation of
HCl1g2 change?
Q6.28 If T is increased at constant total pressure, how will
the degree of dissociation of HCl1g2 change?
Q6.29 Which of the following data provide sufficient information to calculate KP at an arbitrary temperature T? Assume
ideal gas behavior and that the heat capacities of reactants and
products are independent of temperature.
a. ∆ r H ~ 1298.15 K2 and ∆ r G ~ 1298.15 K2
b. ∆ r H ~ 1T2 and ∆ r G ~ 1298.15 K2
c. ∆ r G ~ 1T2
d. ∆ r H ~ 1298.15 K2 and KP1298.15 K2
e. KP1T12 and KP1T22 where T1, T2 ≠ T
Q6.30 Consider a gas-phase reaction for which
∆m1T2 = 15.5 kJ mol-1 and ∆ r H ~ = 21.0 kJ mol-1 at
350. K. If the following statements are true, explain why. If
they are false, rewrite them to make them true.
a. The reaction is spontaneous at 350. K.
b. The reaction becomes spontaneous at temperatures well
below 350. K.
NUMERICAL PROBLEMS
Section 6.1
P6.1 Collagen is the most abundant protein in the mammalian
body. It is a fibrous protein that serves to strengthen and support
tissues. Suppose a collagen fiber can be stretched reversibly
with a force constant of k = 14.0 N m-1 and that the force
F (see Table 2.1) is given by F = -k ∆l. When a collagen
fiber is contracted reversibly, it absorbs heat qrev = 0.015 J.
Calculate the change in the Helmholtz energy, ∆A, as the fiber
contracts isothermally from l = 0.032 m to 0.027 m. Also
calculate the reversible work performed wrev, ∆S, and ∆U.
Assume that the temperature is constant at T = 310. K.
P6.2 Calculate the maximum nonexpansion work that can be
gained from the combustion of benzene(l) and of H21g2 on a
per gram and a per mole basis under standard conditions.
Is it apparent from this calculation why fuel cells based on
H2 oxidation are under development for mobile applications?
C6H61l2 + 15>2O21g2 4 6CO21g2 + 3H2O1l2
P6.3 A hard-working horse can lift a 150.0 kg weight 30. m
in one minute. Assume the horse generates energy to accomplish this work by metabolizing glucose:
C6H12O61s2 + 6O21g2 S 6CO21g2 + 6H2O1l2
Calculate how much glucose a horse must metabolize to sustain this rate of work for one hour. Assume T = 298.15 K.
P6.4 Assume the internal energy of an elastic fiber under
tension is given by dU = T dS - P dV - F d/. Obtain an
expression for 10G>0/2P,T and calculate the maximum
nonexpansion work obtainable when a collagen fiber with a
force constant of k = 15.0 N m-1 contracts from / = 20.0 to
10.0 cm at constant P and T.
Sections 6.2 and 6.3
P6.5 Calculate ∆G for the isothermal expansion of 1.25 mol
of an ideal gas at 285 K from an initial pressure of 16.0 bar to
a final pressure of 1.0 bar.
P6.6 As shown in Example Problem 3.5, 10Um >0V2T = a>V 2m
for a van der Waals gas. In this problem, you will compare the
change in energy with temperature and volume for N2, treating
it as a van der Waals gas.
a. Calculate ∆U per mole of N21g2 at 1 bar pressure and
298 K if the volume is increased by 5.00% at constant T.
Approximate the molar volume with the ideal gas value.
b. Calculate ∆U per mole of N21g2 at 1 bar pressure and 298 K
if the temperature is increased by 5.00% at constant V.
Numerical Problems
c. Calculate the ratio of your results in part (a) to the result in
part (b). What can you conclude about the relative importance of changes in temperature and volume on ∆U?
P6.7 Calculate ∆A for the isothermal compression of 1.25 mol
of an ideal gas at 285 K from an initial volume of 25.0 L to a
final volume of 4.0 L. Does it matter whether the path is reversible or irreversible?
P6.8 Derive an expression for A1V,T2 analogous to that for
G1T,P2 in Equation (6.29).
P6.9 Show that
01A>T2
c
d = U
011>T2 V
Write an expression analogous to Equation (6.32) that would
allow you to relate ∆A at two temperatures.
P6.10 A sample containing 1.70 mol of an ideal gas at 310 K
is expanded from an initial volume of 8.5 L to a final volume
of 25.0 L. Calculate ∆G and ∆A for this process for (a) an
isothermal reversible path and (b) an isothermal expansion
against a constant external pressure of 2.50 bar. Explain why
∆G and ∆A do or do not differ from one another.
Sections 6.4 and 6.5
P6.11 Assume that a sealed vessel at constant pressure
of 1 bar initially contains 2.00 mol of NO21g2. The system is allowed to equilibrate with respect to the reaction
2 NO21g2 N N2O41g2. The number of moles of NO21g2 and
N2O41g2 at equilibrium is 2.00 - 2j and j, respectively,
where j is the extent of reaction.
a. Derive an expression for the entropy of mixing as a function of j.
b. Graphically determine the value of j for which ∆ mixS has
its maximum value.
c. Write an expression for Gpure as a function of j. Use
Equation (6.104) to obtain values of G m~ for NO2 and
N2O4.
d. Plot Gmixture = Gpure + ∆ mixG as a function of j for
T = 298 K and graphically determine the value of j for
which Gmixture has its minimum value. Is this value the
same as for part (b)?
P6.12 Calculate mmixture
1298.15 K, 1 atm2 for nitrogen in air,
N2
assuming that the mole fraction of N2 in air is 0.781.
P6.13 A sample containing 2.85 mol of He (1 bar, 325. K) is
mixed with 3.15 mol of Ne (1 bar, 325. K) and 1.25 mol of Ar
(1 bar, 325. K). Calculate ∆ mixG and ∆ mixS.
P6.14 A gas mixture with 2.0 mol of Ar, x mol of Ne,
and y mol of Xe is prepared at a pressure of 1 bar and a
­temperature of 298 K. The total number of moles in the
­mixture is four times that of Ar. Write an expression for
∆ mixG in terms of x. At what value of x does the magnitude
of ∆ mixG have its minimum value? Calculate ∆ mixG for this
value of x.
P6.15 You have containers of pure O2 and N2 at 298 K and
1 atm pressure. Calculate ∆ mixG relative to the unmixed gases of
a. a mixture of 12 mol of O2 and 6 mol of N2
201
b. a mixture of 12 mol of O2 and 12 mol of N2
c. Calculate ∆ mixG if 10 mol of pure N2 are added to the
mixture of 12 mol of O2 and 12 mol of N2.
Sections 6.6 and 6.7
P6.16 Nitrogen is a vital element for all living systems, but
except for a few types of bacteria, blue-green algae, and some
soil fungi, most organisms cannot utilize N2 from the atmosphere. The formation of “fixed” nitrogen is, therefore, necessary to sustain life, and the simplest form of fixed nitrogen is
ammonia NH3. A possible pathway for ammonia synthesis by
a living system is:
1>2N21g2 + 3>2H2O1l2 S NH31aq2 + 3>2O21g2
where (aq) means the ammonia is dissolved in water.
∆ f G ~ 1NH3, aq2 = -80.3 kJ mol - 1.
a. Calculate ∆ r G ~ for the biological synthesis of ammonia at
298 K.
b. Calculate the equilibrium constant for the biological
synthesis of ammonia.
c. Based on your answer to part (b), does the biological
synthesis of ammonia occur spontaneously?
P6.17 Consider the reaction system
N21g2 + 3H21g2 N 2NH31g2. Assume that initially 3.00 mol
of N21g2, 2.00 mol of H21g2, and 1.00 mol of NH31g2 are
present in a reaction vessel at a constant pressure of 0.50 bar.
a. If z is the extent of reaction, write expressions for the
number of mols and the mole fractions of N21g2, H21g2,
and NH31g2 present at equilibrium.
b. Calculate the partial pressures of all species present at
equilibrium if z = 0.30.
c. What value of z corresponds to the reaction going to
completion?
P6.18 Calculate ∆ r A~ and ∆ r G ~ for the reaction C6H121l2
+ 9O21g2 S 6CO21g2 + 6H2O1l2 at 298 K from the combustion enthalpy of cyclohexane and the entropies of the
reactants and products. All gaseous reactants and products are
treated as ideal gases.
P6.19 In Example Problem 6.11, KP for the reaction
CO1g2 + H2O1l2 S CO21g2 + H21g2 was calculated
to be 3.32 * 103 at 298.15 K. At what temperature does
KP = 8.25 * 103? What is the highest value that KP can have
by changing the temperature? Assume that ∆H R~ is independent of temperature.
Sections 6.8 and 6.9
P6.20 A sample containing 3.50 mol of N2 and 7.50 mol of
H2 is placed in a reaction vessel and brought to equilibrium at
52.0 bar and 555 K in the reaction 1>2 N21g2 + 3>2 H21g2 S
NH31g2.
a. Calculate KP at this temperature.
b. Set up an equation relating KP and the extent of reaction as
in Example Problem 6.12.
c. Using a numerical equation solver, calculate the number of
moles of each species present at equilibrium.
202
CHAPTER 6 Chemical Equilibrium
P6.21 You place 1.65 mol of NOCl1g2 in a ­reaction vessel.
Equilibrium at constant pressure is ­established with respect to
the decomposition reaction NOCl1g2 N NO1g2 + 1>2 Cl21g2
a. Derive an expression for KP in terms of the extent of
­reaction j.
b. Simplify your expression for part (a) in the limit that j is
very small.
c. Calculate j and the degree of dissociation of NOCl in the
limit that j is very small at 350. K and a pressure of 1.50 bar.
d. Solve the expression derived in part (a) using a numerical
equation solver for the conditions stated in the previous
part. What is the relative error in j made using the approximation of part (b)?
P6.22 Calculate the degree of dissociation of N2O4 in the
reaction N2O41g2 S 2NO21g2 at 325 K and a total pressure of
1.00 bar. Do you expect the degree of dissociation to increase
or decrease as the temperature is increased to 550. K? Assume
that ∆ r H ~ is independent of temperature.
P6.23 Calculate KP at 298.15 K and 490 K for the reaction
NO1g2 + 11>22O21g2 S NO21g2 assuming that ∆ r H ~ is
constant over the interval 298–600. K. Do you expect KP to increase or decrease as the temperature is decreased to 273. K?
P6.24 Consider the equilibrium NO21g2 L NO1g2 +
1>2O21g2. One mole of NO21g2 is placed in a vessel and
allowed to come to equilibrium at a total pressure of 1 bar. An
analysis of the contents of the vessel gives the following results:
T
PNO >PNO2
700. K
0.730
800. K
2.11
a. Calculate KP at 700. and 800. K.
b. Calculate ∆ r G ~ and ∆ r H ~ for this reaction at 298.15 K
using only the data in the problem. Assume that ∆ r H ~ is
independent of temperature.
c. Calculate ∆ r G ~ and ∆ r H ~ using the data tables and compare your answer with that obtained in part (b).
P6.25 Consider the equilibrium C2H61g2 L C2H41g2 +
H21g2. At 900 K and a constant total pressure of 1.00 bar,
C2H61g2 is introduced into a reaction vessel. The total pressure is held constant at 1.00 bar, and at equilibrium the composition of the mixture in mole percent is H21g2: 13.427%,
C2H41g2: 13.427%, and C2H61g2: 73.146%.
a. Calculate KP at 900 K.
b. If ∆ r H ~ = 136.4 kJ mol-1, calculate the value of KP at
298.15 K.
c. Calculate ∆ r G ~ for this reaction at 298.15 K.
P6.26 Many biological macromolecules undergo a transition called denaturation. Denaturation is a process whereby a
structured, biological active molecule, called the native form,
unfolds or becomes unstructured and biologically inactive.
The equilibrium is
native1folded2 N denatured1unfolded2
For a protein at pH = 2 and T = 298.15 K the enthalpy
change associated with denaturation is ∆H ~ = 418.0 kJ mol-1,
and the entropy change is ∆S ~ = 1.30 kJ K-1 mol-1.
a. Calculate the Gibbs energy change for the denaturation
of the protein at pH = 2 and T = 310 K. Assume the enthalpy and entropy are temperature independent between
298.15 K and 310 K.
b. Calculate the equilibrium constant for the denaturation of
protein at pH = 2 and T = 310 K.
c. Based on your answer for parts (a) and (b), is the protein
structurally stable at pH = 2 and T = 10 K?
P6.27 Calculate KP at 490 K for the reaction N2O41l2 S
2NO21g2 assuming that ∆ r H ~ is constant over the interval
298.15–725 K.
P6.28 Consider the equilibrium CO1g2 + H2O1g2 L
CO21g2 + H21g2. At 1000. K, the composition of the reaction mixture is
Substance
CO2 1 g2
Mole %
H2 1 g2
24.2
24.2
CO 1 g2
25.8
H2O 1 g2
25.8
a. Calculate KP and ∆ r G at 1150. K.
b. Given the answer to part (a), use the ∆ f H ~ of the reaction
species to calculate ∆ r G ~ at 298.15 K. Assume that ∆ r H ~
is independent of temperature.
P6.29 For the reaction C1graphite2 + H2O1g2 N CO1g2 +
H21g2, ∆ r H ~ = 131.28 kJ mol-1 at 298.15 K. Use the values
of CP,m at 298.15 K in the data tables to calculate ∆ r H ~ at
210.0°C.
~
P6.30 You wish to design an effusion source for Br atoms
from Br21g2. If the source is to operate at a total pressure of
3.00 Torr, what temperature is required to produce a degree
of dissociation of 0.150? What value of the pressure would
increase the degree of dissociation to 0.500 at this temperature?
P6.31 Consider the reaction FeO1s2 + CO1g2 L Fe1s2 +
CO21g2 for which KP is found to have the following values:
T
KP
700.°C
0.688
1200.°C
0.310
a. Calculate ∆ r G ~ , ∆ r S ~ and ∆ r H ~ for this reaction at
700.°C. Assume that ∆ r H ~ is independent of temperature.
b. Calculate the mole fraction of CO21g2 present in the gas
phase at 700.°C.
P6.32 Assuming that ∆ r H ~ is independent of temperature,
calculate ∆ r G ~ for the process 1H2O, g, 298.15 K2 S
1H2O, g, 690.0 K2. Calculate the relative change in ∆ f G ~ for
H2O1g2over the temperature interval.
P6.33 Oxygen reacts with solid glycylglycine C4H8N2O3 to
form urea CH4N2O, carbon dioxide, and water:
3O21g2 + C4H8N2O31s2 S CH4N2O1s2 + 3CO21g2 +
2H2O1l2.
At T = 298 K and 1.00 atm solid glycylglycine has the
following thermodynamic properties:
∆ f G ~ = -491.5 kJ mol-1,
S
~
-1
= 190.0 J K
-1
∆ f H ~ = -746.0 kJ mol-1,
mol .
Calculate ∆ r G ~ at T = 298 K and at T = 310. K. State any
assumptions that you make.
Numerical Problems
P6.34 Calculate ∆ r G ~ for the reaction CO1g2 + 1>2 O21g2 S
CO21g2 at 298.15 K. Calculate ∆ r G ~ at 685. K, assuming
that ∆ r H ~ is constant in the temperature interval of interest.
P6.35 In this problem, you calculate the error in assuming
that ∆ r H ~ is independent of T for the reaction 2CuO 1s2 L
2Cu1s2 + O21g2.
The following data are given at 25°C:
Compound
-1
∆ f H 1kJ mol 2
~
∆ f G ~ 1kJ mol-12
-1
CP,m1J K
CuO 1 s 2
-157
a. From Equation (6.69)
O21g2
-130.
-1
mol 2
Cu 1 s 2
42.3
KP1Tƒ2
L
24.4
Tƒ
d ln KP =
KP1Ti2
29.4
∆r H ~
1
dT.
R L T2
T0
To a good approximation, we can assume that the heat
capacities are independent of temperature over a limited
range in temperature, giving ∆ r H ~ 1T2 = ∆ r H ~ 1Ti2 +
∆CP1T - Ti2 where ∆CP = a ni CP,m1i2. By integrating
i
Equation (6.69), show that
ln KP1T2 = ln KP1Ti2 +
∆ r H ~ 1Ti2 1
1
a - b
R
T
Ti
Ti * ∆CP 1
∆CP T
1
a - b +
ln
R
T
Ti
R
Ti
b. Using the result from part (a), calculate the equilibrium
pressure of oxygen over copper and CuO1s2 at 1100. K.
How is this value related to KP for the reaction
2CuO1s2 N 2Cu1s2 + O21g2?
c. What value would you obtain if you assumed that ∆ r H ~
were constant at its value for 298.15 K up to 1100. K?
P6.36 Consider the equilibrium in the reaction
3O21g2 N 2O31g2, with ∆ r H ~ = 285.4 * 103 J mol-1
at 298 K. Assume that ∆ r H ~ is independent of temperature.
a. Without doing a calculation, predict whether the equilibrium position will shift toward reactants or products as the
pressure is increased.
b. Without doing a calculation, predict whether the equilibrium position will shift toward reactants or products as the
temperature is increased.
c. Calculate KP at 800. and 900. K.
d. Calculate Kx at 800. K and pressures of 1.00 and 2.50 bar.
P6.37 N2O3 dissociates according to the equilibrium
N2O31g2 N NO21g2 + NO1g2. At 275 K and one bar pressure, the degree of dissociation defined as the ratio of moles
of NO1g2 or NO21g2 to the moles of the reactant assuming
that no dissociation occurs is 0.486. Calculate ∆ r G ~ for this
reaction at 275 K.
P6.38 If the reaction Fe 2N1s2 + 3>2H21g2 L
2Fe1s2 + NH31g2 comes to equilibrium at a total pressure
of 1 bar, analysis of the gas shows that at 700. and 800. K,
PNH3 >PH2 = 1.971 and 0.9780, respectively, if only H21g2
203
was initially present in the gas phase and Fe 2N1s2 was
in excess.
a. Calculate KP at 700. and 800. K.
b. Calculate ∆ r S ~ at 700. K and 800. K and ∆ r H ~ , assuming
that it is independent of temperature.
c. Calculate ∆ r G ~ for this reaction at 298.15 K.
P6.39 Under anaerobic conditions, glucose is broken down
in muscle tissue to form lactic acid according to the reaction
C6H12O61s2 S 2CH3CHOHCOOH1s2. Thermodynamic data
at T = 298.15 K for glucose and lactic acid are given in the table.
Glucose(s)
Lactic
acid(s)
𝚫 r H ~ 1kJ mol −1 2 CP,m 1J K−1 mol −1 2 S ~ 1J K−1 mol −1 2
-1273.1
219.2
209.2
-673.6
127.6
192.1
Calculate ∆ r G ~ at T = 298.15 K and T = 310 K using the
data in the table. Assume that all heat capacities are constant
in this temperature interval and that ∆ r H ~ and ∆ r S ~ are constant in this temperature interval.
P6.40 Consider the equilibrium 3O21g2 N 2O31g2.
a. Using the data tables, calculate KP at 298 K.
b. Set up a table as in Example Problem 6.10 describing the
approach to equilibrium. At equilibrium, the number of
moles of O21g2 and O31g2 are n0 - 3deq and 2deq, respectively. Assuming that the extent of reaction at equilibrium
is much less than one, show that the degree of reaction
defined as deq >n0 is given by 11>222KP * P>P ~ .
c. Calculate the degree of reaction at 500. K and a pressure of
5.00 bar.
d. Calculate Kx at 500. K and a pressure of 5.00 bar.
d ln KP
=
P6.41 As shown in Equation (6.64),
dT
∆r H ~
1 d1∆ r G ~ >T2
=
. For most cases, it is reasonable
R
dT
RT 2
to assume that ∆ r H ~ is independent of temperature in calculating the temperature dependence of KP. However, a more
accurate calculation takes the temperature dependence of
∆ r H ~ into account. As listed in the data tables, CP,m for
individual reactants and products can be expressed in the form
CP,m = A112 + A122T + A132T 2 + A142T 3, and the difference in the heat capacity between products can be expressed
as ∆CP,m = ∆A112 + ∆A122T + ∆A132T 2 + ∆A142T 3.
a. Show that ∆ r H ~ 1T2 = ∆A112T +
∆A122T 2
+
2
∆A132T 3
∆A142T 4
+
+ B where B is a constant of
3
4
integration that can be evaluated as ∆ r H ~ can be calculated from standard enthalpies of formation at 298.15 K.
b. Show that ln KP1T2 =
∆A112
∆A122T
ln T +
+
R
2R
∆A132T 2
∆A142T 3
+
+ C where C is a constant of
6R
12R
integration that can be evaluated if KP is known at a given
temperature.
204
CHAPTER 6 Chemical Equilibrium
Section 6.10
P6.42 The shells of marine organisms are constructed of
biologically secreted calcium carbonate CaCO3, largely in a
crystalline form known as calcite. There is a second crystalline form of calcium carbonate known as aragonite. Physical
and thermodynamic properties of calcite and aragonite are
given below.
Properties
1 T = 298 K, P = 1.00 bar2
Calcite
Aragonite
-1206.9
-1207.0
-1128.8
-1127.7
S ~ 1J K-1 mol-12
92.9
88.7
81.9
81.3
Density 1g mL 2
2.710
2.930
∆ f H ~ 1kJ mol-12
-1
∆ f G 1kJ mol 2
~
-1
CP,m1J K
-1
mol 2
-1
a. Based on the thermodynamic data given, would you
expect an isolated sample of calcite at T = 298 K and
P = 1.00 bar to convert to aragonite, given sufficient
time? Explain.
b. Suppose the pressure applied to an isolated sample of
calcite is increased. Can the pressure be increased to the
point that isolated calcite will be converted to aragonite?
Explain.
c. What pressure must be achieved to induce the conversion
of calcite to aragonite at T = 298 K. Assume both calcite
and aragonite are incompressible at T = 298 K.
d. Can calcite be converted to aragonite at P = 1.00 bar if
the temperature is increased? Explain.
P6.43 The pressure dependence of G is quite different for
gases and condensed phases. Calculate ∆Gm for the processes
(C, solid, graphite, 1 bar, 298.15 K) S (C, solid, graphite,
200. bar, 298.15 K) and (He, g, 1 bar, 298.15 K) S (He, g,
200. bar, 298.15 K). By what factor is ∆Gm greater for He
than for graphite?
P6.44 Ca1HCO3221s2 decomposes at elevated temperatures
according to the stoichiometric equation Ca1HCO3221s2 S
CaCO31s2 + H2O1g2 + CO21g2.
a. If pure Ca1HCO3221s2 is put into a sealed vessel, the air
is pumped out, and the vessel and its contents are heated,
the total pressure is 0.150 bar. Determine KP under these
conditions.
b. If the vessel also contains 0.150 bar H2O1g2 at the final
temperature, what is the partial pressure of CO21g2 at
equilibrium?
Section 6.12
P6.45 Use the equation CP,m - CV,m = TVmb2 >kT and the
data tables to determine CV,m for H2O1l2 at 310 K. Calculate
1CP,m - CV,m2>CP,m.
FURTHER READING
Dahl. S., Logadottir, A., Egeberg, R. C., Larsen, J. H.,
Chorkendorff, I., Tornqvist, E., and Norskov, J. K. “Role
of Steps in N2 Activation on Ru(0001).” Physical Review
Letters 83 (1999): 1814–1817.
Ertl, G. “Surface Science and Catalysis—Studies on the
Mechanism of Ammonia Synthesis: The P.H. Emmett
Award Address.” Catalysis Reviews—Science and
Engineering 21 (1980): 201–223.
Gislason, E. A., and Craig, N. C. “Criteria for Spontaneous
Processes Derived from the Global Point of View.”
Journal of Chemical Education 90 (2013): 584–590.
Hwang, R. Q., Schroder, J., Gunther, C., and Behm, R. J.
“Fractal Growth of Two-Dimensional Islands: Au on
Ru(0001).” Physical Review Letters 67 (1991):
3279–3283.
Jacobsen, C. J. H., Dahl, S., Clausen, B. S., Bahn, S.,
Logadottir, A., and Norskov, J. K. “Catalyst Design by
Interpolation in the Periodic Table: Bimetallic Ammonia
Synthesis Catalysts.” Journal of the American Chemical
Society 123 (2001): 8404–8405.
Liphardt, J., Onoa, B., Smith, S., Tinoco, I., and Bustamante,
C. “Reversible Unfolding of Single RNA Molecules by
Mechanical Force.” Science, 292 (2001): 733–737.
Raff, L. M. “Spontaneity and Equilibrium: Why ‘∆G 6 0
Denotes a Spontaneous Process’ and ∆G = 0 Means
the System Is at Equilibrium Are Incorrect.” Journal of
Chemical Education 91 (2014a): 386–395.
Raff, L. M. “Spontaneity and Equilibrium III: A History of
Misinformation.” Journal of Chemical Education
91 (2014b): 2128–2136.
C H A P T E R
7
The Properties of Real Gases
WHY is this material important?
The ideal gas law is only accurate for gases at low values of density. To design production plants that use real gases at high pressures, equations of state valid for gases at
higher densities are needed. Such equations must take into account the finite volume
of a molecule and the intermolecular potential.
7.1
Real Gases and Ideal Gases
7.2
Equations of State for Real
Gases and Their Range of
Applicability
7.3
The Compression Factor
7.4
The Law of Corresponding
States
7.5
Fugacity and the Equilibrium
Constant for Real Gases
WHAT are the most important concepts and results?
Equations of state for real gases accurately describe the P–V relationship of a given
gas at a fixed value of T within their range of validity. These equations use parameters
specific to a given gas. An important consequence of nonideality is that the chemical
potential of a real gas must be expressed in terms of its fugacity rather than its partial
pressure. Fugacities, rather than pressures, must also be used in calculating the thermodynamic equilibrium constant Kp for a real gas at elevated pressures.
WHAT would be helpful for you to review for this chapter?
It would be useful to review the material on ideal gases in Chapter 1.
7.1 REAL GASES AND IDEAL GASES
To this point, the ideal gas equation of state has been assumed to be sufficiently accurate to describe the P–V–T relationship for a real gas. This assumption has allowed
calculations of expansion work and of the equilibrium constant KP in terms of partial
pressures using the ideal gas law. In fact, the ideal gas law provides an accurate description of the P–V–T relationship for many gases, such as He, for a wide range of P,
V, and T values. However, it describes the P–V relationship for water for a wide range
of P and V values within {10, only for T 7 1300 K, as shown in Section 7.4. What
is the explanation for this different behavior of He and H2O? Is it possible to derive a
“universal” equation of state that can be used to describe the P–V relationship for gases
as different as He and H2O?
In Section 1.5, the two main deficiencies in the microscopic model on which the
ideal gas law is based were discussed. The first assumption is that gas molecules are
point masses. However, molecules occupy a finite volume; therefore, a real gas cannot
be compressed to a volume that is less than the total molecular volume. The second assumption that was made is that the molecules in the gas do not interact; but molecules
in a real gas do interact with one another through a potential as depicted in Figure 1.9.
Because the potential has a short range, its effect is negligible at low densities, which
correspond to large distances between molecules. Additionally, at low densities, the
molecular volume is negligible compared with the volume that the gas occupies. Therefore, the P–V–T relationship of a real gas is the same as that for an ideal gas at sufficiently low densities and high temperatures. At higher densities and low temperatures,
molecular interactions cannot be neglected. Because of these interactions, the pressure
Concept
Molecules of real gases have a finite
molecular volume and interact with
one another.
205
206
CHAPTER 7 The Properties of Real Gases
of a real gas can be higher or lower than that for an ideal gas at the same density and
temperature. What determines which of these two cases applies? The questions raised
in this section are the major themes of this chapter.
7.2 EQUATIONS OF STATE FOR REAL GASES
AND THEIR RANGE OF APPLICABILITY
This section discusses several equations of state for real gases and the range of the variables P, V, and T over which they accurately describe a real gas. Such equations of state
must exhibit a limiting P–V–T behavior identical to that for an ideal gas at low density.
They must also correctly model the deviations for ideal gas behavior that real gases
exhibit at moderate and high densities. The first two equations of state considered here
include two parameters, a and b, that must be experimentally determined for a given
gas. The parameter a is a measure of the strength of the attractive part of the intermolecular potential, and b is a measure of the minimum volume that a mole of molecules
can occupy. Real gas equations of state are best viewed as empirical equations whose
functional form has been chosen to fit experimentally determined P–V–T data.
The most widely used equation for real gases is the van der Waals equation of state:
P =
RT
a
nRT
n2a
- 2 =
- 2
Vm - b V m
V - nb
V
(7.1)
A second useful equation of state is the Redlich–Kwong equation of state:
P =
Concept
All equations of state for real gases
have a limited range of P, V, and T
over which they are applicable.
1
1
nRT
n2a
RT
a
=
V - nb
Vm - b
2T V1V + nb2
2T Vm1Vm + b2
(7.2)
Although the same symbols are used for parameters a and b in both equations of state,
they have different values for a given gas.
Figure 7.1 shows that the degree to which the ideal gas, van der Waals, and Redlich–
Kwong equations of state correctly predict the P–V behavior of CO2 depends on P, V,
and T. At 426 K, all three equations of state reproduce the correct P–V behavior reasonably well over the range shown, with the ideal gas law having the largest error.
By contrast, the three equations of state give significantly different results at 310. K.
The ideal gas law gives unacceptably large errors, and the Redlich–Kwong equation
of state is more accurate than is the van der Waals equation. We will have more to say
about the range over which the ideal gas law is reasonably accurate when discussing
the compression factor in Section 7.3.
A third widely used equation of state for real gases is the Beattie–Bridgeman
equation of state. The form of the equation is
P =
RT
c
A
a1 b 1Vm + B2 - 2
V 2m
Vm
VmT 3
A = A0 a1 -
a
b
Vm
and B = B0 a1 -
with
b
b
Vm
(7.3)
This equation uses five experimentally determined parameters to fit P–V–T data.
­ ecause of its complexity, it will not be discussed further.
B
A further important equation of state for real gases has a different form than any of
the previous equations. The virial equation of state is written in the form of a power
series in 1>Vm
P = RT c
B1T2
1
+
+ gd
Vm
V 2m
(7.4)
The power series does not converge at high pressures where Vm becomes small. The
B1T2, C1T2, and so on are called the second, third, and so on virial coefficients. This
equation is more firmly grounded in theory than the previously discussed three equations because a series expansion is always valid in its convergence range. In practical
7.2 Equations of State for Real Gases and Their Range of Applicability
207
Figure 7.1
Comparison of real gas equations of
state to ideal gas equation of state.
Isotherms for CO21g2 are shown at
(a) 426 K and (b) 310 K. The purple curve
is calculated using the van der Waals equation of state, the blue curve is calculated
using the Redlich–Kwong equation of
state, and the red curve is calculated using
the ideal gas equation of state. The black
dots are accurate values taken from the
NIST Chemistry Webbook, which are derived from experimental results. Note that
in (a) the purple curve is mostly obscured
by the blue curve.
140
Pressure (bar)
120
100
T 5 426 K
80
60
40
20
0
0.25
0.50
0.75
1.00
Molar volume (L)
1.25
1.50
1.75
(a)
Pressure (bar)
500
400
300
T 5 310. K
200
100
0
0.1
0.2
0.3
Molar volume (L)
0.4
0.5
0.6
(b)
use, the series is usually terminated after the second virial coefficient because values
for the higher coefficients are not easily obtained from experiments. Table 7.1 (see Appendix A, Data Tables) lists values for the second virial coefficient for selected gases
for different temperatures. If B1T2 is negative, the attractive part of the potential dominates at that value of T. In contrast, if B1T2 is positive, the repulsive part of the potential
dominates at that value of T. Statistical thermodynamics can be used to relate the virial
coefficients with the intermolecular potential function. As you will show in the end-ofchapter problems, B1T2 for a van der Waals gas is given by B1T2 = b - 1a>RT2.
The principal limitation of the ideal gas law is that it does not predict that a gas can
be liquefied under appropriate conditions. Consider the approximate P–V diagram for
CO2 shown in Figure 7.2. (It is approximate because it is based on the van der Waals
equation of state rather than experimental data.) Each of the curves is an isotherm (that
is, each corresponds to a fixed temperature). The behavior predicted by the ideal gas
law is shown for T = 334 K.
To understand the behavior indicated by diagram in Figure 7.2, consider the isotherm for T = 258 K. Starting at large values of Vm, at first as Vm decreases the pressure increases. Then, with a further decrease in Vm, P is constant over a range of values
of Vm. The value of Vm at which P becomes constant depends on T. As Vm decreases
further, the pressure suddenly increases rapidly.
The reason for this unusual dependence of P on Vm becomes clear when the CO2
compression experiment is carried out in the piston and transparent cylinder assembly
shown in Figure 7.3. The system consists of either a single phase or two phases separated
by a sharp interface, depending on the values of T and Vm. For points a, b, c, and d on
208
CHAPTER 7 The Properties of Real Gases
Figure 7.2
140
365 K
120
334 K Ideal gas
334 K
Pressure (bar)
Calculated isotherms for CO2
modeled as a van der Waals gas. The
gas (green) and liquid (blue) regions
and the gas–liquid (yellow) coexistence
region are shown. The dashed curve was
calculated using the ideal gas law. The
isotherm at T = 304.12 K is at the critical temperature and is called the critical
isotherm.
Gas
100
80
60
304.12 K
Liquid
274 K
d
40
c
258 K
b
243 K
20
0
Concept
Ideal gases do not condense to form
liquids or solids.
Concept
Real gases have two-phase
coexistence regions.
Gas 1 Liquid
0.1
0.2
0.3
0.4
Molar volume (L mol–1)
0.5
0.6
the 258 K isotherm of Figure 7.2, the system has the following composition: at point
a: the system consists entirely of CO21g2. However, at points b and c, a sharp interface separates CO21g2 and CO21l2. Along the line linking points b and c, the system
contains CO21g2 and CO21l2 in equilibrium with one another. The proportion of liquid
to gas changes, but the pressure remains constant. The temperature-dependent equilibrium pressure is called the vapor pressure of the liquid. As the volume is decreased
further, the system becomes a single-phase system again, consisting of CO21l2 only
as at point d. The pressure changes slowly with increasing V if Vm 7 0.33 L because
CO21g2 is quite compressible. However, the pressure changes rapidly with decreasing
V if Vm 6 0.06 L because CO21l2 is nearly incompressible. The single-phase regions
and the two-phase gas–liquid coexistence region are shown in Figure 7.2.
If the same experiment is carried out at successively higher temperatures, it is
found that the range of Vm in which two phases are present becomes smaller, as seen in
the 243, 258, and 274 K isotherms of Figure 7.2. The temperature at which the range
of Vm has shrunk to a single value is called the critical temperature, Tc. For CO2,
Tc = 304.12 K. At T = Tc, the isotherm exhibits an inflection point so that
a
W7.1 van der Waals equation of
state isotherms
a
0P
02P
b
= 0 and a 2 b
= 0
0Vm T = Tc
0V m T = Tc
(7.5)
What is the significance of the critical temperature? Although critical behavior will
be discussed in Chapter 8, at this point it is sufficient to know that as the critical point is
approached along the 304.12 K isotherm, the density of CO21l2 decreases and the density of CO21g2 increases, and at T = Tc the densities are equal. At temperatures greater
than Tc, no interface is observed in the experiment depicted in Figure 7.3, and liquid
and gas phases can no longer be distinguished. Instead, the term supercritical fluid
is used to describe the state of CO2. As will be discussed in Chapter 8, Tc and the corresponding values Pc and Vc, which together are called the critical constants, take on
particular significance in describing the phase diagram of a pure substance. The critical
constants for a number of different substances are listed in Table 7.2 (see Appendix A,
Data Tables).
The parameters a and b for the van der Waals and Redlich–Kwong equations of
state are chosen so that the equation of state best represents real gas data. This can be
done by using the values of P, V, and T at the critical point Tc, Pc, and Vc, as shown in
Example Problem 7.1. Parameters a and b calculated from critical constants in this way
are listed in Table 7.4 (see Appendix A, Data Tables).
7.2 Equations of State for Real Gases and Their Range of Applicability
Figure 7.3
Pa
Pb
Pc
Pd
CO2(g)
CO2(g)
CO2(g)
CO2(l)
CO2(l)
Point c
Point d
CO2(l)
Point a
Point b
EXAMPLE PROBLEM 7.1
At T = Tc, 10P>0Vm2T = Tc = 0 and 102P>0V 2m2T = Tc = 0. Use this information to
­determine a and b in the van der Waals equation of state in terms of the experimentally determined values Tc and Pc.
Solution
°
°
0c
02 c
RT
a
- 2d
Vm - b
Vm ¢
RTc
2a
= + 3 = 0
2
0Vm
1Vmc - b2
V mc
T = Tc
RTc
2a
RT
a
0c + 3d
- 2d
2
Vm - b
1V
b2
Vm ¢
Vm ¢
°
m
=
0V
0V 2m
T = Tc
m
T = Tc
=
2RTc
1Vmc - b23
+
6a
= 0
V 4mc
Equating RTC from these two equations gives
RTc =
2a
3a
3
1Vmc - b22 = 4 1Vmc - b23, which simplifies to
1Vmc - b2 = 1
3
2V
V mc
V mc
mc
The solution to this equation is Vmc = 3b. Substituting this result into
10P>0Vm2T = Tc = 0 gives
-
209
RTc
1Vmc - b22
+
RTc
2a
2a
2a 12b22
8a
=
+
=
0
or
T
=
=
c
2
3
3
3 R
27Rb
12b2
V mc
13b2
13b2
Substituting these results for Tc and Vmc in terms of a and b into the van der Waals
equation gives the result Pc = a>27b2. We only need two of the critical constants Tc,
Pc, and Vmc to determine a and b. Because the measurements for Pc and Tc are more
accurate than those for Vmc, we use these constants to obtain expressions for a and b.
These results for Pc and Tc can be used to express a and b in terms of Pc and Tc. The
results are
b =
RTc
8Pc
and a =
27R2T 2c
64Pc
A similar analysis for the Redlich–Kwong equation gives
a =
R2T 5>2
c
9Pc121>3 - 12
and b =
121>3 - 12RTc
3Pc
Schematic diagram of CO2 volume
and phases present in a transparent
cylinder undergoing compression. The
experiment is conducted at a constant
temperature of 258 K, which corresponds
to the brown isotherm in Figure 7.2. Also,
cylinders with labeled points a, b, c, and
d correspond to the same labeled points
in Figure 7.2. The liquid and gas volumes
are not shown to scale. Note that Pb = Pc.
210
CHAPTER 7 The Properties of Real Gases
Concept
Real gases do not exhibit oscillations
predicted by van der Waals equation
of state.
140
Pressure (bar)
120
100
80
60
304 K
274 K
258 K
40
20
0
243 K
0.1 0.2 0.3 0.4 0.5 0.6
Molar volume (L mol21)
Figure 7.4
Van der Waals isotherms and their
­corresponding Maxwell constructions.
The calculated relationship between
pressure and molar volume for CO2 is
shown at the indicated temperatures. The
Maxwell constructions are shown in the
oscillating portion of isotherms. Note that
no oscillations in the calculated isotherms
occur for T Ú Tc.
Concept
The compression factor is a
convenient way to quantify
nonideality.
What is the range of validity of the van der Waals and Redlich–Kwong equations? No
two-parameter equation of state is able to reproduce the isotherms shown in Figure 7.2 for
T 6 Tc because it cannot reproduce either the range in which P is constant or the discontinuity in 10P>0V2T at the ends of this range. This failure is illustrated in Figure 7.4,
in which isotherms calculated using the van der Waals equation of state are plotted for
some of the values of T also used in Figure 7.2. Below Tc, all calculated isotherms have
an oscillating region that is unphysical because V increases as P increases. In a ­Maxwell
construction, the oscillating portion of an isotherm is replaced by the horizontal line
for which the areas above and below the line are equal, as indicated in Figure 7.4.
The Maxwell construction is used in generating the horizontal segments of isotherms
shown in Figure 7.2.
The Maxwell construction can be justified on theoretical grounds, but the equilibrium vapor pressure determined in this way for a given value of T is only in qualitative agreement with experiment. The van der Waals and Redlich–Kwong equations
of state do a good job of reproducing P–V isotherms for real gases only in the singlephase gas region T 7 Tc and for densities well below the critical density, rc = M>Vmc,
where Vmc is the molar volume at the critical point. The Beattie–Bridgeman equation, which has three more adjustable parameters, is accurate above Tc for higher
densities.
7.3 THE COMPRESSION FACTOR
How large is the error in P–V curves if the ideal gas law is used rather than the van der
Waals or Redlich–Kwong equations of state? To address this question, it is useful to
introduce the compression factor, z, defined by
z =
Vm
V ideal
m
=
PVm
RT
(7.6)
For the ideal gas, z = 1 for all values of P and Vm. If z 7 1, the real gas exerts a
greater pressure than the ideal gas, and if z 6 1, the real gas exerts a smaller pressure
than the ideal gas for the same values of T and Vm.
The compression factor for a given gas is a function of temperature, as shown in
Figure 7.5. In this figure, z has been calculated for N2 at two values of T using the van
der Waals and Redlich–Kwong equations of state. The dots are calculated from accurate values of Vm taken from the NIST Chemistry Workbook. Although the results
calculated using these equations of state are not in quantitative agreement with accurate
results for all P and T, the trends in the functional dependence of z on P for different
T values are correct. Because it is inconvenient to rely on tabulated data for individual
gases, we focus on z1P2 curves calculated from real gas equations of state. Both the
1.2
R-K 400. K
Plot of calculated compression factors
for N2 1 g2 as a function of pressure.
The calculated compression factors, z,
are shown for 200. and 400. K using the
Redlich–Kwong (red lines) and van der
Waals (purple lines) equations of state.
For comparison, calculated values of z are
also shown (colored dots and squares) for
the accurate values for Vm taken from the
NIST Chemistry Webbook. The Redlich–
Kwong equation does not give physically
meaningful solutions at P greater than
175 bar for 200. K, for which r = 2.5rc.
Compression factor (z)
Figure 7.5
1.1
vdW 400. K
1.0
50
100
0.9
R-K 200. K
0.8
vdW 200. K
150
200
250
300
Pressure (bar)
7.3 The Compression Factor
van der Waals and Redlich–Kwong equations predict that for T = 200 K, z initially decreases with pressure. The compression factor only becomes greater than the ideal gas
value of one for pressures in excess of 200 bar. For T = 400. K, z increases linearly
with T. This functional dependence is also predicted by the Redlich–Kwong equation.
The van der Waals equation predicts that the initial slope is zero, and z increases slowly
with P. For both temperatures, z S 1 as P S 0. This result shows that the ideal gas law
is obeyed if P is sufficiently small.
To understand why the low-pressure value of the compression factor varies with
temperature for a given gas, we use the van der Waals equation of state. Consider the
variation of the compression factor with P at constant T,
a
0z
0z
1
0z
b = a
b =
a
b
0P T
03 RT>Vm 4 T
RT 03 1>Vm 4 T
for a van der Waals gas in the ideal gas limit as 1>Vm S 0. As shown in Example
­Problem 7.2, the result 10z>0P2T = b>RT - a>1RT22 is obtained.
EXAMPLE PROBLEM 7.2
Show that the slope of z as a function of P as P S 0 is related to the van der Waals
parameters by
lim
a
S
p
0
0z
1
a
b =
ab b
0P T
RT
RT
Solution
Rather than differentiating z with respect to P, we transform the partial derivative to
one involving Vm:
z =
Vm
V ideal
m
=
PVm
=
RT
a
RT
a
- 2 b Vm
Vm - b
Vm
Vm
a
=
RT
Vm - b
RTVm
Vm
0z
0
a
c
d
0Vm
0Vm Vm - b
RTVm
0z
a b = ±
≤ = ±
≤
0P T
0P
0
RT
a
a
- 2b
0Vm T
0Vm Vm - b
Vm
T
Vm
1
a
+
Vm - b
1Vm - b22
RTV 2m
= ±
≤
RT
a
+ 2 3
1Vm - b22
Vm
Vm - b
Vm
a
+
2
2
1Vm - b2
1Vm - b2
RTV 2m
= ±
≤
RT
a
+ 2 3
1Vm - b22
Vm
-b
a
-b
a
+
+
2
2
2
1Vm - b2
RTV m
Vm
RTV 2m
= ±
≤ S±
≤ S
RT
a
RT
+
2
1Vm - b22
V 2m
V 3m
S
1
a
ab b as Vm S ∞
RT
RT
RT
a
RT
+ 2 3 S - 2 as Vm S ∞ because the sec1Vm - b22
Vm
Vm
ond term approaches zero more rapidly than the first term and Vm W b.
In the second to last line, -
W7.2 Quantitative comparison
between van der Waals and ideal
gas equations of state
211
212
CHAPTER 7 The Properties of Real Gases
TABLE 7.3 Boyle Temperatures
of Selected Gases
He
TB1K2
H2
110.
Ne
122
N2
327
CO
352
O2
400.
CH4
510.
Kr
575
Ethene
735
Gas
H2O
From Example Problem 7.2, the van der Waals equation predicts that the initial
slope of the z versus P curve is zero if b = a>RT. The corresponding temperature is
known as the Boyle temperature TB
TB =
23
1250
Source: Calculated from data in Lide, D. R.,
ed. CRC Handbook of Thermophysical and
Thermochemical Data. Boca Raton, FL: CRC
Press, 1994.
Concept
Real gases behave similarly to ideal
gases at the Boyle temperature.
a
Rb
(7.7)
Values for the Boyle temperature of several gases are shown in Table 7.3.
Because the parameters a and b are substance dependent, TB is different for each
gas. Using the van der Waals parameters for N2, TB = 425 K, whereas the experimentally determined value is 327 K. The agreement is qualitative rather than quantitative.
At the Boyle temperature, both z S 1, and 10z>0P2T S 0 as P S 0, which is the behavior exhibited by an ideal gas. It is only at T = TB that a real gas exhibits ideal behavior
as P S 0 with respect to lim
10z>0P2T . Above the Boyle temperature, 10z>0P2T 7 0
PS0
as P S 0, and below the Boyle temperature, 10z>0P2T 6 0 as P S 0. These inequalities provide a criterion to predict whether z increases or decreases with pressure at low
pressures for a given value of T.
Note that the initial slope of a z versus P plot is determined by the relative magnitudes of b and a>RT. Recall that the repulsive interaction is represented through b, and
the attractive part of the potential is represented through a. From the previous discussion,
lim 10z>0P2T is always positive at high temperatures because b - 1a>RT2 S b 7 0 as
PS0
T S ∞. This means that molecules primarily feel the repulsive part of the potential for
T W TB. Conversely, for T V TB, lim
10z>0P2T, will always be negative because the
PS0
molecules primarily feel the attractive part of the potential. For high enough values of
P, z 7 1 for all gases, showing that the repulsive part of the potential dominates at high
gas densities, regardless of the value of T.
Next consider the functional dependence of z on P for different gases at a single
temperature. Calculated values for oxygen, hydrogen, and ethene obtained using the
van der Waals equation at T = 400. K are shown in Figure 7.6, together with accurate results for these gases. It is seen that lim
10z>0P2T at 400. K is positive for
PS0
H2, negative for ethane, and approximately zero for O2. These trends are correctly
predicted by the van der Waals equation. How can this behavior be explained? It is
useful to compare the shape of the curves for H2 and ethene with those in Figure 7.5
for N2 at different temperatures. This comparison suggests that at 400. K the gas temperature is well above TB for H2 so that the repulsive part of the potential dominates.
Figure 7.6
Plot of compression factor as a function
of pressure at T = 400. K for three
different gases. The solid lines have
been calculated using the van der Waals
equation of state. The dots are calculated
from accurate values for Vm taken from
the NIST Chemistry Webbook. Red,
­hydrogen; purple, oxygen; and light blue,
ethene.
Compression factor (z)
1.2
Hydrogen
1.1
Oxygen
1.0
50
100
150
200
250
300
Pressure (bar)
0.9
Ethene
0.8
0.7
7.4 The Law of Corresponding States
By contrast, it suggests that 400. K is well below TB for ethene, and the attractive part
of the potential dominates. Because the initial slope is nearly zero, the data in Figure
7.6 suggest that 400. K is near TB for oxygen. As is seen in Table 7.3, TB = 735 K for
ethene, 400. K for O2, and 110. K for H2. These values are consistent with the previous explanation.
The results shown in Figure 7.6 can be generalized as follows. If lim
10z>0P2T 6 0
PS0
for a particular gas, T 6 TB, and the attractive part of the potential dominates.
If lim
10z>0P2T 7 0 for a particular gas, T 7 TB, and the repulsive part of the
PS0
potential dominates.
7.4 THE LAW OF CORRESPONDING STATES
W7.3 Compression factor for
a van der Waals gas
As shown in Section 7.3, the compression factor is a convenient way to quantify deviations from the ideal gas law. In calculating z using the van der Waals and Redlich–
Kwong equations of state, different parameters must be used for each gas. Is it possible
to find an equation of state for real gases that does not explicitly contain materialdependent parameters? Real gases differ from one another primarily in the value of the
molecular volume and in the magnitude of the attractive potential. Because molecules
that have a stronger attractive interaction exist as liquids to higher temperatures, one
might think that the critical temperature is a measure of the magnitude of the attractive
potential. Similarly, one might think that the critical volume is a measure of the molecular volume. If this is the case, different gases should behave similarly if T, V, and P
are measured relative to their critical values.
These considerations suggest the following hypothesis. Different gases have the
same equation of state if each gas is described by the dimensionless reduced variables
Tr = T>Tc, Pr = P>Pc, and Vmr = Vm >Vmc, rather than by T, P, and Vm. The preceding
statement is known as the law of corresponding states. If two gases have the same
values of Tr, Pr, and Vmr, they are in corresponding states. The values of P, V, and T
can be very different for two gases that are in corresponding states. For example, H2 at
12.93 bar and 32.98 K in a volume of 64.2 * 10-3 L and Br2 at 103 bar and 588 K in a
volume of 127 * 10-3 L are in the same corresponding state.
Next we justify the law of corresponding states and show that it is obeyed by many
gases. The parameters a and b can be eliminated from the van der Waals equation of
state by expressing the equation in terms of the reduced variables Tr, Pr, and Vmr. This
can be seen by writing the van der Waals equation in the form
PrPc =
RTrTc
a
- 2 2
VmrVmc - b
V mrV mc
(7.8)
Next replace Tc, Vmc, and Pc by the relations derived in Example Problem 7.1:
Pc =
a
8a
, Vmc = 3b, and Tc =
2
27Rb
27b
(7.9)
Equation (7.8) becomes
aPr
8aTr
a
- 2 2
27b13bVmr - b2
27b
9b V mr
8Tr
3
Pr =
- 2
3Vmr - 1
V mr
2
=
or
(7.10)
Equation (7.10) relates Tr, Pr and Vmr without reference to the parameters a and b.
­ herefore, it has the character of a universal equation, like the ideal gas equation
T
of state. To a good approximation, the law of corresponding states is obeyed for a
large number of different gases as shown in Figure 7.7, as long as Tr 7 1. However,
­Equation (7.10) is not really universal because the material-dependent quantities enter
through values of Pc, Tc, and Vc rather than through a and b.
213
214
CHAPTER 7 The Properties of Real Gases
Figure 7.7
Concept
1.0
Tr 5 2.00
Tr 5 1.50
Compression factor (z)
Values of compression factor plotted
as a function of the reduced pressure Pr
for seven gases. Compression factors calculated for the six values of reduced temperatures are indicated in the figure. The
solid curves are drawn to guide the eye.
0.8
Tr 5 1.30
Nitrogen
Tr 5 1.20
0.6
Methane
Propane
Tr 5 1.10
Ethane
Butane
0.4
Isopentane
The law of corresponding states
allows the compression factors for
very different gases to be estimated.
Ethene
Tr 5 1.00
0.2
1
4
5
3
Reduced pressure (Pr)
2
6
7
The law of corresponding states implicitly assumes that two parameters are sufficient to describe an intermolecular potential. This assumption is best for molecules that
are nearly spherical because for such molecules the potential is independent of the molecular orientation. It is not nearly as good for dipolar molecules such as HF, for which
the potential is orientation dependent.
The results shown in Figure 7.7 demonstrate the validity of the law of corresponding states. How can this law be applied to a specific gas? The goal is to calculate z for specific values of Pr and Tr and to use these z values to estimate the error in
using the ideal gas law. A convenient way to display these results is in the form of a
graph. Calculated results for z using the van der Waals equation of state as a function
of Pr for different values of Tr are shown in Figure 7.8. For a given gas and specific
P and T values, Pr and Tr can be calculated. A value of z can then be read from the
curves in Figure 7.8.
From Figure 7.8, we see that for Pr 6 5.5, z 6 1, as long as Tr 6 2. This means
that the real gas exerts a smaller pressure than an ideal gas in this range of Tr and Pr.
We conclude that for these values of Tr and Pr, the molecules are more influenced by
the attractive part of the potential than by the repulsive part that arises from the finite
molecular volume. However, z 7 1 for Pr 7 7 for all values of Tr as well as for all
1.6
Compression factor (z)
1.4
Figure 7.8
The compression factor plotted as a
function of Pr for selected values of Tr.
Values of Tr are indicated adjacent to the
curves. The curves were calculated using
the van der Waals equation of state.
1.2
4.0
1.0
2
0.8
1.3
0.6
1.2
1.1
0.4
1.0
2.0
1.7
1.5
1.4
4
6
8
10
Reduced pressure (Pr)
7.4 The Law of Corresponding States
values of Pr if Tr 7 4. Under these conditions, the real gas exerts a larger pressure than
an ideal gas. The molecules are influenced more by the repulsive part of the potential
than by the attractive part.
Using the compression factor, we can define the error in assuming that the pressure
can be calculated using the ideal gas law as
Error = 100, *
z - 1
z
(7.11)
Figure 7.8 shows that the ideal gas law is in error by less than 30% in the range of Tr
and Pr where the repulsive part of the potential dominates if Pr 6 8. However, the
error can be as great as -300% in the range of Tr and Pr where the attractive part of
the potential dominates. The error is greatest near Tr = 1 because at this value of the
reduced temperature, the liquid and gaseous phases can coexist. At or slightly above
the critical temperature, a real gas is much more compressible than an ideal gas. These
curves can be used to estimate the temperature range over which the molar volume for
H2O1g2 predicted by the ideal gas law is within 10% of the result predicted by the van
der Waals equation of state over a wide range of pressure. From Figure 7.8, this is true
only if Tr 7 2.0 or if T 7 1300 K.
EXAMPLE PROBLEM 7.3
Using Figure 7.8 and the data in Table 7.2 (see Appendix A, Data Tables), calculate
the volume occupied by 1.000 kg of CH4 gas at T = 230. K and P = 68.0 bar. Calculate V - Videal and the relative error in V if V were calculated from the ideal gas equation of state.
Solution
From Table 7.2, Tr and Pr can be calculated:
Tr =
230. K
68.0 bar
= 1.21 and Pr =
= 1.48
190.56 K
45.99 bar
From Figure 7.8, z = 0.63.
znRT
V =
=
P
0.63 *
1000. g
16.04 g mol-1
V - Videal
V - Videal
V
* 0.08314 L bar K-1mol-1 * 230. K
68.0 bar
11.0 L
= 11.0 L = -6.5 L
0.63
6.5 L
= = -58,
11.0 L
= 11.0L
Because the critical variables can be expressed in terms of the parameters a and b as
shown in Example Problem 7.1, the compression factor at the critical point can also be
calculated. For the van der Waals equation of state,
zc =
PcVc
1
a
27Rb
3
=
*
* 3b *
=
2
RTc
R
8a
8
27b
(7.12)
Equation (7.12) predicts that the critical compressibility is independent of the parameters a and b and should, therefore, have the same value for all gases. A comparison of
this prediction with the experimentally determined value of zc in Table 7.2 shows qualitative but not quantitative agreement. A similar analysis using the critical parameters
obtained from the Redlich–Kwong equation also predicts that the critical compression
factor should be independent of a and b. In this case, zc = 0.333. This value is in better agreement with the values listed in Table 7.2 than that calculated using the van der
Waals equation.
215
216
CHAPTER 7 The Properties of Real Gases
7.5 FUGACITY AND THE EQUILIBRIUM
CONSTANT FOR REAL GASES
As shown in the previous section, the pressure exerted by a real gas can be greater or
less than that for an ideal gas. We next discuss how this result affects the value of the
equilibrium constant for a mixture of reactive gases. For a pure ideal gas, the chemical
potential as a function of the pressure has the form (see Section 6.3)
mideal1T, P2 = m ~ 1T2 + RT ln
P
P~
(7.13)
To construct an analogous expression for a real gas, we write
mreal1T, p2 = m ~ 1T2 + RT ln
Real gas
Gm– Gm (298.15 K)
Ideal gas
4
2
3
5
Relative Pressure, P/P
\
\
1
f<P
Attractive part
of potential
dominates
f>P
Repulsive part
of potential
dominates
(7.14)
where the quantity ƒ is called the fugacity of the gas. The fugacity can be viewed as
the effective pressure that a real gas exerts. For densities corresponding to the attracideal
tive range of the intermolecular potential, G real
and ƒ 6 P. For densities corm 6 Gm
ideal
responding to the repulsive range of the intermolecular potential, G real
and
m 6 Gm
ƒ 7 P. These relationships are depicted in Figure 7.9.
The fugacity has the limiting behavior that ƒ S P as P S 0. The standard state of fugacity, denoted ƒ ~ , is defined as the value that the fugacity would have if the gas behaved
ideally at 1 bar pressure. This is equivalent to saying that ƒ ~ = P ~ . This standard state
is a hypothetical standard state because a real gas will not exhibit ideal behavior at a pressure of one bar. However, the standard state defined in this way makes Equation (7.14)
become identical to Equation (7.13) in the ideal gas limit ƒ S P as P S 0.
The preceding discussion provides a method to calculate the fugacity for a given gas.
How are the fugacity and the pressure related? For any gas, real or ideal, at constant T,
dGm = VmdP
(7.15)
Therefore,
dmideal = V ideal
m dP
Figure 7.9
The chemical potential for an ideal gas
and a real gas as a function of pressure.
For densities corresponding to the attractive range of the potential, P 6 Pideal.
ideal
Therefore, G real
and ƒ 6 P. The
m 6 Gm
inequalities are reversed for densities corresponding to the repulsive range of the
potential.
ƒ
ƒ~
dmreal = V real
m dP
V ideal
m
(7.16)
V real
m ,
Equation (7.16) shows that because
≠
the chemical potential of a real gas
will change differently with pressure than the chemical potential of an ideal gas. We
form the difference
real
dmreal - dmideal = 1V real
m - V m 2dP
(7.17)
and integrate Equation (7.17) from an initial pressure Pi to a final pressure P:
P
Concept
Fugacity is the effective pressure that
a real gas exerts.
L
Pi
1dmreal - dmideal2 = 3 mreal1P2 - mreal1Pi24 - 3 mideal1P2 - mideal1Pi24
P
=
L
Pi
ideal
1V real
m - V m 2dP′
(7.18)
P′ is used as a dummy variable in the previous equation because P appears in the upper
limit of integration. Equation (7.18) allows us to calculate the difference in chemical
potential between a real and an ideal gas at pressure P. Now let Pi S 0. In this limit,
mreal 1Pi2 = mideal 1Pi2 because all real gases approach ideal gas behavior at a sufficiently low pressure. Equation (7.18) becomes
P
mreal1P2 - mideal1P2 =
L
0
ideal
1V real
m - V m 2dP′
(7.19)
217
7.5 Fugacity and the Equilibrium Constant for Real Gases
Equation (7.19) provides a way to calculate the fugacity of a real gas. Using Equations
(7.13) and (7.14) for mreal1P2 and mideal1P2, P and ƒ are related at the final pressure by
P
1
ln ƒ = ln P +
1V real - V ideal
m 2dP′
RT L m
(7.20)
0
Because tabulated values of the compression factor z of real gases are widely available,
ideal
it is useful to rewrite Equation (7.20) in terms of z by substituting z = V real
m >V m . The
result is
P
P
0
0
z - 1
z - 1
ln ƒ = ln P +
dP′ or ƒ = P expc a
b dP′ d or ƒ = g1P, T2P (7.21)
P′
L
L P′
W7.4 Calculating the fugacity
coefficient from the compression
factor
Equation (7.21) provides a way to calculate the fugacity if z is known as a function of
pressure. Note that ƒ and P are related by the proportionality factor g, which is called
the fugacity coefficient. However, g is not a constant; it depends on both P and T.
EXAMPLE PROBLEM 7.4
For T 7 TB, the equation of state P1Vm - b2 = RT is an improvement over the ideal
gas law because it takes the finite volume of the molecules into account. Derive an expression for the fugacity coefficient for a gas that obeys this equation of state.
Solution
Because z = PVm >RT = 1 + Pb>RT, the fugacity coefficient is given by
P
P
0
0
f
z - 1
d
bP
ln g = ln =
dP′ =
dP′ =
P
P′
RT
RT
L
L
bP>RT
Equivalently, g = e
. Note that g 7 1 for all values of P and T because the
e­ quation of state does not take the attractive part of the intermolecular potential into
account.
1.5
H2
1.4
Fugacity coefficient (g)
Calculated values for g 1P, T2 for H2, N2, and NH3 are shown in Figure 7.10 for
T = 700. K. It can be seen that g S 1 as P S 0. This must be the case because the
fugacity and pressure are equal in the ideal gas limit. The following general statements
can be made about the relationship between g and z: g 1P, T2 7 1 if the integrand of
Equation (7.21) satisfies the condition 1z - 12>P 7 0 for all pressures up to P. Similarly, g 1P, T2 6 1 if 1z - 12>P 7 0 for all pressures up to P.
These predictions about fugacity and the fugacity coefficient can be related to
the Boyle temperature. If T 7 TB, then g1P, T2 7 1 for all pressures, the fugacity is
greater than the pressure, and g 7 1. However, if T 6 TB, then g 1P, T2 6 1, and the
fugacity is smaller than the pressure. The last statement does not hold for very high values of Pr because, as shown in Figure 7.8, z 7 1 for all values of T at such high relative
pressures. Are these conclusions in accord with the curves in Figure 7.10? The Boyle
temperatures for H2, N2, and NH3 are 110, 327, and 995 K, respectively. Because at
700. K, T 7 TB for H2 and N2, but T 6 TB for NH3, we conclude that g 7 1 at all pressures for H2 and N2, and g 6 1 at all but very high pressures for NH3. These conclusions are consistent with the results shown in Figure 7.10.
The fugacity coefficient can also be graphed as a function of Tr and Pr. This is
convenient because it allows g for any gas to be estimated once T and P have been expressed as reduced variables. Graphs of g as functions of Pr and Tr are shown in Figure
7.11. Note that for large values of Tr , g values are close to 1.0 even for large values
of Pr because the attractive potential becomes unimportant in this limit. The curves
have been calculated using Beattie–Bridgeman parameters for N2, and are known to
N2
1.3
1.2
1.1
Pressure (bar)
1.0
100 200 300 400 500 600 700
0.9
NH3
Figure 7.10
Fugacity coefficients for H2 1 g2 , N2 1g2 ,
and NH3 1 g2 plotted as a function of
the partial pressure of the gases for
T = 700. K. The calculations were carried out using the Beattie–Bridgeman
equation of state.
218
CHAPTER 7 The Properties of Real Gases
be accurate for N2 over the indicated range of reduced temperature and pressure. Their
applicability to other gases assumes that the law of corresponding states holds. As
Figure 7.7 shows, this is generally a good assumption.
What are the consequences of the fact that except in the dilute gas limit ƒ ≠ P?
­Because the chemical potential of a gas in a reaction mixture is given by Equation (7.14),
the thermodynamic equilibrium constant for a real gas Kƒ must be expressed in terms of
the fugacities. Therefore, Kƒ for the reaction 3>2H2 + 1>2N2 S NH3 is given by
Kƒ =
a
ƒN2
ƒ~
a
b
ƒNH3
ƒ
1>2
~
a
b
ƒH2
ƒ~
b
3>2
=
a
a
gNH3PNH3
gN2 PN2
= KP
P~
P~
b
1>2
a
b
gH2PH2
P~
gNH3
1>2
1gN22
1gH223>2
b
3>2
(7.22)
We next calculate the error in using Kp rather than Kƒ to calculate an equilibrium constant, using the ammonia synthesis reaction at 700. K and a total pressure of 400. bar
as an example. In the industrial synthesis of ammonia, the partial pressures of H2, N2,
and NH3 are typically 270, 90, and 40 bar, respectively. The calculated fugacity coefficients under these conditions are gH2 = 1.11, gN2 = 1.04, and gNH3 = 0.968 using
the Beattie–Bridgeman equation of state. Therefore,
Kƒ = KP
Fugacities must be used to accurately
calculate the composition of a
reactive mixture of real gases at high
pressures.
11.0421>2 11.1123>2
xNH3 = 1xN221>21xH223>2 a
3.0
2.4
2.0
1.8
1.7
1.6
Fugacity coefficient (g)
1.2
Fugacity coefficient plotted as a
function of the reduced pressure for
the indicated values of the ­reduced
temperature. To avoid overlap,
curves for reduced temperature values
of 1.0–3.0 are shown in (a) and curves
for reduced temperature values of
3.0–24 are shown in (b).
(7.23)
P
b Kƒ
P~
(7.24)
What conclusions can be drawn from this calculation? If KP was used rather than Kƒ,
the calculated mole fraction of ammonia would be in error by -9%, which is a significant error in calculating the economic viability of a production plant. However, for
pressures near 1 bar, activity coefficients for most gases are very close to unity, as can
be seen in Figure 7.11. Therefore, under typical laboratory conditions, the fugacity of
a gas can be set equal to its partial pressure if P, V, and T are not close to their critical
values.
1.4
Figure 7.11
= 0.917 KP
Note that Kƒ is smaller than Kp for the conditions of interest. The NH3 concentration at
equilibrium is proportional to the equilibrium constant, as shown in Section 6.14.
1.0
1.5
1.4
0.8
1.3
0.6
1.2
0.4
1.1
0.2
1.0
3.0
Fugacity coefficient (g)
Concept
10.9682
1.3
6.0
9.0
1.2
12
18
1.1
24
1.0
2 4 6 8 10 12 14
Reduced pressure (Pr)
(a) Reduced temperature values of 1.0–3.0
0
2 4 6 8 10 12 14
Reduced pressure (Pr)
(b) Reduced temperature values of 3.0–24
Conceptual Problems
219
VOCABULARY
Beattie–Bridgeman equation of state
Boyle temperature
compression factor
critical constants
critical temperature
fugacity
fugacity coefficient
law of corresponding states
Maxwell construction
Redlich–Kwong equation of state
supercritical fluid
van der Waals equation of state
vapor pressure
virial equation of state
KEY EQUATIONS
Equation
RT
a
nRT
n2a
- 2 =
- 2
Vm - b V m
V - nb
V
P =
1
1
nRT
n2a
RT
a
=
V1V
+ nb2
V
1V
+
b2
V
nb
Vm - b
m
m
2T
2T
P =
P = RT c
z =
Significance of Equation
Vm
B1T2
1
+
+ gd
Vm
V 2m
V ideal
m
TB =
=
PVm
RT
a
Rb
m1T, p2 = m ~ 1T2 + RT ln
P
g1P, T2 = expc
L
0
a
ƒ
ƒ~
z - 1
bdP′ d
P′
P
P
0
0
z - 1
z - 1
ln ƒ = ln P +
dP′ or ƒ = P expc a
bdP′ d
L P′
L P′
Equation Number
van der Waals equation of state
7.1
Redlich–Kwong equation of state
7.2
Virial equation of state
7.4
Definition of compression factor
7.6
Boyle temperature for van der Waals gas
7.7
Chemical potential of real gas and
­definition of fugacity
7.14
Definition of fugacity coefficient
from 7.21
Calculation of fugacity from
­compressibility
7.21
CONCEPTUAL PROBLEMS
Q7.1 Using the concept of the intermolecular potential,
­explain why two gases in corresponding states can be
­expected to have the same value for z.
Q7.2 Consider the comparison made between accurate results and those based on calculations using the van der Waals
and Redlich–Kwong equations of state in Figures 7.1 and 7.5.
Is it clear that one of these equations of state is better than the
other under all conditions?
Q7.3 Why is the standard state of fugacity, ƒ ~ , equal to the
standard state of pressure, P ~ ?
Q7.4 Explain the significance of the Boyle temperature.
Q7.5 A van der Waals gas undergoes an isothermal revers­
ible expansion under conditions such that z 7 1. Is the work
performed more or less than if the gas followed the ideal
gas law?
Q7.6 For a given set of conditions, the fugacity of a gas is
greater than the pressure. What does this tell you about the
interaction between the molecules of the gas?
Q7.7 What can you conclude about the ratio of fugacity to
pressure for N2, H2, and NH3 at 500 bar using the data in
­Figure 7.10?
Q7.8 Is the ratio of fugacity to pressure greater to or less than
one if the attractive part of the interaction potential between
gas molecules dominates?
Q7.9 A gas is slightly above its Boyle temperature. Do you
expect z to increase or decrease as P increases?
220
CHAPTER 7 The Properties of Real Gases
Q7.10 Explain why the oscillations in the two-phase
­coexistence region using the Redlich–Kwong and van der
Waals equations of state (see Figure 7.4) do not correspond
to reality.
Q7.11 A van der Waals gas undergoes an isothermal reversible expansion under conditions such that z 6 1. Is the work
done more or less than if the gas followed the ideal gas law?
Q7.12 The value of the Boyle temperature increases with the
strength of the attractive interactions between molecules. Arrange the Boyle temperatures of the gases Ar, CH4, and C6H6
in increasing order.
Q7.13 A system containing argon gas is at pressure P1 and
temperature T1. How would you go about estimating the fugacity coefficient of the gas?
Q7.14 By looking at the a and b values for the van der Waals
equation of state, decide whether 1 mol of O2 or H2O has the
higher pressure at the same value of T and V.
Q7.15 Will the fugacity coefficient of a gas at a temperature
greater than the Boyle temperature be less than one at low
pressures?
Q7.16 Show that the van der Waals and Redlich–Kwong
equations of state reduce to the ideal gas law in the limit of
low gas density.
Q7.17 Which of Ne or Ar has the larger van der Waals
­parameter a? Explain your reasoning.
Q7.18 Which of Ne or Ar has the larger van der Waals
­parameter b? Explain your reasoning.
Q7.19 You have calculated the pressure exerted by ethane
using the ideal gas law and the Redlich–Kwong equations of
state. How do you decide if the repulsive or attractive part of
the molecular potential dominates under the given conditions?
Q7.20 Equation 1.20 states that the total pressure in a mixture
of gases is equal to the sum of the partial pressures. Is this equation valid for real gases? If so, under what conditions is it valid?
NUMERICAL PROBLEMS
Section 7.2
P7.1 A van der Waals gas has a value of z = 1.00032 at 298 K
and 1 bar, and the Boyle temperature of the gas is 195 K.
Because the density is low, you can calculate Vm from
the ideal gas law. Use this information and the equation
z ≈ 1 + 1b - a>RT211>Vm2 to estimate a and b.
P7.2 Assume that the equation of state for a gas can be written in the form P1Vm - b1T22 = RT. Derive an expression
for a = 1>V10V>0T2P and kT = -1>V10V>0P2T for such a
gas in terms of b1T2, db1T2>dT, P, and Vm.
P7.3 One mole of Ar initially at 298 K undergoes an
adiabatic expansion against a pressure Pexternal = 0 from a
volume of 4.0 L to a volume of 50.0 L. Calculate the final
­temperature using the ideal gas and van der Waals equations
of state. Assume CV,m = 3>2 R.
P7.4 At 725 K and 280. bar, the experimentally determined
density of N2 is 4.13 mol L- 1. Compare this with values calculated from the ideal and Redlich–Kwong equations of state.
Use a numerical equation solver to solve the Redlich–Kwong
equation for Vm or use an iterative approach starting with Vm
equal to the ideal gas result. Discuss your results.
P7.5 1.35 mol of Ar undergoes an isothermal reversible
­expansion from an initial volume of 1.75 L to a final volume
of 12.00 L at 320. K. Calculate the work done in this process
using the ideal gas and van der Waals equations of state. What
percentage of the work done by the van der Waals gas arises
from the attractive potential?
P7.6 The volume of a spherical molecule can be estimated
as V = b>4NA where b is the van der Waals parameter for the
excluded molar volume and NA is Avogadro’s number. Justify
this relationship by considering a spherical molecule of radius
r, with volume V = 4>3pr 3. What is the volume centered
at the molecule that is excluded for the center of mass of a
second molecule in terms of V? Multiply this volume by NA
and set it equal to b. Apportion this volume equally among
the molecules to arrive at V = b>4NA. Calculate the radius of
a tetrachloromethane molecule from the value of its van der
Waals parameter b.
P7.7 Show that the van der Waals and Redlich–Kwong equations of state reduce to the ideal gas equation of state in the
limit of low density.
P7.8 Another equation of state is the Berthelot equation, Vm = RT>P + b - a>1RT 22. Derive expressions
for a = 1>V10V>0T2P and kT = -1>V10V>0P2T from the
­Berthelot equation in terms of V, T, and P.
0 ln z
b for a real gas where
P7.9 Show that PkT = 1 - P a
0P T
k is the isothermal compressibility.
P7.10 Calculate the density of O21g2 at 310. K and 125. bar
using the ideal gas and the van der Waals equations of state.
Use a numerical equation solver to solve the van der Waals
equation for Vm or use an iterative approach starting with Vm
equal to the ideal gas result. Based on your result, does the
attractive or repulsive contribution to the interaction potential
dominate under these conditions?
P7.11 A sample containing 50.0 g of Ar is enclosed in a
container of volume 0.0750 L at 425 K. Calculate P using the
ideal gas, van der Waals, and Redlich–Kwong equations of
state. Based on your results, does the attractive or repulsive
contribution to the interaction potential dominate under these
conditions?
Web-Based Simulations, Animations, and Problems
Section 7.3
P7.12 For values of z near one, it is a good approximation to
write z1P2 = 1 + 10z>0P2TP. Assume that z = 1.00221 at
298 K and 1 bar, and the Boyle temperature of a gas is 140. K.
Using an equation solver, the relation for z1P2 above, the van
a
der Waals equation of state, and the relation TB =
, calcuRb
late the values of a, b, and Vm for the van der Waals gas.
P7.13 For a gas at a given temperature, the compression
­factor z is described by the empirical equation
P
P 2
z = 1 - 6.40 * 10-3 ~ + 3.20 * 10-5 a ~ b
P
P
where P ~ = 1 bar. Calculate the fugacity coefficient for
P = 100., 200., 300., 400., 500., and 600. bar. For which of
these values is the fugacity coefficient greater than one?
P7.14 The experimentally determined density of O2 at 140. bar
and 298 K is 192 g L- 1. Calculate z and Vm from this information. Compare this result with what you would have estimated
from Figure 7.8. What is the relative error in using Figure 7.8
for this case?
P7.15 Show that the second virial coefficient for a van der
Waals gas is given by
1
0z
a
B1T2 =
= b RT ° 1 ¢
RT
0
Vm T
0 ln z
b for a real gas where
0T P
a is the volumetric thermal expansion coefficient.
P7.16 Show that Ta = 1 + T a
P7.17 At what temperature does the slope of the z versus P
curve as P S 0 have its maximum value for a van der Waals
gas? What is the value of the maximum slope?
P7.18 Show that Ta = 1 + T10 ln z>0T2P and that PkT =
1 - P10 ln z>0P2T .
P7.19 The observed Boyle temperatures of H2, N2, and CH4
are 110., 327, and 510. K, respectively. Compare these values
with those calculated for a van der Waals gas with the appropriate parameters.
221
P7.20 For the Berthelot equation, Vm = 1RT>P2 + b 1a>RT 22, find an expression for the Boyle temperature in
terms of a, b, and R.
P7.21 For a van der Waals gas, z = Vm >1Vm - b2 - a>RTVm.
Expand the first term of this expression in a Taylor series in
the limit Vm W b to obtain z ≈ 1 + 1b - a>RT211>Vm2.
Section 7.4
P7.22
a. Using the relationships derived in Example Problem 7.1
and the values of the critical constants for ethanol from
Table 7.2, calculate values for the van der Waals parameters a, b, and R from zc, Tc, Pc, and Vc. Do your results
agree with those in Tables 1.2 and 7.4 in Appendix A?
b. Calculate the van der Waals parameters a and b using the
critical constants for ethanol and the correct value for R.
Do these results agree with those in Tables 1.2 and 7.4?
P7.23 Calculate the P and T values for which Br21g2 is in a
corresponding state to F21g2 at 400. K and 65.0 bar.
P7.24 Use the law of corresponding states and Figure 7.8
to estimate the molar volume of propane at T = 500. K
and P = 75.0 bar. The experimentally determined value is
0.438 mol L- 1. What is the relative error of your estimate?
P7.25 Calculate the van der Waals parameters of benzene
from the values of the critical constants.
P7.26 Calculate the Redlich–Kwong parameters of water
from the values of the critical constants.
P7.27 Calculate the critical volume for ethane using the data
for Tc and Pc in Table 7.2 (see Appendix A, Data Tables) assuming (a) the ideal gas equation of state and (b) the van der Waals
equation of state. Use an iterative approach to obtain Vc from the
van der Waals equation, starting with the ideal gas result. How
well do the calculations agree with the tabulated values for Vc?
P7.28 The experimental critical constants of O2 are listed in
Table 7.2 (see Appendix A, Data Tables). Use the values of Pc and Tc
to calculate Vc. Assume that at the critical point O2 behaves as (a) an
ideal gas, (b) a van der Waals gas, and (c) a Redlich–Kwong gas. For
parts (b) and (c), use the formulas for the critical compression factor.
Compare your answers with the experimental value. Are any of your
calculated results close to the experimental value in Table 7.2?
WEB-BASED SIMULATIONS, ANIMATIONS, AND PROBLEMS
Simulations, animations, and homework problem worksheets can be accessed on Mastering™ Chemistry.
W7.1 In this problem, the student gains facility in using the
van der Waals equation of state. A set of isotherms will be
generated by varying the temperature and initial volume using
sliders for a given substance. Buttons allow a choice among
more than 20 gases.
a. The student generates a number of P–V curves at
and above the critical temperature for the particular
gas, and explains trends in the ratio PvdW >Pideal as a
­function of V for a given T and as a function of T for a
given V.
b. The compression factor z is calculated for two gases at the
same value of Tr and is graphed versus P and Pr. The degree
to which the law of corresponding states is valid is assessed.
W7.2 A quantitative comparison is made between the ideal gas
law and the van der Waals equation of state for 1 of more than 20
different gases. The temperature is varied using sliders, and Pideal,
PvdW, the relative error PvdW >PvdW - Pideal, and the density of
gas relative to that at the critical point are calculated. The student
is asked to determine the range of pressures and temperatures in
which the ideal gas law gives reasonably accurate results.
222
CHAPTER 7 The Properties of Real Gases
W7.3 The compression factor and molar volume are calculated for an ideal gas and a van der Waals gas as a function of
pressure and temperature. These variables can be varied using
sliders. Buttons allow a choice among more than 20 gases.
The relative error VvdW >VvdW - Videal and the density of gas
relative to that at the critical point are calculated. The student
is asked to determine the range of pressures and temperatures
in which the ideal gas law gives reasonably accurate results
for the molar volume.
W7.4 The fugacity and fugacity coefficient are determined as
a function of pressure and temperature for a model gas. These
variables can be varied using sliders. The student is asked to
determine the Boyle temperature and also the pressure range
in which the fugacity is either more or less than the ideal gas
pressure for the temperature selected.
FURTHER READING
Ben-Amotz, D. Ben, and Levine, R. “Updated Principle of
Corresponding States.” Journal of Chemical Education
81 (2004): 142–146.
David, C. “Fugacity Examples.” Journal of Chemical
Education 81 (2004): 1653–1654.
Eberhart, J. “The Many Faces of van der Waals’ Equation of
State.” Journal of Chemical Education 66 (1989): 906–909.
Rigby, M. “A van der Waals Equation for Nonspherical
Molecules.” Journal of Physical Chemistry 76 (1972):
2014–2016.
Soave, G. “Improvement of the van der Waals Equation
of State.” Chemical Engineering Science 39 (1984):
357–369.
C H A P T E R
8
Phase Diagrams and the
Relative Stability of Solids,
Liquids, and Gases
WHY is this material important?
At room temperature, the most stable form of water is the liquid phase. At t­emperatures
greater than 100°C, the most stable form of H2O is gaseous water vapor. Boiling water
is the manifestation of the transition from liquid water to gaseous water. The process
of boiling occurs at constant temperature. In the elevated pressure of a pressure cooker,
food cooks much more quickly than in a pot on the stove at a pressure of 1 bar. In
this chapter, you will learn how all of this information can be represented in a phase
­diagram.
WHAT are the most important concepts and results?
The occurrence of a single stable phase or the coexistence of several phases is predictable with the application of the principles of thermodynamics. For example, coexistence of two phases of a pure substance occurs if and only if the chemical potential is
equal in both phases. It turns out that no more than three phases of a pure substance
can ­coexist. Furthermore, the vapor pressure of a liquid increases exponentially with
temperature. Moreover, the pressure inside a small liquid droplet can be significantly
greater than that inside a large droplet because of the effects of surface tension. This
observation explains how condensation occurs in clouds.
8.1
What Determines the Relative
Stability of the Solid, Liquid,
and Gas Phases?
8.2
The Pressure–Temperature
Phase Diagram
8.3
The Phase Rule
8.4
Pressure–Volume and Pressure–
Volume–Temperature Phase
Diagrams
8.5
Providing a Theoretical Basis
for the P–T Phase Diagram
8.6
Using the Clausius–Clapeyron
Equation to Calculate Vapor
Pressure as a Function of T
8.7
Dependence of Vapor Pressure
of a Pure Substance on Applied
Pressure
8.8
Surface Tension
8.9
(Supplemental Section)
Chemistry in Supercritical Fluids
8.10 (Supplemental Section) Liquid
Crystal Displays
WHAT would be helpful for you to review for this chapter?
This chapter builds on the concept of chemical potential, introduced in Chapter 6.
Thus, it would be useful to have a firm understanding of this concept.
8.1 WHAT DETERMINES THE RELATIVE STABILITY
OF THE SOLID, LIQUID, AND GAS PHASES?
Substances occur in solid, liquid, and gaseous phases. Phase refers to a form of matter
that is uniform with respect to chemical composition and the state of aggregation on
both microscopic and macroscopic length scales. For example, liquid water in a beaker
is a single-phase system, but a mixture of ice and liquid water consists of two distinct phases, each of which is uniform on microscopic and macroscopic length scales.
A substance may exist in only one gaseous phase. Similarly, most substances have a
223
224
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Concept
The chemical potential determines
which of several phases is the most
stable.
single liquid state, although there are exceptions. For example, helium can exist as a
normal liquid or as a superfluid. In contrast, many substances may exist as several different solid phases. In this section, the conditions under which a pure substance spontaneously forms a solid, liquid, or gas are discussed.
Experience demonstrates that as T is lowered from 300. to 250. K at atmospheric
pressure, liquid water is converted to a solid phase. Similarly, as liquid water is heated
to 400. K at atmospheric pressure, it vaporizes to form a gas. Experience also shows
that if a solid block of carbon dioxide is placed in an open container at 1 bar, it sublimes
over time without passing through the liquid phase. Because of this property, solid CO2
is known as dry ice. These observations can be generalized as follows. Whereas the
solid phase is the most stable state of a substance at sufficiently low temperatures, the
gas phase is the most stable state of a substance at sufficiently high temperatures. At
intermediate temperatures, the liquid state is stable, if it exists at the pressure of interest. What determines which of the solid, liquid, or gas phases exists and which phase is
most stable at a given temperature and pressure? The answers to these questions are the
subject of this chapter.
As discussed in Chapter 6, the criterion for stability at constant temperature and
pressure is that the Gibbs energy, G1T, P, n2, is minimized. For a pure substance, the
chemical potential m is defined as
m = a
03 nGm 4
0G
b
= a
b
= Gm
0n T, P
0n
T, P
where n designates the number of moles of substance in the system, dm = dGm.
We can, therefore, express the differential dm as
dm = -Sm dT + Vm dP
(8.1)
Chemical potential,
From this equation, it is possible to determine how m varies with changes in P and T:
a
Solid
Liquid
Gas
Tm
Tb
Temperature
Figure 8.1
Chemical potential plotted as a function
of temperature and the relative stability of solid, liquid, and gas phases at a
given pressure. The curves for the solid
(green curve), liquid (blue curve), and
gaseous (brown curve) states are shown
for a given value of pressure. The stable
state of the system at any given temperature is that phase with the lowest value of
m, and the temperature range of the most
stable phase is indicated by colored areas.
The substance melts at the temperature Tm,
corresponding to the intersection of the
solid and liquid curves. It boils at the
temperature Tb, corresponding to the intersection of the liquid and gas curves. The
three curves are actually curved slightly
downward but have been approximated
as straight lines.
0m
0m
b = -Sm and a b = Vm
0T P
0P T
(8.2)
Because Sm and Vm are always positive, m decreases as the temperature increases
and increases as the pressure increases. Section 5.4 demonstrated that S varies slowly
with T (approximately as ln T ). Therefore, over a limited range in T, a plot of m versus
T at constant P is a curve with a slowly increasing negative slope, approximating a
straight line.
It is also known from experience that heat is absorbed as a solid melts to form a
liquid and as a liquid vaporizes to form a gas. Both processes are endothermic with
∆H 7 0. Because at the melting and boiling temperatures the entropy increases in
a stepwise fashion by ∆S = ∆ transH > T, the entropy of the three phases follows this
order:
gas
Sm
7 Sliquid
7 Ssolid
m
m
(8.3)
Figure 8.1 graphs the functional relation between m and T for the solid, liquid, and
gas phases at a given value of P. The molar entropy of a phase is the negative of the
slope of the m versus T curve, and the relative entropies of the three phases are given
by Equation (8.3). The stable state of the system at any given temperature is that phase
that has the lowest m.
Assume that the initial state of the system is described by the dot in Figure 8.1.
The most stable phase is the solid phase because m is much larger for the liquid and
gas phases than for the solid. As the temperature is increased, the chemical potential decreases as m remains on the solid curve. However, because the slopes of the
liquid and gas curves are greater than the slope of the solid, each of these m versus
T curves will intersect the solid curve for some value of T. In Figure 8.1, the liquid
curve intersects the solid curve at Tm, which is called the melting temperature. At
this temperature, the solid and liquid phases coexist and are in thermodynamic equilibrium. However, if the temperature is raised by an infinitesimal amount dT, the
solid will melt completely because the liquid phase has the lower chemical potential
at Tm + dT. Similarly, the liquid and gas phases are in thermodynamic equilibrium
8.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phases?
at the boiling temperature Tb. For T 7 Tb, the system is entirely in the gas phase.
Note that the progression of solid S liquid S gas as T increases at this value of
P can be explained with no other information than that 10m > 0T2P = -Sm and that
gas
Sm
7 Sliquid
7 Ssolid
m
m .
If the temperature is changed too quickly, the equilibrium state of the system may
not be reached and a metastable phase may persist. For example, it is possible to form
a superheated liquid in which the liquid phase is metastable at temperatures greater
than Tb. Superheated liquids are dangerous because of the large volume increase that
occurs if the system suddenly converts to the stable vapor phase. Boiling chips are
often used in chemical laboratories to avoid the formation of superheated liquids. Similarly, it is possible to form a supercooled liquid, in which case the liquid is metastable
at temperatures less than Tm. Glass is made by cooling a viscous liquid fast enough
to avoid crystallization. These disordered materials lack the periodicity of crystals but
behave mechanically like solids. Seed crystals can be used to increase the rate of crystallization if the viscosity of a liquid is not too high and the cooling rate is sufficiently
slow. Liquid crystals, which can be viewed as a state of matter intermediate between a
solid and a liquid, are discussed in Section 8.10.
In Figure 8.1, we considered changes of m with T at constant P. How is the relative
stability of the three phases affected if P is changed at constant T ? From Equation
gas
(8.2), 10m > 0P2T = Vm and V m
W V liquid
. For most substances, V liquid
7 V solid
m
m
m .
Therefore, the m versus T curve for the gas changes much more rapidly with pressure
(by a factor of ∼1000) than the liquid and solid curves. This behavior is illustrated in
Figure 8.2, where it can be seen that the temperature at which the solid and the liquid
gas
curves intersect shifts as the pressure is increased. Because V m
W V liquid
7 0, an
m
increase in P always leads to a boiling point elevation. An increase in P leads to
a freezing point elevation if V liquid
7 V solid
and to a freezing point depression if
m
m
liquid
solid
Vm
6 V m . Few substances obey the relation V liquid
6 V solid
m
m , with water being
an important example. The consequences of this unusual behavior for water will be
discussed in Section 8.2.
The m versus T curve for a gas shifts along the T axis much more rapidly with
changes in P than the liquid and solid curves. As a consequence, changes in P can
alter the way in which a system progresses through the phases with increasing T from
the “normal” order solid S liquid S gas shown in Figure 8.1. For example, the sublimation of dry ice at 298 K and 1 bar can be explained using Figure 8.3a. For CO2 at
a given pressure, the m versus T curve for the gas intersects the corresponding curve
for the solid at a lower temperature than the curve for the gas intersects the liquid
curve. Therefore, the solid S liquid transition is energetically unfavorable with respect to the solid S gas transition at this pressure. Under these conditions, the solid
Solid
Either liquid
or gas,
depending on P.
Liquid
DTm
Gas
Either solid
or liquid,
depending on P.
Chemical potential, M
Chemical potential, M
Either solid
or liquid,
depending on P.
Solid
DTm
Either liquid
or gas,
depending on P.
Liquid
DTb
DTb
Tm T'm
Tb
T'm Tm
T'b
Temperature
(a)
Temperature
(b)
Gas
Tb
T'b
225
Concept
Metastable phases do not correspond
to equilibrium states.
Concept
Boiling and freezing temperatures
vary with pressure.
Figure 8.2
The effect of pressure on the relative
stability of solid, liquid, and gas phases.
Chemical potential is plotted as a function of temperature. The diagrams are
for cases in which (a) V liquid
7 V solid
m
m
liquid
solid
and (b) V m
6 V m . The solid curves
show m as a function of temperature for
all three phases at P = P1. The dashed
curves show the same information for
P = P2,where P2 7 P1. The unprimed
temperatures refer to P = P1 and the
primed temperatures refer to P = P2. The
shifts in the solid and liquid curves are
greatly exaggerated. The colored areas indicate the temperature range in which the
phases are most stable.
226
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Figure 8.3
Solid
Gas
Chemical potential, M
At Ttp, all three phases
coexist in equilibrium.
Chemical potential, M
Sublimation and triple point behavior.
The chemical potential of a substance
in the solid (green curve), liquid (blue
curve), and gaseous (brown curve) states
is plotted as a function of temperature for
a fixed value of pressure. (a) Pressure less
than triple point pressure. (b) Triple point
pressure. The colored areas indicate the
temperature range in which the phases are
the most stable. The liquid phase is not
stable at any temperature in part (a), and
is only stable at the single temperature Ttp
in part (b).
Solid
Ts
Temperature
(a) The pressure lies below
the triple point pressure,
and the solid sublimes.
Gas
Ttp
Temperature
(b) The pressure corresponds to
the triple point pressure.
sublimes, and the transition temperature Ts is called the sublimation temperature.
There is also a pressure at which the m versus T curves for all three phases intersect. The P, Vm, T values for this point specify the triple point, so named because all
three phases coexist in equilibrium at this point. This case is shown in Figure 8.3b.
Triple point temperatures for a number of substances are listed in Table 8.1 (see
­Appendix A, Data Tables).
8.2 THE PRESSURE–TEMPERATURE
PHASE DIAGRAM
Concept
A P–T phase diagram summarizes
information on the stability of
different phases as the values of P
and T change.
As shown in the previous section, at a given value of pressure and temperature a system
containing a pure substance may consist of a single phase, two phases in equilibrium,
or three phases in equilibrium. The usefulness of a phase diagram is that it displays
this information graphically. Although any two of the macroscopic system variables
P, V, and T can be used to construct a phase diagram, the P9T diagram is particularly
useful. In this section, the features of a P9T phase diagram that are common to pure
substances are discussed. Phase diagrams must generally be determined experimentally because material-specific forces between atoms determine the temperatures and
pressures at which different phases are stable. Increasingly, calculations have become
sufficiently accurate that major features of phase diagrams can be obtained using microscopic theoretical models. However, as shown in Section 8.4, thermodynamics can
say a great deal about the phase diagram without considering the microscopic properties of the system.
The P–T phase diagram displays stability regions for a pure substance as a function
of pressure and temperature. An example is shown in Figure 8.4. Most P, T points correspond to a single solid, liquid, or gas phase. At the triple point, all three phases coexist. The triple point of water is 273.16 K and 611 Pa. All P, T points for which the same
two phases coexist at equilibrium fall on a curve. Such a curve is called a c­ oexistence
curve. Three separate coexistence curves are shown in Figure 8.4, corresponding to
solid–gas, solid–liquid, and gas–solid coexistence. As shown in ­Section 8.5, the slopes
of the solid–gas and liquid–gas curves are always positive. The slope of the solid–­
liquid curve can be either positive or negative.
The boiling point of a substance is defined as the temperature at which the vapor
pressure of the substance is equal to the external pressure. The standard boiling
temperature is the temperature at which the vapor pressure of the substance is 1 bar.
8.2 The Pressure–Temperature Phase Diagram
Constant temperature
(isothermal) path greater
than triple point
Figure 8.4
Pressure, P (bar)
Critical
point
Solid
1
Liquid
d
Constant pressure
(isobaric) path
for P > Ptp
Gas-liquid coexistence curve
is crossed in the constant
pressure process moving to
lower temperatures.
a
Triple
point
A P9T phase diagram. Shown are
single-phase regions, coexistence curves
for which two phases coexist at equilibrium, and a triple point. The processes
corresponding to paths a, b, c, and the two
processes starting at point d are described
in the text. Two solid–liquid coexistence
curves are shown. For most substances,
the solid curve, which has a positive
slope, is observed. For water or any other
substance for which volume increases in
a transition from the liquid to the solid
phase, the red dashed curve corresponding
to a negative slope is observed.
Gas
b
Constant pressure
(isobaric) path
for P < Ptp
227
c
Tm
Tb
Temperature
The normal boiling temperature is the temperature at which the vapor pressure
of the substance is 1 atm. Values of the normal boiling and freezing temperatures
for a number of substances are shown in Table 8.2. Because 1 bar is slightly less
than 1 atm, the standard boiling temperature is slightly less than the normal boiling
temperature. Along two-phase curves in which one of the coexisting phases is the
gas, P refers to the vapor pressure of the substance. In all regions, P refers to the
external pressure that would be exerted on the pure substance if it were confined in a
piston and cylinder assembly.
The solid–liquid coexistence curve traces out the melting point as a function of
pressure. The magnitude of the slope of this curve is large, as shown in Section 8.5.
Therefore, Tm changes very little with pressure. If the solid is more dense than the
­liquid, the slope of this curve is positive, and the melting temperature increases with
pressure. This is the case for most substances. If the solid is less dense than the liquid, the slope is negative and the melting temperature decreases with pressure. Water
is one of the few substances that exhibits this behavior. Imagine the fate of aquatic
plants and animals in climate zones where the temperature routinely falls below 0°C
in the winter if water behaved “normally.” Lakes would begin to freeze over at the
water–air interface, and the ice formed would fall to the bottom of the lake. This
would lead to more ice formation until the whole lake was full of ice. In cool climate
zones, it is likely that ice at the bottom of the lakes would remain throughout the
summer. The aquatic life that we are familiar with would not survive under such a
scenario.
The slope of the liquid–gas coexistence curve is much smaller than that of the
solid–liquid coexistence curve, as will be shown in Section 8.5. Therefore, the boiling
point is a much stronger function of the pressure than the freezing point. The boiling
point always increases with pressure. This property is utilized in a pressure cooker,
where increasing the pressure by 1 bar increases the boiling temperature of water by
approximately 20°C. The rate of the chemical processes involved in cooking increase
exponentially with T. Therefore, a pressure cooker operating at P = 2 bar can cook
food in 20% to 40% of the time required for cooking at atmospheric pressure. By contrast, a mountain climber in the Himalayas would find that the boiling temperature of
water is reduced by approximately 25°C relative to sea level. Cooking takes significantly longer under these conditions.
Concept
A P–T phase diagram generally
shows gaseous, liquid, and solid
single phase regions, two phase
coexistence curves, and the triple
and critical points.
228
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
TABLE 8.2 Melting and Normal Boiling Temperatures and Enthalpies
of Transition at 1 atm Pressure
Density, r/(g/cm3)
1
Liquid
rc
Gas
0
0
100
200
300
Tc
Temperature, T (°C)
Figure 8.5
Density of the liquid and gaseous
phases of water as a function of
­temperature. At the critical point, the
densities are equal.
Tm1K2
∆ fusH
1kJ mol-12
at Tm
Tb1K2
∆ vapH
1kJ mol-12
at Tb
Substance
Name
Ar
Argon
Cl2
Chlorine
Fe
Iron
1811
H2
Hydrogen
13.81
0.12
20.4
0.90
H2O
Water
273.15
6.010
373.15
40.65
83.8
171.6
1.12
6.41
13.81
0.021
87.3
239.18
3023
4.22
6.43
20.41
349.5
He
Helium
0.95
I2
Iodine
386.8
14.73
457.5
41.57
N2
Nitrogen
63.5
0.71
77.5
5.57
Na
Sodium
370.87
2.60
NO
Nitric oxide
109.5
2.3
121.41
13.83
O2
Oxygen
54.36
0.44
90.7
6.82
SO2
Sulfur dioxide
197.6
7.40
263.1
24.94
Si
Silicon
1687
50.21
2628
359
W
Tungsten
3695
52.31
5933
422.6
1156
0.083
98.0
Xe
Xenon
161.4
1.81
165.11
12.62
CCl4
Carbon
tetrachloride
250
3.28
349.8
29.82
CH4
Methane
90.68
0.94
111.65
8.19
CH3OH
Methanol
175.47
3.18
337.7
35.21
CO
Carbon
monoxide
0.83
81.6
6.04
C2H4
Ethene
3.35
169.38
13.53
C2H6
Ethane
90.3
2.86
184.5
14.69
C2H5OH
Ethanol
159.0
5.02
351.44
38.56
C3H8
Propane
85.46
3.53
231.08
19.04
C5H5N
Pyridine
231.65
8.28
388.38
35.09
C6H6
Benzene
278.68
9.95
353.24
30.72
C6H5OH
Phenol
314.0
455.02
45.69
C6H5CH3
Toluene
178.16
6.85
383.78
33.18
C10H8
Naphthalene
353.3
17.87
491.14
43.18
68
103.95
11.3
Sources: Data from Lide, D. R., ed. Handbook of Chemistry and Physics; 83rd ed. Boca Raton, FL: CRC
Press, 2002; Lide, D. R., ed. CRC Handbook of Thermophysical and Thermochemical Data. Boca Raton,
FL: CRC Press, 1994; and Blachnik, R., ed. D'Ans Lax Taschenbuch für Chemiker und Physiker, 4th ed.
Berlin: Springer, 1998.
Concept
Above the critical pressure and
temperature, liquid and gas phases
are not distinguishable.
The termination of the coexistence curves is different for each curve. Specifically,
the solid–gas coexistence curve ends at the triple point, the liquid–solid coexistence
curve extends indefinitely, and the liquid–gas curve ends at the critical point. The
critical point is characterized by T = Tc and P = Pc. For T 7 Tc and P 7 Pc, the liquid and gas phases have the same density as shown in Figure 8.5, so it is not meaningful to refer to distinct phases. Because the liquid and gas phases are indistinguishable
at the critical point, ∆ vapH approaches zero as the critical point is reached, as shown
in Figure 8.6.
229
8.2 The Pressure–Temperature Phase Diagram
DsubH
Temperature
50
DvapH /(kJ mol–1)
40
30
20
10
0
0
100
200
300
Tc
Temperature, T (°C)
Figure 8.6
The enthalpy of vaporization for water
plotted as a function of temperature. At
the critical point, ∆ vapH approaches zero.
Solid
Temperature
Substances for which T 7 Tc and P 7 Pc are called supercritical fluids. As discussed in Section 8.9, supercritical fluids have unusual properties that make them
useful in chemical technologies.
Each of the paths labeled a, b, c, and d in Figure 8.4 corresponds to a process that
demonstrates the usefulness of the P9T phase diagram. In the following, each process
is considered individually. Process a follows a constant pressure (isobaric) path. An
example of this path is heating one mole of ice. The system is initially in the solid
single-phase region. Assume that heat is added to the system at a constant rate using
current flow through a resistive heater. Because the pressure is constant, qP = ∆H.
Furthermore, as discussed in Section 2.4, ∆H ≈ CP ∆T in a single-phase region.
Combining these equations, ∆T = qP > C solid
P . Along path a, the temperature increases
linearly with qp in the single-phase solid region, as shown in Figure 8.7. At the melting
temperature Tm, heat continues to be absorbed by the system as the solid is transformed
into a liquid. During melting, the system consists of two distinct phases, solid and
liquid. As heat continues to flow into the system, the temperature will not increase until
the system consists entirely of liquid. The heat taken up per mole of the system at the
constant temperature Tm is ∆ fusH. As soon as the last amount of solid melts, the temperature again increases linearly with qp and ∆T = qP > C liquid
until the boiling point is
P
reached. At this temperature, the system consists of two phases, liquid and gas. During
boiling, the temperature remains constant until all the liquid has been converted into
gas. The heat taken up per mole of the system at the constant temperature Tb is ∆ vapH.
Finally, the system enters the single-phase gas region.
Along path b, the pressure is less than the triple point pressure. Therefore, the
liquid phase is not stable, and the solid is converted directly into the gaseous form.
As Figure 8.4 shows, there is only one two-phase interval along this path. A diagram
illustrating the relationship between temperature and heat flow is shown in Figure 8.8.
Note that the initial and final states in process b can be reached by an alternative
route described by the vertical dashed arrows in Figure 8.4. The pressure of the system
in process b is first increased to the value of the initial state of process a at constant
temperature. The process then follows that described for process a, after which the
pressure is returned to the final pressure of process b. Invoking Hess’s law, the enthalpy change for this pathway and for pathway b are equal. Now imagine that the constant pressures for processes a and b differ only by an infinitesimal amount, a­ lthough
that for a is higher than the triple point pressure and that for b is lower than the triple
point pressure. We can express this mathematically by setting the pressure for process a equal to Ptp + dP and that for process b equal to Ptp - dP. We examine the
limit dP S 0. In this limit, ∆H S 0 for the two steps in the process indicated by
Solid 1 Gas
Gas
Tm
Tb
DsubH
Tm
DfusH
DvapH
Heat flow at constant pressure, qP
Figure 8.8
Solid
Liquid 1 Solid
Liquid
Liquid 1 Gas
Gas
Heat flow at constant pressure, qP
Figure 8.7
Trends in a temperature versus heat curve for solid u liquid u gas. This schematic
diagram illustrates the process corresponding to path a in Figure 8.4. Temperature rises
linearly with qp in single-phase regions and remains constant along the two-phase curves
as the relative amounts of the two phases in equilibrium change (not to scale).
Trends in a temperature versus heat
curve for solid u gas. This schematic
diagram illustrates the process corresponding to path b in Figure 8.4. Temperature rises linearly with qp in single-phase
regions and remains constant along the
two-phase curves as the relative amounts
of the two phases in equilibrium change
(not to scale).
230
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
the dashed arrows because dP S 0. Therefore, ∆H for the transformation solid S
liquid S gas in process a and for the transformation solid S gas in process b must be
identical. We conclude that
∆H = ∆ subH = ∆ fusH + ∆ vapH
(8.4)
This statement is strictly true at the triple point. Far from the triple point, the
­temperature-dependent heat capacities of the different phases and the temperature
­dependence of the transition enthalpies must be taken into account.
Path c indicates an isothermal process in which the pressure is increased. The
initial state of the system is the single-phase gas region. As the gas is compressed, it
is liquefied as it crosses the gas–liquid coexistence curve. As the pressure is increased
further, the sample freezes as it crosses the liquid–solid coexistence curve. Freezing
is exothermic, and heat must flow to the surroundings as the liquid solidifies. If the
process is reversed, heat must flow into the system to keep T constant as the solid
melts.
If T is less than the triple point temperature, the liquid exists at equilibrium only
if the slope of the liquid–solid coexistence curve is negative, as is the case for water.
Solid water below the triple point temperature can melt at constant T if the pressure
is increased sufficiently to cross the liquid–solid coexistence curve. In an example
of such a process, a thin wire to which a heavy weight is attached on each end is
stretched over a block of ice. With time, it is observed that the wire lies within the
ice block and eventually passes through the block. There is no visible evidence of
the passage of the wire in the form of a narrow trench. What happens in this process?
Because the wire is thin, the force on the wire results in a high pressure in the area
of the ice block immediately below the wire. This high pressure causes local melting of the ice below the wire. Melting allows the wire to displace the liquid water,
which flows to occupy the volume immediately above the wire. Because water in this
region no longer experiences a high pressure, it freezes again and hides the passage
of the wire.
The consequences of having a critical point in the gas–liquid coexistence curve
are illustrated in Figure 8.4. The gas–liquid coexistence curve is crossed in the
constant pressure process starting at point d and moving to lower temperatures. A
clearly visible interface will be observed as the two-phase gas–liquid coexistence
curve is crossed. However, the same overall process can be carried out in four steps
indicated by the blue arrows. In this case, a two-phase coexistence is not observed
because the gas–liquid coexistence curve ends at Pcritical. The overall transition is the
same along both paths; namely, gas is transformed into liquid. However, no interface will be observed if P for the higher pressure constant pressure process is greater
than Pcritical.
EXAMPLE PROBLEM 8.1
Draw a generic P9T phase diagram like that shown in Figure 8.4. Draw pathways in
the diagram that correspond to the processes described here:
a. You hang washed clothing out to dry for a temperature below the triple point
value. Initially, the water in the wet clothing is frozen. However, after a few hours
in the sun, the clothing is warmer, dry, and soft.
b. A small amount of ethanol is contained in a thermos bottle. A test tube is inserted
into the neck of the thermos bottle through a rubber stopper. A few minutes after
filling the test tube with liquid nitrogen, the ethanol is no longer visible at the
­bottom of the bottle.
c. A transparent cylinder and piston assembly contains only a pure liquid in equilibrium with its vapor phase. An interface is clearly visible between the two phases.
When you increase the temperature by a small amount, the interface disappears.
8.2 The Pressure–Temperature Phase Diagram
Solution
The phase diagram with the paths is shown here:
c
Critical
point
Liquid
Pressure
Solid
b
Gas
Triple
point
a
Temperature
Paths a, b, and c are not unique. Path a must occur at a pressure lower than the
triple point pressure. Process b must occur at a pressure greater than the triple point
pressure and originate on the liquid–gas coexistence curve. Path c will lie on the
liquid–gas coexistence curve up to the critical point, but it can deviate once T 7 Tc
and P 7 Pc.
A P9T phase diagram for water at high P values is shown in Figure 8.9. Because
the logarithm of pressure is plotted on the y axis, the curvature of the solid–liquid and
liquid–gas coexistence curves decreases with increasing temperature rather than increases
as shown in Figure 8.4. Water has a number of solid phases that are stable in different
pressure ranges because they have different densities. Eleven different crystalline forms
of ice have been identified up to a pressure of 1012 Pa. Note, for example, in Figure 8.9
that ice VI does not melt until the temperature is raised to ∼300 K for P ≈ 1000 MPa
X
VII
VIII
109
VI
II
V
Pressure, P (Pa)
IX
Liquid
III
Critical point
106
I
XI
103
Gas
Triple point
1
0
50
100
150
200
250
300
350
Temperature, T (K)
400
450
500
Figure 8.9
The P–T phase diagram for H2O for pressures to P = 1012 Pa and T = Tc.
Source: Adapted from “Water Structure and Science,” http://www.lsbu.ac.uk/water/phase.html, courtesy of Martin Chaplin.
550
600
231
232
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
(a) Hexagonal ice is the stable form of
ice under atmospheric conditions.
It has a relatively open structure
and a density less than that of
liquid water.
and that ice VII and ice X are stable to very high temperatures (though at very high
pressures). For a comprehensive collection of material on the phase diagram of water,
see http://www.lsbu.ac.uk/water/phase.html.
Hexagonal ice (ice I) is the normal form of ice and snow. The structure shown in
Figure 8.10 may be thought of as consisting of a set of parallel sheets connected to
one another through hydrogen bonding. Hexagonal ice has a moderately open structure
with a density of 0.931 g cm-3 near the triple point. Figure 8.10 also shows the crystal
structure of ice VI. All water molecules in this structure are hydrogen bonded to four
other molecules. Ice VII is much more closely packed than hexagonal ice and does not
float on liquid water.
As shown in Figure 8.9, phase diagrams can be quite complex for simple substances
because a number of solid phases can exist at different values of P and T. Another example is sulfur, which can also exist in several different solid phases. A portion of the
phase diagram for sulfur is shown in Figure 8.11, and the solid phases are described
by the symmetry of their unit cells. Note that several points correspond to three-phase
equilibria. By contrast, the CO2 phase diagram shown in Figure 8.12 is simpler. It is
similar in structure to that of water, but the solid–liquid coexistence curve has a positive slope. Several of the end-of-chapter problems and questions refer to the phase diagrams in Figures 8.11, 8.12, and 8.13.
Liquid
Pressure, P (atm)
1
Monoclinic
solid
153°C,
1420 atm
115.21°C,
1 atm
95.39°C,
1 atm
444.6°C,
1 atm
Rhombic
solid
(b) Ice VI has a much more compact
arrangement of water molecules
and thus a density much greater
than ice I (~1.3 g cm23).
95.31°C,
5.1 3 1026 atm
Figure 8.10
Crystal structures for two different
forms of ice. (a) Hexagonal ice, (b) Ice
VI. Ice VI is only stable at elevated pressures, as shown in the phase diagram
of Figure 8.9. The dotted lines indicate
­hydrogen bonds.
115.18°C,
3.2 3 1025 atm
Gas
Temperature
Figure 8.11
The P–T phase diagram for sulfur. The diagram is not to scale.
10000
Solid
Pressure, P (atm)
1000
Solid
Liquid
100
10
uid
1 Liq
Sublimation point
278.5°C at 1 atm
as
Critical point
d1G
Liqui
Triple point
256.6°C at 5.11 atm
1
Figure 8.12
The P–T phase diagram for CO2.
The diagram is not to scale.
So
0.01
lid
1
Ga
s
0.1
Gas
0.001
2140 2120 2100 280
260 240 220
0
20
Temperature, T (°C)
40
60
80
100
8.4 Pressure–Volume and Pressure–Volume–Temperature Phase Diagrams
233
8.3 THE PHASE RULE
As we have seen in Section 8.1, the coexistence of two phases, a and b, of a substance
requires that the chemical potentials are equal:
ma1T, P2 = mb1T, P2
(8.5)
ma1T,P2 = mb1T,P2 = mg1T,P2
(8.6)
The state variables T and P, which are independent in a single-phase region, are linked
through Equation (8.5) for a two-phase coexistence region. Consequently, only one of
T and P is needed to describe the system, and the other variable is determined by Equation (8.5). For this reason, the two-phase coexistence regions in Figures 8.4, 8.9, 8.11,
and 8.12 are curves that have the functional form P1T2 or T1P2. If three phases, a, b,
and g, coexist in equilibrium, then
Equation (8.6) imposes another requirement on the variables T and P. Now only one
combination of T and P satisfies the requirement posed by Equation (8.6), and threephase coexistence for a pure substance occurs only at a single point in the P, T phase
diagram called the triple point. These restrictions on the number of independent variables accessible to a system can be summarized as follows. Because T and P can be
varied independently in a single-phase region, a system of a pure substance has two
degrees of freedom. The degrees of freedom are the number of independent variables
needed to describe the physical state of the system. Because only one of T and P can be
varied independently along a two-phase coexistence curve, the system has one degree
of freedom. Finally, because neither T nor P can be varied at a triple point, such a system has zero degrees of freedom.
The American chemist J. W. Gibbs (1839–1903) derived the phase rule, which
links the number of degrees of freedom to the number of phases in a system at equilibrium. For a pure substance, the phase rule takes the form
F = 3 - p
(8.7)
Concept
The phase rule for a pure system
relates the number of phases that
can coexist at equilibrium and the
number of degrees of freedom
accessible to the system.
where F is the number of degrees of freedom and p is the number of phases. Gibbs
proved that no more than three phases of a pure substance can be in equilibrium as F
cannot be a negative number. As we will see in Chapter 9, the number of degrees of
freedom increases if a system contains several chemically independent species, for example, in a system consisting of ethanol and water.
8.4 PRESSURE–VOLUME AND PRESSURE–
Liquid
Solid + Liquid
Solid
In Section 8.2, the regions of stability and equilibrium among the solid, liquid, and gas
phases were described using a P9T phase diagram. Any phase diagram that includes
only two of the three state variables is limited because it does not contain information
on the third variable. We first complement the information contained in the P9T phase
diagram with a P–V phase diagram, and then we combine these two representations
into a P9V9T phase diagram. Figure 8.13 shows a P9V phase diagram for a substance
for which V liquid
7 V solid
m
m .
Significant differences are seen in the way that two- and three-phase coexistences
are represented in the two-phase diagrams. The two-phase coexistence curves of the
P9T phase diagram become two-phase regions in the P9V phase diagram because the
volume of a system in which two phases coexist varies with the relative amounts of
the material in each phase. For pressures well below the critical point, the range in V
over which the gas and liquid coexist is large compared to the range in V over which
gas
the solid and liquid coexist because V solid
6 V liquid
V Vm
. Therefore, the gas–liquid
m
m
coexistence region is broader than the solid–liquid coexistence region. Note that the triple point in the P9T phase diagram becomes a triple line in the P9V diagram. Although
P and T have unique values at the triple point, V can range between a maximum value
Pressure
VOLUME–TEMPERATURE PHASE DIAGRAMS
Critical
point
Liquid+Gas
Gas
a
Triple line
Solid + Gas
b
Volume
Figure 8.13
A P–V phase diagram. In this type of
diagram, single-phase regions, two-phase
coexistence regions, a critical point, and
a triple line are displayed.
234
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Concept
A P–V phase diagram displays the
information contained in a P–T phase
diagram in a different manner.
Concept
A P–V–T phase diagram is three
dimensional and displays all
information contained in P–T and
P–V phase diagrams.
for which the system consists almost entirely of gas with trace amounts of the liquid
and solid phases and a minimum value for which the system consists almost entirely of
solid with trace amounts of the liquid and gas phases.
The usefulness of the P9V diagram is illustrated by tracing several processes in
Figure 8.13. In process a, a solid is converted to a gas by increasing the temperature
in an isobaric process for which P is greater than the triple point pressure. This same
process was depicted as path a in Figure 8.4. In the P9V phase diagram, it is clear that
this process involves large changes in volume in the two-phase coexistence region over
which the temperature remains constant, which was not obvious in Figure 8.4. Process b
shows an isobaric transition from solid to gas for P below the triple point pressure, for
which the system has only one two-phase coexistence region (see path b in Figure 8.4).
Process b in Figure 8.13 is known as freeze drying. Assume that the food to be
freeze dried is placed in a vessel at -10°C, and the system is allowed to equilibrate at
this temperature. The partial pressure of H2O1g2 in equilibrium with the ice crystals in
the food under these conditions is 260 Pa. A vacuum pump is now started, and the gas
phase is pumped away. The temperature reaches a steady-state value determined by
heat conduction into the sample and heat loss through sublimation. The solid has an
equilibrium vapor pressure determined by the steady-state temperature. As the pump
removes water from the gas phase, ice spontaneously sublimes in order to keep the gasphase water partial pressure at its equilibrium value. After the sublimation is complete,
the food has been freeze dried. The advantage of freeze drying over boiling off water
at 1 atm is that food can be dehydrated at low temperatures, and thus the food is not
cooked.
All the information on the values of P, V, and T corresponding to single-phase regions, two-phase regions, and the triple point displayed in the P–T and P–V phase diagrams is best displayed in a three-dimensional P–V–T phase diagram. Such a diagram
is shown in Figure 8.14 for an ideal gas, which does not exist in the form of condensed
phases. It is easy to obtain the P–T and P–V phase diagrams from the P–V–T phase
diagram. The P–T phase diagram is a projection of the three-dimensional surface on the
P–T plane, and the P–V phase diagram is a projection of the three-dimensional surface
on the P–V plane.
Figure 8.15 shows a P–V–T diagram for a substance that expands upon melting. The
usefulness of the P–V–T phase diagram can be illustrated by revisiting the isobaric conversion of a solid to a gas at a temperature above the triple point shown as process a in
Figure 8.4. This process is shown as the path a S b S c S d S e S f in Figure 8.15.
We can see now that the temperature increases along the segments a S b, c S d, and
V1,V2,V3
P
V1
P
V2
V3
T1,T2,T3,T4
T1
T2
T3
T4
Pressure
T
V
Figure 8.14
A P–V–T diagram for an ideal gas.
Constant pressure (gray curves), constant
volume (red curves), and constant temperature (blue curves) paths are shown.
Vo
lum
e
tur
e
Te
m
ra
pe
8.5 PROVIDING A THEORETICAL BASIS FOR THE P–T PHASE DIAGRAM
Figure 8.15
P
m
P
Solid
Liquid
Critical
point
T
Gas
m
l
q
a
Pressure
g
Triple
point
A P–V–T phase diagram for a ­substance
that expands upon melting. The indicated processes are discussed in the text.
Solid–Liquid
f
Solid
b
f
Li
Ga
s
as
g
o
n
e
V
G
e
lid
–
Gas
d
Lin
So
lum
Critical
point
qu
i Gas id e
Tri
h
ple
p
Vo
Liquid
c
a
Liq
Ga uid
s
So
lid
–G
as
0
k
Solid–Liqu
id
Liq
uid
Critical
point
Solid
T2
T1
T3Tc
T4
ure
rat
e
mp
235
Te
e S f , all of which lie within single-phase regions, and it remains constant along the
segments b S c and d S e, which lie within two-phase regions. Similarly, process c in
Figure 8.4 is shown as the path g S h S i S k S l S m in Figure 8.15.
8.5 PROVIDING A THEORETICAL BASIS
FOR THE P–T PHASE DIAGRAM
In this section, a theoretical basis is provided for the coexistence curves that separate
different single-phase regions in the P–T phase diagram. Along the coexistence curves,
two phases are in equilibrium. From Section 5.6, we know that if two phases, a and b,
are in equilibrium at a pressure P and temperature T, their chemical potentials must be
equal:
ma1P, T2 = mb1P, T2
(8.8)
ma1P, T2 + dma = mb1P, T2 + dmb
(8.9)
dma = dmb
(8.10)
If the macroscopic variables are changed by a small amount, P, T S P + dP, T + dT,
such that the system pressure and temperature still lie on the coexistence curve, then
In order for the two phases to remain in equilibrium,
Because dm can be expressed in terms of dT and dP,
dma = -Sma dT + Vma dP and dmb = -Smb dT + Vmb dP
(8.11)
The expressions for m can be equated, giving
-Sma dT + Vma dP = -Smb dT + Vmb dP or
1Smb - Sma2 dT = 1Vmb - Vma2 dP
(8.12)
∆Sm
dP
=
dT
∆Vm
(8.13)
Assume that as P, T S P + dP, T + dT an incremental amount of phase a is transformed to phase b. In this case, ∆Sm = Smb - Sma and ∆Vm = Vmb - Vma. Rearranging Equation (8.12) gives the Clapeyron equation:
236
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
The Clapeyron equation allows us to calculate the slope of the coexistence curves in a
P–T phase diagram if ∆Sm and ∆Vm for the transition are known. This information is
necessary when constructing a phase diagram.
We next use the Clapeyron equation to estimate the slope of the solid–liquid coexistence curve. At the melting temperature,
∆ fusG = ∆ fus H - T∆ fus S = 0
(8.14)
Therefore, the ∆Sm values for the fusion transition can be calculated from the enthalpy
of fusion and the fusion temperature. Values of the normal fusion and vaporization
temperatures, as well as ∆Hm for fusion and vaporization at 1 atm pressure, are shown
in Table 8.2 for a number of different elements and compounds. Although there is a
significant variation in these values, for our purposes it is sufficient to use the average
value of ∆ fusS = 22 J mol-1 K-1 calculated from the data in Table 8.2 in order to estimate the slope of the solid–liquid coexistence curve.
For the fusion transition, ∆V is small because the densities of the solid and liquid states are quite similar. The average ∆ fusV for Ag, AgCl, Ca, CaCl2 K, KCl, Na,
NaCl, and H2O is +4.0 * 10-6 m3 mol-1. Of these substances, only H2O has a negative value for ∆ fusV. We next use the average values of ∆ fusS and ∆ fusV to estimate the
slope of the solid–liquid coexistence curve:
a
∆ fusS
dP
22 J mol-1 K-1
b =
≈
dT fus
∆ fusV
{4.0 * 10-6 m3mol-1
= {5.5 * 106 Pa K-1 = {55 bar K-1
(8.15)
Inverting this result, we see that 1dT > dP2fus ≈ {0.018 K bar . An increase of P by
∼50 bar is required to change the melting temperature by one degree. This result explains the very steep solid–liquid coexistence curve shown in Figure 8.4.
The same analysis applies to the liquid–gas coexistence curve. Because ∆ vapH
gas
and ∆ vapV = V m
- V liquid
are always positive, 1dP > dT2vap is always positive. The
m
average of ∆ vapS for the substances shown in Table 8.2 is 95 J mol-1 K-1. This value is
in accord with Trouton’s rule, which states that ∆ vapS ≈ 90 J mol-1 K-1 for liquids.
The rule fails for liquids in which there are strong interactions between molecules such
as -OH or -NH2 groups capable of forming hydrogen bonds. For these substances,
∆ vapS 7 90 J mol-1 K-1.
The molar volume of an ideal gas is approximately 20. L mol-1 in the temperature
gas
range in which many liquids boil. Because V m
W V liquid
, ∆ vapV ≈ 20.*10-3 m3mol-1,
m
the slope of the liquid–gas coexistence curve is given by
-1
Concept
Trouton’s rule is useful in predicting
∆ vapS for liquids.
a
∆ vapS
dP
95 J mol-1 K-1
b
=
≈
≈ 4.8 * 103 Pa K-1
dT vap
∆ vapV
2.0 * 10-2 m3 mol-1
= 4.8 * 10-2 bar K-1
3
(8.16)
This slope is a factor of 10 smaller than the slope for the liquid–solid coexistence
curve. Inverting this result, we find that 1dT > dP2vap ≈ 21 K bar -1. This result shows
that it takes only a modest increase in the pressure to increase the boiling point of a
liquid by a significant amount. For this reason, a pressure cooker or an automobile
radiator does not need to be able to withstand high pressures. Note that the slope of the
liquid–gas coexistence curve in Figure 8.4 is much less than that of the solid–liquid coexistence curve. The slope of both curves increases with T because ∆S increases with
increasing T and ∆V decreases with increasing P.
The solid–gas coexistence curve can also be analyzed using the Clapeyron equation. Because entropy is a state function, the entropy change for the processes solid
1P, T2 S gas1P, T2 and solid 1P, T2 S liquid 1P, T2 S gas 1P, T2 must be the same.
Therefore, ∆ subS = ∆ fusS + ∆ vapS 7 ∆Svap. However, because the molar volume of
the gas is so much larger than that of the solid or liquid, ∆ subV ≈ ∆ vapV. We conclude
that 1dP > dT2sub 7 1dP > dT2vap. Therefore, the slope of the solid–gas coexistence
curve will be greater than that of the liquid–gas coexistence curve. Because this comparison applies to a common value of the temperature, it is best made for temperatures
just above and just below the triple point temperature. This difference in slope of these
two coexistence curves is exaggerated in Figure 8.4.
8.6 USING THE CLAUSIUS–CLAPEYRON EQUATION TO CALCULATE VAPOR PRESSURE AS A FUNCTION OF T
8.6 USING THE CLAUSIUS–CLAPEYRON
EQUATION TO CALCULATE VAPOR
PRESSURE AS A FUNCTION OF T
From watching a pot of water as it is heated on a stove, it is clear that the vapor pressure of a liquid increases rapidly with increasing temperature. The same conclusion
holds for a solid below the triple point. To calculate the vapor pressure at different
temperatures, the Clapeyron equation must be integrated. Consider the solid–liquid
­coexistence curve:
Pf
Tf
Tf
Tf
Pi
Ti
Ti
Ti
∆ fusH dT
∆ fusS
∆ fusH dT
dP =
dT =
≈
∆
V
∆
V
T
∆ fusV L T
L
L fus
L fus
(8.17)
where the integration is along the solid–liquid coexistence curve. In the last step, it has
been assumed that ∆ fusH and ∆ fusV are independent of T over the temperature range
of interest. Assuming that 1Tf - Ti2>Ti is small, we can simplify the previous equation
to give
Pf - Pi =
∆ fusH Tf
∆ fusH Ti + ∆T
∆ fusH ∆T
ln =
ln
≈
∆ fusV Ti
∆ fusV
Ti
∆ fusV Ti
(8.18)
The last step uses the result ln11 + x2 ≈ x for x V 1, obtained by expanding
ln11 + x2 in a Taylor series about x = 0. We see that the vapor pressure of a solid
­varies linearly with ∆T in this limit. (The value of the slope dP>dT was discussed in
the previous section.)
For the liquid–gas coexistence curve, we have a different result because ∆V ≈ V gas.
Assuming that the ideal gas law holds, we obtain the Clausius–Clapeyron equation,
∆ vapS
∆ vapH
P∆ vapH
dP
=
≈
gas =
dT
∆ vapV
TV m
RT 2
∆ vapH dT
dP
=
P
R T2
(8.19)
Assuming that ∆ vapH remains constant over the range of temperature of interest, the
variation of the vapor pressure of the liquid with temperature is given by
Pf
Tf
Pi
Ti
∆ vapH
dP
dT
=
*
2
P
R
L
LT
ln
Pf
Pi
= -
∆ vapH
R
* a
1
1
- b
Tf
Ti
(8.20)
We see that the vapor pressure of a liquid rises exponentially with temperature. The
same procedure is followed for the solid–gas coexistence curve. The result is the same
as Equation (8.20), with ∆ subH substituted for ∆ vapH. Equation (8.20) provides a way
to determine the enthalpy of vaporization for a liquid by measuring its vapor pressure
as a function of temperature, as shown in Example Problem 8.2. In this discussion, it
has been assumed that ∆ vapH is independent of temperature. More accurate values of
the vapor pressure as a function of temperature can be obtained by fitting experimental
data. This leads to an expression for the vapor pressure as a function of temperature.
These functions for selected liquids and solids are listed in Tables 8.3 and 8.4 (see
­Appendix A, Data Tables).
Concept
The vapor pressure of a liquid
increases exponentially with
temperature.
237
238
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
EXAMPLE PROBLEM 8.2
The normal boiling temperature of benzene is 353.24 K, and the vapor pressure of
­liquid benzene is 1.19 * 104 Pa at 20.0°C. The enthalpy of fusion is 9.95 kJ mol-1,
and the vapor pressure of solid benzene is 137 Pa at -44.3°C. Calculate the following:
a. ∆ vapH
b. ∆ vapS
c. Triple point temperature and pressure
Solution
a. We can calculate ∆ vapH using the Clausius–Clapeyron equation because we know
the vapor pressure at two different temperatures:
ln
Pf
Pi
= -
∆ vapH 1
1
a - b
R
Tf
Ti
Pf
101,325 Pa
8.314 J mol-1 K-1 * ln
Pi
1.19 * 104 Pa
∆ vapH = = 1
1
1
1
a - b
a
b
Tf
Ti
353.24 K
273.15 K + 20.0 K
R ln
= 30.7 kJ mol-1
b. ∆ vapS =
∆ vapH
Tb
30.7 * 103 J mol-1
= 86.9 J mol-1 K-1
353.24 K
=
c. At the triple point, the vapor pressures of the solid and liquid are equal:
ln
P liquid
tp
P
ln
ln
R aln
8.314 J K-1 mol-1
= ln
∆ vapH 1
P liquid
1
i
a
- liquid b
~
P
R
Ttp
Ti
P solid
∆ subH 1
P solid
1
tb
i
=
ln
a
- solid b
~
~
P
P
R
Ttp
Ti
∆ vapH
∆ vapH - ∆ subH
∆ subH
P liquid
P solid
i
i
ln
+
=
~
~
solid
liquid
P
P
RTtp
RT i
RT i
∆ vapH - ∆ subH
Ttp =
=
~
∆ vapH
∆ subH
P liquid
P solid
i
i
- ln
+
b
~
~
solid
P
P
RT i
RT liquid
i
-9.95 * 103J mol-1
130.7 * 103 + 9.95 * 1032 J mol-1
1.19 * 104 Pa
137 Pa
ln
ln
8.314 J K-1 mol-1 * 228.9 K
105 Pa
105 Pa
* ±
≤
30.7 * 103 mol-1
+
8.314 J K-1mol-1 * 293.15 K
= 277 K
We calculate the triple point pressure using the Clapeyron equation:
ln
ln
ln
Pf
Pi
Ptp
P
~
Ptp
= = -
∆ vapH 1
1
a - b
R
Tf
Ti
30.7 * 103 J mol-1
1
1
* a
b
-1 -1
278
K
353.24
K
8.314 J mol K
= 8.70483
P~
Ptp = 6.03 * 103 Pa
8.7 Dependence of Vapor Pressure of a Pure Substance on Applied Pressure
239
8.7 DEPENDENCE OF VAPOR PRESSURE OF A
PURE SUBSTANCE ON APPLIED PRESSURE
To evaluate the effects of applied pressure on vapor pressure, consider the piston and
cylinder assembly shown in Figure 8.16. If the cylinder contains only water at 25°C,
the equilibrium vapor pressure of water at this temperature is P * = 3.16 * 103 Pa,
or 0.0316 bar. Therefore, if the weightless piston is loaded with a mass sufficient to
generate a pressure of 1 bar, the system is in the single-phase liquid region of the phase
diagram in Figure 8.4. This state of the system is shown in Figure 8.16a.
Next, the mass is reduced so that the pressure is exactly equal to the vapor pressure
of water. The system now lies in the two-phase liquid–gas region of the phase diagram
described by the liquid–gas coexistence curve. The piston can be pulled outward or
pushed inward while maintaining this pressure. This action leads to a larger or smaller
volume for the gas phase, but the pressure will remain constant at 3.16 * 103 Pa as
long as the temperature of the system remains constant. This state of the system is
shown in Figure 8.16b.
Finally, keeping the temperature constant, enough argon gas is introduced into the
cylinder such that the sum of the argon and H2O partial pressures is 1 bar. This state of
the system is shown in Figure 8.16c. What is the vapor pressure of water in this case,
and does it differ from that for the system shown in Figure 8.16b? The vapor pressure P
is used to denote the partial pressure of water in the gas phase, and P is used to denote
the sum of the argon and water partial pressures.
To calculate the partial pressure of water in the argon–water mixture, the following
equilibrium condition holds:
mliquid1T, P2 = mgas1T, P2
(8.21)
Mass
Piston
H2O(l)
Pext 5 1.00 bar
(a) Pressure is greater than
the vapor pressure, and
the contents consists of
only liquid water.
Mass
Piston
H2O(g)
H2O(l)
Pext 5 0.0316 bar
(b) Pressure is equal to the
vapor pressure, and the
contents consist of liquid
and gaseous water.
Differentiating this expression with respect to P, we obtain
a
0mliquid1T, P2
0P
b = a
T
0mgas1T, P2
0P
b a
T
0P
b
0P T
Mass
(8.22)
Piston
H2O(g)1Ar(g)
Because dm = -Sm dT + Vm dP, 1dm>dP2T = Vm, the previous equation becomes
gas
V liquid
= Vm
a
m
V liquid
0P
0P
m
b
or a b =
0P T
0P T
V gas
m
(8.23)
This equation shows that the vapor pressure P increases if the total pressure P increases.
gas
However, the rate of increase is small because V liquid
>V m
V 1. It is reasonable to rem
gas
place V m with the ideal gas value RT>P in Equation (8.23). This leads to the equation
P
P
RT
dP′
dP = V liquid
d P or RT
= V liquid
d P′
m
m
P
L* P′
L*
P
(8.24)
P
H2O(l)
Pext 5 1.00 bar
(c) Pressure is 1 bar, and the
contents consist of a
mixture of argon and
liquid and gaseous water.
Figure 8.16
A piston and cylinder assembly
­containing water at 298 K.
Integrating Equation (8.24) gives
RT ln a
P
P*
b = V liquid
1P - P *2
m
(8.25)
For the specific case of water, P * = 0.0316 bar, P = 1 bar, and V liquid
=
m
1.81 * 10-5 m3 mol-1:
ln a
P
P
b =
*
=
V liquid
1P - P *2
m
RT
1.81 * 10-5m3 mol-1 * 11 - 0.03162 * 105 Pa
8.314 J mol-1 K-1 * 298 K
= 7.04 * 10-4
P = 1.0007 P * ≈ 0.0316 bar
For an external pressure of 1 bar, the effect is negligible. However, for P = 100. bar,
P = 0.0339 bar, amounting to an increase in the vapor pressure of 7%.
Concept
The vapor pressure of a pure
substance depends on the applied
pressure.
240
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
8.8 SURFACE TENSION
Concept
Surface tension arises at an atomic
level because atoms at the surface
have fewer nearest neighbors than
those in the interior of a liquid or solid.
In discussing the liquid phase, the effect of the boundary surface on the properties of
the liquid has thus far been neglected. In the absence of a gravitational field, a liquid
droplet will assume a spherical shape because in this geometry the maximum number
of molecules is surrounded by neighboring molecules. Because the interaction between
molecules in a liquid is attractive, minimizing the surface-to-volume ratio minimizes
the energy. Therefore, increasing the area of a liquid droplet requires work to be done
on the droplet. How does the energy of the droplet depend on its surface area? Starting with the equilibrium spherical shape, assume that the droplet is distorted to create
more area while keeping the volume constant. The work associated with the creation of
­additional surface area ds at constant V and T is
(8.26)
dw = dA = gds
The proportionality constant, g, linking dw and ds in Equation 8.26 is called the
s­ urface tension. It has the units of energy/area or J m-2, which is equivalent to N m-1.
Equation (8.26) predicts that a liquid, a bubble, or a liquid film suspended in a wire
frame will tend to minimize its surface area because dA 6 0 for a spontaneous process
at constant V and T.
Consider the spherical droplet depicted in Figure 8.17. There must be a force acting on the droplet in the radially inward direction for the liquid to assume a spherical
shape. An expression for the force can be generated as follows. If the radius of the
droplet is increased from r to r + dr, the area increases by ds.
s = 4pr 2 so ds = 8pr dr
(8.27)
From Equation (8.26), the work done in the expansion of the droplet is 8pgr dr. The
force, which is normal to the surface of the droplet, is the work divided by the distance or
(8.28)
F = 8pgr
Figure 8.17
The forces acting on a spherical droplet
that arise from surface tension.
The net effect of this force is to generate a pressure differential across the droplet surface. At equilibrium, there is a balance between the inward and outward acting forces.
The inward acting force is the sum of the force exerted by the external pressure and the
force arising from the surface tension, whereas the outward acting force arises solely
from the pressure in the liquid:
4pr 2Pouter + 8pgr = 4pr 2Pinner or
2g
Pinner = Pouter +
(8.29)
r
Note that Pinner - Pouter S 0 as r S ∞ . Therefore, the pressure differential exists only
for a curved surface. From the geometry in Figure 8.17, it is apparent that the larger
pressure is always on the concave side of the interface. Values for the surface tension
for a number of liquids are listed in Table 8.5.
TABLE 8.5 Surface Tension of Selected Liquids at 298 K
Formula
Name
Br2
Bromine
H2O
Water
Hg
Mercury
CCl4
Carbon
tetrachloride
CH3OH
Methanol
g 1mN m-12
g 1mN m-12
Formula
Name
40.95
CS2
Carbon
disulfide
71.99
C2H5OH
Ethanol
21.97
C6H5N
Pyridine
36.56
26.43
C6H6
Benzene
28.22
22.07
C8H18
Octane
21.14
485.5
31.58
Source: Data from Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL:
CRC Press, 2002.
8.8 Surface Tension
Equation (8.29) has interesting implications for the relative stabilities of bubbles
with different curvatures 1>r. Consider two air-filled bubbles of a liquid with the same
surface tension g, one with a large radius R1 and one with a smaller radius R2. Assume
there is a uniform pressure outside both bubbles. From Equation (8.29) we obtain the
difference between the pressures P1 and P2 within each bubble:
P1 - P2 =
2g
2g
1
1
= 2g a
b
R1
R2
R1
R2
2g
rgr
r2
r1
(8.30)
Now suppose the two bubbles come into contact as shown at the top of Figure 8.18.
Because R1 7 R2, from Equation (8.30) the pressure P2 inside bubble 2 (with respect
to some common exterior pressure) is greater than the pressure P1 in bubble 1. Thus,
air will flow from bubble 2 to bubble 1 until the smaller bubble disappears entirely, as
shown at the bottom of Figure 8.18. In foams this process is called coarsening, and in
crystals it is called Ostwald ripening.
A biologically relevant issue that follows from Equations (8.29) and (8.30) is the
stability of lung tissue. Lung tissue is composed of small water-lined, air-filled chambers called alveoli. According to Equation (8.29), the pressure difference across the
surface of a spherical air-filled cavity is proportional to the surface tension and inversely proportional to the radius of the cavity. Therefore, if two air-filled alveoli,
­approximated as spheres with radii r and R and for which r 6 R are interconnected,
the smaller alveolus will collapse and the larger alveolus will expand because the
pressure within the smaller alveolus is greater than the pressure within the larger alveolus. In theory, because of the surface tension of water that lines the alveoli, these
cavities will coarsen (as defined above) as smaller alveoli collapse, leaving an ever
diminishing number of alveoli of increasing size. Coarsening of the alveoli according
to ­Equations (8.29) will result eventually in collapse of the lungs.
In the alveoli, a lipid surfactant called phosphatidylcholine lowers the surface tension of the water lining the alveoli, thus stabilizing the lungs. But in persons lacking
sufficient surfactant in the lining of their lungs, such as prematurely born infants and
heavy smokers, the surface tension of the water in the alveoli is not lowered, and as a
consequence lung tissue is destabilized by coarsening.
Another consequence of the pressure differential across a curved surface is that
the vapor pressure of a droplet depends on its radius. By substituting numbers in
­Equation (8.29) and using Equation (8.25) to calculate the vapor pressure, we find that
the vapor pressure of a 10-7 m water droplet is increased by 1%, that of a 10-8 m
droplet is increased by 11%, and that of a 10-9 m droplet is increased by 270% relative to a droplet with zero surface tension. [At such a small diameter, the application
of Equation (8.29) is questionable because the size of an individual water molecule is
comparable to the droplet diameter. Therefore, a microscopic theory is needed to describe the forces within the droplet.] This effect plays a role in the formation of liquid
droplets in a condensing gas such as fog. Small droplets evaporate more rapidly than
large droplets, and the vapor condenses on the larger droplets, allowing them to grow at
the expense of small droplets.
Capillary rise and capillary depression are other consequences of the pressure differential across a curved surface. Assume that a capillary of radius r is partially immersed
in a liquid. When the liquid comes in contact with a solid surface, there is a natural tendency to minimize the energy of the system. If the surface tension of the liquid is lower
than that of the solid, the liquid will wet the surface, as shown in Figure 8.19a. However,
if the surface tension of the liquid is higher than that of the solid, the liquid will avoid
the surface, as shown in Figure 8.19b. In either case, there is a pressure differential in the
capillary across the gas–liquid interface because the interface is curved. If we assume
that the liquid–gas interface is tangent to the interior wall of the capillary at the solid–­
liquid interface, the radius of curvature of the interface is equal to the capillary radius.
The difference in the pressure across the curved interface 2g>r is balanced by the
weight of the column in the gravitational field rgh. Therefore, 2g>r = rgh, and the
capillary rise or depression is given by
h =
241
(8.31)
Airflow
Figure 8.18
Interaction of two bubbles with
­unequal radii. In the top diagram,
bubbles with radii R1 and R2 make contact
and interact. Because the pressure within
a bubble varies inversely with the bubble
radius, air flows from the smaller into the
larger bubble (middle diagram) until a
single large bubble remains (bottom).
242
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Capillary rise
Pouter
Pouter
In the preceding discussion, it was assumed that either (1) the liquid completely
wets the interior surface of the capillary, in which case the liquid coats the capillary
walls but does not fill the core, or (2) the liquid is completely nonwetting, in which
case the liquid does not coat the capillary walls but fills the core. In a more realistic
model, the interaction is intermediate between these two extremes. In this case, the
liquid–­surface is characterized by the contact angle u, as shown in Figure 8.20.
Complete wetting corresponds to u = 0°, and complete nonwetting corresponds
to u = 180°. For intermediate cases,
Pinner = Pouter +
Pinner 1 rgh
h
2g cos u
2g cos u
and h =
r
rgr
(8.32)
Measurement of the contact angle is one of the main experimental methods used to
measure the difference in surface tension at the solid–liquid interface.
EXAMPLE PROBLEM 8.3
The six-legged water strider supports itself on the surface of a pond on four of its legs.
Each of these legs causes a depression to be formed in the pond surface. Assume that
each depression can be approximated as a hemisphere of radius 1.2 * 10-4 m and that
u (as in Figure 8.20) is 0°. Calculate the force that one of the insect’s legs exerts on the
pond.
(a) If the liquid wets the interior wall
of the capillary, a capillary rise is
observed. The combination
Pyrex–water exhibits this behavior.
Solution
2g cos u
2 * 71.99 * 10-3 N m-1
∆P =
=
= 1.20 * 103 Pa
r
1.2 * 10-4 m
F = PA = P * pr 2 = 1.20 * 103 Pa * p11.2 * 10-4 m22 = 5.4 * 10-5 N
EXAMPLE PROBLEM 8.4
Water is transported upward in trees through channels in the trunk called xylem.
­Although the diameter of the xylem channels varies from species to species, a typical
value is 2.0 * 10-5 m. Is capillary rise sufficient to transport water to the top of a redwood tree that is 100 m high? Assume complete wetting of the xylem channels.
Pouter
Solution
From Equation (8.31),
Pouter
h
Capillary
depression
Pouter 1rgh
(b) If the liquid does not wet the capillary
surface, a capillary depression is
observed. The combination
Pyrex–mercury exhibits this behavior.
Figure 8.19
Capillary rise and depression in a
capillary partially i­mmersed in a liquid.
(a) Capillary rise, (b) capillary depression.
h =
2g
2 * 71.99 * 10-3 N m-1
=
= 0.74 m
rgr cos u
997 kg m-3 * 9.81 m s-2 * 2.0 * 10-5 m * 1
No, capillary rise is not sufficient to account for water supply to the top of a
redwood tree.
As Example Problem 8.4 shows, capillary rise is insufficient to account for water
transport to the leaves in all but the smallest plants. The property of water that accounts for water supply to the top of a redwood is its large tensile strength. Imagine
pulling on a piston in a cylinder containing only liquid water to create a negative pressure. How hard can a person pull on the water without “breaking” the water column?
The answer to this question depends on whether bubbles are nucleated in the liquid.
This phenomenon is called cavitation. If cavitation occurs, the bubbles will grow rapidly as the piston is pulled outward. The bubble pressure is given by Equations (8.31),
where Pext is the vapor pressure of water. The height of the water column in this case
is limited to about 9.7 m. However, bubble nucleation is a kinetic phenomenon initiated at certain sites at the wall surrounding the water, and under the conditions present
in xylem tubes it is largely suppressed. In the absence of bubble nucleation, theoretical calculations predict that negative pressure in excess of 1000 atm can be generated.
8.9 Chemistry in Supercritical Fluids
The pressure is negative because the water is under tension rather than compression.
­Experiments on water inclusions in very small cracks in natural rocks have verified
these estimates. However, bubble nucleation occurs at much lower negative pressures
in capillaries similar in d­ iameter to xylem tubes. Even in these capillaries, negative
pressures of more than 50 atm have been observed.
How does the high tensile strength of water explain the transport of water to the top
of a redwood? If one cuts into a tree near its base, the sap oozes rather than spurts out,
showing that the pressure in the xylem tubes is ∼1 atm at the base of a tree, whereas
the pressure at the base of a tall water column would be much greater than 1 atm.
Imagine the redwood in its infancy as a seedling. Capillary rise is sufficient to fill the
xylem tubes to the top of the plant. As the tree grows taller, the water is pulled upward
because of its large tensile strength. As the height of the tree increases, the pressure at
the top becomes increasingly negative. As long as cavitation does not occur, the water
column remains intact. As water evaporates from the leaves, it is resupplied from the
roots through the pressure gradient in the xylem tubes that arises from the weight of the
column. If the tree (and each xylem tube) grows to a height of ∼100 m, and P = 1 atm
at the base, the pressure at the top must be ∼ -9 atm, from ∆P = rgh. We again encounter a negative pressure because the water is under tension. If water did not have
a sufficiently high tensile strength, gas bubbles would form in the xylem tubes. This
would disrupt the flow of sap, and tall trees could not exist.
S U PPL EM EN TAL
243
u
Figure 8.20
The shape of the liquid–air interface in a
capillary. For cases intermediate between
wetting and nonwetting, the contact angle u
lies in the range 0° 6 u 6 180°.
SEC TI O N
8.9 CHEMISTRY IN SUPERCRITICAL FLUIDS
Chemical reactions can take place in the gas phase, in solution (homogeneous reactions), or on the surfaces of solids (heterogeneous reactions). Solvents with suitable
properties can influence both the yield and the selectivity of reactions in solution. It
has been found that the use of supercritical fluids as solvents increases the number
of parameters that chemists have at their disposal to tune a reaction system to meet
their needs.
Supercritical fluids (SCFs) near the critical point with reduced temperatures and
pressures (see Section 7.4) in the range Tr ∼ 1.091.1 and Pr ∼ 192 have a density that is
an appreciable fraction of the liquid-phase density. They are unique in that they exhibit
favorable properties of liquids and gases. Because the density is high, the solubility of
solid substances is quite high, but the diffusion of solutes in the fluid is greater than
that in the normal liquid. This is the case because the density of the SCF is lower than
that in normal liquids. For similar reasons, the viscosity of SCFs is lower than that in
normal liquids. As a result, mass transfer is faster, and the overall reaction rate can be
increased. Because an SCF is more gas-like than a liquid, the solubility of gases can
be much higher in the SCF. This property is of particular usefulness in enhancing the
reactivity of reactions in which one of the reactants is a gas, such as oxidation or hydrogenation. For example, hydrogen generally has a low solubility in organic solvents but
is quite soluble in organic SCFs.
Carbon dioxide and water exhibit the unusual properties of SCFs. Because the values for the critical pressure and temperature of CO2 1Pc = 73.74 bar and Tc = 304 K2
are easily attainable, it is possible to use chemically inert supercritical CO2 to replace
toxic organic solvents in the dry cleaning industry. Although supercritical CO2 is a
good solvent for nonpolar molecules, it must be mixed with other substances to achieve
a required minimum solubility for polar substances. This can be done without increasing the toxicity of the process greatly. Supercritical CO2 is also used in the decaffeination of coffee and tea.
The critical constants of H2O are considerably higher 1Pc = 220.64 bar and
Tc = 647 K2, so the demands on the vessel used to contain the supercritical fluid are
greater than for CO2 However, as supercritical H2O is formed, many of the hydrogen
bonds in the normal liquid are broken. As a result, the dielectric constant can be varied
between the normal value for the liquid of 80 and a value of 5. At the lower end of
Concept
Supercritical fluids have unique
properties than can be used to carry
out reactions that would not be
possible under “normal” conditions.
244
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
this range, supercritical water acts like a nonpolar solvent and is effective in dissolving organic materials. Supercritical water has considerable potential for use at the high
temperatures at which organic solvents begin to decompose. For example, it can be
used to destroy toxic substances through oxidation in the remediation of contaminated
groundwater. One challenge in using supercritical water is that it is highly corrosive,
placing demands on the materials used to construct a practical facility.
SU P P LE ME NTAL
SE CTION
8.10 LIQUID CRYSTAL DISPLAYS
CN
O2
N1
H3CO
OCH3
N
Figure 8.21
Examples of molecules that form liquid
crystals. Liquid crystals are generally
formed from polar organic molecules with
a rod-like shape.
Liquid crystals are an exception to the general statement (superfluid He is another) that
there are three equilibrium states of matter, namely, solids, liquids, and gases. Glass is
a liquid of such high viscosity that it cannot achieve equilibrium on the timescale of a
human lifetime. Glasses are a commonly encountered nonequilibrium state of matter.
An example is SiO2 in the form of window glass. The properties of liquid ­crystals
are intermediate between liquids and solids. Molecules that form liquid crystals typically contain rod-shaped structural elements. Several such molecules are shown in
Figure 8.21.
How do liquid crystals differ from the other states of matter? The ordering in
solid, liquid, and liquid crystal phases of such molecules is shown schematically in
­Figure 8.22. Whereas the crystalline solid phase is perfectly ordered and the liquid
phase has no residual order, the liquid crystal phase retains some order. The long axes
of all molecules deviate only by a small amount from their value in the liquid crystal
structure (Figure 8.22c).
The twisted nematic phase shown in Figure 8.23 is of particular importance because liquid crystals with this structure are the basis of the multibillion-dollar liquid
crystal display (LCD) industry and also the basis for sensor strips that change color
with temperature. The twisted nematic phase consists of parallel planes in which the
angle of the preferred orientation direction varies in a well-defined manner as the number of layers increases. If light is incident on such a crystal, it is partially reflected and
partially transmitted from a number of layers. A maximum in the reflection occurs if
the angle of incidence and the spacing between layers are such that constructive interference occurs in the light reflected from successive layers. Because this occurs only
in a narrow range of wavelengths, the film appears colored, with the color determined
by the wavelength. As temperature increases or decreases, the layer spacing and, therefore, the color of the reflected light change. For this reason, the color of the liquid crystal film changes with temperature.
The way in which an LCD display functions is illustrated in Figure 8.24. A twisted
nematic liquid crystal film is sandwiched between two transparent conducting
Figure 8.23
Schematic molecular diagram of a
twisted nematic phase. The direction
of preferential ­orientation of the rod-like
molecules varies in a well-­defined manner
from one plane to the next.
(a) Solid
(b) Liquid
(c) Liquid crystal
Figure 8.22
Schematic molecular structures of (a) solid, (b) liquid, and (c) liquid crystal phases.
Vocabulary
Figure 8.24
Unpolarized light
Schematic diagram of an LCD display.
The display consists of a twisted nematic
liquid film enclosed between parallel
transparent conducting electrodes. Polarizers whose transmission direction is
rotated 90° with respect to one another are
mounted on the electrodes. (a) Polarized
light transmitted, (b) no light transmitted,
(c) resulting display.
Unpolarized light
Polarizer
Polarizer
E50
Electrodes
245
Electrodes
(voltage on)
E
Analyzer
Analyzer
Polarized light
(a) Light passing through the
first polarizer is transmitted
by the second polarizer
because the plane of
polarization of the light
is rotated by the crystal.
No light
(b) The orientational ordering
of the twisted nematic
phase is destroyed by
application of the electric
field. No light is passed.
(c) Arrangement of
electrodes in an
LCD alphanumeric
display.
electrodes. The thickness of the film is such that the orientational direction rotates by
90° from the bottom to the top of the film. Ambient light is incident on an upper polarizing filter that allows only one direction of polarization to pass. The rod-like molecules act as a waveguide and rotate the plane of polarization as the light passes through
the liquid crystal film. Therefore, light passes through the lower polarizer, which is
rotated by 90° with respect to the upper polarizer.
If an electric field is applied to the electrodes, the rotation of the orientational direction from plane to plane no longer occurs because the polar molecules align along
the electric field. Therefore, the plane of polarization of the light transmitted by the
upper polarizer is not rotated as it passes through the film. As a result, the light is not
able to pass through the second polarizer. Now imagine the lower electrode to be in
the form of a mirror. In the absence of the electric field (Figure 8.24a), the display will
appear bright because the plane of polarization of the light reflected from the mirror is
again rotated 90° as it passes back through the film. Consequently, it passes through the
upper polarizer. By contrast, no light is reflected if the field is on (Figure 8.24b), and
the display appears dark. Now imagine each of the electrodes to be patterned as shown
in Figure 8.24c, and it will be clear why, in a liquid crystal watch display, dark numbers are observed on a light background.
Concept
Liquid crystal displays are possible
because of the ways that liquid
crystals transmit polarized light.
VOCABULARY
boiling point elevation
capillary depression
capillary rise
Clapeyron equation
Clausius–Clapeyron equation
coexistence curve
contact angle
critical point
degrees of freedom
freeze drying
freezing point depression
freezing point elevation
glass
LCD display
liquid crystals
metastable phase
nonwetting
normal boiling temperature
phase
phase diagram
phase rule
P–T phase diagram
P–V phase diagram
P–V–T phase diagram
standard boiling temperature
sublimation temperature
supercritical fluids
surface tension
tensile strength
triple point
Trouton’s rule
twisted nematic phase
vapor pressure
wetting
246
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
KEY EQUATIONS
Equation
Significance of Equation
liquid
Sgas
7 Ssolid
m 7 Sm
m
Order of states of matter based on entropy values
8.3
∆H = ∆ sub H = ∆ fus H + ∆ vap H
Relation between enthalpies of fusion, vaporization, and sublimation
8.4
ma1T,P2 = mb1T,P2 = mg1T,P2
Condition for coexistence of three phases
8.6
F = 3 - p
Phase rule for a pure substance
8.7
∆Sm
dP
=
dT
∆Vm
Clapeyron equation allows calculation of slope of coexistence curves
8.13
Change in vapor pressure of liquid with temperature
8.20
Change in vapor pressure of liquid with applied pressure
8.25
Difference in pressure on liquid across a curved interface due to
­surface tension
8.29
Capillary rise as a function of contact angle and curvature of the
­interface
8.32
ln
Pf
Pi
= -
RT ln a
P
P*
∆ vap H
R
1
1
- b
Tf Ti
b = V liquid
1P - P *2
m
Pinner = Pouter +
h =
* a
2g cos u
rgr
2g
r
Equation Number
CONCEPTUAL PROBLEMS
Q8.1 Why is it reasonable to show the m versus T segments
for the three phases as straight lines as is done in Figure 8.1?
More realistic curves would have some curvature. Is the curvature upward or downward on a m versus T plot? Explain
your answer.
Q8.2 Why do the temperature versus heat curves in the solid,
liquid, and gas regions of Figure 8.7 have different slopes?
Q8.3 Figure 8.7 is not drawn to scale. What would be the
relative lengths on the qP axis of the liquid + solid, liquid,
and liquid + gas segments for water if the drawing were to
scale and the system consisted of H2O?
Q8.4 Show the paths n S o S p S q and a S b S c S
d S e S f of the P–V–T phase diagram of Figure 8.15 in the
P–T phase diagram of Figure 8.4.
Q8.5 At a given temperature, a liquid can coexist with its
vapor at a single value of pressure. However, you can measure the presence of H2O1g2 above the surface of a lake, and
H2O1g2 is still there if the barometric pressure rises or falls at
constant temperature. How is this possible?
Q8.6 Why are the triple point temperature and the normal
freezing point very close in temperature for most substances?
Q8.7 Give a molecular-level explanation as to why the
­surface tension of Hg(l) is not zero.
Q8.8 A vessel containing a liquid is opened inside an evacuated chamber. Will you see a liquid–gas interface if the volume of the initially evacuated chamber is (a) less than the
critical volume, (b) much larger than the critical volume, and
(c) slightly larger than the critical volume.
Q8.9 Why are there no points in the phase diagram for sulfur
in Figure 8.11 that show rhombic and monoclinic solid phases
in equilibrium with liquid and gaseous sulfur?
Q8.10 What is the physical origin of the pressure difference
across a curved liquid–gas interface?
Q8.11 A triple point refers to a point in a P–T phase diagram
for which three phases are in equilibrium. Do all triple points
correspond to a gas–liquid–solid equilibrium?
Q8.12 Why does the liquid–gas coexistence curve in a
P–T phase diagram end at the critical point?
Q8.13 How can you get a P–T phase diagram from a
P–V–T phase diagram?
Q8.14 Why does the triple point in a P–T phase diagram
­become a triple line in a P–V phase diagram?
Q8.15 Why does water have several different solid phases
but only one liquid and one gaseous phase?
Q8.16 As the pressure is increased at -45°C, ice I is converted to ice II. Which of these phases has the lower density?
Q8.17 Why is ∆ subHm = ∆ fusHm + ∆ fusHm?
Q8.18 What can you say about the density of liquid and
­gaseous water as the pressure approaches the critical value
from lower (gas) and higher (liquid) values? Assume that the
temperature is constant at the critical value.
Numerical Problems
Q8.19 What can you say about ∆ vapH of a liquid as the
t­emperature approaches the critical temperature?
Q8.20 Is the following statement correct? Because dry ice
sublimes, carbon dioxide has no liquid phase. Explain your
answer.
Q8.21 Make a drawing indicating the three-step process d of
Figure 8.4 in Figure 8.13.
Q8.22 Consider the phase diagram of Figure 8.4 and answer
the following questions.
a. Which phase is present if P is increased at constant T
­starting at the triple point?
247
b. Which phase is present if P is increased at constant T
­starting at the normal boiling point?
c. Which phase is present if T is decreased at constant P
starting at the critical point?
d. Are any of your answers different if the substance under
consideration is water?
Q8.23 Use Trouton’s rule to predict the enthalpy of
vaporization for H2O1l2, CCl41l2, CH3OH1l2, C3H81l2, and
C6H61l2 using boiling temperatures from Table 8.2. Are your
estimated enthalpies high or low? Explain the trends in your
results.
NUMERICAL PROBLEMS
Section 8.2
P8.1 Within what range can you restrict the values of P
and T if the following information is known about CO2? Use
Figure 8.12 to answer this question.
a. As temperature is increased, the solid is first converted to
liquid and subsequently to the gaseous state.
b. As pressure on a cylinder containing pure CO2 is increased
from 65 to 80. atm, no interface delineating liquid and gaseous phases is observed.
c. Solid, liquid, and gas phases coexist at equilibrium.
d. An increase in pressure from 10. to 50. atm converts the
liquid to the solid.
e. An increase in temperature from -80.° to 20.°C converts a
solid to a gas with no intermediate liquid phase.
P8.2 A P–T phase diagram for potassium is shown in the
­following diagram.
POTASSIUM
fcc (II)
12
8
Diamond
Liquid
bcc (I)
4
Liquid
Pressure
Pressure, P (GPa)
16
P8.3 It has been suggested that the surface melting of
ice plays a role in enabling speed skaters to achieve peak
performance. Carry out the following calculation to test
this hypothesis. At 1 atm pressure, ice melts at 273.15 K,
∆ fus H = 6010 J mol-1, the density of ice is 920. kg m-3, and
the density of liquid water is 997 kg m-3.
a. What pressure is required to lower the melting temperature
by 4.40°C?
b. Assume that the width of the skate in contact with the ice
has been reduced by sharpening to 18.0 * 10-3 cm and
that the length of the contact area is 20.0 cm. If a skater
of mass 60 kg is balanced on one skate, what pressure is
­exerted at the interface of the skate and the ice?
c. What is the melting point of ice under this pressure?
d. If the temperature of the ice is -4.40°C, do you expect
melting of the ice at the ice-skate interface to occur?
P8.4 Answer the following questions using the P–T phase
diagram for carbon sketched in the following diagram.
Graphite
0
Vapor
0
200
400
600
Temperature, T (K)
800
a. Which phase has a greater density, the fcc or the bcc
phase? Explain your answer.
b. Indicate the range of P and T in the phase diagram for
which fcc and liquid potassium are in equilibrium. Does
fcc potassium float on or sink in liquid potassium? Explain
your answer.
c. Extend P and T on this diagram and indicate where you
might expect to find the vapor phase. Explain how you
choose the slope of your liquid–vapor coexistence curve.
Temperature
a. Which substance is denser, graphite or diamond? Explain
your answer.
b. Which phase is denser, graphite or liquid carbon? Explain
your answer.
c. Why does the phase diagram have two triple points?
­Explain your answer.
248
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
Liquid
Solid
Pressure
Pressure
P8.5 Are the two P–T phase diagrams shown in this problem
likely to be observed for a pure substance? If not, explain all
features of the diagram that will not be observed.
Vapor
Liquid
Solid
Vapor
Temperature
Temperature
(a)
(b)
P8.6 The phase diagram of NH3 can be characterized by the
following information. The normal melting and boiling temperatures are 195.2 and 239.82 K, respectively, whereas the
triple point pressure and temperature are 6077 Pa and 195.41
K, respectively. The critical point parameters are 112.8 * 105
Pa and 405.5 K. Make a sketch of the P–T phase diagram (not
necessarily to scale) for NH3. Place a point in the phase diagram for the following conditions. State which and how many
phases are present.
a. 195.41 K, 1050 Pa
b. 195.41 K, 6077 Pa
c. 237.51 K, 101325 Pa
d. 420. K, 130 * 105 Pa
e. 190. K, 6077 Pa
P8.7 Prove that a substance for which the solid–liquid coexistence curve has a positive slope expands upon melting.
P8.8 Are the two P–T phase diagrams shown here likely to
be observed for a pure substance? If not, explain all features
of the diagram that will not be observed.
Solid
II
Vapor
Solid
Pressure
Pressure
Liquid
Solid
I
Vapor
Temperature
Temperature
(a)
Liquid
(b)
P8.9 Within what range can you restrict the values of P
and/or T if the following information is known about sulfur?
Use Figure 8.11 to answer this problem.
a. Only the monoclinic solid phase is observed for
P = 1 atm.
b. When the pressure on the vapor is increased, the liquid
phase is formed.
c. Solid, liquid, and gas phases coexist at equilibrium.
d. As temperature is increased, the rhombic solid phase is
converted to the liquid directly.
e. As temperature is increased at 1 atm, the monoclinic solid
phase is converted to the liquid directly.
P8.10 The normal melting point of H2O is 273.15 K, and
∆ fusH = 6010 J mol-1. Calculate the change in the normal
freezing point at 20. bar and 100. bar, assuming that the
­density of the liquid and solid phases remains constant at 997
and 917 kg m-3, respectively. Explain why your answer is
positive or negative.
P8.11 The densities of a given solid and liquid of molar mass
122.5 g mol-1 at its normal melting temperature of 432.5 K
are 1075 and 1012 kg m-3, respectively. If the pressure is
increased to 150. bar, the melting temperature increases to
437.0 K. Calculate ∆ fusH and ∆ fusS for this substance.
P8.12 In this problem, you will calculate the differences in
the chemical potentials of ice and supercooled water and of
steam and superheated water, all at 1 atm pressure, shown
schematically for an arbitrary substance in Figure 8.1. For
this problem, S ~ 1H2O, s2 = 48.0 J mol-1 K-1, S ~ 1H2O, l2
= 70.0 J mol-1 K-1, and S ~ 1H2O, g2 = 188.8 J mol-1 K-1.
a. By what amount does the chemical potential of water
­exceed that of ice at -1.95°C?
b. By what amount does the chemical potential of water
­exceed that of steam at 101.95°C?
c. Which is the most stable phase in parts (a) and (b)?
P8.13 15.0 g of water is in a container of 25.0 L at 298.15 K.
The vapor pressure of water at this temperature is 0.0313 atm.
a. What phases are present?
b. At what volume would only the gas phase be present?
c. At what volume would only the liquid phase be present?
P8.14 Consider the transition between two forms of solid tin,
Sn1s,gray2 N Sn1s,white2. The two phases are in equilibrium
at 1 bar and 18°C. The densities for gray and white tin are
5750 and 7280 kg m-3, respectively, and the molar entropies for gray and white tin are 44.14 and 51.18 J K-1 mol-1,
­respectively. Calculate the temperature at which the two
phases are in equilibrium at 425 bar.
Section 8.6
P8.15 Use the vapor pressures of CCl4(l) given in the following table to calculate the enthalpy of vaporization using a
graphical method or a least squares fitting routine.
T 1K2
280.
290.
300.
310.
P1 Pa2
6440.
10540
16580
25190
T 1K2
320.
330.
340
P1 Pa2
37130.
53250.
74520.
P8.16 The vapor pressure of benzene(l) is given by
ln a
P
2.7738 * 103
b = 20.767 Pa
T
- 53.08
K
a. Calculate the standard boiling temperature.
b. Calculate ∆ vap H at 298 K and at the standard boiling
­temperature.
Numerical Problems
P8.17 Use the vapor pressures for tetrachloromethane given
in the following table to estimate the temperature and pressure of the triple point as well as the enthalpies of fusion,
vaporization, and sublimation.
T 1K2
Phase
Solid
Solid
Liquid
Liquid
230.
250.
280.
340.
T 1K2
Solid
Solid
Liquid
Liquid
160
170
275
305
P1 Pa2
0.0185
0.204
667
2690
P8.20 Autoclaves that are used to sterilize surgical tools require
a temperature of 135°C to kill some bacteria. If water is used
for this purpose, at what pressure must the autoclave operate?
P8.21 You have collected a tissue specimen that you
would like to preserve by freeze drying. To ensure the
­integrity of the specimen, the temperature should not exceed
-5.50°C. The vapor pressure of ice at 273.16 K is 624 Pa;
∆ fusH = 6.010 kJ mol-1 and ∆ vap H = 40.656 kJ mol-1.
What is the maximum pressure at which the freeze drying can
be carried out?
P8.22 Use the vapor pressures of ice given here to calculate
the enthalpy of sublimation using a graphical method or a
least squares fitting routine.
T 1K2
200.
210.
220.
230.
240.
250.
260.
P1 Torr2
0.1676
0.7233
2.732
9.195
27.97
77.82
200.2
P8.23 The vapor pressure of a liquid can be written in the
empirical form known as the Antoine equation, where A112,
A122, and A132 are constants determined from measurements:
P1T2
A122
= A112 Pa
T
+ A132
K
Starting with this equation, derive an equation giving ∆ vap H
as a function of temperature.
ln
T 1K2
100.
120.
140.
160.
180.
200.
220.
225.7
905
6440
62501
P8.19 Use the vapor pressures for hexane given in the following
table to estimate the temperature and pressure of the triple point
as well as the enthalpies of fusion, vaporization, and sublimation.
Phase
P8.24 Use the following vapor pressures of propane to calculate the enthalpy of vaporization using a graphical method or
a least squares fitting routine.
P1 Pa2
P8.18 You have a compound dissolved in benzene and need
to remove the solvent by distillation. Because the compound
is heat sensitive, you hesitate to raise the temperature above
8.50°C and decide on vacuum distillation. What is the maximum pressure at which the distillation takes place? Use the
data in Table 8.3 (see Appendix A, Data Tables).
249
P1 Pa2
0.01114
2.317
73.91
838.0
5054
2.016 * 104
6.046 * 104
P8.25 The vapor pressure of liquid toluene is 5333 Pa at
308.15 K and ∆ vap H = 38.06 kJ mol-1. Calculate the normal
and standard boiling points. Does your result for the normal
boiling point agree with that in Table 8.3 (see Appendix A,
Data Tables)? If not, suggest a possible cause.
P8.26 Benzene(l) has a vapor pressure of 0.48423 bar at
331.06 K and an enthalpy of vaporization of 30.72 kJ mol-1.
Assume that CP,m of the vapor and liquid phases at that
temperature have their 298.15 K values of 82.4 and
136.0 J K-1 mol-1, respectively. Calculate the vapor pressure
of C6H61l2 at 349.0 K assuming that
a. the enthalpy of vaporization does not change with
­temperature.
b. the enthalpy of vaporization at temperature T can be
­calculated from the equation ∆ subH1T2 = ∆ subH1T02 +
∆CP1T - T02, assuming that ∆CP does not change with
temperature.
c. the experimentally determined value is 0.88717 bar. If
your calculated values differ from this value, provide a
possible explanation.
P8.27 Use the values for ∆ fG ~ 1H2O, g2 and ∆ fG ~ 1H2O, l2
from Appendix A (Data Tables) to calculate the vapor pressure of H2O at 298.15 K.
P8.28 The vapor pressure of an unknown solid is approximately
given by ln1P>Torr2 = 20.34 - 1980.1K>T2, and the
vapor pressure of the liquid phase of the same substance is
approximately given by ln1P>Torr2 = 16.56 - 1530.1K>T2.
a. Calculate ∆ vap H and ∆ sub H.
b. Calculate ∆ fus H.
c. Calculate the triple point temperature and pressure.
P8.29 For water, ∆ vap H is 40.65 kJ mol-1, and the normal
boiling point is 373.15 K. Calculate the boiling point for water
at the summit of Mount Kilimanjaro (elevation 5895 m),
where the normal barometric pressure is 51530 Pa.
P8.30 Use the vapor pressures of SO21l2 given in the
­following table to calculate the enthalpy of vaporization
using a graphical method or a least squares fitting routine.
T 1K2
190.
200.
210.
220.
P1 Pa2
824.1
2050.
4591
9421
T 1K2
230.
240.
250.
260.
P1 Pa2
17950
32110
54410
87930
250
CHAPTER 8 Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases
P8.31 The vapor pressure of carbon tetrachloride(l) is
15.28 * 103 Pa at 298.15 K. Use this value to calculate
∆ f G ~ 1CCl4, g2 - ∆ f G ~ 1CCl4, l2. Compare your result
with those in Table 4.2 (see Appendix A, Data Tables).
P8.32 Carbon tetrachloride melts at 250. K. The vapor pressure of the liquid is 10,539 Pa at 290. K and 74,518 Pa at
340. K. The vapor pressure of the solid is 270. Pa at 232 K
and 1092 Pa at 250. K.
a. Calculate ∆ vap H and ∆ sub H.
b. Calculate ∆ fusH.
c. Calculate the normal boiling point and ∆ vap S at the boiling
point.
d. Calculate the triple point pressure and temperature.
P8.33 In Equation (8.16), 1dP>dT2vap was calculated by
gas
­assuming that V m
W V liquid
. In this problem, you will test
m
the validity of this approximation. For water at its normal
boiling point of 373.13 K, ∆ vap H = 40.65 * 103 J mol-1,
rliquid = 958.66 kg m-3, and rgas = 0.58958 kg m-3.
­Compare the calculated values for 1dP>dT2vap with and
without the approximation of Equation (8.16). What is
the relative error in making the approximation?
P8.34 The variations of the vapor pressure of the liquid
and solid forms of a pure substance near the triple
Psolid
K
point are given by ln
= -7225. + 28.70 and
Pa
T
Pliquid
K
ln
= -3400. + 17.50. Calculate the temperature
Pa
T
and pressure at the triple point.
P8.35 Use the vapor pressures of n-butane given in the
­following table to calculate the enthalpy of vaporization using
a graphical method or a least squares fitting routine.
T 1 °C 2
-134.3
-121
-104
P1 Pa2
1.00
10.00
100.00
T 1 °C 2
-81.1
-49.1
-0.800
P1 Pa2
1000.
10000.
100000.
P8.36 At 298.15 K, ∆ f G ~ 1C2H5OH, g2 = -167.9 kJ mol-1
and ∆ f G ~ 1C2H5OH, l2 = -174.8 kJ mol-1. Calculate the
vapor pressure of ethanol at this temperature.
P8.37 Solid iodine, I21s2, at 25.0°C has an enthalpy of sublimation of 56.30 kJ mol-1. The CP, m of the vapor and solid
phases at that temperature are 36.9 and 54.4 J K-1 mol-1,
respectively. The sublimation pressure at 25.00°C is
0.30844 Torr. Calculate the sublimation pressure of the solid
at the melting point 1113.6°C2 assuming that
a. the enthalpy of sublimation and the heat capacities do not
change with temperature.
b. the enthalpy of sublimation at temperature T can be
calculated from the equation ∆ sub H1T2 ∆ sub H1T02 +
∆CP1T - T02
P8.38 A reasonable approximation to the vapor pressure of
krypton is given by log101P>Torr2 = b - 0.052231a>T2
For solid krypton, a = 10065 and b = 7.1770. For liquid
krypton, a = 9377.0 and b = 6.92387. Use this formula and
these constants to estimate the triple point temperature and
pressure as well as the enthalpies of vaporization, fusion, and
sublimation of krypton.
Section 8.7
P8.39 Calculate the vapor pressure of CH3OH1l2 at 310. K if
He is added to the gas phase at a partial pressure of 200. bar.
By what factor does the vapor pressure change? (Hint: You
need to calculate the vapor pressure of methanol at this temperature and 1 bar using the data in Table 8.3.)
P8.40 23.5 g of H2O1s2 at 260. K are transformed to H2O1l2
and brought to a temperature of 283 K, with an increase in
­entropy of 34.5 J K-1. Use this information and the heat
­capacities in Table 2.4 (see Appendix A, Data Tables) to
­calculate ∆ fusH for water.
P8.41 At temperatures below the triple point, H2O1s2 can
sublime to form H2O1g2. Calculate ∆S if 24.0 g of H2O1s2
sublimes at 273 K and is subsequently heated to 298 K.
­Assume that the heat capacity of H2O1g2 is constant at its
298 K value over the temperature interval of interest. What
fraction of ∆S comes from heating the gas? (Hint: Use
Equation (8.4) and an analogy of Equation (5.16) to calculate
the entropy of sublimation.)
Section 8.8
P8.42 In Section 8.8, it is stated that the maximum height of
a water column in which cavitation does not occur is ∼9.7 m.
Show that this is the case at 298 K.
P8.43 A cell is roughly spherical with a radius of
1.40 * 10-5 m. If the cell grows to double its volume,
­calculate the magnitude of the work required to expand the
cell surface. Assume the cell is surrounded by pure water and
that T = 298.15 K.
P8.44 Calculate the vapor pressure for a mist of
­spherical water droplets of radius (a) 1.75 * 10-8 m and
(b) 1.75 * 10-6 m surrounded by water vapor at 298 K. The
vapor pressure of water at this temperature is 25.2 Torr.
P8.45 Calculate the vapor pressure of water droplets of radius
1.65 * 10-8 m at 360. K in equilibrium with water vapor.
Use the tabulated value of the density and the surface tension
at 298 K from Appendix A (Data Tables) for this problem.
(Hint: You need to calculate the vapor pressure of water at this
temperature using the data in Table 8.3.)
P8.46 Calculate the difference in pressure across the liquid–
air interface for a droplet of radius 85.0 nm for (a) mercury
and (b) methanol.
P8.47 Calculate the vapor pressure of a droplet of benzene
of radius 1.00 * 10-8 m at 15.0°C in equilibrium with its
vapor. Use the tabulated value of the density and the surface
tension at 298 K from Appendix A (Data Tables) for this
problem. (Hint: You need to calculate the vapor pressure of
benzene at this temperature using the data in Table 8.3.)
Further Reading
FURTHER READING
Alper, J. S. “The Phase Rule Revisited: Interrelationships
between Components and Phases.” Journal of Chemical
Education 76 (1999): 1567–1569.
Brooks, S. D., Gonzales, M., and Farias, R. “Using
Surface Tension Measurements to Understand How
Pollution Can Influence Cloud Formation, Fog, and
Precipitation.” Journal of Chemical Education
86 (2009): 838–841.
Jensen, W. B. “Generalizing the Phase Rule.” Journal of
Chemical Education 78 (2001): 1369–1370.
Lin, F., Liu, D., Das, S., Prempeh, N., Hua, Y., and Lu,
J. “Recent Progress in Heavy Metal Extraction by
Supercritical CO2 Fluids.” Industrial and Chemical
Engineering Research 53 (2014): 1866–1877.
Prasad, R. “On Capillary Rise and Nucleation.” Journal of
Chemical Education 85 (2008): 1389–1391.
251
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C H A P T E R
9
Ideal and Real Solutions
9.1
Defining the Ideal Solution
9.2
The Chemical Potential of a
Component in the Gas and
Solution Phases
9.3
Applying the Ideal Solution
Model to Binary Solutions
Solutions are commonly encountered in chemistry. Because the components of a solution interact strongly, there is no equivalent to the ideal gas equation of state for liquid
solutions. The ideal solution model and ideal dilute solution model discussed in this
chapter allow calculations to be carried out for real solutions.
9.4
The Temperature–Composition
Diagram and Fractional
Distillation
9.5
The Gibbs–Duhem Equation
9.6
Colligative Properties
WHAT are the most important concepts and results?
9.7
Freezing Point Depression
and Boiling Point Elevation
9.8
Osmotic Pressure
9.9
Deviations from Raoult’s Law
in Real Solutions
WHY is this material important?
For ideal solutions, the partial pressure of a component in the vapor phase above a
solution is simply related to the mole fraction in the liquid phase by Raoult’s law. For
solutions characterized by the ideal dilute solution model, the solvent is best described
by Raoult’s law and the solute is best described by Henry’s law. Osmotic pressure,
freezing point depression, and boiling point elevation can all be described in terms of
the ideal and ideal dilute solution models.
9.13 Chemical Equilibrium
in Solutions
9.1 DEFINING THE IDEAL SOLUTION
9.14 Solutions Formed from Partially
Miscible Liquids
In an ideal gas, the atoms or molecules do not interact with one another. Clearly, this is
not a good model for a liquid, in which the atoms and molecules clearly do interact. In
fact, without attractive interactions, a liquid would transform into a gas. The attractive
interaction in liquids varies greatly, as shown by the large variation in boiling points
among the elements; for instance, helium has a normal boiling point of 4.2 K, whereas
hafnium has a boiling point of 5400 K.
In developing a model for solutions, the vapor phase in equilibrium with the solution must be taken into account. Consider pure liquid benzene in a beaker placed in
a closed room. Because the liquid is in equilibrium with the vapor phase above it,
there is a nonzero partial pressure of benzene in the air surrounding the beaker. This
pressure is called the vapor pressure of benzene at the temperature of the liquid. What
happens when toluene is added to the beaker? It is observed that the partial pressure of
benzene is reduced, and the vapor phase now contains both benzene and toluene. For
this particular mixture, the partial pressure of each component (i) above the liquid is
given by
i = 1, 2
Respect to Standard States
of Gases in a Solvent
It would be helpful to review chemical equilibrium discussed in Chapter 6.
Pi =
9.11 Activities Are Defined with
9.12 Henry’s Law and the Solubility
WHAT would be helpful for you to review for this chapter?
xiP *i
9.10 The Ideal Dilute Solution
(9.1)
9.15 Solid–Solution Equilibrium
Concept
The ideal solution is defined by
Raoult’s law.
where xi is the mole fraction of that component in the liquid and P *i is the vapor pressure of the pure component. This equation states that the partial pressure of each of the
two components is directly proportional to P *i and that the proportionality constant is xi.
Equation (9.1) is known as Raoult’s law and is the definition of an ideal solution.
253
254
CHAPTER 9 Ideal and Real Solutions
Solvent
Solute
Figure 9.1
Schematic model of a solution. The
white and black spheres represent solvent
and solute molecules, respectively.
Raoult’s law holds for each substance in an ideal solution over the range from
0 … xi … 1. In binary solutions, one refers to the component that has the higher value
of xi as the solvent and the component that has the lower value of xi as the solute.
Few solutions satisfy Raoult’s law. However, it is useful to study the thermodynamics of ideal solutions and to introduce departures from ideal behavior later. Why
is Raoult’s law not generally obeyed over the whole concentration range of a binary
solution consisting of molecules A and B? Equation (9.1) is only obeyed if the A–A,
B–B, and A–B interactions are all equally strong. This criterion is satisfied for a mixture of benzene and toluene because the two molecules are similar in size, shape, and
chemical properties. However, it is not satisfied for arbitrary molecules. Raoult’s law
is an example of a limiting law; the solvent in a real solution obeys Raoult’s law as the
solution becomes highly dilute.
Raoult’s law is derived in Example Problem 9.1 and can be rationalized using the
model depicted in Figure 9.1. In the solution, molecules of solute are distributed in the
solvent. The solution is in equilibrium with the gas phase, and the gas-phase composition is determined by a dynamic balance between evaporation from the solution and
condensation from the gas phase, as indicated for one solvent and one solute molecule
in Figure 9.1.
EXAMPLE PROBLEM 9.1
Assume that the rates of evaporation Revap and condensation Rcond of the solvent from
the surface of pure liquid solvent are given by the expressions
Revap = Akevap
Rcond = Akcond P *solvent
where A is the surface area of the liquid and kevap and kcond are the rate constants for
evaporation and condensation, respectively. Derive a relationship between the vapor
pressure of the solvent above the pure solvent and above a solution.
Solution
For the pure solvent, the equilibrium vapor pressure is found by setting the rates of
evaporation and condensation equal:
Revap = Rcond
Akevap = Akcond P *solvent
P *solvent =
kevap
kcond
The superscript * is used to indicate a pure substance.
Next, consider the ideal solution. In this case, the rate of evaporation is reduced by
the factor xsolvent because only that fraction of the surface is available for evaporation
of the solvent.
Revap = Akevap xsolvent
Rcond = Akcond Psolvent
and at equilibrium
Revap = Rcond
Akevap xsolvent = Akcond Psolvent
Psolvent = xsolvent
The derived relationship is Raoult’s law.
kevap
kcond
= xsolvent P *solvent
9.2 The Chemical Potential of a Component in the Gas and Solution Phases
255
9.2 THE CHEMICAL POTENTIAL OF A
COMPONENT IN THE GAS AND
SOLUTION PHASES
If the liquid and vapor phases are in equilibrium, the following equation holds for each
component of the solution, where mi is the chemical potential of species i:
msolution
= mivapor
i
(9.2)
Recall from Section 6.3 that the chemical potential of a substance in the gas phase is
related to its partial pressure Pi by
Pi
mivapor = mi~ + RT ln ~
(9.3)
P
where mi~ is the chemical potential of pure component i in the gas phase at the
standard state pressure P ~ = 1 bar. Because at equilibrium msolution
= mivapor = mi,
i
Equation (9.3) can be written in the form
mi = mi~ + RT ln
Pi
P~
(9.4)
For pure liquid i in equilibrium with its vapor, m*i 1liquid2 = m*i 1vapor2 = m*i . Therefore, the chemical potential of the pure liquid is given by
m*i = mi~ + RT ln
P *i
P~
(9.5)
Subtracting Equation (9.5) from (9.4) gives
mi = m*i + RT ln
Pi
(9.6)
P *i
For an ideal solution, Pi = xiP *i . Combining Equations (9.6) and (9.1), the central
equation describing ideal solutions is obtained:
mi = m*i + RT ln xi
(9.7)
This equation relates the chemical potential of a component in an ideal solution to the
chemical potential of the pure liquid form of component i and the mole fraction of that
component in the solution. This equation is most useful in describing the thermodynamics of solutions in which all components are volatile and miscible in all proportions.
Keeping in mind that mi = ∆ f Gi, the form of Equation (9.7) is identical to that
derived for the Gibbs energy of a mixture of gases in Section 6.5. Therefore, one can
derive relations for the thermodynamics of mixing to form ideal solutions as was done
for ideal gases in Section 6.5. These relations are shown in Equation (9.8). Note in particular that ∆ mixH = ∆ mixV = 0 for an ideal solution:
∆ mixG = nRT a xi ln xi
i
∆ mixS = -a
∆ mixV = -a
0∆ mixG
b
= -nR a xi ln xi
0T
P, n1, n2
i
0∆ mixG
b
= 0 and
0P
T, n1, n2
∆ mixH = ∆ mixG + T∆ mixS
= nRT a xi ln xi - T anR a xi ln xi b = 0
i
i
A calculation for ∆ mixG and ∆ mixS is shown in Example Problem 9.2.
(9.8)
Concept
In an ideal solution, the chemical
potential of a component is
determined by the chemical potential
of the pure component and the mole
fraction of the component.
256
CHAPTER 9 Ideal and Real Solutions
EXAMPLE PROBLEM 9.2
An ideal solution is made from 5.00 mol of benzene and 3.25 mol of toluene. Calculate ∆ mixG and ∆ mixS at 298 K and 1 bar pressure. Is mixing a spontaneous process?
Solution
The mole fractions of the components in the solution are xbenzene = 0.606 and
xtoluene = 0.394.
∆ mixG = nRT a xi ln xi
i
= 8.25 mol * 8.314 J mol-1 K-1
* 298 K * 10.606 ln 0.606 + 0.394 ln 0.3942
= -13.7 * 103 J
∆ mixS = -nR a xi ln xi
i
= -8.25 mol * 8.314 J mol-1 K-1 * 10.606 ln 0.606 + 0.394 ln 0.3942
= 46.0 J K-1
Mixing is spontaneous because ∆ mixG 6 0 for an ideal solution. If two liquids are
miscible, it is always true that ∆ mixG 6 0.
9.3 APPLYING THE IDEAL SOLUTION MODEL
TO BINARY SOLUTIONS
Although the ideal solution model can be applied to any number of components, to
simplify the mathematics, the focus in this chapter is on binary solutions, which consist of only two components. Because Raoult’s law holds for both components of the
mixture, P1 = x1P *1 and P2 = x2P *2 = 11 - x12P *2. The total pressure in the gas phase
varies linearly with the mole fraction of each of its components in the liquid:
Ptot = P1 + P2 = x1P *1 + 11 - x12P *2 = P *2 + 1P *1 - P *22x1
(9.9)
The individual partial pressures as well as Ptot above a benzene–1,2 dichloroethane
(C2H4Cl22 solution are shown in Figure 9.2. Small deviations from Raoult’s law are
seen. Such deviations are typical because few solutions obey Raoult’s law exactly.
250
Total gas pressure
200
Comparison of gas pressures calculated
from Raoult’s law with experimental
results for a binary solution. The
vapor pressure of benzene (blue), 1,2
dichloroethane (red), and the total pressure (purple) above the solution is shown
as a function of the mole fraction of 1,2
dichloroethane. The triangles, squares,
and circles are experimental data points
[Zawidski, J. V. Zeitschrift für Physikalische Chemie 35 (1900): 129]. The solid
curves are polynomial fits to the data.
The dashed lines are calculated using
Raoult’s law.
Pressure, P (Torr)
Figure 9.2
Benzene
150
Dichloroethane
100
50
0
0.2
0.4
0.6
Mole fraction of 1,2 dichloroethane
0.8
1
257
9.3 Applying the Ideal Solution Model to Binary Solutions
P1
= *
Ptot
P2 +
y1 =
x1P *1
(P *1 -
(9.10)
P *22x1
To obtain the pressure in the vapor phase as a function of y1, we first solve
Equation (9.10) for x1:
x1 =
y1P *2
+
(P *2
-
(9.12)
P *12y1
(9.13)
Ptot(P *1 - P *22
The variation of the total pressure with x1 and y1 is not the same, as is seen in
Figure 9.3. In Figure 9.3a, the system consists of a single-phase liquid for pressures
above the curve and of a two-phase vapor–liquid mixture for points lying on the curve
in the P - x1 diagram. Only points lying on or above the curve are meaningful because
points lying below the curve do not correspond to equilibrium states at which liquid is
present. In Figure 9.3b, the system consists of a single-phase vapor for pressures below
the curve and of a two-phase vapor–liquid mixture for points lying on the curve in the
P - y1 diagram. Points lying above the curve do not correspond to equilibrium states
at which vapor is present. The excess vapor would condense to form liquid.
Note that pressure is plotted as a function of different variables in the two parts of
Figure 9.3. To compare the gas phase and liquid composition at a given total pressure,
both are graphed as a function of the average composition of benzene, denoted by
Zbenzene, in the whole system in Figure 9.4. The average composition Z is defined by
Zbenzene =
nliquid
toluene
+
nvapor
toluene
+
nliquid
benzene
+
nvapor
benzene
nbenzene
=
ntotal
In the region labeled Liquid in Figure 9.4, the system consists entirely of a liquid phase
and Zbenzene = xbenzene. In the region labeled Vapor, the system consists entirely of a
gaseous phase and Zbenzene = ybenzene. The area separating the single-phase liquid and
vapor regions corresponds to the two-phase liquid–vapor coexistence region.
We next apply the phase rule to the benzene–toluene ideal solution for which there
are two components (benzene and toluene) and three independent state variables,
namely, P, T, and Zbenzene (or Ztoluene). Because Zbenzene = 1 - Ztoluene, these two variables are not independent. For a system containing C components, the phase rule discussed in Section 8.3 takes the form
F = C - p + 2
40
Liquid + vapor
(9.14)
For the benzene–toluene ideal solution, C = 2 and F = 4 - p. Therefore, in the singlephase regions labeled Liquid and Vapor in Figure 9.4, the system has three degrees of
freedom. One of these degrees of freedom is already fixed because the temperature is
held constant in Figure 9.4. Therefore, the two remaining degrees of freedom are P and
Zbenzene. In the two-phase region labeled Liquid + Vapor, the system has only one degree of freedom. Once P is fixed, the Z values at the ends of the horizontal line determine
xbenzene and ybenzene and, therefore, Zbenzene. We note that the number of components
0.2
0.4 0.6
xbenzene
0.8
1
(a)
100
Pressure, P (Torr)
P *1Ptot - P *1P *2
vapor
nliquid
benzene + nbenzene
Liquid
20
Equation (9.12) can be rearranged to give an equation for y1 in terms of the vapor
pressures of the pure components and the total pressure:
y1 =
60
0
P *1P *2
P *1
80
(9.11)
P *1 + (P *2 - P *12y1
and obtain Ptot from Ptot = P *2 + (P *1 - P *22x1
Ptot =
100
Pressure, P (Torr)
Nonideal solutions, which generally exhibit large deviations from Raoult’s law, are
discussed in Section 9.9.
The concentration unit used in Equation (9.9) is the mole fraction of each component in the liquid phase. The mole fraction of each component in the gas phase can also
be calculated. Using the definition of the partial pressure and the symbols y1 and y2 to
denote the gas-phase mole fractions, we can write
80
Liquid + vapor
60
40
Vapor
20
0
0.2
0.4 0.6
ybenzene
0.8
1
(b)
Figure 9.3
Total pressure above a solution as a
function of gas and liquid composition.
For a benzene–toluene ideal solution, the
total gas-phase pressure is shown for different values of (a) the mole fraction of
benzene in the solution and (b) the mole
fraction of benzene in the vapor. Points
on the curves correspond to vapor–liquid
coexistence. Only the curves and the
darkly shaded areas are of physical
significance, as explained in the text.
258
CHAPTER 9 Ideal and Real Solutions
Figure 9.4
100
a
Liquid
80
Pressure, P (Torr)
A pressure–composition phase diagram
for a benzene–toluene ideal solution. Z
is the average composition of a component of interest, in this case benzene. The
upper curve shows the vapor pressure as
a function of xbenzene. The lower curve
shows the vapor pressure as a function of
ybenzene. Above the two curves, the system
is totally in the liquid phase, whereas
below the two curves, the system is totally
in the vapor phase. The elongate-shaped
area between the two curves is the liquid–
vapor coexistence region. The horizontal
lines connecting the curves are tie lines.
60
b
Vapor pressure as a
function of xbenzene
c
Liquid + vapor
40
Vapor
d
20
Vapor pressure as a
function of ybenzene
0
0.2
0.4
0.6
0.8
1
Zbenzene
Concept
The vapor phase above a solution
is enriched in the most volatile
component relative to the solution.
can be less than the number of chemical species. For example, a solution containing
the three species A, B, and C linked by the chemical equilibrium A + B N C contains
only two independent components because only two of the three concentrations are
independent variables.
To demonstrate the usefulness of this pressure–average composition (P–Z)
diagram, consider a constant temperature process in which the pressure of the system is
decreased from the value corresponding to point a in Figure 9.4. Because P 7 Ptot, initially the system is entirely in the liquid phase. As the pressure is decreased at constant
composition, the system remains entirely in the liquid phase until the constant composition line intersects the P versus xbenzene curve. At this point, the system enters the twophase vapor–liquid coexistence region. As point b is reached, what are the values for
xbenzene and ybenzene, the mole fractions of benzene in the liquid and vapor phases? These
values can be determined by constructing a tie line in the two-phase coexistence region.
The values for xbenzene and ybenzene are given by the Z values at the ends of the tie line.
A tie line graphically connects two coexisting phases on a phase diagram.
In Figure 9.4, tie lines are horizontal lines (two are shown) at the pressure of interest
that connects the P versus xbenzene and P versus ybenzene curves. Only the values at the
ends of the tie line have a physical meaning. Note that for all values of the pressure,
ybenzene is greater than xbenzene, showing that the vapor phase is always enriched in the
more volatile or higher vapor pressure component in comparison with the liquid phase.
Calculations illustrating these concepts are shown in Example Problems 9.3 and 9.4.
EXAMPLE PROBLEM 9.3
An ideal solution of 5.00 mol of benzene and 3.25 mol of toluene is placed in a
­piston and cylinder assembly. At 298 K, the vapor pressure of the pure substances are
P *benzene = 96.4 Torr and P *toluene = 28.9 Torr.
a. The pressure above this solution is reduced from 760. Torr. At what pressure does
the vapor phase first appear?
b. What is the composition of the vapor under these conditions?
Solution
a. The mole fractions of the components in the solution are xbenzene = 0.606 and
xtoluene = 0.394. The vapor pressure above this solution is
Ptot = xbenzene P *benzene + xtoluene P *toluene
= 0.606 * 96.4 Torr + 0.394 * 28.9 Torr
= 69.8 Torr
No vapor will be formed until the pressure has been reduced to this value.
9.3 Applying the Ideal Solution Model to Binary Solutions
259
b. The composition of the vapor at a total pressure of 69.8 Torr is given by
ybenzene =
Pbenzene
Ptot
=
xbenzeneP *benzene
Ptot
=
0.606 * 96.4 Torr
= 0.837
69.8 Torr
Liquid
l
ytoluene = 1 - ybenzene = 0.163
Note that the vapor is enriched relative to the liquid in the more volatile component, which has the higher vapor pressure and lower boiling temperature.
To calculate the relative amount of material in each of the two phases in a coexistence region, we will derive the lever rule for a binary solution of the components A
and B. Figure 9.5 shows a magnified portion of Figure 9.4 centered at the tie line that
passes through point b. We derive the lever rule using the following geometrical argument. The lengths of the line segments lb and bn are given by
lb = ZB - xB =
ntot
B
tot
bv = yB - ZB =
nvapor
ntot
B
B
ntot
ntot
vapor
n
-
nliq
B
ntot
liq
(9.15)
(9.16)
n
b
Liquid + vapor
Vapor
Figure 9.5
An enlarged portion of the two-phase
coexistence region of Figure 9.4. The
vertical line through point b indicates a
constant system composition. The lever
rule (see text) is used to determine the
fraction of the liquid and vapor phases
present in the system.
The superscripts on n indicate whether we are referring to the moles of component B
in the vapor or liquid phases or the total number of moles of B in the system. If Equatot
tion (9.15) is multiplied by ntot
liq, Equation (9.16) is multiplied by nvapor, and after the
modified Equation (9.16) is subtracted from the modified Equation (9.15), we find that
tot
lb ntot
liq - bv nvapor =
We conclude that
ntot
B
liq
vapor
tot
1ntot + ntot
2 = ntot
vapor2 - 1nB + nB
B - nB = 0
ntot liq
ntot
liq
ntot
vap
=
bv
lb
(9.17)
It is convenient to restate this result as
tot
ntot
liq1ZB - xB2 = nvapor1yB - ZB2
(9.18)
Equation (9.18) is called the lever rule by analogy with the torques acting on a lever
of length lb + bv with the fulcrum positioned at point b. For the specific case shown
tot
in Figure 9.5, ntot
liq >nvapor = 2.36. Therefore, 70.1% of the total number of moles in the
system is in the liquid phase, and 29.9% is in the vapor phase.
EXAMPLE PROBLEM 9.4
For the benzene–toluene solution of Example Problem 9.3, calculate
a. the total pressure
b. the liquid composition
c. the vapor composition
when 1.50 mol of the solution has been converted to vapor.
Solution
The lever rule relates the average composition, Zbenzene = 0.606, and the liquid and
vapor compositions:
tot
ntot
vapor 1ybenzene - Zbenzene2 = nliq 1Zbenzene - xbenzene2
Entering the parameters of the problem, this equation simplifies to
6.75xbenzene + 1.50ybenzene = 5.00
Concept
The lever rule allows the proportions
of a component between different
phases to be calculated.
260
CHAPTER 9 Ideal and Real Solutions
The total pressure is given by
Ptot = xbenzene P *benzene + 11 - xbenzene2P *toluene
= 3 96.4xbenzene + 28.911 - xbenzene24 Torr
and the vapor composition is given by
ybenzene =
P *benzenePtotal - P *benzeneP *toluene
Ptotal(P *benzene - P *toluene2
Ptot
- 2786
Torr
¥
Ptotal
67.5
Torr
96.4
= ≥
These three equations in three unknowns can be solved by using an equation solver or
by eliminating the variables by combining equations. For example, the first equation
can be used to express ybenzene in terms of xbenzene. This result can be substituted into
the second and third equations to give two equations in terms of xbenzene and Ptot. The
solution for xbenzene obtained from these two equations can be substituted in the first
equation to give ybenzene. The relevant answers are xbenzene = 0.561, ybenzene = 0.810,
and Ptot = 66.8 Torr.
9.4 THE TEMPERATURE–COMPOSITION
DIAGRAM AND FRACTIONAL DISTILLATION
The enrichment of the vapor phase above a solution in the more volatile component is
the basis for fractional distillation, an important separation technique in chemistry and
in the chemical industry. It is more convenient to discuss fractional distillation using a
temperature–composition diagram than using the pressure–composition ­
diagram
discussed in the previous section. The temperature–composition diagram gives the
temperature of the solution as a function of the average system composition for a predetermined total vapor pressure, Ptot. It is convenient to use the value Ptot = 1 atm
so the vertical axis is the normal boiling point of the solution. Figure 9.6 shows a
boiling temperature–composition diagram for a benzene–toluene solution. Neither the
Tb versus xbenzene nor the Tb versus ybenzene curves are linear in a temperature–composition
diagram. Note that because the more volatile component has the lower boiling point,
the vapor and liquid regions are inverted when compared with the pressure–composition
diagram.
Figure 9.6
Temperature–composition diagram
for the benzene–toluene system. The
boiling temperature for an ideal solution
of benzene and toluene is plotted against
the average system composition, Zbenzene.
The upper curve shows the boiling temperature as a function of ybenzene, and the
lower curve shows the boiling temperature as a function of xbenzene. The area
intermediate between the two curves shows
the vapor–liquid coexistence region.
Boiling Temperature, T (K)
380
b
a
Vapor
375
c
d
Boiling temperature
as a function of ybenzene
370
Liquid
365
Boiling temperature
as a function of xbenzene
360
f
e
h
g
i
355
j
k
0
0.2
0.4
0.6
Zbenzene
0.8
1
9.4 The Temperature–Composition Diagram and Fractional Distillation
The principle of fractional distillation can be illustrated using the sequence of
lines labeled a through k in Figure 9.6. The vapor above the solution at point a is enriched in benzene and has a composition given by point b. If this vapor is separated
from the original solution and condensed by lowering the temperature, the resulting
liquid, indicated by point c, will have a composition whose mole fraction of benzene
is greater than the original solution, as indicated by point a. As for the original solution, the vapor above this separately collected liquid is enriched in benzene. As this
process is repeated, the successively collected vapor samples become more enriched in
benzene. In the limit of a very large number of steps, the last condensed samples are essentially pure benzene. The multistep procedure described earlier is very cumbersome
because it requires the collection and evaporation of many different samples. In practice, the separation into pure benzene and toluene is accomplished using a distillation
column, shown schematically in Figure 9.7.
Rather than have the whole system at a uniform temperature, a distillation column
operates with a temperature gradient so that the top part of the column is at a lower
temperature than the boiling solution. For such a distillation column, the horizontal
segments ab, cd, ef, and so on in Figure 9.6 correspond to successively higher (cooler)
portions of the column. Only the most volatile component of the solution initially
passes through the column and is collected as a liquid. As the most volatile component
is depleted from the solution, the boiling temperature increases, and the next most volatile component can be separated and collected, and so on.
Distillation plays a major role in the separation of crude oil into its various components. The lowest boiling liquid is gasoline, followed by kerosene, diesel fuel and
heating oil, and gas oil, which is further broken down into lower-molecular-weight
components by catalytic cracking. It is important in distillation to keep the temperature
of the boiling liquid as low as possible to avoid the occurrence of thermally induced
reactions. A useful way to reduce the boiling temperature is to distill the liquid under
a partial vacuum, which is done by pumping away the gas phase. The boiling point of
a typical liquid mixture can be reduced by approximately 100°C if the pressure is reduced from 760 to 20 Torr.
Although the principle of fractional distillation is the same for real solutions, it is
not possible to separate a binary solution into its pure components if the nonideality
is strong enough. If the A–B interactions are more attractive than the A–A and B–B
interactions, the boiling point of the solution will go through a maximum at a concentration intermediate between xA = 0 and xA = 1. An example of such a case is an
acetone–chloroform mixture. The hydrogen on the chloroform forms a hydrogen bond
with the oxygen in acetone, as shown in Figure 9.8, leading to stronger A–B than A–A
and B–B interactions. A boiling point diagram for this case is shown in Figure 9.9a. At
the maximum boiling temperature, the liquid and vapor composition lines coincide and
are tangent to one another.
Fractional distillation of a solution with a maximum boiling temperature is shown
schematically in Figure 9.9a. Beginning with an initial value of xA greater than that corresponding to the maximum boiling point, the component with the lowest boiling point
will initially emerge at the top of the distillation column. In this case, it will be pure
component A. However, the liquid left in the heated flask will not be pure component
B, but rather the solution corresponding to the concentration at which the maximum
boiling point is reached. Continued boiling of the solution at this composition will lead
to evaporation at constant composition. Such a mixture is called an azeotrope, and because Tb, azeotrope 7 Tb, A, Tb, B, it is called a maximum boiling azeotrope. An example
for a maximum boiling azeotrope is a mixture of H2O 1Tb = 100°C2 and HCOOH
1Tb = 100°C2 at xH2O = 0.427, which boils at 107.2°C. Other commonly occurring
azeotropic mixtures are listed in Table 9.1.
If the A–B interactions are less attractive than the A–A and B–B interactions, a minimum boiling azeotrope is formed (see Figure 9.9b). Fractional distillation of such a solution beginning with an initial value of xA less than that corresponding to the minimum
boiling point leads to a liquid with the azeotropic composition initially emerging at the top
of the distillation column. An example for a minimum boiling azeotrope is a mixture of
CS2 (Tb = 46.3°C) and acetone (Tb = 56.2°C) at xCS2 = 0.608, which boils at 39.3°C.
261
Resistive
heater
Initial solution
Figure 9.7
Schematic diagram of a fractional distillation column. The initial solution is
introduced at the bottom of the column.
The resistive heater provides the energy
needed to vaporize the liquid. It is assumed that the liquid and vapor are at
equilibrium at each level of the column.
The equilibrium temperature decreases
from the bottom to the top of the column.
H3C
C
O
H
CCI3
H3C
Figure 9.8
Hydrogen bond formation between
acetone and chloroform. The relatively
strong bonding between species leads
to the formation of a maximum boiling
azeotrope.
Concept
The different components in an
ideal solution can be separated by
fractional distillation.
262
CHAPTER 9 Ideal and Real Solutions
TABLE 9.1 Composition and Boiling Temperatures of Selected Azeotropes
Boiling Temperature
Vapor
Water–ethanol
Boiling Temperature
of Components 1°C2
100. and 78.5
0.096
Azeotrope Boiling
Point 1°C2
Water–trichloromethane
100. and 61.2
0.160
56.1
Water–benzene
100. and 80.2
0.295
69.3
Water–toluene
100. and 111
0.444
84.1
Ethanol–hexane
78.5 and 68.8
0.332
58.7
Ethanol–benzene
78.5 and 80.2
0.440
67.9
Ethyl acetate–hexane
78.5 and 68.8
0.394
65.2
Carbon disulfide–acetone
46.3 and 56.2
0.608
39.3
Toluene-acetic acid
111 and 118
0.625
100.7
Azeotropic Mixture
Liquid
xA (initial)
1
0
xA
(a) Maximum boiling point azeotrope
78.2
Source: Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002.
Vapor
Boiling Temperature
Mole Fraction of
First Component
It is still possible to collect one component of an azeotropic mixture using the
property that the azeotropic composition is a function of the total pressure, as shown
in Figure 9.10. The mixture is first distilled at atmospheric pressure, and the volatile
distillate is collected. This vapor is condensed and subsequently distilled at a reduced
pressure for which the azeotrope contains less A (more B). If this mixture is distilled,
the azeotrope evaporates, leaving some pure B behind.
Liquid
xA (initial)
1
0
xA
(b) Minimum boiling point azeotrope
Figure 9.9
Boiling behavior of azeotroic binary
solutions. (a) Maximum boiling point
azeotrope and (b) minimum boiling point
azeotrope. The dashed lines indicate the
initial composition of the solution and
the composition of the azeotrope. The
sequence of horizontal segments with
­arrows corresponds to successively higher
(cooler) portions of the column.
9.5 THE GIBBS–DUHEM EQUATION
In this section, we will show that the chemical potentials of the two components in a
binary solution are not independent. This is an important result because it allows the
chemical potential of a nonvolatile solute such as sucrose in a volatile solvent such as
water to be determined. Such a solute has no measurable vapor pressure; therefore, its
chemical potential cannot be measured directly. As shown here, its chemical potential
can be determined if only the chemical potential of the solvent as a function of concentration is known.
From Chapter 6, the differential form of the Gibbs energy is given by
dG = -S dT + V dP + a mi dni
i
so that at constant T and P,
dG = a mi dni
i
Concept
The Gibbs–Duhem equation states
that not all chemical potentials can
be varied independently.
(9.19)
(9.20)
However, from Equation (6.37) we have
G = a nimi and
i
dG = a ni dmi + a mi dni
i
i
(9.21)
Because both of the preceding equations for dG are correct, we conclude that
a ni dmi = 0
i
(9.22)
which is known as the Gibbs–Duhem equation. This equation states that not all chemical potentials can be varied independently.
9.6 Colligative Properties
263
For a binary solution at constant T and P, the Gibbs–Duhem equation reduces to
n1dm1 + n2dm2 = 0 or x1dm1 + x2dm2 = 0
Atmospheric pressure
(9.23)
dm2 = -
n1dm1
n2
(9.24)
The use of the Gibbs–Duhem equation is illustrated in Example Problem 9.5.
EXAMPLE PROBLEM 9.5
Boiling Temperature
If the change in the chemical potential of the first component is dm1, the change of the
chemical potential of the second component is given by
Reduced pressure
One component in a solution follows Raoult’s law, m1 = m*1 + RT ln x1, over the
­entire composition range 0 … x1 … 1. Using the Gibbs–Duhem equation, show that
the second component must also follow Raoult’s law.
Solution
From Equation (9.24),
x1dm1
n1
x1 dx1
dm2 = = - d1m*1 + RT ln x12 = -RT
x2
n2
x2 x1
Because x1 + x2 = 1, then dx2 = -dx1 and dm2 = RT dx2 >x2. Integrating this equation, one obtains m2 = RT ln x2 + C, where C is a constant of integration. This constant can be evaluated by examining the limit x2 S 1. This limit corresponds to the
pure substance 2 for which m2 = m*2 = RT ln 1 + C. We conclude that C = m*2 and,
therefore, m2 = m*2 + RT ln x2.
9.6 COLLIGATIVE PROPERTIES
Many solutions consist of nonvolatile solutes that have limited solubility in a volatile solvent. Examples are solutions of sucrose or sodium chloride in water. Important
properties of these solutions, including boiling point elevation, freezing point depression, and osmotic pressure only depend on the solute concentration—not on the nature
of the solute. These properties are called colligative properties. In this section, colligative properties are discussed using the ideal solution model. Corrections for nonideality
will be made in Section 9.13.
As discussed in Section 9.1, the vapor pressure above a solution containing a solute
is lowered with respect to the pure solvent. For the case of dissolution of a nonvolatile
solute in which the solute does not crystallize out with the solvent during freezing (because the solute cannot be easily integrated into the solvent crystal structure), the effect
on the solvent chemical potential is shown in Figure 9.11a. The chemical potential of
the solid is unaffected because of the assumption that the solute does not crystallize out
with the solvent. Furthermore, although the gas pressure above the solution is lowered
by the addition of the solute, the chemical potential of the gas is unaffected. Only the
chemical potential of the liquid is affected through formation of the solution. As shown
in Figure 9.11a, the melting temperature Tm, defined as the intersection of the solid and
liquid m versus T curves, is lowered. Similarly, the boiling temperature Tb, defined as
the intersection of the gas and liquid m versus T curves, is raised by dissolution of a
nonvolatile solute in the solvent.
The same information is shown in a P–T phase diagram in Figure 9.11b. Because
the vapor pressure above the solution is lowered by the addition of the solute, the liquid–
gas coexistence curve intersects the solid–gas coexistence curve at a lower temperature
than for the pure solvent. This intersection defines the triple point, and it must also be
the origin of the solid–liquid coexistence curve. Therefore, the solid–liquid coexistence
curve is shifted to lower temperatures through the dissolution of the nonvolatile solute.
The overall effect of the shifts in the solid–gas and solid–liquid coexistence curves is a
0
xA
Figure 9.10
A minimum boiling point diagram at
two different pressures. Because the
azeotropic composition depends on the
total pressure, pure B can be recovered
from the A–B mixture by first distilling
the mixture under atmospheric pressure
and, subsequently, under a reduced pressure. Note that the boiling temperature is
lowered as pressure is reduced.
Concept
Colligative properties depend on the
solute concentration and not on the
identity of the solute.
1
264
CHAPTER 9 Ideal and Real Solutions
Pure
solid
Chemical Potential, M
Pure liquid
Solution
freezing point depression and a boiling point elevation, both of which arise through
lowering the vapor pressure of the solvent. This effect depends only on the concentration, and not on the identity, of the nonvolatile solute. The preceding discussion is
qualitative in nature. In the next two sections, quantitative relationships are developed
between the colligative properties and the concentration of the nonvolatile solute.
9.7 FREEZING POINT DEPRESSION
AND BOILING POINT ELEVATION
DTm
If a solution is in equilibrium with a pure solid solvent, the following relation must be
satisfied:
msolution = m*solid
(9.25)
Pure
gas
Recall that msolution refers to the chemical potential of the solvent in the solution, and
m*solid refers to the chemical potential of the pure solvent in solid form. From Equation
(9.7), we can express msolution in terms of the chemical potential of the pure solvent and
its concentration and rewrite Equation (9.25) in the form
DTb
T¿m T¿m
Tb T¿b
Temperature
(a)
m*solvent + RT ln xsolvent = m*solid
(9.26)
This equation can be solved for ln xsolvent:
ln xsolvent =
1 bar
(9.27)
m*solid - m*solvent = ∆ fG solid - ∆ fG solvent =
The difference in chemical potentials
∆ fusG so that
Pressure
Solution
Pure substance
m*solid - m*solvent
RT
ln xsolvent =
- ∆ fusG
RT
(9.28)
Because we are interested in how the freezing temperature is related to xsolvent at constant pressure, we need the partial derivative 10T>0xsolvent2P. This quantity can be
­obtained by differentiating Equation (9.28) with respect to xsolvent:
T¿m Tm
Tb T¿b
Temperature
(b)
Figure 9.11
Illustration of boiling point elevation
and freezing point depression. Addition
of a nonvolatile solute affects the chemical
potential of a solution but has essentially
no effect on the chemical potentials of
the solid and gas phases. This difference
in behavior leads to an expansion of the
liquid stability field, which in turn produces
elevation of the boiling point, T S T′b, and
depression of the freezing point, T S T′m.
These effects are illustrated on plots of
(a) chemical potential versus temperature
and (b) pressure versus temperature.
0ln xsolvent
1
1 °
a
b =
= xsolvent
0xsolvent P
R
0
∆ fusG
T ¢
0T
a
b
0T
P 0xsolvent P
(9.29)
The first partial derivative on the right side of Equation (9.29) can be simplified using
the Gibbs–Helmholtz equation (see Section 6.3), giving
1
xsolvent
=
∆ fusH
RT
2
a
0T
0xsolvent
b
P
or
dxsolvent
∆ fusH dT
= d ln xsolvent =
(constant P)
xsolvent
R T2
(9.30)
This equation can be integrated between the limits given by the pure solvent
(xsolvent = 1), for which the fusion temperature is Tfus, and an arbitrary small concentration of solute, for which the fusion temperature is T:
xsolvent
T
∆ fusH dT′
dx
=
2
L R T′
L x
1
(9.31)
Tfus
For xsolvent, not very different from 1, ∆ fusH is assumed to be independent of T, and
Equation (9.31) simplifies to
R ln xsolvent
1
1
=
T
Tfus
∆ fusH
(9.32)
9.7 Freezing Point Depression and Boiling Point Elevation
In discussing solutions, it is more convenient to use molality of the solute expressed in moles of solute per kg of solvent as the concentration unit,
rather than the mole fraction of the solvent. For dilute solutions, ln xsolvent =
ln1nsolvent >1nsolvent + msolute + Msolvent nsolvent22 = -ln11 + Msolvent msolute2,
and
Equation (9.32) can be rewritten in terms of the molality (m) rather than the mole fraction. The result is
∆Tf = -
RMsolventT 2fus
msolute = -Kf msolute
∆ fusH
(9.33)
where Msolvent is the molar mass of the solvent. In recasting concentration units
from Equation (9.32) to Equation (9.33), we have made the approximation
ln11 + Msolvent msolute2 ≈ Msolvent msolute and 1>T - 1>Tfus ≈ - ∆Tf >T 2fus. Note that
Kf depends only on the properties of the solvent and is primarily determined by the
molecular mass and the enthalpy of fusion. For most solvents, the magnitude of Kf
lies between 1.5 and 10, as shown in Table 9.2. However, Kf can reach unusually high
values—for example, 30. 1K kg2>mol for carbon tetrachloride and 40. 1K kg2>mol for
camphor.
The boiling point elevation can be calculated using the same argument with ∆ vapG
and ∆ vapH substituted for ∆ fusG and ∆ fusH. The result is
and
a
RMsolvent T 2vap
0T
b
=
0msolute P, m S 0
∆ vapH
∆Tb =
RMsolventT 2vap
∆ vapH
(9.34)
(9.35)
msolute = Kbmsolute
Because ∆ vapH 7 ∆ fusH, it follows that Kf 7 Kb. Typically, Kb values range between
0.5 and 5.0 K > 1mol kg -12, as shown in Table 9.2. Note also that, by convention, both
Kf and Kb are positive; hence, the negative sign in Equation (9.33) is replaced by a positive sign in Equation (9.35).
Example Problem 9.6 illustrates calculations for the freezing point depression and
determination of the molecular mass.
TABLE 9.2 Freezing Point Depression and Boiling Point
Elevation Constants
Standard
Boiling
Point (K)
Substance
Standard
Freezing
Point (K)
Acetic acid
289.6
Kf 1K kg mol-12
3.59
391.2
Kb1K kg mol-12
Benzene
278.6
5.12
353.3
2.53
Camphor
449
482.3
5.95
Carbon disulfide
161
319.2
2.40
Carbon tetrachloride
250.3
30.
349.8
4.95
Cyclohexane
279.6
20.0
353.9
2.79
Ethanol
158.8
2.0
351.5
1.07
Phenol
314
7.27
455.0
3.04
Water
273.15
1.86
373.15
0.51
40.
3.8
3.08
Source: Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002.
265
Concept
Freezing point depression and boiling
point elevation arise because of the
lowering of the vapor pressure of the
solvent above the solution.
266
CHAPTER 9 Ideal and Real Solutions
EXAMPLE PROBLEM 9.6
In this example, 4.50 g of a substance dissolved in 125 g of CCl4 leads to an elevation of the boiling point of 0.650 K. Calculate the freezing point depression, the molar
mass of the substance, and the factor by which the vapor pressure of CCl4 is lowered.
Solution
∆Tf = a
Kf
Kb
b ∆Tb = -
30. K>1mol kg -12
4.95 K>1mol kg -12
* 0.650 K = -3.9 K
To avoid confusion, we use the symbol m for molality and m for mass. We solve for
the molar mass Msolute using Equation (9.35):
∆Tb = Kb msolute = Kb * a
msolute >Msolute
b
msolvent
Msolute =
Kbmsolute
msolvent ∆Tb
Msolute =
4.95 K kg mol-1 * 4.50 g
= 274 g mol-1
0.125 kg * 0.650 K
We solve for the factor by which the vapor pressure of the solvent is reduced by using
Raoult’s law:
Psolvent
P *solvent
= xsolvent = 1 - xsolute = 1 -
nsolute
nsolute + nsolvent
4.50 g
274 g mol-1
= 1 a
4.50 g
274 g mol-1
b + a
= 0.980
125 g
153.8 g mol-1
b
9.8 OSMOTIC PRESSURE
P
P1P
Solution
Pure
solvent
Some membranes allow the passage of small molecules such as water, yet do not
allow larger molecules such as sucrose to pass through them. Such a semipermeable
membrane is an essential component in medical technologies, including kidney dialysis,
which is described below. Consider a sac of such a membrane that contains a solute
that cannot pass through the membrane. If the sac is immersed in a beaker filled with
pure solvent, then initially the solvent diffuses into the sac. Diffusion ceases when
equilibrium is attained, and at equilibrium the pressure is higher in the sac than in the
surrounding solvent. This result is shown schematically in Figure 9.12. The process in
which solvent diffuses through a membrane and dilutes a solution is known as osmosis.
The increase in pressure in the solution is known as the osmotic pressure.
To understand the origin of the osmotic pressure, denoted by p, the equilibrium
condition is applied to the contents of the sac and the surrounding solvent:
*
msolution
solvent 1T, P + p, xsolvent2 = msolvent 1T, P2
(9.36)
*
msolution
solvent 1T, P + p, xsolvent2 = msolvent 1T, P + p2 + RT ln xsolvent
(9.37)
Using Raoult’s law to express the concentration dependence of msolvent,
Semipermeable
membrane
Figure 9.12
Osmotic pressure experiment. An
­osmotic pressure arises if a solution containing a solute that cannot pass through
the membrane boundary is immersed in
the pure solvent.
Because m for the solvent is lower in the solution than in the pure solvent, only an
increased pressure in the solution can raise its m sufficiently to achieve equilibrium
with the pure solvent. The dependence of m on pressure and temperature is given by
dm = dGm = VmdP - SmdT. At constant T we can write
P+p
m*solvent1T,
P + p, xsolvent2 -
m*solvent1T,
P2 =
L
P
V *mdP′
(9.38)
9.8 Osmotic Pressure
where V *m is the molar volume of the pure solvent and P is the pressure in the solvent
outside the sac. Because a liquid is nearly incompressible, it is reasonable to assume
that V *m is independent of P to evaluate the integral in the previous equation. Therefore,
m*solvent 1T, P + p, xsolvent2 - m*solvent 1T, P2 = V *m p, and Equation (9.37) reduces to
pV *m + RT ln xsolvent = 0
(9.39)
For a dilute solution, nsolvent W nsolute, and
ln xsolvent = ln(1 - xsolute2 ≈ -xsolute = -
nsolute
nsolute
≈ nsolvent
nsolute + nsolvent
(9.40)
Equation (9.39) can be simplified further by recognizing that for a dilute solution,
V ≈ nsolventV *m. With this substitution, Equation (9.39) becomes
p =
nsolute RT
V
(9.41)
Note the similarity in form between this equation and the ideal gas law.
An important application of the selective diffusion of the components of a solution through a membrane is dialysis. In healthy individuals, the kidneys remove waste
products from the bloodstream, whereas individuals with damaged kidneys use a dialysis machine for this purpose. Blood from the patient is shunted through tubes made
of a selectively porous membrane surrounded by a flowing sterile solution made up of
water, sugars, and other components. Blood cells and other vital components of blood
are too large to fit through the pores in the membranes, but urea and salt flow out of
the bloodstream through membranes into the sterile solution and are removed as waste.
Example Problem 9.7 illustrates a calculation of the osmotic pressure.
EXAMPLE PROBLEM 9.7
Calculate the osmotic pressure generated at 298 K if a cell with a total solute concentration of 0.500 mol L-1 is immersed in pure water. The cell wall is permeable to water
molecules but not to the solute molecules.
Solution
nsolute RT
= 0.500 mol L-1 * 8.206 * 10-2 L atm K-1 mol-1 * 298 K
V
= 12.2 atm
p =
As this calculation shows, the osmotic pressure generated for moderate solute concentrations can be quite high. Hospital patients have died after pure water has accidentally
been injected into their blood vessels because the osmotic pressure is sufficient to
burst the walls of blood cells.
Plants use osmotic pressure to achieve mechanical stability in the following way.
A plant cell is bounded by a cellulose cell wall that is permeable to most components
of the aqueous solutions that it encounters. A semipermeable cell membrane through
which water, but not solute molecules, can pass is located just inside the cell wall. When
the plant has sufficient water, the cell membrane expands, pushing against the cell wall,
which gives the plant stalk a high rigidity. However, if there is a drought, the cell membrane is not totally filled with water and it moves away from the cell wall. As a result,
only the cell wall contributes to the rigidity, and plants droop under these conditions.
Another important application involving osmotic pressure is the desalination of seawater using reverse osmosis. Seawater is typically 1.1 molar in NaCl. Equation (9.41)
shows that an osmotic pressure of 27 bar is needed to separate the solvated Na+ and
Cl- ions from the water. If the seawater side of the membrane is subjected to a pressure
greater than 27 bar, H2O from the seawater will flow through the membrane, resulting
in a separation of pure water from the seawater. This process is called reverse osmosis.
The challenge in carrying out reverse osmosis on the industrial scale needed to provide
Concept
Osmotic pressure arises as a
consequence of the selective
diffusion of solution components
through a membrane.
267
268
CHAPTER 9 Ideal and Real Solutions
a coastal city with potable water is to produce robust membranes that accommodate
the necessary flow rates without getting fouled by algae and also effectively separate
the ions from the water. The mechanism that prevents passage of the ions through suitable membranes is not fully understood. It is not based on the size of pores within the
membrane alone, and it involves charged surfaces within the hydrated pores. These
membrane-anchored ions repel the mobile Na+ and Cl- ions, while allowing the passage of the neutral water molecule.
9.9 DEVIATIONS FROM RAOULT’S LAW
IN REAL SOLUTIONS
In Sections 9.1 through 9.8, the discussion was limited to ideal solutions. However, in
general, if two volatile and miscible liquids are combined to form a solution, Raoult’s
law is not obeyed. This is the case because the A–A, B–B, and A–B interactions in a
binary solution of A and B are in general not equal. If the A–B interactions are less
attractive than the A–A and B–B interactions, positive deviations from Raoult’s law
will be observed. In contrast, if the A–B interactions are more attractive than the A–A
and B–B interactions, negative deviations from Raoult’s law will be observed. An example of a binary solution with positive deviations from Raoult’s law is CS2 9acetone.
Experimental data for this system are shown in Table 9.3, and the data are plotted in
Figure 9.13. How can a thermodynamic framework analogous to that presented for the
TABLE 9.3 Partial and Total Pressures above a CS2–Acetone Solution
xCS2
0
PCS2
(Torr)
0
Pacetone
(Torr)
Ptot
(Torr)
xCS2
PCS2
(Torr)
Pacetone
(Torr)
Ptot
(Torr)
343.8
343.8
0.4974
404.1
242.1
646.2
0.0624
110.7
331.0
441.7
0.5702
419.4
232.6
652.0
0.0670
119.7
327.8
447.5
0.5730
420.3
232.2
652.5
0.0711
123.1
328.8
451.9
0.6124
426.9
227.0
653.9
0.1212
191.7
313.5
505.2
0.6146
427.7
225.9
653.6
0.1330
206.5
308.3
514.8
0.6161
428.1
225.5
653.6
0.1857
258.4
295.4
553.8
0.6713
438.0
217.0
655.0
0.1991
271.9
290.6
562.5
0.6713
437.3
217.6
654.9
0.2085
283.9
283.4
567.3
0.7220
446.9
207.7
654.6
0.2761
323.3
275.2
598.5
0.7197
447.5
207.1
654.6
0.2869
328.7
274.2
602.9
0.8280
464.9
180.2
645.1
0.3502
358.3
263.9
622.2
0.9191
490.7
123.4
614.1
0.3551
361.3
262.1
623.4
0.9242
490.0
120.3
610.3
0.4058
379.6
254.5
634.1
0.9350
491.9
109.4
601.3
0.4141
382.1
253.0
635.1
0.9407
492.0
103.5
595.5
0.4474
390.4
250.2
640.6
0.9549
496.2
85.9
582.1
0.4530
394.2
247.6
641.8
0.9620
500.8
73.4
574.2
0.4933
403.2
242.8
646.0
0.9692
502.0
62.0
564.0
1
512.3
Source: Zawidski, J. V. Zeitschrift für Physikalische Chemie 35 (1900): 129.
0
512.3
269
9.9 Deviations from Raoult’s Law in Real Solutions
Figure 9.13
Deviations from Raoult’s law. The
data in Table 9.3 are plotted versus xCS2.
Dashed lines show the expected behavior
if Raoult’s law were obeyed.
Ptotal
600
Pressure, P (Torr)
500
400
300
200
0
0.2
0.4
xCS2
0.6
0.8
1
ideal solution in Section 9.2 be developed for real solutions? This issue is addressed
throughout the rest of this chapter.
Figure 9.13 shows that the partial and total pressures above a real solution can differ substantially from the behavior predicted by Raoult’s law. Another way that ideal
and real solutions differ is that the set of equations denoted in Equation (9.8), which
describes the change in volume, entropy, enthalpy, and Gibbs energy that results from
mixing, are not applicable to real solutions. For real solutions, these equations can only
be written in a much less explicit form. Assuming that A and B are miscible,
∆ mixG
∆ mixS
∆ mixV
∆ mixH
6 0
7 0
≠ 0
≠ 0
=
xAV *m, A
+ 11 -
Real solutions can show appreciable
deviations from Raoult’s law.
(9.42)
Whereas ∆ mixG 6 0 and ∆ mixS 7 0 always hold for miscible liquids, ∆ mixV and
∆ mixH can be positive or negative, depending on the nature of the A–B interaction in
the solution.
As indicated in Equation (9.42), the volume change upon mixing is not generally 0.
Therefore, the volume of a solution will not be given by
V ideal
m
Concept
xA2V *m, B
V *m, A
(9.43)
V *m, B
as expected for 1 mol of an ideal solution, where
and
are the molar volideal
umes of the pure substances A and B. Figure 9.14 shows ∆Vm = V real
for an
m - Vm
acetone–chloroform solution as a function of xchloroform. Note that ∆Vm is positive for
some values of xchloroform but negative for other values of xchloroform. The deviations
from ideality are small but clearly evident.
The deviation of the volume from ideal behavior can best be understood by defining the concept of partial molar quantities. One can form partial molar quantities for
any extensive property of a system (for example, U, H, G, A, and S). Partial molar
quantities (other than the chemical potential, which is the partial molar Gibbs energy)
are usually denoted by the appropriate symbol topped by a horizontal bar, such as Vi.
The concept of partial molar quantities can be illustrated by discussing the partial
molar volume. The volume of 1 mol of pure water at 25°C is 18.1 cm3. However, if 1
mol of water is added to a large volume of an ethanol–water solution with xH2O = 0.75,
the volume of the solution increases by only 16 cm3. This is the case because the
local structure around a water molecule in the solution is more compact than that in
pure water. The partial molar volume of a component in a solution is defined as the
volume by which the solution changes if 1 mol of the component is added to such a
0.2
0.4
0.6
0.8
1
0
Volume Change DVsol, (cm3)
100
Pacetone
PCS2
20.025
x chloroform
20.050
20.075
20.100
20.125
20.150
20.175
Figure 9.14
Deviations of measured molar volume
for the acetone–chloroform system
relative to that for ideal solution behavior. Ideal solution volume calculated from
Equation 9.43. Molar volume deviations
plotted as a function of the mole fraction
of chloroform.
CHAPTER 9 Ideal and Real Solutions
Figure 9.15
82.0
75.0
Acetone
74.5
81.5
74.0
81.0
73.5
80.5
73.0
72.5
80.0
Chloroform
72.0
79.5
71.5
0.2
0.4
xchloroform
0.6
0.8
Partial molar volume of acetone (cm3 mol21)
The partial molar volumes in a binary
system. Chloroform (blue curve) and
­acetone (red curve) partial molar volumes
in a chloroform–acetone binary solution
are shown as a function of xchloroform.
The arrows indicate the appropriate y-axis
labels for the given curve.
Partial molar volume of chloroform (cm3 mol21)
270
1
large volume that the solution composition can be assumed constant. This statement is
expressed mathematically in the following form:
V11P, T, n1, n22 = a
0V
b
0n1 P, T, n2
(9.44)
With this definition, the volume of a binary solution is given
Concept
The Gibbs–Duhem equation also
applies to partial molar volumes
in real solutions.
V = n1V11P, T, n1, n22 + n2V21P, T, n1, n22
(9.45)
Note that because the partial molar volumes depend on the concentration of all components, the same is true of the total volume.
Partial molar volume is a function of P, T, n1, and n2, and Vi can be greater than
or less than the molar volume of the pure component. Therefore, the volume of a solution of two miscible liquids can be greater than or less than the sum of the volumes of
the pure components of the solution. Figure 9.15 shows data for the partial volumes
of acetone and chloroform in an acetone–chloroform binary solution at 298 K. Note
that the changes in the partial molar volumes with concentration are small, but not
negligible.
In Figure 9.15, we can see that V1 increases if V2 decreases and vice versa. This is
the case because partial molar volumes are related in the same way as the chemical
potentials are related in the Gibbs–Duhem equation [Equation (9.23)]. In terms of the
partial molar volumes, the Gibbs–Duhem equation takes the form
x2
x1 dV1 + x2 dV2 = 0 or dV1 = - dV2
(9.46)
x1
Therefore, as seen in Figure 9.15, if V2 changes by an amount dV2 over a small concentration interval, V1 will change in the opposite direction. The Gibbs–Duhem equation is
applicable to both ideal and real solutions.
9.10 THE IDEAL DILUTE SOLUTION
Although no simple model exists that describes all real solutions, there is a limiting law
that merits further discussion. In this section, we describe the model of the ideal dilute
solution, which provides a useful basis for describing the properties of real solutions if
they are dilute. Just as for an ideal solution, at equilibrium the chemical potentials of
a component in the gas and solution phase of a real solution are equal. As for an ideal
solution, the chemical potential of a component in a real solution can be separated into
9.10 The Ideal Dilute Solution
271
two terms, a standard state chemical potential and a term that depends on the partial
pressure.
mi = m*i + RT ln
Pi
(9.47)
P *i
Recall that for a pure substance m*i (vapor) = m*i (liquid) = m*i . Because the solution is
not ideal, Pi ≠ xi P *i .
First consider only the solvent in a dilute binary solution. To arrive at an equation
for mi that is similar to Equation (9.7) for an ideal solution, we define the dimensionless
activity, asolvent, of the solvent by
asolvent =
Psolvent
(9.48)
P *solvent
Note that asolvent = xsolvent for an ideal solution. For a nonideal solution, the activity
and the mole fraction are related through the activity coefficient gsolvent, defined by
gsolvent =
asolvent
xsolvent
(9.49)
The activity coefficient quantifies the degree to which the solution is nonideal. The
activity plays the same role for a component of a solution that the fugacity plays for a
real gas in expressing deviations from ideal behavior. In both cases, ideal behavior is
observed in the appropriate limit, namely, P S 0 for the gas, and xsolute S 0 for the solution. To the extent that there is no atomic-scale model that tells us how to calculate g,
it should be regarded as a correction factor that exposes the inadequacy of our model,
rather than as a fundamental quantity. As will be discussed in Chapter 10, there is such
an atomic-scale model for dilute electrolyte solutions.
How is the chemical potential of a component related to its activity? Combining
Equations (9.47) and (9.48), one obtains a relation that holds for all components of a
real solution:
mi = m*i + RT ln ai
(9.50)
Equation (9.50) is a central equation in describing real solutions. It is the starting point
for the discussion in the rest of this chapter.
To this point in the chapter, we have focused on the solvent in a dilute solution. However, the ideal dilute solution is defined by the conditions xsolute S 0 and
xsolvent S 1. Because the solvent and solute are considered in different limits, we use
different expressions to relate the mole fraction of a component and the partial pressure
of the component above the solution.
Consider the partial pressure of acetone as a function of xCS2 shown in Figure 9.16.
Although Raoult’s law is not obeyed over the whole concentration range, it is obeyed
in the limit that xacetone S 1 and xCS2 S 0. In this limit, the average acetone molecule
at the surface of the solution is surrounded by acetone molecules. Therefore, to a good
approximation, Pacetone = xacetone P *acetone as xacetone S 1. Because the majority species
Pressure, P (Torr)
Henry’s law
Pacetone
300
200
Raoult’s law
100
Figure 9.16
0
0.2
0.4
0.6
xCS2
0.8
1
Partial pressure of acetone plotted as a
function of xCS2. Note that Pacetone follows
Raoult’s law as xCS2 S 0 and Henry’s law
as xCS2 S 1. Data from Table 9.3.
272
CHAPTER 9 Ideal and Real Solutions
TABLE 9.4 Henry’s Law Constants for Aqueous Solutions Near 298 K
Substance
kH (Torr)
kH (bar)
Ar
2.80 * 107
3.72 * 104
C2H6
2.30 * 107
3.06 * 104
CH4
3.07 * 107
4.08 * 104
CO
4.40 * 106
5.84 * 103
CO2
1.24 * 106
1.65 * 103
H2S
4.27 * 105
5.68 * 102
He
1.12 * 108
1.49 * 106
N2
6.80 * 107
9.04 * 104
O2
3.27 * 107
4.95 * 104
Source: Alberty, R. A., and Silbey, R. S. Physical Chemistry. New York: John Wiley & Sons, 1992.
Concept
In an ideal dilute solution, the solvent
follows Raoult’s law and the solute
follows Henry’s law at low solute
concentrations.
is defined to be the solvent, we see that Raoult’s law is obeyed for the solvent in a dilute solution. This limiting behavior is also observed for CS2 in the limit in which it is
the solvent, as seen in Figure 9.13.
Consider the opposite limit in which xacetone S 0. In this case, the average acetone molecule at the surface of the solution is surrounded by CS2 molecules. Therefore, the molecule experiences very different interactions with its neighbors than if it
were surrounded by acetone molecules. For this reason, Pacetone ≠ xacetone P *acetone as
xacetone S 0. However, it is apparent from Figure 9.16 that Pacetone also varies linearly
with xacetone in this limit. This behavior is described by the following equation:
Pacetone = xacetone k H
acetone as xacetone S 0
(9.51)
This relationship is known as Henry’s law, and the constant kH is known as the
Henry’s law constant. The value of the constant depends on the nature of the solute
and solvent and quantifies the degree to which deviations from Raoult’s law occur. As
*
the solution approaches ideal behavior, k H
i S P i . For the data shown in Figure 9.13,
H
H
k CS2 = 1750 Torr and k acetone = 1950 Torr. Note that these values are substantially
greater than the vapor pressures of the pure substances, which are 512.3 and 343.8 Torr,
respectively. The Henry’s law constants are less than the vapor pressures of the pure
substances if the system exhibits negative deviations from Raoult’s law. Henry’s law
constants for aqueous solutions are listed for a number of solutes in Table 9.4.
Based on these results, the ideal dilute solution is defined as a solution in which
the solvent is described using Raoult’s law and the solute is described using Henry’s
law. As shown by the data in Figure 9.13, the partial pressures above the CS2-acetone
mixture are consistent with this model in either of two limits, xacetone S 1 or xCS2 S 1.
In the first of these limits, we consider acetone to be the solvent and CS2 to be the
solute. In the second limit, acetone is the solute and CS2 is the solvent.
9.11 ACTIVITIES ARE DEFINED WITH RESPECT
TO STANDARD STATES
Predictions based on the ideal dilute solution model that Raoult’s law is obeyed for the
solvent and Henry’s law is obeyed for the solute are not valid over a wide range of concentration. The concept of the activity coefficient introduced in Section 9.10 is used to
quantify these deviations. In doing so, it is useful to define the activities in such a way
that the solution approaches ideal behavior in the limit of interest, which is generally
9.11 Activities Are Defined with Respect to Standard States
xA S 0, or xA S 1. With this choice, the activity approaches the concentration, and it
is reasonable to set the activity coefficient equal to 1. Specifically, ai S xi as xi S 1
for the solvent, and ai S xi as xi S 0 for the solute. The reason for this choice is that
numerical values for activity coefficients are generally not known. Choosing the standard state in this way ensures that the concentration (divided by the unit concentration
to make it dimensionless), which is easily measured, is a good approximation of the
activity.
In Section 9.10, the activity and activity coefficient for the solvent in a dilute solution (xsolvent S 1) were defined by the relations
ai =
Pi
P *i
and gi =
ai
xi
(9.52)
As shown in Figure 9.13, the activity approaches unity as xsolvent S 1. We refer to an
activity calculated using Equation (9.52) as being based on a Raoult’s law standard
state. The standard state chemical potential based on Raoult’s law is m*solvent, which is
the chemical potential of the pure solvent.
However, this definition of the activity and choice of a standard state is not optimal
for the solute at a low concentration because the solute obeys Henry’s law rather than
Raoult’s law; therefore, the value of the activity coefficient will differ appreciably from
one. In this case,
*
msolution
solute = msolute + RT ln
kH
solute xsolute
P *solute
= m*H
solute + RT ln xsolute as xsolute S 0 (9.53)
The standard state chemical potential is the value of the chemical potential when
xi = 1. We see that the Henry’s law standard state chemical potential is given by
*
m*H
solute = msolute + RT ln
kH
solute
P *solute
(9.54)
The activity and activity coefficient based on Henry’s law are defined, respectively, by
ai =
Pi
kH
i
and gi =
ai
xi
(9.55)
Note that Henry’s law is still obeyed for a solute that has such a small vapor pressure
that we refer to it as a nonvolatile solute.
The Henry’s law standard state is a state in which the pure solute would have
a vapor pressure kH,solute rather than its actual value P *solute. It is a hypothetical state
that does not exist. Recall that the value k H
solute is obtained by extrapolation from the
low coverage range in which Henry’s law is obeyed. Although this definition may
seem peculiar, only in this way can we ensure that asolute S xsolute and gsolute S 1 as
xsolute S 0. We reiterate the reason for this choice. If the preceding conditions are satisfied, the concentration (divided by the unit concentration to make it dimensionless)
is, to a good approximation, equal to the activity. Therefore, equilibrium constants
for reactions in solution can be calculated without knowing the values of the activity
coefficients. We emphasize the difference in the Raoult’s law and Henry’s law standard states because it is the standard state chemical potentials m*solute and m*H
solute that
are used to calculate ∆G ~ and the thermodynamic equilibrium constant K. Different
choices for standard states will result in different numerical values for K. The standard
chemical potential m*H
solute refers to the hypothetical standard state in which xsolute = 1,
and each solute species is in an environment characteristic of the infinitely dilute
solution.
We now consider a less well-defined situation. For solutions in which the components are miscible in all proportions, such as the CS2 9acetone system, either a Raoult’s
law or a Henry’s law standard state can be defined, as we show with sample calculations in Example Problems 9.8 and 9.9. In this case, there is no unique choice for the
standard state over the entire concentration range in such a system. Numerical values
for the activities and activity coefficients will differ, depending on whether the Raoult’s
law or the Henry’s law standard state is used.
273
Concept
Standard states for real solutions
are chosen to make the value of the
activity coefficient approach one
if possible.
274
CHAPTER 9 Ideal and Real Solutions
1.0
EXAMPLE PROBLEM 9.8
Calculate the activity and activity coefficient for CS2 at xCS2 = 0.3502 using data
from Table 9.3. Assume a Raoult’s law standard state.
CS2 activity (RL)
0.8
Solution
0.6
aRCS2 =
0.4
g RCS2
0.2
0
0.2
0.4
0.6
xCS
0.8
1
2
Activity coefficient CS2 (RL)
4.0
3.5
3.0
2.5
2.0
=
PCS2
P *CS2
aRCS2
xCS2
=
358.3 Torr
= 0.6994
512.3 Torr
=
0.6994
= 1.997
0.3502
The activity and activity coefficients for CS2 are concentration dependent. Calculations of activity and activity coefficients using a Raoult’s law standard state and the
same methods as in this Example Problem are shown in Figure 9.17 as a function of
xCS2. For this solution, g RCS2 7 1 for all values of the concentration for which xCS2 6 1.
Note that g RCS2 S 1 as xCS2 S 1, as the model requires. The activity and activity coefficients for CS2 using a Henry’s law standard state are shown in Figure 9.18 as a function of xCS2. For this solution, g H
CS2 6 1 for all values of the concentration for which
xCS2 7 0. Note that g H
CS2 S 1 as xCS2 S 0 as the model requires. Which of these two
possible standard states should be chosen? There is a good answer to this question only
in the limits xCS2 S 0 or xCS2 S 1. For intermediate concentrations, either standard
state can be used.
EXAMPLE PROBLEM 9.9
1.5
1.0
Calculate the activity and activity coefficient for CS2 at xCS2 = 0.3502 using data
from Table 9.3. Assume a Henry’s law standard state.
0.5
Solution
0
0.2
0.4
0.6
0.8
aH
CS2 =
1
xCS2
gH
CS2 =
Figure 9.17
Activity and activity coefficient for CS2
in a CS2 9acetone solution based on a
Raoult’s law standard state. Activity
and activity coefficient are plotted as a
function of xCS2. RL = Raoult’s law.
PCS2
kH
CS2
aH
CS2
xCS2
=
358.3 Torr
= 0.204
1750 Torr
=
0.204
= 0.584
0.3502
The Henry’s law standard state is defined with respect to concentration measured in
units of mole fraction. This is not a particularly convenient scale, and either the molarity or molality concentration scales are generally used in laboratory experiments. The
mole fraction of the solute can be converted to the molality scale by dividing the numerator and denominator of the first expression in Equation (9.56) by nsolvent Msolvent:
xsolute =
nsolute
=
nsolvent + nsolute
msolute
1
Msolvent
(9.56)
+ msolute
where msolute is the molality of the solute, and Msolvent is the molar mass of the solvent
in kg mol-1. We see that msolute S xsolute >Msolvent as xsolute S 0. Using molality as the
concentration unit, we define the activity and activity coefficient of the solute by
amolality
solute =
g molality
solute =
Psolute
k molality
H
amolality
solute
msolute
with amolality
solute S msolute as msolute S 0 and
(9.57)
with g molality
solute S 1 as msolute S 0
(9.58)
The Henry’s law constants and activity coefficients determined on the mole fraction scale must be recalculated to describe a solution characterized by its molality or
275
9.11 Activities Are Defined with Respect to Standard States
0.25
0.20
CS2 activity (HL)
molarity. Example Problem 9.10 shows the conversion from mole fraction to molarity;
similar conversions can be made if the concentration of the solution is expressed in
terms of molality.
What is the standard state if concentrations rather than mole fractions are used? The
standard state in this case is the hypothetical state in which Henry’s law is obeyed by a
solution that is 1.0 molar (or 1.0 molal) in the solute concentration. It is a hypothetical
state because at this concentration, substantial deviations from Henry’s law will be observed. Although this definition may seem peculiar at first reading, only in this way can
we ensure that the activity becomes equal to the molarity (or molality), and the activity
coefficient approaches 1, as the solute concentration approaches 0.
0.15
0.10
0.05
EXAMPLE PROBLEM 9.10
a. Derive a general equation, valid for dilute solutions, relating the Henry’s law
­constants for the solute on the mole fraction and molarity scales.
b. Determine the Henry’s law constants for acetone on the molarity scale.
Use the results for the Henry’s law constants on the mole fraction scale cited in
Section 9.10. The density of acetone is 789.9 g L-1.
a. We use the symbol csolute to designate the solute molarity, and c ~ to indicate a
1 molar concentration.
dxsolute
dP
dP
=
csolute
csolute
dxsolute
da ~ b
da ~ b
c
c
To evaluate this equation, we must determine
xsolute =
dxsolute
csolute
da ~ b
c
1
0.2
0.4 0.6
xCS2
0.8
1
0.6
0.4
0.2
Activity and activity coefficient for CS2
in a CS2 9acetone solution based on a
Henry’s law standard state. Activity and
activity coefficient are plotted as a function of xCS2. HL = Henry’s law.
c ~ Msolvent dP
dP
=
rsolution dxsolute
csolute
da ~ b
c
1 mol L-1 * 58.08 g mol-1
dP
=
* 1950 Torr = 143.4 Torr
csolute
789.9 g L-1
da ~ b
c
The colligative properties for an ideal solution discussed in Sections 9.7 and 9.8
refer to the properties of the solvent in a dilute solution. The Raoult’s law standard state
applies to this case and in an ideal dilute solution, Equations (9.33), (9.35), and (9.41)
can be used with activities replacing concentrations:
∆Tf = -Kf gmsolute
∆Tb = Kbgmsolute
p = gcsolute RT
0.8
Figure 9.18
Therefore,
b.
0.4 0.6
xCS2
0.8
0
nsolute Msolvent
Msolvent
nsolute
nsolute
≈
=
= csolute
rsolution
nsolvent
nsolute + nsolvent
Vsolution rsolution
0.2
1.0
Activity coefficent CS2 (HL)
Solution
0
(9.59)
The activity coefficients are defined with respect to molality for the boiling point
­elevation ∆Tb and freezing point depression ∆Tf, and with respect to molarity for the
osmotic pressure p. Equations (9.59) provide a useful way to determine activity coefficients, as shown in Example Problem 9.11.
276
CHAPTER 9 Ideal and Real Solutions
EXAMPLE PROBLEM 9.11
In 500. g of water, 24.0 g of a nonvolatile solute of molecular weight 241 g mol-1 is
dissolved. The observed freezing point depression is 0.359°C. Calculate the activity
coefficient of the solute.
Solution
∆Tƒ = -Kƒ gmsolute; g = g =
1.86 K kg mol-1
∆Tƒ
Kƒ msolute
0.359 K
= 0.969
24.0
*
mol kg -1
241 * 0.500
9.12 HENRY’S LAW AND THE SOLUBILITY
OF GASES IN A SOLVENT
The ideal dilute solution model can be applied to the solubility of gases in liquid solvents. An example for this type of solution equilibrium is the amount of N2 absorbed
by water at sea level, which is considered in Example Problem 9.12. In this case, one
of the components of the solution is a liquid and the other is a gas. The equilibrium of
interest between the solution and the vapor phase is
Concept
Henry’s law is widely used to describe
the solubility of gases in liquids.
N21aqueous2 N N21vapor2
(9.60)
The chemical potential of the dissolved N2 is given by
msolution
= m*H
N2
N2 1vapor2 + RT ln asolute
(9.61)
In this case, a Henry’s law standard state is the appropriate choice because the nitrogen
is sparingly soluble in water. The mole fraction of N2 in solution, xN2, is given by
xN2 =
nN2
nN2 + nH2O
≈
nN2
n H 2O
(9.62)
The amount of dissolved gas is given by
nN2 = nH2O xN2 = nH2O
PN2
kH
N2
(9.63)
Example Problem 9.12 shows how Equation (9.63) is used to model the dissolution of
a gas in a liquid.
EXAMPLE PROBLEM 9.12
The average human with a body weight of 70. kg has a blood volume of 5.00 L. The
Henry’s law constant for the solubility of N2 in H2O is 9.04 * 104 bar at 298 K.
­Assume that this is also the value of the Henry’s law constant for blood and that the
density of blood is 1.00 kg L-1.
a. Calculate the number of moles of nitrogen absorbed in this amount of blood in air
of composition 80.% N2 at sea level, where the pressure is 1 bar, and at a pressure
of 50. bar.
b. Assume that a diver accustomed to breathing compressed air at a pressure of
50. bar is suddenly brought to sea level. What volume of N2 gas is released as
bubbles in the diver’s bloodstream?
9.13 Chemical Equilibrium in Solutions
277
Solution
nN2 = nH2O
a. =
PN2
kH
N2
5.0 * 103g
18.02 g mol
-1
*
0.80 bar
9.04 * 104 bar
= 2.5 * 10-3 mol at 1 bar total pressure
At 50. bar, nN2 = 50.*2.5 * 10-3 mol = 0.13 mol.
b. V =
nRT
P
(0.13 mol - 2.5 * 10-3 mol) * 8.314 * 10-2 L bar mol-1 K-1 * 300. K
1.00 bar
= 3.2 L
=
The symptoms induced by the release of air into the bloodstream are known to divers
as the bends. The volume of N2 just calculated is far more than is needed to cause
the formation of arterial blocks due to gas-bubble embolisms.
9.13 CHEMICAL EQUILIBRIUM IN SOLUTIONS
The concept of activity can be used to express the thermodynamic equilibrium constant
in terms of activities for real solutions. Consider a reaction between solutes in a solution. At equilibrium, the following relation must hold:
a a vjmj1solution2 b
j
equilibrium
= 0
(9.64)
where the subscript states that the individual chemical potentials must be evaluated
under equilibrium conditions. Each of the chemical potentials in Equation (9.64) can be
expressed in terms of a standard state chemical potential and a concentration-dependent
term. Assume a Henry’s law standard state for each solute. Equation (9.64) then takes
the form
*H
eq vj
a vjmj 1solution2 + RT a ln 1ai 2 = 0
j
j
(9.65)
Using the relation between the Gibbs energy and the chemical potential, we can write
the previous equation in the form
vj
∆ rG ~ = -RT a ln (aeq
i 2 = -RT ln K
j
(9.66)
The equilibrium constant in terms of activities is given by
eq vj
vj
eq vj ci
K = q 1aeq
2
=
1g
2
a
b
i
q i
c~
i
i
(9.67)
where the symbol Π indicates that the terms following the symbol are multiplied with
one another. It can be viewed as a generalization of KP, defined in Equation (6.60), and
it can be applied to equilibria involving gases, liquids, dissolved species, and solids.
For gases, the fugacities divided by ƒ ~ (see Section 7.5) are the activities.
To obtain a numerical value for K, the standard state Gibbs reaction energy ∆ rG ~
must be known. In general, the ∆ rG ~ must be determined experimentally. This can be
done by measuring the individual activities of the species in solution and calculating K
from these results. After a series of ∆ rG ~ for different reactions has been determined,
they can be combined to calculate the ∆ rG ~ for other reactions, as discussed for reaction enthalpies in Chapter 4. Because of the significant interactions between the solutes
and the solvent, K values depend on the nature of the solvent. Furthermore, for electrolyte solutions K values depend on the ionic strength, as will be discussed in Chapter 10.
Concept
Chemical equilibrium in solutions is
defined in terms of activities rather
than concentrations.
278
CHAPTER 9 Ideal and Real Solutions
An equilibrium constant in terms of molarities or molalities can also be defined
starting from Equation (9.67) and setting all activity coefficients equal to 1. This
is most appropriate for a dilute solution of a nonelectrolyte, using a Henry’s law
standard state:
eq vj
vj
ceq
i
vj ci
K = q (g eq
i 2 a ~ b ≈ qa ~ b
c
c
i
i
(9.68)
Example Problem 9.13 illustrates an application of Equation (9.68).
EXAMPLE PROBLEM 9.13
a. Write the equilibrium constant for the reaction N21aq, m2 N N21g, P2 in terms of
activities at 25°C, where m is the molality of N21aq2.
b. By making suitable approximations, convert the equilibrium constant of part (a)
into one in terms of pressure and molality only.
Solution
a. K =
eq nj
q 1ai 2
i
=
eq nj
q 1g i 2 a c ~
i
ceq
i
b
nj
=
a
a
P
gN2, g a P ~ b
=
gN2, aq m
a ~b
m
gN2, g P
P~
b
gN2, aq m
m~
b
b. Using a Henry’s law standard state for dissolved N2, gN2, aq ≈ 1, because the
concentration is very low. Similarly, because N2 behaves like an ideal gas to quite
high pressures at 25°C, gN2, g ≈ 1. Therefore,
P
b
P~
K ≈
m
a ~b
m
Note that in this case, the equilibrium constant is simply the Henry’s law constant
in terms of molality.
a
The numerical values for the dimensionless thermodynamic equilibrium constant
depend on the choice of the standard states for the components involved in the reaction.
The same is true for ∆ rG ~ . Therefore, it is essential to know which standard state has
been assumed before an equilibrium constant is used. The activity coefficients of most
neutral solutes are close to 1 with the appropriate choice of standard state. Therefore,
example calculations of chemical equilibria using activities will be deferred until electrolyte solutions are discussed in Chapter 10. For such solutions, gsolute differs substantially from 1, even for dilute solutions.
An important example of equilibrium processes in solution is the physical interaction of a molecule with a binding site. Such binding equilibria processes are ubiquitous in chemistry and biochemistry, being central to the function of oxygen transport
proteins such as hemoglobin or myoglobin, the action of enzymes, and the adsorption
of molecules to surfaces. All of these processes involve equilibrium between the free
molecule (R) and the molecule absorbed or bound to another species (protein, solid,
etc.) that contains a binding site (M). In the limit where there is a single binding site per
species, the equilibrium can be written as
R + M N RM
(9.69)
9.13 Chemical Equilibrium in Solutions
The equilibrium constant for this reaction in terms of the concentrations of the species
of interest is
cRM
cR * cM
(9.71)
Expressing this fraction in terms of the equilibrium constant yields
v =
1>1cRcM2
cRM
KcR
K
a
b =
=
cM + cRM 1>1cRcM2
1>cR + K
1 + KcR
(9.72)
This expression demonstrates that v depends on both the molecular concentration and
the value of the equilibrium constant. The variations of v with c R for different values of
K are presented in Figure 9.19. With an increase in K (consistent with the bound form
of the molecule being lower in Gibbs energy relative to the free form), the average
number of bound molecules per species approaches unity (that is, the binding sites are
completely occupied) even at low values of cR.
To this point in this section, the development of a theory for chemical equilibrium
in a solution was limited to a single binding site for a given molecule. To consider molecules with multiple binding sites, two issues must be considered. First, binding to one
site may have an effect on binding at other sites. This effect is important, for example,
in the binding of oxygen to hemoglobin. However, in the current theory development,
we will restrict ourselves to independent site binding, where the binding at one site
has no effect on binding at other sites. In this case, the average number of molecules
bound to a specific site (i) is
vi =
KicR
1 + KicR
K5100
0.8
0.6
K55
0.4
0
0
Figure 9.19
Average number of bound molecules
per species plotted against the concentration of free molecules for different
equilibrium constant values. The average number of bound molecules per species, v; concentration of free species, cR;
equilibrium constant, K. Values for K of
5, 10, 30, and 100 are shown.
500
400
(9.74)
(9.75)
This expression is almost identical to our previous expression for v for a single binding
site, but with the addition of a factor of N. Therefore, at large values of cR, the average
number of bound molecules per species will approach N instead of 1. To determine K
and N for specific binding equilibrium, the expression for v is recast in a linear form
referred to as the Scatchard equation:
v =
NKcR
1 + KcR
v
= -K v + NK
cR
0.1 0.2 0.3 0.4 0.5 0.6 0.7
Concentration of
free species, CR
(9.73)
300
n
CR
The second issue to address is the equivalency of the binding sites. Assuming that all of
the binding sites are equivalent, v becomes
N
KicR
NKcR
v = a
=
1
+
K
c
1
+ KcR
i R
i=1
K510
0.2
If there are N binding sites, then the average number of molecules adsorbed per species
is simply the sum of the average values for each specific binding site:
N
N
KicR
v = a vi = a
1
+
KicR
i=1
i=1
K530
1
(9.70)
(In biochemistry, K is commonly expressed in terms of concentrations and, therefore,
has units.) Assuming that there is a single binding site for the molecule, then the average number of bound molecules per species (v2 is given by
cRM
v =
cM + cRM
1.2
Average number of bound
–
molecules per species, n
K =
279
200
100
0
1
2
3
4
5
0
Average number of bound
molecules per species, n
Figure 9.20
(9.76)
The Scatchard equation demonstrates that a plot of v>cR versus v should yield a
straight line having a slope of -K and a y-intercept of NK from which N can be readily determined. A Scatchard plot for K = 100 and N = 4 is presented in Figure 9.20.
Scatchard plot of independent binding
for K = 100 and N = 4. K, equilibrium
constant; N, number of binding sites for
the molecule of interest. Note that the y
intercept is K * N = 400, the slope is
- K = -100, and the x ­intercept is 4.
280
CHAPTER 9 Ideal and Real Solutions
Deviations from linear behavior in a Scatchard plot can occur if binding is not independent or if the binding sites are not equivalent.
Example Problem 9.14 illustrates an application of Equation 9.76.
EXAMPLE PROBLEM 9.14
A variety of molecules physically bind to DNA by intercalation or insertion between
the base pairs. The intercalation of chlorobenzylidine (CB) to calf thymus DNA was
studied by circular dichroism spectroscopy [Zhong, W., Yu, J.-S., Liang, Y., Fan, K.,
and Lai, L., Spectrochimica Acta, Part A, 60A (2004): 2985]. Using the following
­representative binding data, determine the binding constant (K) and the number of
binding sites per base (N).
cCB 1 :106 M 2
v
1.0
2.0
3.0
5.0
0.0063
0.0123
0.0180
0.0285
10.0
0.0515
Solution
Plot of v>cCB versus v. Using the provided binding data allows for a determination of
K and N. Using the data, we can construct the following Scatchard plot:
–
21
n/C
CB (M )
6500
6000
5500
5000
Tuc 5
65.85°C
Temperature
a9
L1
T1
l1
L2
l2
a
L1 + L2
0
x1
x2
xphenol
x3 1
Figure 9.21
A temperature–composition (T–x)
phase diagram with an upper consolute
temperature for the partially miscible
liquids water and phenol. Pressure is
held constant at 1 bar. The beige-colored
region labeled L 1 + L 2 is a two-phase
region. The blue-colored balance of the
diagram is a one-phase region.
0
0.02
0.04
–
Average number of bound molecules per species, n
0.06
A best fit to a straight line results in a slope of -26,000 M-1 and a y intercept of
6500. Comparison to the Scatchard equation reveals that the slope is equal to -K,
so that K = 26,000 M -1. In addition, the y intercept is equal to the product of N and
K; therefore, N = 0.25. This value of N is consistent with a binding site composed
of four bases, or two base pairs, which is consistent with intercalation between
base pairs.
9.14 SOLUTIONS FORMED FROM PARTIALLY
MISCIBLE LIQUIDS
Two liquids need not be miscible in all proportions. Over a limited range of relative concentration, a homogeneous solution can be formed, and outside of this range, two liquid
phases separated by a phase boundary can be observed. An example is the water–phenol
system for which a temperature–composition (T–x) phase diagram is shown in
Figure 9.21. The pressure is assumed to be sufficiently high that no gas phase is present.
As the phenol concentration increases from 0 at the temperature T1, a single, waterrich liquid phase of variable composition is formed. However, at x1, the appearance
of a second, phenol-rich liquid phase is observed, separated from the first phase by a
9.14 Solutions Formed from Partially Miscible Liquids
210
l2
a0
Temperature, T (°C)
Temperature, T (°C)
l1
L1 + L2
L1
18.5
a9
281
L2
L1
L1 1 L2
L2
61
a
0
0
1
1
xnicotine
xtriethylamine
Figure 9.22
Figure 9.23
A T–x phase diagram with a lower consolute temperature for the partially miscible liquids water and triethylamine.
Pressure is held constant at 1 bar. The
beige-colored region labeled L 1 + L 2 is a
two-phase region. The blue-colored balance of the diagram is a one-phase region.
A T–x phase diagram with an upper
and lower consolute temperature for
the partially miscible liquids water
and nicotine. Pressure is held constant at
1 bar. The beige-colored region labeled
L 1 + L 2 is a two-phase region. The
blue-colored balance of the diagram is a
one-phase region.
horizontal boundary with the denser phase below the less dense phase. If we start with
pure phenol and add water at T1, separation into two phases is observed at x3. In the
phenol–water system, solubility increases with temperature so that the extent in x of the
two-phase region decreases with increasing T, and phase separation is observed only
at a single composition at the upper consolute temperature Tuc. For T 7 Tuc, phenol
and water are completely miscible.
The composition of the two phases in equilibrium for the phenol concentration
given by x2 at a given temperature is given by the x values at the ends of the tie line l1l2,
and the relative amount of the two phases present is given by the lever rule:
x3 - x2
moles L 1
al2
=
=
x2 - x1
moles L 2
al1
(9.77)
Although less common, liquid–liquid systems are known in which the solubility decreases as the temperature increases (at least for part of the temperature range). In this
case, a lower consolute temperature is observed, below which the two liquids are completely miscible. An example is the triethylamine–water system shown in Figure 9.22.
The water–nicotine system has both a lower and an upper consolute temperature, and
phase separation is only observed between 61°C and 210°C, as shown in Figure 9.23.
We next consider solutions of partially miscible liquids at pressures for which a
vapor phase is in equilibrium with one or more liquid phases. The composition of the
vapor for a given value of the average composition of the liquid depends on the relative
values of the upper consolute temperature and the minimum boiling temperature for
the azeotrope. For the system shown in Figure 9.24a, the vapor phase is only present
at temperatures greater than the upper consolute temperature. In this case, the vapor
and liquid composition is the same as that described in Section 9.4 for minimum boiling point azeotropes. However, if the upper consolute temperature is greater than the
boiling temperature of the minimum boiling azeotrope, the T–x diagram shows new
features, as illustrated for the water–butanol system in Figure 9.24b.
First, consider the system at point a in Figure 9.24b. At this temperature, the system
consists of a single liquid phase enriched in butanol. If the temperature is increased to Tb
at constant pressure, the gas-phase composition is given by point b. If the vapor of this
composition is removed from the system and cooled to the original temperature Ta, it is
at point c in the phase diagram, which corresponds to two liquid phases of composition
d and e. This system consists of two components and, therefore, F = 3 - p = 1.
Concept
Phase diagrams are useful in
describing partially miscible liquids.
282
CHAPTER 9 Ideal and Real Solutions
Figure 9.24
Vapor
L1V
Temperature
T0a
c
a
b
Solid
0
1
xB
(a)
Temperature
Liquid
T0a
g
Te
h
f
e
T0b
d
L1 1 L2
Z1
xB
100
l
k
Tj
L2 1 V
b
Tb
j
L1 1 V
94
L1
g
d
i
f
h
c
L2
Ta
e a
L1 1 L2
0
ZB
1
0
1
Z butanol
(b)
Next, consider heating the liquid described by point ƒ in the two-liquid-phase
region at constant pressure. Vapor first appears when the temperature is increased to
94°C and has the composition given by point i, corresponding to a minimum boiling
point azeotrope. The vapor is in equilibrium with two liquid phases that have the composition given by points g and h. Because the pressure is constant in Figure 9.24b and
p = 3 at point i, the system has zero degrees of freedom. Distillation of the solution
under these conditions results in two separate liquid phases of fixed composition g and
h. As more heat is put into the system, liquid is converted to vapor, but the composition
of the two liquid phases and the vapor phase remains constant as described by points g,
h, and i. Because the vapor is enriched in butanol with respect to the average composition of the system, the butanol-rich phase L 2 is converted to vapor before the water-rich
phase L 1. At the point when the L 2 phase disappears, a degree of freedom is gained, the
temperature increases beyond 94°C, and the vapor composition changes along the i–j
curve. As the vapor approaches the composition corresponding to j, the entire system is
converted to vapor and the last drop of L 1, which disappears at Tj, has the composition
given by point k. As the last drop of liquid is converted to vapor, an additional degree
of freedom is gained, and further heat input results in an increase in the temperature of
the vapor, with no change in its composition.
To summarize, continued input of heat into a system of composition ƒ at 94°C
increases the amount of material in the vapor phase. As long as the two phases of
composition g and h are present, the temperature is constant and the amount of vapor
increases, but its composition is fixed at i. After the last drop of L 2 of composition h
boils off, the vapor composition continuously changes along the curve i–j. As the last
drop of L 1 of composition k evaporates, the vapor composition approaches the average
composition of the system given by ƒ, j, and l.
9.15 SOLID–SOLUTION EQUILIBRIUM
Solid
0
L1V
Liquid
(a)
Liquid
118
Vapor
Temperature, T (°C)
Temperature
T–x composition phase diagrams for
partially miscible liquids. (a) A generic
phase diagram in which the upper consolute temperature is at a temperature lower
than the azeotrope boiling temperature.
(b) The system water–butanol. The upper
consolute temperature is at a temperature
higher than the minimum azeotrope boiling
temperature. For both diagrams, pressure is
held constant at 1 bar and the beige-colored
regions labeled L 1 + L 2 and L + V are
two-phase regions, whereas the bluecolored regions are one-phase regions.
ZB
1
(b)
Figure 9.25
T–x phase diagrams showing solid–
liquid equilibria in a two-component
system. Pressure is held constant at 1 bar.
(a) Component B is assumed to be the
solute. (b) Full-phase diagram, including
eutectic point for all possible concentrations of A and B.
Consider the binary system A–B. At low temperatures, the components separate into
pure solid phases A and B. At higher temperatures, this system consists of a single
liquid phase, as A and B are miscible for all proportions of A and B. The transition
between two immiscible solid phases and a single liquid phase exhibits interesting behavior, which we will now discuss. If a liquid solution of B in A (A is the solvent and,
therefore, xB 6 xA2 is cooled sufficiently, pure solvent A will crystallize out in a solid
phase, as discussed in Section 9.7 in connection with the freezing point depression. For
example, point a in Figure 9.25a represents a liquid solution of composition b in equilibrium with pure solid A (point c). The relative number of moles of solution and solid can
be found using the lever rule. The temperature T at which freezing occurs is given by
R ln xA
1
1
=
T
Tfus1A2
∆ fusH1A2
(9.78)
Vocabulary
Concept
A minimum in the melting
temperature with concentration can
be observed for solid-solid solutions.
1500
1300
Temperature, T (°C)
where Tfus1A2and ∆ fus H1A2 are, respectively, the freezing temperature and the enthalpy
of fusion of pure solvent A. The temperature T at which freezing occurs is shown as
a function of xB by the dark curve in Figure 9.25a, and the dashed range indicates the
region beyond which Equation (9.78) is unlikely to be valid.
The same argument holds for a solution of A in B whereby we have interchanged
the solvent and solute roles. Therefore, the appropriate phase diagram for all values of
xB is that shown in Figure 9.25b. For the solid, we use the variables ZA and ZB rather
than xA and xB because the solid consists of an inhomogeneous mixture of crystallites
of pure A and pure B. In Figure 9.25b, point d represents pure solid B in equilibrium
with a solution of composition ƒ. Point g represents pure solid A in equilibrium with the
solution of composition h. Point e is a special point called the eutectic point that is
the minimum melting point observed in the A–B system. The temperature Te is called
the eutectic temperature (from the Greek word meaning easily melted).
Because the tie line at Te extends across the entire phase diagram, a solution at Te
is in equilibrium with both pure A (point i) and pure B (point j). As more heat is put
into the system, the volume of the solution phase increases, but its composition remains
constant at Ze until one of the pure solids is completely melted because, while the three
phases are present, F = 0.
Eutectic materials are widely used in making plumbing and electrical connections
with a solder. The benefit of such materials is that the components to be joined do not
need to be heated to high temperatures. Until the 1980s, copper plumbing components
were connected using Pb–Sn solders. Now that it is recognized that even small amounts
of lead present a health hazard, Pb–Sn solders have been replaced by lead-free solders,
which may contain tin, copper, silver, indium, zinc, antimony, or bismuth.
Another important application of eutectic materials is in the manufacture of electrical connections to silicon chips in the microelectronics industry. Gold is deposited onto
the Si surface using masks to establish an appropriate pattern. Subsequently, the chip
is heated to nearly 400°C. As the temperature increases, gold atoms diffuse rapidly
into the silicon, and a very thin liquid film forms at the interface once enough gold has
diffused to reach the eutectic composition. As more heat is put into the chip, a larger
volume of eutectic alloy is produced, which is ultimately the electrical connection after
the chip is cooled rapidly to room temperature, freezing in the high-temperature equilibrium state. The Au–Si phase diagram is shown in Figure 9.26; it is seen that the
eutectic temperature is approximately 700°C below the melting point of pure gold and
950°C below the melting point of pure Si.
The eutectic mixture is useful in bonding a gold wire to a silicon chip to form an
electrical connection. Imagine that a gold wire is placed in contact with a Si chip on
which Au has been deposited. If the chip is heated, the eutectic liquid will form before
the Si or the Au melts, allowing the eutectic mixture to form a connection between the
gold wire and the chip without melting the Au or the Si.
1100
Liquid
900
700
Liquid + Solid
500
363°C
300
18.6%
100
0
Solid
20
40
60
80 100
Atomic percent silicon
Figure 9.26
The T–x phase diagram for the gold–
silicon system. Note that the eutectic
melting point is much lower than the
melting points of the pure substances
Au and Si. Pressure is constant at 1 bar.
VOCABULARY
activity
activity coefficient
average composition
binding equilibria
boiling point elevation
colligative properties
eutectic temperature
fractional distillation
freezing point depression
Gibbs–Duhem equation
Henry’s law
Henry’s law constant
Henry’s law standard state
ideal dilute solution
ideal solution
independent site binding
lever rule
limiting law
lower and upper consolute temperature
osmosis
osmotic pressure
partial molar volume
283
pressure–average composition
(P–Z) diagram
Raoult’s law
Raoult’s law standard state
Scatchard equation
Scatchard plot
semipermeable membrane
solute
solvent
temperature–composition diagram
tie line
284
CHAPTER 9 Ideal and Real Solutions
KEY EQUATIONS
Equation
Significance of Equation
Equation Number
Pi = xiP *i i = 1, 2
Definition of ideal solution
9.1
msolution
= mvapor
i
i
Definition of equilibrium between liquid and vapor phases
9.2
Chemical potential of pure liquid in equilibrium with its vapor
9.6
Central equation describing ideal solutions
9.7
Pressure in vapor above ideal binary solution in terms of mole
fraction in liquid phase
9.9
Mole fraction of component in vapor phase above binary solution
9.10
Pressure in vapor above ideal binary solution in terms of mole
fraction in vapor phase
9.12
Phase rule
9.14
Lever rule for calculating amounts of vapor and liquid
in ­coexistence region
9.18
Gibbs–Duhem equation for binary solution relates chemical
­potential of solute and solvent
9.23
Freezing point depression
9.33
Boiling point elevation
9.35
van’t Hoff equation for osmotic pressure
9.41
Definition of partial volume
9.44
Volume of binary solution in terms of partial molar volumes
9.45
Gibbs–Duhem equation relating partial molar volumes for binary
solution
9.46
mi = m*i + RT ln
Pi
P *i
*
mi = mi + RT ln xi
Ptot = P *2 + 1P *1 - P *22x1
y1 =
P1
x1P *1
= *
Ptot
P 2 + 1P *1 - P *22x1
P *1P *2
Ptot =
P *1
+
1P *2
-
F = C - p + 2
P *12y1
tot
ntot
liq1ZB - xB2 = nvapor1yB - ZB2
n1dm1 + n2dm2 = 0 or x1dm1 + x2dm2 = 0
∆Tƒ = ∆Tb =
p =
RMsolventT 2fus
msolute = - Kƒ msolute
∆ fusH
RMsolventT 2vap
∆ vapH
msolute = Kb msolute
nsolute RT
V
V1(P, T, n1, n22 = a
0V
b
0n1 P, T, n2
V = n1V1(P, T, n1, n22 + n2V21P, T, n1, n22
x1 dV1 + x2 dV2 = 0 or dV1 = -
asolvent =
ai =
Pi
kH
i
Psolvent
P *solvent
; gsolvent =
and gi =
x2
dV
x1 2
asolvent
xsolvent
ai
xi
Definition of activity and activity coefficient for ideal dilute
­solution
9.48, 9.49
Definition of activity and activity coefficient for ideal dilute
­solution based on Henry’s law
9.55
CONCEPTUAL PROBLEMS
Q9.1 Why is the magnitude of the boiling point elevation less
than that of the freezing point depression?
Q9.2 Fractional distillation of a particular binary liquid mixture leaves a residual liquid that consists of both components,
the composition of which does not change as the liquid is
boiled off. Is this behavior characteristic of a maximum or a
minimum boiling point azeotrope?
Q9.3 In the description of Figure 9.24, the following sentence appears: “At the point when the L 2 phase disappears,
the temperature increases beyond 94°C and the vapor composition changes along the i–j curve.” Why does the vapor
­composition change along the i–j curve?
Numerical Problems
Q9.4 Explain why chemists conducting quantitative work
using liquid solutions prefer to express concentration in terms
of molality rather than molarity.
Q9.5 Explain the usefulness of a tie line on a P–Z phase
­diagram such as that of Figure 9.4.
Q9.6 For a pure substance, the liquid and gaseous phases
can only coexist for a single value of pressure at a given
temperature. Is this also the case for an ideal solution of two
volatile liquids?
Q9.7 Using the differential form of G, dG = VdP - SdT, show
that if ∆ mixG = nRT a i xi ln xi, then ∆ mixH = ∆ mixV = 0.
Q9.8 What information can be obtained from a tie line in a
P–Z phase diagram?
Q9.9 You boil an ethanol–benzene mixture with
xethanol = 0.35. What is the composition of the vapor phase
that first appears? What is the composition of the last liquid
left in the vessel?
Q9.10 What can you say about the composition of the solid at
temperatures less than the eutectic temperature in Figure 9.26
on a microscopic scale?
285
Q9.11 Is a whale likely to get the bends when it dives deep
into the ocean and resurfaces? Answer this question by
­considering the likelihood of a diver getting the bends if he or
she dives and resurfaces on one lung full of air as ­opposed to
breathing air for a long time at the deepest point of the dive.
Q9.12 Why is the preferred standard state for the solvent in
an ideal dilute solution the Raoult’s law standard state? Why is
the preferred standard state for the solute in an ideal dilute solution the Henry’s law standard state? Is there a preferred standard state for the solution in which xsolvent = xsolute = 0.5?
Q9.13 The entropy of two liquids is lowered if they mix.
How can immiscibility be explained in terms of thermodynamic state functions?
Q9.14 Explain why colligative properties depend only on the
concentration, and not on the identity of the solute molecules.
Q9.15 The statement “The boiling point of a typical liquid
mixture can be reduced by approximately 100°C if the pressure is reduced from 760 to 20 Torr” is found in Section 9.4.
What figure or figures in Chapter 9 can you identify to support this statement in a qualitative sense?
NUMERICAL PROBLEMS
Section 9.3
P9.1 At 325 K, the vapor pressure of benzene is 295 Torr and
that of hexane is 439 Torr. Calculate the vapor pressure of a
solution for which xbenzene = 0.65 assuming ideal behavior.
P9.2 An ideal solution is made up of the volatile liquids A
and B, for which P *A = 115. Torr and P *B = 65.0 Torr.
Initially, the pressure is high enough that only the liquid phase
is present. As pressure is reduced, the first vapor is observed
at a total pressure of 94.5 Torr. Calculate xA.
P9.3 At 280. K, the vapor pressure of ethyl bromide is
0.295 bar, and that of ethyl chloride is 0.823 bar. Assume that
the solution is ideal. For the case in which there is only a
trace of liquid present and the mole fraction of ethyl chloride
in the vapor is 0.750, answer these questions:
a. What are the total pressure and the mole fraction of ethyl
chloride in the liquid?
b. If there are 4.25 mol of liquid and 2.75 mol of vapor present at the same pressure as in part (a), what is the overall
composition of the system?
P9.4 A and B form an ideal solution at 298 K. With xA = 0.230,
P *A = 84.0 Torr and P *B = 37.5 Torr.
a. Calculate the partial pressures of A and B in the gas phase.
b. A portion of the gas phase is removed and condensed in a
separate container.
Calculate the partial pressures of A and B in equilibrium with
this liquid sample at 298 K.
P9.5 Given the vapor pressures of the pure liquids and the
overall composition of the system, what are the upper and
lower limits of pressure between which liquid and vapor
­coexist in an ideal solution?
P9.6 A and B form an ideal solution. At a total pressure of
0.650 bar, yA = 0.485 and xA = 0.390. Use this information
to calculate the vapor pressure of pure A and pure B.
P9.7 An ideal solution is formed by mixing liquids A and B
at 298 K. The vapor pressure of pure A is 160 Torr, and that
of pure B is 95 Torr. If the mole fraction of A in the vapor is
0.570, what is the mole fraction of A in the solution?
P9.8 A solution is prepared by dissolving 54.0 g of a nonvolatile solute in 150 g of water. The vapor pressure above the
solution is 22.97 Torr, and the vapor pressure of pure water is
23.76 Torr at this temperature. What is the molecular mass of
the solute?
P9.9 A volume of 4.90 L of air is bubbled through liquid
toluene at 360. K, thus reducing the mass of toluene in the
beaker by 7.48 g. Assuming that the air emerging from the
beaker is saturated with toluene, determine the vapor pressure
of toluene at this temperature.
P9.10 The vapor pressures of 1-bromobutane and
1-chlorobutane can be expressed in the form
ln
Pbromo
1584.8
= 17.076 Pa
T
- 111.88
K
and
ln
Pchloro
2688.1
= 20.612 Pa
T
- 55.725
K
Assuming ideal solution behavior, calculate xbromo and ybromo
at 280. K and a total pressure of 4000. Pa.
286
CHAPTER 9 Ideal and Real Solutions
P9.11 In an ideal solution of A and B, 2.25 mol is in the liquid phase and 4.25 mol is in the gaseous phase. The overall
composition of the system is ZA = 0.150 and xA = 0.225.
Calculate yA.
P9.12 Assume that 1-bromobutane and 1-chlorobutane
form an ideal solution. At 285 K, P *chloro = 7242 Pa and
P *bromo = 2756 Pa. When only a trace of liquid is present at
285 K, ychloro = 0.800.
a. Calculate the total pressure above the solution.
b. Calculate the mole fraction of 1-chlorobutane in the s­ olution.
c. Calculate Zchloro and Zbromo.
P9.13 An ideal dilute solution is formed by dissolving the
solute A in the solvent B. Write expressions equivalent to
Equations (9.9) through (9.13) for this case.
P9.14 At 325 K, pure toluene and hexane have vapor pressures of 1.42 * 104 Pa and 5.77 * 104 Pa, respectively.
a. Calculate the mole fraction of hexane in the liquid mixture
that boils at a pressure of 0.375 atm.
b. Calculate the mole fraction of hexane in the vapor that is in
equilibrium with the liquid of part (a).
P9.15 You wish to remediate soil that has been contaminated
by 1-chlorobutane by passing air through it at a total pressure of 1.00 bar. (a) Assuming that the air leaving the soil is
saturated with 1-chlorobutane, calculate how many liters of
air are required to remove 10.0 g of 1-chlorobutane at 298 K.
(b) Would raising the temperature of the soil make the process
more efficient? Answer this question by calculating how many
liters of air are required to remove 10.0 g of 1-chlorobutane
at 320. K using the data in Table 8.3.
Section 9.7
P9.16 The dissolution of 22.0 g of a substance in 650. g of
benzene at 298 K raises the boiling point by 0.668°C. Note
that Kƒ = 5.12 K kg mol-1, Kb = 2.53 K kg mol-1, and the
density of benzene is 876.6 kg m-3. Calculate the freezing
point depression, the ratio of the vapor pressure above the
solution to that of the pure solvent, the osmotic pressure
and the molar mass of the solute.
P9.17 The heat of fusion of ice is 6.008 * 103 J mol-1 at
its normal melting point of 273.15 K. Calculate the freezing
point depression constant Kƒ.
Section 9.8
P9.18 The osmotic pressure of an unknown substance is measured at 298 K. Determine the molar mass if the concentration
of this substance is 23.7 kg m-3 and the osmotic pressure is
2.50 * 104 Pa. The density of the solution is 997 kg m-3.
P9.19 A sample of glucose (C6H12O62 of mass 9.75 g is
placed in a test tube of radius 2.00 cm. The bottom of the test
tube is a membrane that is semipermeable to water. The tube
is partially immersed in a beaker of water at 298 K, so that the
bottom of the test tube is only slightly below the level of the
water in the beaker. The density of water at this temperature
is 997 kg m-3. After equilibrium is reached, how high is the
level of the water in the tube above that in the beaker? What is
the value of the osmotic pressure? You may find the approximation ln 11>11 + x22 ≈ -x useful.
Section 9.9
P9.20 The partial molar volumes of water and ethanol
in a solution with xH2O = 0.45 at 25°C are 17.0 and
57.0 cm3 mol-1, respectively. Calculate the volume change
upon mixing sufficient ethanol with 4.50 mol of water to
give this concentration. The densities of water and ethanol
are 0.997 and 0.7893 g cm-3, respectively, at this
temperature.
P9.21 A solution is made up of 258.7 g of ethanol and
151.7 g of H2O. If the volume of the solution is 468.1 cm3
and the partial molar volume of H2O is 17.0 cm3 mol-1,
what is the partial molar volume of ethanol under these
conditions?
P9.22 The densities of pure water and ethanol are 997 and
789 kg m-3, respectively. The partial molar volumes of
­ethanol and water are 55.2 and 17.8 * 10-3 L mol-1, respectively. Calculate the change in volume relative to the pure
components when 3.25 L of a solution with xethanol = 0.540
is prepared.
Section 9.10
P9.23 The partial pressures of Br2 above a solution containing CCl4 as the solvent at 25°C have the values listed, as a
function of the mole fraction of Br2 in solution, in the following table [G. N. Lewis and H. Storch, J. American Chemical
Society 39 (1917), 2544]. Use these data and a graphical
method to determine the Henry’s law constant for Br2 in
CCl4 at 25°C.
xBr2
0.00394
0.00420
0.00599
0.0102
PBr2 (Torr)
1.52
1.60
2.39
4.27
xBr2
PBr2 (Torr)
0.0130
0.0236
0.0238
0.0250
5.43
9.57
9.83
10.27
P9.24 The data from Problem P9.23 can be expressed in
terms of the molality rather than the mole fraction of Br2.
Use the data from the following table and a graphical method
to determine the Henry’s law constant for Br2 in CCl4 at 25°C
in terms of molality.
mBr2
1mol kg −1 2
0.026
0.028
0.039
0.067
PBr2 (Torr)
1.52
1.60
2.39
4.27
mBr2
1 mol kg −1 2
0.086
0.157
0.158
0.167
PBr2 (Torr)
5.43
9.57
9.83
10.27
P9.25 (a) How many grams of CO2 are dissolved in a 1.00 L
bottle of carbonated water at 293.15 K if the pressure used
in the carbonation process was 2.20 bar? (b) What volume would the dissolved CO2 occupy as a gas at 293.15 K
and 1.00 bar? The density of water at this temperature is
998 kg m-3.
P9.26 Calculate the concentration of O2 dissolved in water
in equilibrium with air at 298 K and a total pressure of
1.00 bar. The vapor pressure of water at this temperature is
3.17 * 103 Pa, and the mole fraction of O2 in air is 0.2095.
Numerical Problems
Section 9.11
P9.27 G. Ratcliffe and K. Chao [Canadian Journal of Chemical Engineering 47 (1969), 148] obtained the following
tabulated results for the variation of the total pressure above
a solution of isopropanol (P *1 = 1008 Torr) and n-decane
(P *2 = 48.3 Torr) as a function of the mole fraction of the
n-decane in the solution and vapor phases. Using these data,
calculate the activity coefficients for both components using a
Raoult’s law standard state.
P (Torr)
x2
y2
942.6
909.6
883.3
868.4
830.2
786.8
758.7
0.1312
0.2040
0.2714
0.3360
0.4425
0.5578
0.6036
0.0243
0.0300
0.0342
0.0362
0.0411
0.0451
0.0489
P9.28 At a given temperature, a nonideal solution of the
volatile components A and B has a vapor pressure of 275. Torr.
For this solution, yA = 0.320. In addition, xA = 0.210,
P *A = 585. Torr, and P *B = 465 Torr. Calculate the activity
and activity coefficient of A and B.
P9.29 At 39.9°C, a solution of ethanol (x1 = 0.9006,
P *1 = 130.4 Torr) and isooctane (P *2 = 43.9 Torr) forms
a vapor phase with y1 = 0.6667 at a total pressure of
185.9 Torr.
a. Calculate the activity and activity coefficient of each
­component.
b. Calculate the total pressure that the solution would have
if it were ideal.
P9.30 Calculate the activity and activity coefficient for
CS2 at xCS2 = 0.7220 using the data in Table 9.3 for both a
Raoult’s law and a Henry’s law standard state.
Section 9.12
P9.31 At high altitudes, mountain climbers are unable to
absorb a sufficient amount of O2 into their bloodstreams to
maintain a high activity level. At a pressure of 1 bar, blood
is typically 95% saturated with O2, but near 29,000 ft (the
altitude of the summit of Mount Everest), where the pressure
is about 0.300 bar, the corresponding degree of saturation is
63.%. Assuming that the Henry’s law constant for blood is the
same as that for water, calculate the mass of O2 dissolved in
1.00 L of blood for pressures of 1 bar and 0.300 bar. Air contains 20.99% O2 by volume. Assume that the density of blood
is 998 kg m-3.
P9.32
a. Calculate the solubility of N2 in 1 L of water at 298 K if
its pressure above the solution is 12.75 bar. The density of
water is 997 kg m-3.
b. If the pressure were suddenly reduced to 1.00 bar, as could
happen in a diving accident, what volume of N2 is released
per liter of blood, assuming that the solubility of N2 is the
same as that in water?
287
Section 9.13
P9.33 The binding of NADH to human liver mitochondrial
isozyme was studied [Biochemistry 28 (1989) 5367] and
it was determined that only a single binding site is present
with K = 2.0 * 107 M-1. What concentration of NADH is
required to occupy 8% of the binding sites?
P9.34 DNA is capable of forming complex helical structures.
An unusual triple-helix structure of poly(dA).2poly(dT) DNA
was studied by P. V. Scaria and R. H. Shafer [Journal of Biological Chemistry, 266 (1991) 5417] where the intercalation
of ethidium bromide was studied using UV absorption and
circular dichroism spectroscopy. The following representative
data were obtained using the results of this study:
cEth 1 MM 2
N
0.63
1.7
5.0
10.0
14.7
0.007
0.017
0.039
0.059
0.070
Using the above data, determine K and N for the binding of
ethidium bromide to the DNA triple-helical structure.
Section 9.14
P9.35 Two liquids, A and B, are immiscible for T 6 75.0°C
and for T 7 45.0°C and are completely miscible outside of
this temperature range. Sketch the phase diagram, showing as
much information as you can from these observations.
P9.36 Describe the changes you would observe as the temperature of a mixture of phenol and water at point a in
Figure 9.21 is increased until the system is at point a′. How
does the relative amount of separate phases of phenol-rich
liquid L 2 and water-rich liquid L 1 change along this path?
P9.37 Describe what you would observe if you heated the
liquid mixture at the composition corresponding to point i in
Figure 9.24b from a temperature below Ta to 118°C.
P9.38 Two liquids, A and B, are immiscible for
xA = xB = 0.5, for T = 75.0°C, and completely miscible
for T 7 75.0°C. Sketch the phase diagram, showing as much
information as you can from these observations.
P9.39 Describe the changes you would observe as the temperature of a mixture of triethylamine and water at point a in
Figure 9.22 is increased until the system is at point a″. How
does the relative amount of separate phases of triethylamine
and water change along this path?
P9.40 Describe the changes in a beaker containing water and
butanol that you would observe along the path a S b S c in
Figure 9.24b. How would you calculate the relative amounts
of different phases present along the path?
P9.41 Describe the changes in a beaker containing water and
butanol that you would observe along the path ƒ S j S l in
Figure 9.24b. How would you calculate the relative amounts
of different phases present along the path?
P9.42 Describe the changes in a beaker containing water and
butanol that you would observe along the path ƒ S j S k in
Figure 9.24b. How would you calculate the relative amounts
of different phases present along the path?
288
CHAPTER 9 Ideal and Real Solutions
b. The vertical dashed line shows a cooling process, starting
at point a which corresponds to a single phase liquid
region. Describe which phases and components, and the
relative amounts of the components are present at the
points ­labeled a–e.
a
b
c
Temperature
Section 9.15
P9.43 Describe what you would observe if you heated the
solid at the composition 40. atomic percent Si in Figure 9.26
from 300.°C to 1300.°C.
P9.44 Describe the system at points d and g in Figure 19.25b.
How would you calculate the relative amounts of different
phases present at these points?
P9.45 A phase diagram for an alloy, the two components of
which are completely soluble in the liquid phase and completely insoluble in the solid phase, is shown below.
a. Label the four regions of the phase diagram indicating
what components (A and/or B, where A and B denote
­solids) and phases are present.
b. The vertical dashed line shows a cooling process, starting
at point a. Describe which components and the relative
amounts of the components that are present at the points
labeled a–e.
d
e
a
0
b
P9.47 A phase diagram for an alloy for which an intermetallic compound, AB, is formed is shown here.
a. Label the different regions of the phase diagram indicating
what components (A and/or B and/or AB where A, B, and
AB denote solids) and phases are present.
b. The vertical dashed line shows a cooling process, ­starting at
point a. Describe which components and the ­relative amounts
of the components are present at the points ­labeled a–d.
c
Temperature
1
xA
d
e
a
xA
1
P9.46 A phase diagram for an alloy, the two components of
which are completely soluble in the liquid phase and partially
soluble in the solid phase, is shown here.
a. Label the different regions of the phase diagram, indicating
what components (A and/or B where A and B denote
components) and phases are present.
b
Temperature
0
c
d
AB
FURTHER READING
Castellan, G. “Phase Equilibria and the Chemical Potential.”
Journal of Chemical Education 32 (1955): 424–427.
Chinard, F. P. “The Definition of Osmotic Pressure.” Journal
of Chemical Education 31 (1954): 66–69.
McInnes, D. M., and Kampbell, D. “Bubble Stripping to
Determine Hydrogen Concentrations in Ground Water:
A Practical Application of Henry’s Law.” Journal of
Chemical Education 80 (2003): 516–519.
0
xa
xA
1
Nikitas, P. “Applications of the Gibbs–Duhem Equation.”
Journal of Chemical Education, 78 (2001): 1070–1075.
Rosenberg, R. M., and Peticolas, W. L. “Henry’s Law: A
Retrospective.” Journal of Chemical Education 81 (2004):
1647–1652.
C H A P T E R
10
Electrolyte Solutions
10.1 Enthalpy, Entropy, and Gibbs
Energy of Ion Formation
in Solutions
10.2 Understanding the
Thermodynamics of Ion
Formation and Solvation
WHY is this material important?
Electrolyte solutions occur all around us, with seawater being the most obvious example.
These solutions are quite different from the ideal and real solutions of neutral solutes
discussed in Chapter 9. The fundamental reason for this difference is that solutes in
electrolyte solutions exist as solvated positive and negative ions.
10.3 Activities and Activity
Coefficients for Electrolyte
Solutions
10.4 Calculating g { Using the
Debye–Hückel Theory
10.5 Chemical Equilibrium
in Electrolyte Solutions
WHAT are the most important concepts and results?
Electrolyte and nonelectrolyte solutions are so different because the Coulomb interactions between ions in an electrolyte solution are of much longer range than the interactions between neutral solutes. For this reason, electrolyte solutions deviate from
ideal behavior at much lower concentrations than do solutions of neutral solutes. The
Debye–Hückel limiting law provides a useful way to calculate activity coefficients for
dilute electrolyte solutions.
WHAT would be helpful for you to review for this chapter?
It would be helpful to review the material on solutions in Chapter 9.
10.1 ENTHALPY, ENTROPY, AND GIBBS ENERGY
OF ION FORMATION IN SOLUTIONS
This chapter discusses substances called electrolytes, which dissociate into positively
and negatively charged mobile solvated ions when dissolved in an appropriate solvent.
Consider the following overall reaction in water:
1 > 2 H21g2 + 1 > 2 Cl21g2 S H +1aq2 + Cl-1aq2
(10.1)
∆ r H ~ = ∆ f H ~ 1H +, aq2 + ∆ f H ~ 1Cl-, aq2
(10.2)
in which H +1aq2 and Cl-1aq2 represent mobile solvated ions. Although similar in structure, Equation (10.1) represents a reaction that is quite different from the gas-phase
dissociation of an HCl molecule to give H +1g2 and Cl-1g2. For the reaction in solution,
∆ r H ~ is -167.2 kJ mol-1. The shorthand notation H +1aq2 and Cl-1aq2 refers to positive and negative ions together with their associated solvation shell. A solvation shell
around an ion consists of those solvent molecules that are bound to the ion by electrostatic forces and do not diffuse freely in the solution. The solvation shell is essential
in lowering the energy of the ions, thereby making the previous reaction spontaneous.
Although energy flow into the system is required to dissociate and ionize hydrogen and
chlorine, even more energy is gained in the reorientation of the dipolar water molecules
around the ions in the solvation shell. Therefore, the reaction is exothermic.
The standard state enthalpy for this reaction can be written in terms of formation
enthalpies:
289
290
CHAPTER 10 Electrolyte Solutions
Concept
The reference zero of energy for
electrolyte solutions is set by the
equation ∆ f G ~ 1H +, aq2 = 0 for all T.
There is no contribution of H21g2 and Cl21g2 to ∆ r H ~ in Equation (10.2) because
∆ f H ~ for a pure element in its standard state is zero.
Unfortunately, no direct calorimetric experiment can measure only the heat of formation of the solvated anion or cation. This is the case because the solution remains
electrically neutral; any dissociation reaction of a neutral solute must produce both
anions and cations. As we have seen in Chapter 4, tabulated values of formation enthalpies, entropies, and Gibbs energies for various chemical species are very useful. How
can this information be obtained for individual solvated cations and anions?
The discussion in the rest of this chapter is restricted to aqueous solutions, for
which water is the solvent but can be generalized to other solvents. As was discussed
in Chapters 2 and 4, there is no natural choice for the zero of energy. Because only differences of energy can be measured, we can choose the zero of energy arbitrarily, as long
as we use the same reference point for all substances. Therefore, numerical values of
thermodynamic functions for anions and cations in aqueous solutions can be obtained by
making an appropriate choice for the zero of ∆ f H ~ , ∆ f G ~ , and Sm~ . By convention, the
formation Gibbs energy for H +1aq2 at unit activity is set equal to zero at all temperatures:
With this choice,
∆ f G ~ 1H +, aq2 = 0 for all T
Sm~ 1H +, aq2 = - a
0∆ f G ~ 1H +, aq2
0T
b
P
(10.3)
= 0 and
∆ f H ~ 1H +, aq2 = ∆ f G ~ 1H +, aq2 + TSm~ 1H +, aq2 = 0
(10.4)
Using the convention of Equation (10.3), which has the consequences shown in Equation (10.4), the values of ∆ f H ~ , ∆ f G ~ , and Sm~ for an individual ion can be assigned
numerical values, as shown next.
As discussed earlier, ∆ r H ~ for the reaction 1>2 H21g2 + 1>2 Cl21g2 S H +1aq2 +
Cl 1aq2 can be directly measured. The value of ∆ r G ~ can be determined from
∆ rG ~ = -RT ln K by measuring the degree of dissociation in the reaction, using the
solution conductivity, and ∆ r S ~ can be determined from the relation
∆r S ~ =
∆ r H ~ - ∆ rG ~
T
Using the conventions stated in Equations (10.3) and (10.4), together with the conventions regarding ∆ f H ~ and ∆ f G ~ for pure elements, ∆ r H ~ = ∆ f H ~ 1Cl-, aq2,
∆ rG ~ = ∆ f G ~ 1Cl-, aq2, and
∆ r S ~ = Sm~ 1Cl-, aq2 -
1 ~
1
Sm1H2, g2 - Sm~ 1Cl2, g2
2
2
for the reaction of Equation (10.1). In this way, the numerical values ∆ f H ~ 1Cl-, aq2 =
-167.2 kJ mol-1, Sm~ 1Cl-, aq2 = 56.5 J K-1 mol-1, and ∆ f G ~ 1Cl-, aq2 = -131.2 kJ
mol-1 can be obtained.
These values can be used to determine the formation functions of other ions. To
illustrate how this is done, consider the following reaction:
NaCl1s2 S Na+1aq2 + Cl-1aq2
(10.5)
-1
for which ∆ r H is found to be +3.09 kJ mol . For this reaction,
~
∆ r H ~ = ∆ f H ~ 1Cl-, aq2 + ∆ f H ~ 1Na+, aq2 - ∆ f H ~ 1NaCl, s2
-1
(10.6)
We use the tabulated value of ∆ f H 1NaCl, s2 = -411.2 kJ mol and the value for
∆ f H ~ 1Cl-, aq2 just calculated to determine that ∆ f H ~ 1Na+, aq2 = -240.1 kJ mol-1.
Proceeding to other reactions that involve either Na+1aq2 or Cl-1aq2, we can determine
the enthalpies of formation of the counter ions. This procedure can be extended to include other ions. Values for ∆ f G ~ and Sm~ can be determined in a similar fashion. Values
for ∆ f H ~ , ∆ f G ~ , and Sm~ for aqueous ionic species are tabulated in Table 10.1. These
thermodynamic quantities are called conventional formation enthalpies, conventional Gibbs energies of formation, and conventional formation entropies because
of the convention described earlier.
~
10.2 Understanding the Thermodynamics of Ion Formation and Solvation
291
TABLE 10.1 Conventional Formation Enthalpies, Gibbs Energies,
and Entropies of Selected Aqueous Anions and Cations
Ion
Ag +1aq2
∆ f H ~ 1kJ mol-12
105.6
∆ f G ~ 1kJ mol-12
77.1
Sm~ 1J K-1 mol-12
72.7
Br -1aq2
-121.6
-104.0
82.4
2+
Ca 1aq2
-542.8
-553.6
-53.1
-
Cl 1aq2
-167.2
-131.2
56.5
+
Cs 1aq2
-258.3
-292.0
133.1
+
Cu 1aq2
71.7
50.0
40.6
64.8
65.5
-99.6
-332.6
-278.8
-13.8
Cu2 + 1aq2
F -1aq2
+
H 1aq2
-
I 1aq2
K+1aq2
+
Li 1aq2
2+
1aq2
0
0
0
-55.2
-51.6
111.3
-252.4
-283.3
102.5
-278.5
-293.3
13.4
-466.9
-454.8
-138.1
NO3-1aq2
+
-207.4
-111.3
146.4
-240.1
-261.9
59.0
OH 1aq2
-230.0
-157.2
-10.9
- 1277.4
-1018.7
-220.5
-909.3
-744.5
20.1
-153.9
-147.1
-112.1
Mg
Na 1aq2
-
PO34 1aq2
SO24 1aq2
2+
Zn 1aq2
Source: Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002.
Note that ∆ f H ~ , ∆ f G ~ , and Sm~ for ions are defined relative to H +1aq2. Negative
values for ∆ f H ~ indicate that the formation of the solvated ion is more exothermic than
the formation of H +1aq2. A similar statement can be made for ∆ f G ~ . Generally speaking, ∆ f H ~ for multiply charged ions is more negative than that of singly charged ions,
and ∆ f H ~ for a given charge is more negative for smaller ions because of the stronger
electrostatic attraction between the multiply charged or smaller ion and the water in the
solvation shell.
Recall from Section 5.8 that the entropy of an atom or a molecule was shown to
be always positive. This is not the case for solvated ions because the entropy is measured relative to H +1aq2. The entropy of solvated ions decreases with formation of the
hydration shell because liquid water molecules are converted to relatively immobile
molecules. Ions with a negative value for the conventional standard entropy such as
+
Mg 2+1aq2, Zn2+1aq2, and PO34 1aq2 have a larger charge-to-size ratio than H 1aq2.
For this reason, the solvation shell is more tightly bound. Conversely, ions with a positive value for the standard entropy such as Na+1aq2, Cs+1aq2, and NO3-1aq2 have a
smaller charge-to-size ratio than H +1aq2 and a less tightly bound solvation shell.
10.2 UNDERSTANDING THE THERMODYNAMICS
OF ION FORMATION AND SOLVATION
As discussed in the preceding section, ∆ f H ~ , ∆ f G ~ , and Sm~ can be determined for a
formula unit but not for an individual ion in a calorimetric experiment. However, as
we will next see, values for thermodynamic functions associated with individual ions
can be calculated with a reasonable level of confidence using a thermodynamic model.
This result allows the conventional values of ∆ f H ~ , ∆ f G ~ , and Sm~ to be converted to
absolute values for individual ions. In the following discussion, the focus is on∆ f G ~ .
Concept
In electrolyte solutions, ∆ f H ~ , ∆ f G ~ ,
~
and S m
for ions are defined relative
+
to H 1aq2.
292
CHAPTER 10 Electrolyte Solutions
Figure 10.1
Infinite separation
1
2 H2(g) 1
1
2 Cl2(g)
2131.2
Cl2(aq) 1 H1(aq)
DsolvG (Cl2,aq)
[
Gibbs energy of formation of aqueous
hydrogen and chloride ions. ∆ r G ~ is
shown pictorially for two different paths
starting with 1>2 H21g2 and 1>2 Cl21g2
and ending with H +1aq2 + Cl-1aq2.
The units for the numbers are kJ mol-1.
Because ∆G is the same for both paths,
∆ solvG ~ 1H +, aq2 can be expressed in
terms of gas-phase dissociation and
­ionization energies.
105.7
203.3
[
DsolvG (H1,aq)
Cl(g)
2349
2(g)
Cl
1312
H(g)
H1(g)
We first discuss the individual contributions to ∆ f G ~ , and we will do so by
analyzing the following sequence of steps that describe the formation of H +1aq2 and
Cl-1aq2:
1>2 H21g2 S H1g2
∆ rG ~ = 203.3 kJ mol-1
∆ rG ~ = 105.7 kJ mol-1
1>2 Cl21g2 S Cl1g2
+
H1g2 S H 1g2 + e
-
∆ rG ~ = 1312 kJ mol-1
Cl1g2 + e- S Cl-1g2
Cl-1g2 S Cl-1aq2
∆ rG ~ = - 349 kJ mol-1
1>2 H21g2 + 1>2 Cl21g2 S H +1aq2 + Cl-1aq2
∆ rG ~ = - 131.2 kJ mol-1
+
∆ rG ~ = ∆ solvG ~ 1Cl-, aq2
+
∆ rG ~ = ∆ solvG ~ 1H +, aq2
H 1g2 S H 1aq2
This pathway is shown pictorially in Figure 10.1. Because G is a state function, both
the black and red paths must have the same ∆G value. The first two reactions in this
sequence are the dissociation of the molecules in the gas phase, and the second two
reactions are the formation of gas-phase ions from the neutral gas-phase atoms. ∆ rG ~
can be determined experimentally for these four reactions. Substituting the known values for ∆ rG ~ for these four reactions in ∆ rG ~ for the overall process,
∆ rG ~ = ∆ solvG ~ 1Cl-, aq2 + ∆ solvG ~ 1H +, aq2 + 1272 kJ mol-1
(10.7)
Equation 10.7 allows us to relate the ∆ rG ~ ∆ solvG ~ of the H + and Cl- ions with ∆ rG ~
for the overall reaction.
As Equation (10.7) shows, ∆ solvG ~ plays a critical role in determining the Gibbs
energies of ion formation. Although ∆ solvG ~ of an individual cation or anion cannot be
determined experimentally, it can be estimated using a simplified model of the solution
developed by Max Born in which the solvent is treated as a uniform fluid with the
appropriate dielectric constant, and the ion is treated as a charged sphere. How can
∆ solvG ~ be calculated with these assumptions? At constant T and P, the nonexpansion
work for a reversible process equals ∆G for the process. Therefore, if the reversible
work associated with solvation can be calculated, ∆ solvG ~ for the process is known.
Imagine a process in which a neutral atom A gains the charge Q, first in a vacuum and
second in a uniform dielectric medium. The value of ∆ solvG ~ of an ion with a charge
q is the reversible work for the process 3 A1g2 S AQ1aq24 solvent minus that for the
reversible process 3 A1g2 S AQ1aq24 vacuum.
The electrical potential around a sphere of radius r with the charge Q′ is given by
f = Q′>4per. From electrostatics, we know that the work in charging the sphere by
the additional amount dQ is f dQ. Therefore, the work in charging a neutral sphere in
vacuum to the charge Q is
Q
Q
Q′dQ′
Q2
1
w =
=
Q′dQ′ =
4pe0r L
8pe0r
L 4pe0r
0
0
(10.8)
10.2 Understanding the Thermodynamics of Ion Formation and Solvation
293
where e0 is the permittivity of free space. The work of the same process in a solvent is
Q2 >8pe0 er r, where er is the relative permittivity (dielectric constant) of the solvent.
Consequently, ∆ solvG ~ for an ion of charge Q = ze is given by
∆ solvG ~ =
z 2e2NA 1
a - 1b
8pe0 r er
(10.9)
Because er 7 1, ∆ solvG ~ 6 0, showing that solvation is a spontaneous process. Values
for er for a number of solvents are listed in Table 10.2 (see Appendix A, Data Tables).
To test the Born model, we need to compare absolute values of ∆ solvG ~ for ions of
different radii with the functional form proposed in Equation (10.9). However, experimentally determined values of ∆ solvG ~ are referenced to H +1aq2. Therefore, we need to
know ∆ solvG ~ 1H +, aq2 to convert experimental values referenced to H +1aq2 to absolute
values. It turns out that ∆ solvG ~ 1H +, aq2, ∆ solvH ~ 1H +, aq2, and ∆ solvS ~ 1H +, aq2 can
be calculated. Because the calculation is involved, the results are simply stated here.
Concept
∆ solvG ~ for an ion can be calculated
using the Born model.
∆ solvG ~ 1H +, aq2 ≈ -1090 kJ mol-1
∆ solvG ~ 1H +, aq2 ≈ -1050 kJ mol-1
∆ solvS ~ 1H +, aq2 ≈ -130 J mol-1 K-1
(10.10)
The values listed in Equation (10.10) can be used to calculate absolute values of
∆ solvH ~ , ∆ solvG ~ , and ∆ solvS ~ for other ions from the conventional values referenced
to H +1aq2. These calculated absolute values can be used to test the validity of the Born
model. If the model is valid, a plot of ∆ solvG ~ versus z 2 >r will give a straight line as
shown by Equation (10.9), and the data points for individual ions should lie on the line.
The results are shown in Figure 10.2, where r is the ionic radius obtained from crystal
structure determinations.
The first and second clusters of data points in Figure 10.2 are for singly and doubly charged ions, respectively. The data are compared with the result predicted by
Equation (10.9) in Figure 10.2a. As can be seen from the figure, the trends are reproduced, but there is no quantitative agreement. The agreement can be considerably
improved by using an effective radius for the solvated ion rather than the ionic radius
from crystal structure determinations. The effective radius is defined as the distance
from the center of the ion to the center of charge in the dipolar water molecule in
the solvation shell. Latimer, Pitzer, and Slansky [J. Chemical Physics, 7 (1939): 109]
found the best agreement with the Born equation by adding 0.085 nm to the crystal
radius of positive ions and 0.100 nm to the crystal radius for negative ions to account
for the fact that the H2O molecule is not a point dipole. This difference is explained by
the fact that the center of charge in the water molecule is closer to positive ions than
to negative ions. Figure 10.2b shows that the agreement obtained between the predictions of Equation (10.9) and experimental values is very good if this correction to the
ionic radii is made.
9 z2/r
(m21)
100
150
10
\
DsolvG /(kJ mol21)
22000
24000
26000
Li1
Ba21
0
0
200
La31 Cr31
Fe31 31
Al
28000
210000
21000
Ba21
Mg21
22000
23000
La31
24000
212000
214000
(a) Born model without solvation shell
Li1
Figure 10.2
Mg21
\
0
50
DsolvG /(kJ mol21)
0
109 z2/reff (m21)
10 20 30 40 50 60 70
Fe31
31
Cr31 Al
25000
(b) Born model with solvation shell
Evaluation of the Born model of
­solvation. (a) The solvation energy
calculated using the Born model is
shown as a function of z 2 >r. (b) The
same results are shown as a function
of z 2 >reff , where estimated values of
the solvation shell have been added
to account for the fact that the H2O
molecule is not a point dipole, as
discussed in the text. The dashed line
shows the behavior predicted by
Equation (10.9).
294
CHAPTER 10 Electrolyte Solutions
Figure 10.2b shows good agreement between the predictions of the Born model
and calculated values for ∆ solvG ~ , and justifies the approach used to calculate absolute
enthalpies, Gibbs energies, and entropies for solvated ions. However, because of uncertainties about the numerical values of the ionic radii and the dielectric constant of the
immobilized water in the solvation shell of an ion, the uncertainty is {50 kJ mol-1 for
the absolute solvation enthalpy and Gibbs energy and {10 JK1 mol-1 for the absolute
solvation entropy. Because these uncertainties are large compared to the uncertainty of
the thermodynamic functions using the convention described in Equations (10.3) and
(10.4), conventional rather than absolute values of ∆ f H ~ , ∆ f G ~ , and Sm~ for solvated
ions in aqueous solutions are generally used by chemists.
10.3 ACTIVITIES AND ACTIVITY COEFFICIENTS
FOR ELECTROLYTE SOLUTIONS
The ideal dilute solution model presented for the activity and activity coefficient of
components of real solutions in Chapter 9 is not valid for electrolyte solutions. This is
the case because solute–solute interactions are dominated by the long-range electrostatic forces present between ions in electrolyte solutions. For example, the reaction
that occurs when NaCl is dissolved in water is
NaCl1s2 + H2O1l2 S Na+1aq2 + Cl-1aq2
(10.11)
G = nsolvent msolvent + nsolute msolute
(10.12)
In dilute solutions, NaCl is completely dissociated. Therefore, all solute–solute interactions are electrostatic.
Although the concepts of activity and activity coefficients introduced in Chapter 9
are applicable to electrolytes, these concepts must be formulated differently for electrolytes to include the Coulombic interactions among ions. We will next discuss a model
for electrolyte solutions stated in terms of the chemical potentials, activities, and activity
coefficients of the individual ionic species. However, only mean activities and activity
coefficients (to be defined later) can be determined through experiments.
Consider the Gibbs energy of the solution, which can be written as
For the general electrolyte Av+ Bv- that dissociates completely, one can also write an
equivalent expression for G:
Concept
Because it is not possible to make
a solution that contains only anions
or cations, a mean ionic chemical
potential, m{ , is defined for an
electrolyte solution.
G = nsolvent msolvent + n + m + + n - m - = nsolvent msolvent + nsolute1v + m + + v - m - 2
(10.13)
where v + and v - are the stoichiometric coefficients of the cations and anions, respectively, produced upon dissociation of the electrolyte. In shorthand notation, an electrolyte is called a 1–1 electrolyte if v + = 1 and v - = 1. Similarly, for a 2–3 electrolyte,
v + = 2 and v - = 3. Because Equations (10.12) and (10.13) describe the same solution, they are equivalent. Therefore,
msolute = v + m + + v - m -
(10.14)
Although this equation is formally correct for a strong electrolyte, it is not possible
to make a solution of either cations or anions alone because all solutions are electrically neutral. Therefore, it is useful to define a mean ionic chemical potential m{
for the solute
msolute
v + m+ + v - mm{ =
=
(10.15)
v
v
where v = v+ + v - . The reason for defining the mean chemical potential is that m{
can be determined experimentally, whereas m + and m - are not accessible through
experiments.
The next task is to relate the chemical potentials of the solute and its individual ions
to the activities of these species. For the individual ions,
m+ = m+~ + RT ln a +
and m- = m-~ + RT ln a -
(10.16)
10.3 Activities and Activity Coefficients for Electrolyte Solutions
295
where the standard chemical potentials of the ions, m+~ and m-~ , are based on a Henry’s
law standard state. Substituting Equation (10.16) in (10.15), an equation for the mean
ionic chemical potential is obtained that is similar in structure to the expressions that
we derived for the ideal dilute solution:
~
m{ = m{
+ RT ln a {
(10.17)
The mean ionic activity a { is related to the individual ion activities by
av{ = av++ av--
or a { = 1av++ av-- 21>v
(10.18)
EXAMPLE PROBLEM 10.1
Write the mean ionic activities of NaCl, K2SO4, and H3PO4 in terms of the ionic
activities of the individual anions and cations. Assume complete dissociation.
Solution
a2NaCl = aNa+ aCl-
aNaCl = 2aNa+ aCl+
or
a3K2SO4 = a2K+ aSO2or aK2SO4 = 1a2K+ aSO221>3
4
4
a4H3PO4 = a3H+ aPO3or aH3PO4 = 1a3H+ aPO321>4
4
4
If the ionic activities are referenced to the concentration units of molality, then
a+ =
m+
g
m~ +
and a - =
mg
m~ -
(10.19)
where m + = v + m and m - = v - m. Because the activity is unitless, the molality must
be referenced to a standard state concentration chosen to be m ~ = 1 mol kg -1. As in
Chapter 9, a hypothetical standard state based on molality is defined. In this standard
state, Henry’s law, which is valid in the limit m S 0, is obeyed up to a concentration of
m = 1 molal. Substitution of Equation (10.19) in Equation (10.18) shows that
av{ = a
m+ v+ m- v- v+ vb a ~ b g+ gm~
m
(10.20)
To simplify this notation, we define the mean ionic molality m { and mean ionic
activity coefficient g{ by
mv{ = mv++ mv-m { = 1v v++ vv-- 21>v m and
g v{ = g v++ g v--
g{ = 1g v++ g v-- 21>v
(10.21)
With these definitions, the mean ionic activity is related to the mean ionic activity
coefficient and mean ionic molality as follows:
av{ = a
m{ v v
b g{
m~
or a { = a
m{
bg
m~ {
(10.22)
Equations (10.19) through (10.22) relate the activities, activity coefficients, and
molalities of the individual ionic species to mean ionic quantities and measurable properties of the system such as the molality and activity of the solute. With these definitions, Equation (10.23) defines the chemical potential of the electrolyte solute in terms
of its mean ionic activity:
~
msolute = msolute
+ RT ln av{
(10.23)
Concept
The mean ionic molality and mean
ionic activity coefficient are defined
by Equation (10.21).
296
CHAPTER 10 Electrolyte Solutions
Equations (10.15), (10.21), (10.22), and (10.23) can be used to express the chemical potential of the solute in terms of measurable or easily accessible quantities:
~
msolute = vm{ = 3 vm{
+ RT ln1v v++ v v-- 24 + vRT ln a
m
b + vRT ln g{
m~
(10.24)
The first term in the square bracket is defined by the “normal” standard state, which is
usually taken to be a Henry’s law standard state. The second term is obtained from the
chemical formula for the solute. These two terms can be combined to create a new stan~ ~
dard state m{
defined by the terms in the square brackets in Equation (10.24):
~ ~
msolute = vm{ = m{
+ vRT ln a
m
b + vRT ln g{
m~
(10.25)
The first two terms in Equation (10.25) correspond to the “ideal” ionic solution, which
is associated with g{ = 1.
The last term in Equation (10.25), which is the most important term in this discussion, contains the deviations from ideal behavior. The mean activity coefficient
g{ can be obtained through experiment. For example, the activity coefficient of the
solvent can be determined by measuring the boiling point elevation, the freezing
point depression, or the lowering of the vapor pressure above the solution upon solution formation. The activity of the solute is obtained from that of the solvent using
the Gibbs–Duhem equation. As shown in Section 11.8, g{ can also be determined
through measurements on electrochemical cells. In addition, a very useful theoretical model allows g{ to be calculated for dilute electrolytic solutions. This model is
discussed in the next section.
10.4 CALCULATING Gt USING THE
DEBYE–HÜCKEL THEORY
Concept
The Debye–Hückel limiting law allows
the mean ionic activity coefficient
to be calculated for an electrolyte
solution.
No model adequately explains the deviations from ideality for the solutions of neutral
solutes discussed in Sections 9.1 through 9.4. The lack of a good model reflects the fact
that the deviations arise through A–A, B–B, and A–B interactions that are specific to
components A and B. This precludes a general model that holds for arbitrary components A and B. However, the situation for solutions of electrolytes is different.
Deviations from ideal solution behavior occur at a much lower concentration for
electrolytes than for nonelectrolytes because the dominant interaction between the ions
in an electrolyte is a long-range electrostatic Coulomb interaction rather than a shortrange van der Waals or chemical interaction. Because of its long range, the Coulomb
interaction among the ions cannot be neglected even for very dilute solutions of electrolytes. The Coulomb interaction allows a model of electrolyte solutions to be formulated independent of the identity of the solute for the following reason. The attractive or
repulsive interaction of two ions depends only on their charge and their separation, and
not on their chemical identity. Therefore, the solute–solute interactions can be modeled
knowing only the charge on the ions, and the model becomes independent of the identity of the solute species.
Measurements of activity coefficients in electrolyte solutions show that g{ S 1
for dilute solutions in the limit m S 0. Because g{ 6 1, the chemical potential of the
solute in a dilute solution is lower than that for a solution of uncharged solute species.
Why is this the case? The lowering of msolute arises because the net electrostatic interaction among the ions surrounding an arbitrarily chosen central ion is attractive rather
than repulsive. The model that describes the lowering of the energy of electrolytic solutions is due to Peter Debye (1884–1966) and Erich Hückel (1896–1980). Rather than
derive their results, we describe the essential features of their model next.
The solute ions in the solvent give rise to a spatially dependent electrostatic potential, f, which can be calculated if the spatial distribution of ions is known. In dilute
electrolyte solutions, the energy increase or decrease experienced by an ion of charge
297
10.4 CALCULATING g{ USING THE DEBYE–HÜCKEL THEORY
1
Figure 10.3
c50
Ratio of the electrostatic potential in
a dilute electrolyte solution to that for
an isolated ion as a function of radial
distance. Computed values are shown
for three different molarities of a 1–1
electrolyte such as NaCl. Note that the
potential falls off much more rapidly in
the electrolyte solution than in the uniform
dielectric medium corresponding to the
c = 0 curve.
Fsolution(r)
Fisolated ion(r)
0.8
c 5 0.001
0.6
0.4
c 5 0.01
0.2
c 5 0.1
2 3 1029
4 3 1029
6 3 1029
r(m)
8 3 1029
1 3 1028
{ze that arises from the potential f is small compared to the thermal energy. This condition can be expressed in the form
0 {zef 0 V kBT
(10.26)
In Equation (10.26), e is the charge on a proton, and k is the Boltzmann constant. In
this limit, the dependence of f on the spatial coordinates and the spatial distribution of
the ions around an arbitrary central ion can be calculated. In contrast to the potential
around an isolated ion in a dielectric medium, which is described by
f isolated ion1r2 =
{ze
4per e0r
(10.27)
the potential in the dilute electrolyte solution has the form
f solution1r2 =
{ze
exp1-kr2
4per e0r
Concept
An individual ion in an electrolyte
solution experiences a screened
potential from other ions.
(10.28)
In Equations (10.27) and (10.28), e0 and er are the permittivity of free space and the
relative permittivity (dielectric constant) of the dielectric medium or solvent, respectively. Because of the exponential term, f solution1r2 falls off much more rapidly with
r than f isolated ion1r2. We say that an individual ion experiences a screened potential
from the other ions.
The Debye–Hückel theory shows that k is related to the individual charges on the
ions and to the solute molality m by
v+z 2+ + v- z 2b rsolvent
k = e NA11000 L m 2 m a
e 0e r k BT
2
2
-3
2
2
1
1
2
(10.29)
From this formula, we can see that screening becomes more effective as the concentration of the ionic species increases. Screening is also more effective for multiply
charged ions and for larger values of v+ and v- .
The ratio
f solution1r2
= e-kr
f isolated ion1r2
1
is plotted in Figure 10.3 for different values of m for an aqueous solution of a 1–1 electrolyte. Note that the potential falls off much more rapidly with the radial distance r in
the electrolyte solution than in the uniform dielectric medium. Note also that the potential falls off more rapidly with increasing concentration of the electrolyte. The origin
of this effect is that ions of sign opposite to the central ion are more likely to be found
close to the central ion. These surrounding ions form a diffuse ion cloud around the
central ion, as shown pictorially in Figure 10.4. If a spherical surface is drawn centered
1
2
2
1
1
2
Figure 10.4
Pictorial rendering of the arrangement
of ions about an arbitrary ion in an
electrolyte solution. The central ion is
more likely to have oppositely charged
ions as neighbors than other ions. The
large circle represents a sphere of radius
r ∼ 8>k. From a point outside of this
sphere, the charge on the central ion is
essentially totally screened.
298
CHAPTER 10 Electrolyte Solutions
at the central ion, the net charge within the surface can be calculated. The results show
that the net charge has the same sign as the central charge, falls off rapidly with distance, and is close to zero for kr ∼ 8. For larger values of kr, the central ion is completely screened by the diffuse ion cloud, meaning that the net charge in the sphere
around the central ion is zero. The net effect of the diffuse ion cloud is to screen the
central ion from the rest of the solution, and the quantity 1>k is known as the Debye–
Hückel screening length. Larger values of k correspond to a smaller diffuse cloud and
a more effective screening.
It is convenient to combine the concentration-dependent terms that contribute to k
in the ionic strength I, which is defined by
I =
m
1
1vi + z 2i + + vi - z 2i - 2 = a 1mi + z 2i + + mi - z 2i - 2
a
2 i
2 i
(10.30)
EXAMPLE PROBLEM 10.2
Calculate I for (a) a 0.050 molal solution of NaCl and for (b) a Na 2SO4 solution of the
same molality.
Solution
a. INaCl =
0.050 mol kg -1
m
1v + z 2+ + v - z 2- 2 =
* 11 + 12 = 0.050 mol kg -1
2
2
b. INa2SO4 =
0.050 mol kg -1
m
1v + z 2+ + v - z 2- 2 =
* 12 + 42 = 0.15 mol kg -1
2
2
Using the definition of the ionic strength, we can write Equation (10.29) in the form
k =
= 2.91 * 1010
I
0.1
0.2
0.3
0.4
0.5
0
20.25
NaCl
20.50
lng6
20.75
CaCl2
21.00
21.25
2e2NA
I
b 11000L m-32 a b rsolvent
e
B e 0k BT
B r
a
LaCl3
21.50
21.75
Figure 10.5
Decrease in the Debye–Hückel mean
activity coefficient with the square root
of ionic strength. Plots for a 1–1, a 1–2,
and a 1–3 electrolyte solution are shown.
All curves correspond the same molality
in the solute.
I>mol kg -1 rsolvent
B er
kg L-1
m-1 at 298 K
(10.31)
The first factor in this equation contains temperature and fundamental constants that
are independent of the solvent and solute. The second factor contains the ionic strength
of the solution and the unitless relative permittivity of the solvent. For the more conventional units of mol L-1, and for water, for which er = 78.5, k = 3.29 * 109 2I m-1,
at 298 K.
By calculating the charge distribution of the ions around the central ion and the
work needed to charge these ions up to their charges z+ and z- from an initially neutral
state, Debye and Hückel were able to obtain an expression for the mean ionic activity
coefficient. It is given by
ln g{ = - 0 z + z - 0
e2k 2I
8pe0er kBT
(10.32)
This equation is known as the Debye–Hückel limiting law. It is called a limiting law
because Equation (10.32) is only obeyed for small values of the ionic strength. Note
that because of the negative sign in Equation (10.32), g{ 6 1. From the concentration
dependence of k shown in Equation (10.31), the model predicts that ln g{ decreases
with the ionic strength as 1I. This dependence is shown in Figure 10.5. Although all
three solutions have the same solute concentration, they have different values for z +
and z -. For this reason, the three lines have a different slope.
Equation (10.32) can be simplified for a particular choice of solvent and temperature. For aqueous solutions at 298.15 K, the result is
log g{ = -0.5092 0 z + z - 0 2I or ln g{ = -1.173 0 z + z - 0 2I
(10.33)
10.4 CALCULATING g{ USING THE DEBYE–HÜCKEL THEORY
I
0.2
0.4
Figure 10.6
I
0.6
0.8
1
0.2
0
0.4
0.6
0.8
1
0
20.2
20.5
20.4
Experimentally determined activity
coefficients as a function of the square
root of the ionic strength. (a) AgNO3
and (b) CaCl2. Experimentally measured
values shown as solid dots. The solid lines
were generated according to predictions
of the Debye–Hückel theory.
21.0
lng6
lng6
299
20.6
AgNO3
21.5
20.8
21.0
CaCl2
22.0
21.2
(a)
(b)
How well does the Debye–Hückel limiting law agree with experimental data? Figure 10.6
shows a comparison of the model with data for aqueous solutions of AgNO3 and CaCl2.
In each case, ln g{ is plotted versus 1I. The Debye–Hückel limiting law predicts that
the data will fall on the line indicated in each figure. The data points deviate from the
predicted behavior above 1I = 0.1 for AgNO31m = 0.012, and above 1I = 0.006
for CaCl21m = 0.0042. In the limit that I S 0, the limiting law is obeyed. However,
the deviations are significant at a concentration for which a neutral solute would exhibit
ideal behavior.
The deviations continue to increase with increasing ionic strength. Figure 10.7
shows experimental data for ZnBr2 out to 1I = 5.5, corresponding to m = 10. Note
that, although the Debye–Hückel limiting law is obeyed as I S 0, ln g{ goes through
a minimum and begins to increase with increasing ionic strength. At the highest value
of the ionic strength, g{ = 2.32, which is significantly greater than one. Although the
deviations from ideal behavior are less pronounced in Figure 10.6, the trend is the
same for all the solutes. The mean ionic activity coefficient g{ falls off more slowly
with the ionic strength than predicted by the Debye–Hückel limiting law. The behavior
shown in Figure 10.7 is typical for most electrolytes; after passing through a minimum,
g{ rises with increasing ionic strength, and for high values of I, g{ 7 1. Experimental
values for g{ for a number of solutes at different concentrations in aqueous solution
are listed in Table 10.3 (see Appendix A, Data Tables).
The experimental values of g{ differ at high ionic strength from those calculated
from the Debye–Hückel limiting law for a number of reasons. They mainly involve
1.0
0.75
0.50
Figure 10.7
lng6
0.25
0
20.25
20.50
20.75
21.0
1
2
3
I
4
5
Experimentally determined values for
the mean activity coefficient for ZnBr2
as a function of the square root of the
ionic strength. The solid line is calculated
from the predictions of the Debye–Hückel
theory. Note that after passing through a
minimum, g{ increases with increasing
ionic strength. This behavior is typical for
electrolytes.
300
CHAPTER 10 Electrolyte Solutions
Concept
The Davies equation can be used
to estimate the mean ionic activity
coefficient at high ionic strengths.
0.2
I
0.4
0.6
0.8
0
20.1
1–1 electrolyte
lng6
20.2
20.3
20.4
1–2 electrolyte
the assumptions made in the model. It has been assumed that the ions can be treated
as point charges with zero volume, whereas ions and their associated solvation shells
occupy a finite volume. As a result, there is an increase in the repulsive interaction
among ions in an electrolyte over that predicted for point charges, which becomes more
important as the concentration increases. Repulsive interactions raise the energy of the
solution and, therefore, increase g{. The Debye–Hückel model also assumes that the
solvent can be treated as a structureless dielectric medium. However, the ion is surrounded by a relatively ordered primary solvation shell, as well as by more loosely
bound water molecules. The atomic-level structure of the solvation shell is not adequately represented by using the dielectric strength of bulk solvent. Another factor that
has not been taken into account is that, as the concentration increases, some ion pairing
occurs such that the concentration of ionic species is less than would be calculated
assuming complete dissociation.
Additionally, consider the fact that the water molecules in the solvation shell have
effectively been removed from the solvent. For example, in an aqueous solution of
H2SO4, approximately nine H2O molecules are tightly bound per dissolved H2SO4
formula unit. Therefore, the number of moles of H2O as solvent in 1 L of a one molar
H2SO4 solution is reduced from 55 for pure H2O to 46 in the solution. Consequently,
the actual solute molarity is larger than that calculated by assuming that all the H2O is
in the form of solvent. Because the activity increases linearly with the actual molarity, g{ increases as the solute concentration increases. If there were no change in the
enthalpy of solvation with concentration, all the H2O molecules would be removed
from the solvent at a concentration of six molar H2SO4. Clearly, this assumption is
unreasonable. What actually happens is that solvation becomes energetically less
favorable as the H2SO4 concentration increases. This corresponds to a less negative
value of ln g{ , or, equivalently, to an increase in g{ . Summing up, many factors
explain why the Debye–Hückel limiting law is only valid for small concentrations.
Because of the complexity of these different factors, there is no simple formula based
on theory that can replace the Debye–Hückel limiting law. However, the main trends
exhibited in Figures 10.6 and 10.7 are reproduced in more sophisticated theories of
electrolyte solutions.
Because none of the usual models are valid at high concentrations, empirical models
that “improve” on the Debye–Hückel model by predicting an increase in g{ for high
concentrations are in widespread use. An empirical modification of the Debye–Hückel
limiting law that has the form
20.5
20.6
1–3 electrolyte
Figure 10.8
Comparison between predictions from
the Debye–Hückel limiting law and
the Davies equation. Predictions for the
Debye–Hückel limiting law are shown as
dashed curves and those for the Davies
equation as solid curves. Electrolytes are
as follows: for 1–1 (red), 1–2 (purple),
and 1–3 (blue).
log10 g{ = -0.51 0 z + z - 0 ≥
a
I 1>2
b
m~
I
1 + a ~b
m
1>2
- 0.30 a
I
b¥
m~
(10.34)
is known as the Davies equation. As seen in Figure 10.8, this equation for g{ shows
the correct limiting behavior for low I values, and the trend at higher values of I
is in better agreement with the experimental results shown in Figures 10.6 and 10.7.
However, unlike the Debye–Hückel limiting law, there is no theoretical basis for the
Davies equation.
10.5 CHEMICAL EQUILIBRIUM
IN ELECTROLYTE SOLUTIONS
As discussed in Section 9.13, the equilibrium constant in terms of activities is given by
Equation (9.67):
vj
K = q 1aeq
i 2
i
(10.35)
10.5 Chemical Equilibrium in Electrolyte Solutions
TABLE 10.4 Solubility Product Constants (Molarity Based)
for Selected Salts
Salt
Ksp
-13
Salt
Ksp
CaSO4
4.9 * 10-6
AgBr
4.9 * 10
AgCl
1.8 * 10-10
Mg1OH22
5.6 * 10-11
AgI
8.5 * 10-17
Mn1OH22
1.9 * 10-13
Ba1OH22
5.0 * 10-3
PbCl2
1.6 * 10-5
BaSO4
1.1 * 10-10
Pb SO4
1.8 * 10-8
ZnS
1.6 * 10-23
3.4 * 10
CaCO3
-9
Source: Lide, D. R., ed. Handbook of Chemistry and Physics. 83rd ed. Boca Raton, FL: CRC Press, 2002.
It is convenient to define the activity of a species relative to its molarity. In this case,
a i = gi
ci
c~
(10.36)
where gi is the activity coefficient of species i. We specifically consider chemical equilibrium in electrolyte solutions, illustrating that activities rather than concentrations
must be taken into account to accurately model equilibrium concentrations. We restrict
our considerations to the range of ionic strengths for which the Debye–Hückel limiting law is valid. As an example, we calculate the degree of dissociation of MgF2 in
water. The equilibrium constant in terms of molarities for ionic salts is usually given
the symbol Ksp, where the subscript refers to the solubility product. The equilibrium
constant Ksp is unitless and has the value of 6.4 * 10-9 for the reaction shown in Equation (10.37). Values for Ksp are generally tabulated for reduced concentration units of
molarity 1c>c ~ 2 rather than molality 1m>m ~ 2, and values for selected substances are
listed in Table 10.4. Because the mass of 1 L of water is 0.998 kg, the numerical value
of the concentration is the same on both scales for dilute solutions.
Now consider dissociation of MgF2 in an aqueous solution:
MgF21s2 S Mg 2 + 1aq2 + 2F -1aq2
(10.37)
Because the activity of the pure solid can be set equal to one,
Ksp = aMg2 + a2F - = a
cMg2 +
c~
ba
cF - 2 3
b g { = 6.4 * 10-9
c~
(10.38)
From the stoichiometry of the overall equation, we know that cF - = 2cMg2 + , but
Equation (10.38) still contains two unknowns, g{ and cF - , that we solve for iteratively,
as shown in Example Problem 10.3.
EXAMPLE PROBLEM 10.3
Calculate the solubility of MgF2 in an aqueous solution. Use the Debye–Hückel limiting
law to obtain g{. Ksp = 6.4 * 10-9.
Solution
The equilibrium expression is stated in Equation 10.38.
Ksp = aMg2 + a2F - = a
cMg2 +
c~
ba
cF - 2 3
b g { = 6.4 * 10-9
c~
First, assume that g{ = 1, and solve Equation (10.38) for cMg2 + .
Ksp = aMg2 + a2F - = a
cMg2 +
c
4x 3 = 6.4 * 10-9; x =
~
ba
cMg2 +
c~
cF - 2
b = 6.4 * 10-9
c~
= 1.17 * 10-3 mol L-1 = 1.17 * 10-3 mol kg -1
301
302
CHAPTER 10 Electrolyte Solutions
We next calculate the ionic strength of the solution using the concentrations just
obtained.
IMgF2 =
1.17 * 10-3 mol kg -1
m
1v + z 2+ + v - z 2- 2 =
* 122 + 22
2
2
= 3.51 * 10-3 mol kg -1
We next calculate g{ from the Debye–Hückel limiting law of Equation (10.33).
ln g{ = -1.173 0 z + z - 0 2I
ln g{ = -1.173 * 2 * 23.57 * 10-3 = -0.1390
g{ = 0.870
We use this value of g{ in the equilibrium expression and recalculate the solubility.
Ksp = aMg2 + a2F - = a
4x 3 =
cMg2 +
c
~
ba
cF - 2 3
b g { = 6.4 * 10-9
c~
cMg2 +
6.4 * 10-9
;
x
=
= 1.34 * 10-3 mol L-1 = 1.34 * 10-3 mol kg -1
c~
10.87023
A second iteration gives g{ = 0.862 and cMg2 + = 1.36 * 10-3 mol L-1, and a third
iteration gives g{ = 0.861 and cMg2 + = 1.36 * 10-3 mol L-1, showing that the iteration has converged. These results show that assuming that g{ = 1 leads to an unacceptably large error of 14% in cMg2 + .
Concept
The ionic strength can have an
appreciable effect on solubility.
Another effect of the ionic strength on solubility is described by the terms salting in
and salting out. The behavior shown in Figure 10.7, in which the activity coefficient first
decreases and subsequently increases with concentration, affects the solubility of a salt
in the following way. For a salt such as MgF2, the product 3 Mg 2 + 4 3 F - 4 2 g 3{ = Ksp
is constant as the concentration varies for constant T because the thermodynamic equilibrium constant K depends only on T. Therefore, the concentrations 3 Mg 2 + 4 and
3 F - 4 change in an opposite way to g{. At small values of the ionic strength, g{ 6 1,
and the solubility increases as g{ decreases with concentration until the minimum in
a plot of g{ versus I is reached. This effect is known as salting in. For high values of
the ionic strength, g{ 7 1, and the solubility is less than that at low values of I. This
effect is known as salting out. Salting out can be used to purify proteins by selective
precipitation from solution. For example, the blood-clotting protein fibrinogen is precipitated in a 0.8 molar ammonium sulfate solution, whereas a 2.4 molar concentration
of ammonium sulfate is required to precipitate albumin.
We next consider equilibria in electrolyte solutions at concentrations where the
Debye–Hückel limiting law is no longer valid. In this case, activity coefficients must
be estimated as there is no theoretical model for calculating them. The Davies equation
[Equation (10.34)] can be used in such a case. An example is a buffer solution that consists of a weak acid and its conjugate base.
EXAMPLE PROBLEM 10.4
Calculate the pH of a buffer solution that is 0.100 molar in CH3COOH and 0.100 molar
in CH3COONa. The molality-based equilibrium constant K for the dissociation of
acetic acid is 1.75 * 10-5. Compare your value with what you would have calculated
assuming g{ = 1.
Vocabulary
Solution
We write the equilibrium constant in the form
K =
1.75 * 10-5 =
m1H3O+2 =
g{m1H3O+2g{m1CH3COO-2
m1CH3COOH2
1g{22x * 10.100 + x2
≈ 1g{22 x
10.100 - x2
1.75 * 10-5
1g{22
pH = -log5 m1H3O+26
To calculate g{ from the Davies equation, we must first calculate the ionic strength
of the solution. The small value of K tells us that the degree of ionization of acetic
acid is small, so that the ionic strength can be calculated from the concentration of
CH3COONa alone.
I =
0.100 mol kg -1
m
1v + z 2+ + v - z 2- 2 =
* 11 + 12 = 0.100 mol kg -1
2
2
Using the Davies equation
log10 g{ = -0.51 0 z + z - 0 ≥
= -0.51c
g{ = 0.781
m1H3O+2 =
a
I 1>2
b
m~
1 + a
10.10021>2
I
b
m~
1 + 10.10021>2
1>2
- 0.30 a
I
b¥
m~
- 0.30 * 0.100 d = -0.1072
1.75 * 10-5
= 2.87 * 10-5 mol kg -1
10.78122
pH = -log12.87 * 10-52 = 4.54
If we had assumed g{ = 1
m1H3O+2 =
1.75 * 10-5
= 1.75 * 10-5
1g{22
This value for m1H3O+2 gives a pH value of 4.76, which is significantly different
from the value calculated earlier.
VOCABULARY
conventional formation enthalpies
conventional formation entropies
conventional Gibbs energies of formation
Davies equation
Debye–Hückel limiting law
Debye–Hückel screening length
electrolyte
ionic strength
mean ionic activity
mean ionic activity coefficient
mean ionic chemical potential
mean ionic molality
salting in
salting out
screened potential
solvation shell
303
304
CHAPTER 10 Electrolyte Solutions
KEY EQUATIONS
Equation
Significance of Equation
∆ f G ~ 1H +, aq2 = Sm~ 1H +, aq2 = ∆ f H ~ 1H +, aq2 = 0 for all T
Conventions used in calculating G, H, and S
for electrolyte solutions
∆ solvG ~ =
m{ =
z 2e2NA 1
a - 1b
8pe0r er
msolute
v+m+ + v-m=
v
v
vv+ v- 1>v
av{ = av+
+ a - or a { = 1a + a - 2
mv{ = mv++ mv--
m{ =
g v{ =
g{ =
1v v++v v-- 21>vm
g v++ g v-1g v++ g v-- 21>v
and
~
msolute = msolute
+ RT ln av{
I =
m
1
1vi + z 2i + + vi - z 2i - 2 = a 1mi + z 2i + + mi - z 2i - 2
2a
2 i
i
ln g{ = - 0 z + z- 0
e2k2I
8pe0erkBT
log g{ = - 0.5092 0 z + z- 0 2I or ln g{ = -1.173 0 z + z- 0 2I
log10 g{ = -0.51 0 z + z- 0 ≥
a
I 1>2
b
m~
1 + a
I
b
m~
1>2
- 0.30 a
I
b¥
m~
Equation Number
10.3 and 10.4
Solvation Gibbs energy of ion
10.9
Definition of mean ionic chemical potential
10.15
Definition of mean ionic activity, a {
10.18
Definition of mean ionic molality m { and
mean ionic activity coefficient g{
10.21
Chemical potential of electrolyte solute in
terms of its activity
10.23
Definition of ionic strength
10.30
Debye–Hückel limiting law
10.32
Debye–Hückel limiting law for aqueous
solutions at 298.15 K
10.33
Davies equation gives approximate values for
activity coefficients at high concentrations
10.34
CONCEPTUAL PROBLEMS
Q10.1 Discuss how the Debye–Hückel screening length
changes as the (a) temperature, (b) solvent dielectric constant,
and (c) ionic strength of an electrolyte solution are increased.
Q10.2 Why is it not possible to measure the Gibbs energy of
solvation of Cl - directly?
Q10.3 Why are activity coefficients calculated using the
Debye–Hückel limiting law always less than one?
Q10.4 How is the mean ionic chemical potential of a solute
related to the chemical potentials of the anion and cation produced when the solute is dissolved in water?
Q10.5 How is the chemical potential of a solute related to its
activity?
Q10.6 Tabulated values of standard entropies of some
aqueous ionic species are negative. Why is this statement not
inconsistent with the third law of thermodynamics?
Q10.7 Why is it not possible to measure the activity coefficient of Na + 1aq2?
Q10.8 Why is it possible to formulate a general theory for
the activity coefficient for electrolyte solutions but not for
nonelectrolyte solutions?
Q10.9 Why does an increase in the ionic strength in the range
where the Debye–Hückel law is valid lead to an increase in
the solubility of a weakly soluble salt?
Q10.10 What is the correct order of the following inert electrolytes in their ability to increase the degree of dissociation
of acetic acid?
a. 0.001 m NaCl
b. 0.001 m KBr
c. 0.10 m CuCl2
Q10.11 How does salting in affect solubility?
Q10.12 Why is it not appropriate to use ionic radii from
crystal structures to calculate ∆ solvG ~ of ions using the
Born model?
Numerical Problems
Q10.13 Why do deviations from ideal behavior occur at lower
concentrations for electrolyte solutions than for solutions in
which the solute species are uncharged?
Q10.14 Why is the value for the dielectric constant for water
in the solvation shell around ions less than that for bulk water?
Q10.15 What can you conclude about the interaction between
ions in an electrolyte solution if the mean ionic activity coefficient is greater than one?
Q10.16 Why is the inequality g{ 6 1 always satisfied in dilute electrolyte solutions?
305
Q10.17 Under what conditions does g{ S 1 for electrolyte
solutions?
Q10.18 How do you expect Sm~ for an ion in solution to
change as the ionic radius increases at constant charge?
Q10.19 How do you expect Sm~ for an ion in solution to
change as the charge increases at constant ionic radius?
Q10.20 It takes considerable energy to dissociate NaCl in
the gas phase. Why does this process occur spontaneously in
an aqueous solution? Why does it not occur spontaneously
in CCl4?
NUMERICAL PROBLEMS
Section 10.2
P10.1 Calculate ∆ r S ~ for the reaction Ba1OH221aq2 +
2 NaCl1aq2 S BaCl21s2 + 2 NaOH1aq2.
P10.2 Calculate ∆ r S ~ for the reaction CuSO41aq2 +
2KCl1aq2 S CuCl21s2 + K2SO41aq2.
P10.3 Calculate ∆ r H ~ and ∆ rG ~ for the reaction
MgCl21aq2 + 2NaOH1aq2 S Mg1OH221s2 + 2NaC11aq2.
P10.4 Calculate ∆ solvG ~ in an aqueous solution for Mg 2 + 1aq2
using the Born model. The radius of the Mg + ion is 72 pm.
P10.5 Calculate ∆ r H ~ and ∆ rG ~ for the reaction
AgNO31aq2 + KCl1aq2 S AgCl1s2 + KNO31aq2.
Section 10.3
P10.6 Calculate the mean ionic molality, m {, in 0.0225 m
solutions of (a) Ca1NO322, (b) NaOH, (c) MgSO4, and
(d) AlCl3.
P10.7 Calculate the mean ionic activity of a 0.0250 m K2SO4
solution for which the mean activity coefficient is 0.745.
P10.8 From the data in Table 10.3 (see Appendix A, Data
Tables), calculate the activity of the electrolyte in 0.300 m
solutions of
a. AgNO3
b. CuSO4
c. BaCl2
P10.9 Express a { in terms of a + and a - for (a) Ba(NO3)2,
(b) LiF, (c) K2CO3, and (d) K3Fe1CN26. Assume complete
dissociation.
P10.10 Express m{ in terms of m+ and m- for (a) NaCl,
(b) MgBr2, (c) Li3PO4, and (d) Ca1NO322. Assume complete
dissociation.
P10.11 Express g{ in terms of g+ and g- for (a) CuSO4,
(b) CaI2, (c) Na3PO4, and (d) AgNO3. Assume complete dissociation.
P10.12 Calculate the mean ionic molality and mean ionic
­activity of a 0.150 m K2MnO4 solution for which the mean
ionic activity coefficient is 0.145.
P10.13 Calculate the value of m { in 3.25 * 10-3 molal
solutions of (a) NaBr, (b) Ba(NO3)2, and (c) CuSO4. Assume
complete dissociation.
Section 10.4
P10.14 Using the Debye–Hückel limiting law, calculate
the value of g{ in a 3.5 * 10-3 m solution of (a) KCl,
(b) BaCl2, and (c) CaHPO4. Assume complete dissociation.
P10.15 Calculate the ionic strength in a solution that is
0.0625 m in K2SO4, 0.0125 m in Na 3PO4, and 0.0100 m
in MgCl2 .
P10.16 Calculate I, g{, and a { for a 0.0225 m solution of
Na3PO4 at 298 K. Assume complete dissociation.
P10.17 In the Debye–Hückel theory, the countercharge in a
spherical shell of radius r and thickness dr around the central
ion of charge +q is given by -qk2re-krdr. Calculate the
radius at which the counter charge has its maximum value,
rmax, from this expression. Evaluate rmax for a 0.0750 m
aqueous solution of K2SO4 at 298 K.
P10.18 Calculate the probability of finding an ion at a distance greater than 1>k from the central ion.
P10.19 Calculate the Debye–Hückel screening length 1>k
at 298 K in a 0.00625 m solution of K3PO4.
P10.20 Calculate I, g{, and a { for a 0.0150 m solution of
K4Fe1CN26 at 298 K. How confident are you that your calculated results will agree with experimental results?
P10.21 Calculate I, g{, and a { for a 0.0225 m solution of
Na 3PO4 at 298 K. Assume complete dissociation. How confident are you that your calculated results will agree with
experimental results?
P10.22 Calculate the ionic strength of each of the solutions
in Problem P10.6.
P10.23 Use the Davies equation to calculate g{ for a
0.100 m solution of BaCl2. Calculate the percent error of your
result compared to the value in Table 10.3 (see Appendix A,
Data Tables).
306
CHAPTER 10 Electrolyte Solutions
Section 10.5
P10.24 A weak acid has a dissociation constant of
Ka = 2.50 * 10-2. (a) Calculate the degree of dissociation
for a 0.0750 m solution of this acid using the Debye–Hückel
limiting law. (b) Calculate the degree of dissociation for a
0.0750 m solution of this acid that is also 0.200m in KCl from
the Debye–Hückel limiting law using an iterative calculation
until the answer is constant to within {2 in the first decimal
place. (c) Repeat the calculation in (b) using the mean activity
coefficient for KCl in Table 10.3. Do you need to repeat the
iterative calculation of (a) to solve (b) and (c)?
P10.25 At 25°C, the equilibrium constant for the dissociation
of acetic acid, Ka, is 1.75 * 10-5. Using the Debye–Hückel
limiting law, calculate the degree of dissociation in 0.0150 m
and 0.150 m solutions using an iterative calculation until the
answer is constant to within {2 in the first decimal place.
Compare these values with what you would obtain if the ionic
interactions had been ignored.
P10.26 Estimate the degree of dissociation of a 0.300 m solution of nitrous acid 1Ka = 4.00 * 10-42 that is also 0.300 m
in the strong electrolyte given in parts (a)–(c). Use the data
tables to obtain g{, as the electrolyte concentration is too
high to use the Debye–Hückel limiting law.
a. BaCl2
b. KOH
c. AgNO3
Compare your results with the degree of dissociation of the
acid assuming g{ = 1.
P10.27 Calculate the solubility of CaCO31Ksp = 3.4 * 10-92
(a) in pure H2O and (b) in an aqueous solution with
I = 0.0375 mol kg -1. For part (a), do an iterative calculation
of g{ and the solubility until the answer is constant to within
{2 in the first decimal place. Do you need to repeat this
procedure in part (b)?
P10.28 The base dissociation constant for dimethylamine,
1CH322NH1aq2 + H2O1l2 S CH3NH3+ 1aq2 + OH - 1aq2
is 5.12 * 10-4. Calculate the extent of hydrolysis for (a) a
0.125 m solution of 1CH322NH in water using an iterative
calculation until the answer is constant to within {2 in the
first decimal place and (b) a solution that is also 0.400 m
in NaNO3.
P10.29 Dichloroacetic acid has a dissociation constant of
Ka = 3.32 * 10-2. Calculate the degree of dissociation for
a 0.0750 m solution of this acid (a) from the Debye–Hückel
limiting law using an iterative calculation until the answer is
constant to within {2 in the first decimal place and (b) assuming that the mean ionic activity coefficient is one.
P10.30 Calculate the pH of a buffer solution that is 0.225
molal in CH3COOH and 0.375 molal in CH3COONa using
the Davies equation to calculate g{. What pH value would
you have calculated if you had assumed that g{ = 1?
FURTHER READING
Davies, M., and Debye, Peter J. W. Journal of Chemical
Education 45 (1968): 467–473.
Morel, J. P. “Debye’s Activity Coefficient: In Which
Concentration Scale?” Journal of Chemical Education
57 (1980): 246.
Naiman, B. “The Debye–Hückel Theory and Its Application
in the Teaching of Quantitative Analysis.” Journal of
Chemical Education 26 (1949): 280–282.
Neidig, H. A., and Yingling, R. T. “Relationship of Enthalpy
of Solution, Solvation Energy, and Crystal Energy.”
Journal of Chemical Education 42 (1965): 473.
Wright, M. R., Patterson, I. L. J., and Harris, K. D. M.
“Non-Ideality and Ion Association in Aqueous Electrolyte
Solutions: Overview and a Simple Experimental
Approach.” Journal of Chemical Education 75 (1998):
352–357.
C H A P T E R
11
Electrochemical Cells,
Batteries, and Fuel Cells
WHY is this material important?
In electrochemical cells, spontaneity is governed by the electrochemical potential
rather than the chemical potential. By applying a voltage, the electrochemical potential
of reactants and products can easily be changed, allowing the direction of spontaneous change to be reversed. Batteries are electrochemical cells in which energy can be
stored and one of the most important applications of electrochemistry is the development of batteries with a high enough energy density to power cars, trucks, and boats.
WHAT are the most important concepts and results?
Reaction Gibbs energies, reaction entropies, equilibrium constants, and activity coefficients can all be determined from experiments carried out in electrochemical cells.
Batteries and fuel cells are efficient energy sources because electrical work is directly
converted to mechanical work. They are not subject to the limit imposed on combustion engines by the second law of thermodynamics.
11.1
The Effect of an Electrical
Potential on the Chemical
Potential of Charged Species
11.2
Conventions and Standard
States in Electrochemistry
11.3
Measurement of the Reversible
Cell Potential
11.4
Chemical Reactions in
Electrochemical Cells and the
Nernst Equation
11.5
Combining Standard Electrode
Potentials to Determine the
Cell Potential
11.6
Obtaining Reaction Gibbs
Energies and Reaction
Entropies from Cell Potentials
11.7
Relationship between the
Cell EMF and the Equilibrium
Constant
11.8
Determination of E ~ and
Activity Coefficients Using
an Electrochemical Cell
11.9
Cell Nomenclature and Types
of Electrochemical Cells
WHAT would be helpful for you to review for this chapter?
11.10 The Electrochemical Series
It would be helpful to review the material in Chapter 6 in which a framework for the
spontaneity of chemical reactions was discussed.
11.11 Thermodynamics of Batteries
and Fuel Cells
11.12 Electrochemistry of Commonly
Used Batteries
11.1 THE EFFECT OF AN ELECTRICAL POTENTIAL
ON THE CHEMICAL POTENTIAL OF
CHARGED SPECIES
11.14 (Supplemental Section)
Electrochemistry at the Atomic
Scale
If a Zn electrode is partially immersed in an aqueous solution of ZnSO4, an equilibrium
is established between Zn1s2 and Zn2+1aq2 as a small amount of the Zn enters the solution as Zn2+1aq2 ions, as depicted in Figure 11.1. However, the electrons remain on the
Zn electrode. Therefore, a negative charge builds up on the Zn electrode, and a corresponding positive charge builds up in the surrounding solution. This charging leads to a
difference in the electrical potential f between the electrode and the solution, which we
call the half-cell potential. As we will see in Section 11.2, two half-cells can be combined
to form an electrochemical cell. The charge separation in the system arises through the
dissociation equilibrium
Zn1s2 ∆ Zn2+1aq2 + 2e-
11.13 Fuel Cells
11.15 (Supplemental Section) Using
Electrochemistry for Nanoscale
Machining
(11.1)
The equilibrium position in this reaction lies far toward Zn1s2. At equilibrium, fewer
than 10-14 mol of the Zn1s2 dissolves in 1 liter of solution to form Zn2+1aq2. However, this minuscule amount of charge transfer between the electrode and the solution is
307
308
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Zn(s)
Zn21
SO422
Zn21
solution
2e2
Zn
Zn21
Zn21
Figure 11.1
Electrochemistry at a zinc electrode.
When a Zn electrode is immersed in an
aqueous solution containing Zn2+1aq2
ions, a very small amount of the Zn enters
the solution as Zn2+1aq2, leaving behind
two electrons for each ion formed on the
Zn electrode.
sufficient to create a difference of approximately 1 V in the electrical potential between
the Zn electrode and the electrolyte solution. A similar dissociation equilibrium is established for other metal electrodes. Because the value of the equilibrium constant depends on
∆ f G ~ of the solvated metal ion, the equilibrium constant for the dissociation reaction and
f depends on the identity of the metal.
Can f be measured directly? Let us assume that we can carry out the measurement using two chemically inert Pt wires as probes. One Pt wire is placed on the Zn
electrode, and the second Pt wire is placed in the ZnSO4 solution. However, the measured voltage is the difference in electrical potential between a Pt electrode connected
to a Zn electrode in a ZnSO4 solution and a Pt electrode in a ZnSO4 solution, and not the
electrical potential between the Zn electrode and the electrolyte solution. The measurement problem arises because a difference in electrical potential can only be measured
between one phase and a second phase of identical composition. For example, the difference in electrical potential across a resistor is measured by contacting the metal wire
at each end of the resistor, with two metal probes of identical composition connected
to the terminals of a voltmeter. Although we cannot measure the half-cell potential f
directly, half-cell potentials can be determined relative to a reference half-cell, as will
be shown in Sections 11.2 and 11.4.
How are chemical species affected by the electrical potential f? To a very good
approximation, a neutral atom or molecule is not affected if a small electrical potential
is applied to the environment containing the species. However, this is not the case for a
charged species such as a Na+ ion in an electrolyte solution that experiences a force that
causes it to move. The work required to transfer dn moles of charge reversibly from a
chemically uniform phase at an electrical potential f 1 to a second, otherwise identical
phase at an electrical potential f 2 is equal to the product of the difference in the electrical potential between the two locations and the charge:
dw = 1f 2 - f 12 dQ
(11.2)
In this equation, dQ = -z F dn is the charge transferred through the potential, z is
the charge in units of the electron charge 1+1, -1, +2, -2, c2, and the Faraday
constant F is the absolute magnitude of the charge associated with 1 mol of a singly
charged species. The Faraday constant has the numerical value F = 96,485 Coulombs
mole-1 1C mol-12.
Because the work being carried out in this reversible process is nonexpansion work,
∼ of the charged
dwrev = dG, which is the difference in the electrochemical potential m
particle in the two phases:
∼ dn - m
∼ dn
dG = m
(11.3)
2
1
The electrochemical potential is a generalization of the chemical potential to include
the effect of an electrical potential on a charged particle. It is the sum of the normal
chemical potential m and a term that results from the nonzero value of the electrical
potential:
∼ = m + zfF
m
(11.4)
Concept
Electrochemical potential is a
generalization of chemical potential
and includes the effect of an electrical
potential on a charged particle.
∼ S m as f S 0.
Note that with this definition m
Combining Equations (11.2) and (11.4) gives
∼ - m
∼ = + z1f - f )F or m
∼ = m
∼ + z1f - f 2F
m
2
1
2
1
2
1
2
1
(11.5)
Because only the difference in the electrical potential between two points can be
measured, one can set f 1 = 0 in Equation (11.5) to obtain the result
∼ = m
∼ + zfF
m
(11.6)
2
1
This result shows that charged particles in two otherwise identical phases have different values for the electrochemical potential if the phases are at different electrical
potentials. Because the particles will flow in a direction that decreases their electrochemical potential, the flow of negatively charged particles in a conducting phase is
toward a region of more positive electric potential. The opposite is true for positively
charged particles.
11.2 Conventions and Standard States in Electrochemistry
Equation (11.6) is the basis for understanding all electrochemical reactions. In an
electrochemical environment, the equilibrium condition is a ni ∼
mi = 0
i
∼ =
∼
∆m
a ni mi = 0, rather than ∆m = a ni mi = 0
i
i
(11.7)
309
Concept
In an electrochemical environment, the
equilibrium condition is a ni ∼
mi = 0.
i
In this equation, the sum is over reactants and products with appropriate signs for the
stoichiometric coefficients. Chemists have a limited ability to change ∆m by varying
P, T, and concentration. However, because the electrochemical potential can be varied
∼ can be changed easily. The term
through the application of an electrical potential, ∆m
zifF can be larger in magnitude than a nimi and can have either the same or the oppoi
site sign. Because a change in the electrical potential can lead to a change in the sign of
∼, the direction of spontaneous change in an electrochemical reaction system can be
∆m
changed simply by applying suitable electrical potentials within the system. A practical
∼ is the recharging of a battery by the application of
example of changing the sign of ∆m
potential external to the battery.
11.2 CONVENTIONS AND STANDARD STATES
IN ELECTROCHEMISTRY
In order to use Equation (11.7) to describe equilibrium in an electrochemical cell, we need
∼ for the individual species in a half-cell. Consider the equito be able to assign values to m
2+
librium Zn1s2 ∆ Zn 1aq2 + 2e-. To do so, we need to define standard states for
the various species that take part in an electrochemical reaction. Because Zn1s2 is in its
∼1Zn1s22 = m1Zn1s22 = ∆ G1Zn1s22 = 0.
standard reference state and is uncharged, m
ƒ
Next, let us consider an ion in solution. Presently, we will consider only the case that
the overall chemical reaction in an electrochemical cell takes place in a single phase, in
this case the solution between the electrodes. Therefore, f has the same value for all solvated ionic reactants and products. For this scenario, the equilibrium condition becomes
∼ =
∼
∆m
(11.7a)
a nimi = a nimi + Fw a nizi = a nimi = 0
because a nizi = 0 for an electrically neutral solution. We see that the value of f does
i
not influence the equilibrium if the overall reaction takes place in a single phase. Therefore, we can set f = 0 and assign the standard state
∼ = m 1ions in solution2
m
(11.8)
i
i
i
i
i
i
Adopting this standard state simplifies calculations because mi can be calculated
from the solute concentration at low concentrations using the Debye–Hückel limiting
law discussed in Section 10.4 or at higher concentrations if the activity coefficients are
known.
Next, consider the appropriate standard state for an electron in a metal electrode. We
cannot set f = 0 as we did for the ions in solution because there is a potential difference
between the solution and the electrode. As shown in Equation (11.6), the electrochemical
potential consists of two parts, a chemical component and a component that depends on
the electrical potential. For an electron in a metal, there is no way to determine the relative magnitude of the two components. Therefore, it is convenient to choose the standard
state mi 1electrons in electrode2 = 0 so that
∼ - = - fF 1electrons in metal electrode2
m
(11.9)
e
These choices for standard states do not affect the values of thermodynamic state functions obtained in experiments that use electrochemical cells because both half-cells are
affected equally.
Having discussed the standard states for neutral species, charged ions, and electrons
in half-cells, we next combine two half-cells to form an electrochemical cell. As we
discussed earlier, half-cell potentials cannot be measured directly. They are measured
relative to one another rather than absolutely. To understand how this is done, it is useful to consider an electrochemical cell, which consists of two half-cells, such as the one
Concept
Standard states for ions in solutions,
electrons in an electrode, and neutral
atoms differ.
310
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Figure 11.2
Digital
voltmeter
Schematic diagram of the Daniell cell.
Zn2+ >Zn and Cu2+ >Cu half-cells are
connected through a salt bridge in the
internal circuit. A voltmeter is shown in
the external circuit. The inset shows the
atomic-level processes that occur at each
electrode.
e2
e2
Salt bridge
Anode
Cathode
Zn(s)
Cu(s)
2e2
2e2
Zn
Zn21
Zn21
Concept
The standard hydrogen electrode
is the half-cell against which the
potentials of all other half-cells are
measured.
Cu
Cu21
Cu21
shown in Figure 11.2. This particular cell is known as the Daniell cell, after its inventor
John Frederic Daniell (1790–1845). On the left, a Zn electrode is immersed in a solution of ZnSO4. The solute is completely dissociated to form Zn2+1aq2 and SO24 1aq2.
On the right, a Cu electrode is immersed in a solution of CuSO4, which is completely
dissociated to form Cu2+ 1aq2 and SO24 1aq2. The two half-cells are connected by an
ionic conductor known as a salt bridge. The salt bridge consists of an electrolyte such
as KCl suspended in a gel. A salt bridge allows current to flow between the half-cells
while preventing the mixing of the solutions. A metal wire fastened to each electrode
allows the electron current to flow through the external part of the circuit. Note that
because the wire is connected on one end to a Cu electrode and on the other end to a Zn
electrode, the two phases between which we are measuring the electrical potential are
not identical.
Using the experimental setup of Figure 11.2, we can measure the electrical potential difference between two half-cells, rather than the absolute electrical potential of
each half-cell. However, as we will see in the subsequent discussion, we need to know
values for potentials of individual half-cells to be able to discuss electrochemical reactions. Therefore, it is convenient to choose one half-cell as a reference and arbitrarily
assign an electrical potential of zero to this half-cell. Once this is done, the electrical
potential associated with any other half-cell can be determined by combining it with
the reference half-cell. The measured potential difference across the full cell is associated with the half-cell of interest. It is next shown that the standard hydrogen electrode
fulfills the role of a reference half-cell of zero potential. The measurement of electrical
potentials is discussed using this cell. The reaction in the standard hydrogen electrode is
H +1aq2 + e- S
1
H 1g2
2 2
(11.10)
and the equilibrium in the half-cell is described by
∼ - = 1m
mH +1aq2 + m
e
2 H21g2
(11.11)
11.2 Conventions and Standard States in Electrochemistry
311
Note that the standard state convention for ions in solution [Equation (11.8)] allows
us to write Equation (11.11) in terms of the chemical potential rather than the electrochemical potential. Half-cell reactions such as Equation (11.10) are generally written as
reduction reactions by convention, as is done in Tables 11.1 and 11.2 (see Appendix A,
Data Tables). It is useful to separate mH2 1g2 and mH+1aq2 into a standard state portion
and a portion that depends on the activity and use Equation (11.9) for the electrochemical potential of the electron. The preceding equation then takes the form
mH~ + + RT ln aH + - Ff H +>H2 =
1 ~
1
m + RT ln ƒH2
2 H2
2
where ƒH2 is the fugacity of the hydrogen gas. Solving Equation (11.12) for f H
f H +>H2
1
mH~ + - mH~ 2
1>2
2
RT ƒH2
=
ln
aH +
F
F
(11.12)
H2 out
+
H2 in
>H2
(11.13)
For unit activities of all species, the cell has its standard state potential, designated
fH~ +>H2. Because mH~ 2 = ∆ f G1H2, g2 = 0,
fH~ +>H2 =
mH~ +
F
(11.14)
In Section 10.1 the convention that ∆ f G1H + , aq2 = mH~ + = 0 was introduced.
Therefore, we find that
(11.15)
fH~ +>H2 = 0
We have shown that the standard hydrogen electrode is a convenient reference electrode
with zero potential against which the potentials of all other half-cells can be measured.
A schematic drawing of this electrode is shown in Figure 11.3. To achieve equilibrium
on a short timescale, this reaction H +1aq2 + e- ∆ 1>2 H21g2 is carried out over a
Pt catalyst electrode. It is also necessary to establish a standard state for the activity of
H +1aq2. It is customary to use a Henry’s law standard state based on molarity. Therefore,
ai S ci and gi S 1 as ci S 0. The standard state is a (hypothetical) aqueous solution of
H +1aq2 that shows ideal solution behavior at a concentration of c ~ = 1 mol L-1.
The usefulness of the result fH~ + >H2 = 0 is that values for the electrical potential can
be assigned to individual half-cells by measuring their potential relative to the H + >H2
half-cell. For example, the cell potential of the electrochemical cell in Figure 11.4 is
assigned to the Zn>Zn2+ half-cell if the H +1aq2 and H21g2 activities both have the value 1.
A
Switch
H1(aq)
Pt
Figure 11.3
The standard hydrogen electrode. This
electrode consists of a solution of an acid,
H2 gas, and a Pt catalyst electrode that
allows the equilibrium in the half-cell
reaction to be established rapidly. The
activities of H2 and H + are equal to one.
Resistor
V
H2 out
Zn
H2 in
Salt bridge
Figure 11.4
Zn
21(aq)
1(aq)
H
Pt
Voltage determination using a standard
hydrogen electrode. In a cell consisting
of a half-cell and the standard hydrogen
electrode, the entire cell voltage is assigned
to the half-cell.
312
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
2
1
dc source
Although not directly measurable, absolute values for half-cell potentials can be determined indirectly. For more details, see the articles by Donald and Williams (2014) and
Gomer and Tryson (1977) cited in Further Reading. These authors have shown that the
currently accepted value for the absolute potential of the standard hydrogen electrode is
-4.44 { 0.02 V. Chemists generally use half-cell potentials relative to the standard
hydrogen electrode, assuming that fH~ +>H2 = 0.
11.3 MEASUREMENT OF THE REVERSIBLE
CELL POTENTIAL
I
Figure 11.5
Schematic diagram showing how the
reversible cell potential is measured.
The cell potential measured under reversible conditions is directly related to the state
functions G, H, and S. The reversible cell potential, also called electromotive force
(emf), is determined in an experiment depicted in Figure 11.5. The dc source provides
a voltage to a potentiometer circuit with a sliding contact. The sliding contact is
attached to the positive cell terminal as shown, and the slider is adjusted until the currentsensing device labeled I shows a null current. At this position of the potentiometer,
the voltage applied through the potentiometer exactly opposes the cell potential. The
voltage measured in this way is the reversible cell potential. If the sliding contact is
moved to a position slightly to the left of this position, the electron current will flow
through the external circuit in one direction. However, if the sliding contact is moved to
a position slightly to the right of this position, the electron current will flow through
the external circuit in the opposite direction, showing that the direction of the cell
reaction has been reversed. Because a small variation of the applied voltage can reverse
the direction of spontaneous change, the criterion for reversibility is established. This
discussion also demonstrates that the direction of spontaneous change in the cell can be
reversed by changing the electrochemical potential of the electrons in one of the electrodes relative to that in the other electrode using an external voltage source.
11.4 CHEMICAL REACTIONS IN
ELECTROCHEMICAL CELLS
AND THE NERNST EQUATION
What reactions occur in the Daniell cell shown in Figure 11.2? If the half-cells are connected through the external circuit, Zn atoms leave the Zn electrode to form Zn2+ in
solution, and Cu2+ ions are deposited as Cu atoms on the Cu electrode. In the external
circuit, it is observed that electrons flow through the wires and the resistor in the direction from the Zn electrode to the Cu electrode. These observations are consistent with
the following electrochemical reactions:
Left half-cell: Zn1s2 S Zn2+1aq2 + 2e-
(11.16)
-
2+
Right half-cell: Cu 1aq2 + 2e S Cu1s2
(11.17)
Overall: Zn1s2 + Cu2+1aq2 ∆ Zn2+1aq2 + Cu1s2
(11.18)
2+
In the left half-cell, Zn is being oxidized to Zn , and in the right half-cell, Cu2+ is
being reduced to Cu. By convention, the electrode at which oxidation occurs is called
the anode, and the electrode at which reduction occurs is called the cathode. Each
half-cell in an electrochemical cell must contain a species that can exist in an oxidized
and a reduced form. For a general redox reaction, the reactions at the anode and cathode and the overall reaction can be written as follows:
+
Anode: Red1 S Ox1 n 1 + n1e-
(11.19)
n -2
Cathode: Ox2 + n2e- S Red2
Overall: n2Red1 + n1Ox2 ∆
(11.20)
+
n2Ox1n 1
+
n1Red2n 2
(11.21)
313
11.4 Chemical Reactions in Electrochemical Cells and the Nernst Equation
Note that electrons do not appear in the overall reaction because the electrons produced
at the anode are consumed at the cathode.
How are the cell voltage and ∆ rG for the overall reaction related? This important
relationship can be determined from the electrochemical potentials of the species
involved in the overall reaction of the Daniell cell:
∼ = m
∼ 2+ + m
∼ - m
∼ 2+ - m
∼ = m ~ 2 + - m ~ 2 + + RT ln aZn2 +
∆m
Zn
Cu
Cu
Zn
Zn
Cu
aCu2 +
= ∆ rG ~ + RT ln
aZn2 +
aCu2 +
(11.22)
If this reaction is carried out reversibly, the electrical work done is equal to the product
of the charge and the potential difference through which the charge is moved. However,
the reversible work at constant pressure is also equal to ∆G. Therefore, we can write the
following equation:
∆ rG = -nF∆f = -nFE
Concept
The Gibbs energy for the overall
reaction in an electrochemical cell is
proportional to the cell potential.
(11.23)
In Equation (11.23), ∆f is the measured potential difference generated by the spontaneous chemical reaction for particular values of aZn2+ and aCu2+, and n is the number
of moles of electrons involved in the redox reaction. The measured cell voltage is
directly proportional to ∆ rG. For a reversible reaction, the symbol E is used in place
of ∆f, and E is referred to as the electromotive force. Using this definition, we rewrite
Equation (11.23) as follows:
-2FE = ∆ rG ~ + RT ln
aZn2 +
aCu2 +
(11.24)
For standard state conditions, aZn2+ = aCu2+ = 1, and Equation (11.24) takes the form
∆ rG ~ = -2FE ~ . This definition of E ~ allows Equation (11.24) to be rewritten as
E = E~ -
RT aZn2 +
ln
2F aCu2 +
Concept
The Nernst equation relates the
potential of an electrochemical cell to
its standard state potential and to the
activities of ionic species in solution.
(11.25)
For a general overall electrochemical reaction involving the transfer of n moles of
electrons,
(11.26)
where Q is the familiar reaction quotient. The preceding equation is known as the
Nernst equation. At 298.15 K, the Nernst equation can be written in the form
E = E~ -
0.05916 V
log10 Q
n
n51
2
n52
n53
[
RT
ln Q
nF
(E2E )
RT/F
E = E~ -
4
0
(11.27)
22
This function is graphed in Figure 11.6. The Nernst equation allows the emf for an
electrochemical cell to be calculated if the activity is known for each species and if E ~
is known.
The Nernst equation has been derived on the basis of the overall cell reaction. For
a half-cell, an equation of a similar form can be derived. The equilibrium condition for
the half-cell reaction
Oxn+ + ne- ∆ Red
(11.28)
∼- = m
mOxn+ + nm
e
Red
(11.29)
is given by
24
24
22
0
log10 Q
2
Figure 11.6
Plots for Nernst equation solutions
for various numbers of transferred
electrons. The cell potential E varies
linearly with log Q. The slope of a plot
of 1E - E ~ 2>1RT>F2 is inversely
proportional to the number of electrons
transferred in the redox reaction.
4
314
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Using the standard state for the electrochemical potential of an electron in a metal electrode [Equation (11.9)], Equation (11.29) can be written in the form
~
~
n+ + RT ln a
mOx
Oxn+ - nFf Ox>Red = mRed + RT ln aRed
f Ox>Red = -
~
~
n+ - m
mOx
aRed
RT
Red
ln
nF
nF aOxn +
~
EOx>Red = E Ox>Red
-
aRed
RT
ln
nF aOxn+
(11.30)
The last line in Equation (11.30) has the same form as the Nernst equation. Note that
the activity of the electrons does not appear in Q. The application of Equation (11.30)
to a half-cell reaction is illustrated in Example Problem 11.1.
EXAMPLE PROBLEM 11.1
Calculate the potential of the H + > H2 half-cell when aH+ = 0.770 and ƒH2 = 1.13.
Solution
E = E~ -
2ƒH2
0.05916 V
0.05916 V
21.13
log10
= 0 log10
= -0.0083 V
+
n
aH
1
0.770
11.5 COMBINING STANDARD ELECTRODE
POTENTIALS TO DETERMINE THE CELL
POTENTIAL
A representative set of standard potentials is listed in Tables 11.1 and 11.2 (see
Appendix A, Data Tables). By convention, half-cell emfs are always tabulated as
reduction potentials. Because ∆ rG = -nFE, and because ∆ rG for the oxidation and
reduction reactions of a given half-cell are equal in magnitude and opposite in sign,
~
~
E red
= -E ox
Concept
The standard potential of an
electrochemical cell is determined
by the standard potentials of its
half-cells.
(11.31)
How is the cell potential related to the potentials of the half-cells? The standard
potential of a half-cell is given by E ~ = - ∆ rG ~ >nF and is an intensive property because although both ∆ rG ~ and n are extensive quantities, the ratio ∆ rG ~ >n is an intensive quantity. In particular, E ~ is not changed if all the stoichiometric coefficients are
multiplied by any integer because both ∆ rG ~ and n are changed by the same factor.
Therefore,
~
~
~
E cell
= E red
+ E ox
(11.32)
even if the balanced reaction for the overall cell is multiplied by an arbitrary number. In
Equation (11.32), the standard potentials on the right refer to the half-cells. However,
whether the reduction or the oxidation reaction is spontaneous in a half-cell of an electrochemical cell is determined by the relative emfs of the two half-cells that constitute
the cell. Because ∆ rG = -nFE, the spontaneous direction of the overall reaction in
the cell is the direction for which E 7 0, as illustrated in Example Problem 11.2.
EXAMPLE PROBLEM 11.2
An electrochemical cell is constructed using a half-cell for which the reduction reaction is given by
Fe1OH)21s2 + 2e- S Fe1s2 + 2 OH -1aq2 E ~ = -0.877 V
It is combined with a half-cell for which the reduction reaction is given by
a. Al3+1aq2 + 3e- S Al1s2
E ~ = -1.66 V
b. AgBr1s2 + e- S Ag1s2 + Br -1aq2
E ~ = +0.071 V
11.5 Combining Standard Electrode Potentials to Determine the Cell Potential
315
The value of the activity of all species is one for both reactions. Write the overall reaction for the cells in the direction of spontaneous change. Is the Fe reduced or oxidized
in the spontaneous reaction?
Solution
The emf for the cell is the sum of the emfs for the half-cells, one of which is written as
an oxidation reaction and the other of which is written as a reduction reaction. For the
~
~
direction of spontaneous change, E ~ = E reduction
+ E oxidation
7 0. Note that the two
half-cell reactions are combined with the appropriate stoichiometry, so that electrons
do not appear in the overall equation. Recall that the half-cell emfs are not changed in
multiplying the overall reactions by the integers necessary to eliminate the electrons in
the overall equation because E ~ is an intensive quantity.
a. 3 Fe1OH221s2 + 2 Al1s2 ∆ 3 Fe1s2 + 6 OH -1aq2 + 2 Al3+1aq2
E ~ = -0.877 V + 1.66 V = +0.783 V
In this cell, Fe is reduced.
b. Fe1s2 + 2 OH -1aq2 + 2 AgBr1s2 ∆ Fe1OH)21s2 + 2 Ag1s2 + 2 Br -1aq2
E ~ = 0.071 V + 0.877 V = +0.95 V
In this cell, Fe is oxidized.
Because ∆G is a state function, the cell potential for a third half-cell can be obtained
from the cell potentials of two half-cells if they have an oxidation or reduction reaction
in common. The procedure is analogous to the use of Hess’s law in Chapter 4 and is
illustrated in Example Problem 11.3.
EXAMPLE PROBLEM 11.3
You are given the following reduction reactions and E ~ values:
Fe3+1aq2 + e- S Fe2+1aq2
Fe2+1aq2 + 2e- S Fe1s2
E ~ = +0.771 V
E ~ = -0.447 V
Calculate E ~ for the half-cell reaction Fe3+1aq2 + 3e- S Fe1s2.
Solution
We calculate the desired value of E ~ by converting the given E ~ values to ∆G ~ values,
and combining these reduction reactions to obtain the desired equation.
∆ rG ~ = -nFE ~
∆ rG ~ = -nFE ~
Fe3+1aq2 + e- S Fe2+1aq2
= -1 * 96485 C mol-1 * 0.771 V = -74.39 kJ mol-1
Fe2+1aq2 + 2e- S Fe1s2
= -2 * 96485 C mol-1 * 1-0.447 V2 = 86.26 kJ mol-1
We next add the two equations as well as their ∆ rG ~ to obtain
Fe3+1aq2 + 3e- S Fe1s2
∆ rG ~ = -74.39 kJ mol-1 + 86.26 kJ mol-1 = 11.87 kJ mol-1
∆ rG ~
-11.87 * 103 J mol-1
~
3+
E Fe
=
= -0.041 V
>Fe = nF
3 * 96485 C mol-1
The E ~ values cannot be combined directly because they are intensive rather than
extensive quantities.
The preceding calculation can be generalized as follows. Assume that n1 electrons
~
are transferred in the reaction with the potential E A>B
, and n2 electrons are transferred
~
in the reaction with the potential E B>C. If n3 electrons are transferred in the reaction
~
~
~
~
with the potential E A>C
, then n3E A>C
= n1E A>B
+ n2E B>C
.
Concept
If the number of electrons transferred
in two connected half-cells differ, the
cell potential is not the sum of the
half-cell potentials.
316
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
11.6 OBTAINING REACTION GIBBS ENERGIES
AND REACTION ENTROPIES FROM CELL
POTENTIALS
In Section 11.4, we demonstrated that ∆ rG = -nFE. Therefore, if the cell potential is
measured under standard conditions,
Concept
Gibbs energies and reaction
entropies can be determined from
the temperature dependence of the
cell potential.
∆ rG ~ = -nFE ~
(11.33)
If E ~ is known, ∆ rG ~ can be determined using Equation (11.33). For example, E ~ for
the Daniell cell is +1.10 V. Therefore, ∆ rG ~ for the reaction Zn1s2 + Cu2+1aq2 ∆
Zn2+1aq2 + Cu1s2 is
∆ rG ~ = -nFE ~ = -2 * 96,485 C mol-1 * 1.10 V = -212 kJ mol-1
(11.34)
The reaction entropy is related to ∆G R~ by
∆ rS ~ = -a
0∆ rG ~
0E ~
b = nF a
b
0T
0T P
P
(11.35)
Therefore, a measurement of the temperature dependence of E ~ can be used to determine ∆ r S ~ , as shown in Example Problem 11.4.
EXAMPLE PROBLEM 11.4
The standard potential of the cell formed by combining the Cl21g2 > Cl-1aq2 half-cell with
the standard hydrogen electrode is +1.36 V, and 10E ~ >0T)P = -1.20 * 10-3 V K-1.
Calculate ∆ r S ~ for the reaction H21g2 + Cl21g2 ∆ 2 H +1aq2 + 2 Cl-1aq2. Compare
your result with the value that you obtain by using the values of ∆ r S ~ in the data tables
from Appendix A.
Solution
From Equation (11.35),
∆r S ~ = - a
0∆ rG ~
0E ~
b
b = nF a
0T P
0T
P
= 2 * 96485 C mol-1 * 1-1.2 * 10-3V K-12
= -2.3 * 102J K-1 mol-1
From Table 11.1 (see Appendix A, Data Tables),
∆ rS ~ = 2Sm~ 1H +, aq2 + 2Sm~ 1Cl-, aq2 - Sm~ 1H2, g2 - Sm~ 1Cl2, g2
= 2 * 0 + 2 * 56.5 J K-1mol-1 - 130.7 J K-1mol-1 - 223.1 J K-1mol-1
= -2.408 * 102 J K-1mol-1
The limited precision of the temperature dependence of E ~ limits the precision in the
experimental determination of ∆ r S ~ .
11.7 RELATIONSHIP BETWEEN THE CELL EMF
AND THE EQUILIBRIUM CONSTANT
If the redox reaction is allowed to proceed until equilibrium is reached, ∆ rG = 0, so
that E = 0. For the equilibrium state, the reaction quotient Q = K. Therefore, from
Equation (11.26)
E~ =
RT
ln K
nF
(11.36)
11.7 Relationship between the Cell EMF and the Equilibrium Constant
Equation (11.36) shows that a measurement of the standard state cell potential, for
which ai = 1 for all species in the redox reaction, allows K for the overall reaction to be determined. Although this statement is true, it is not practical to adjust all
activities to the value 1. The experimental determination of E ~ is discussed in the
next section.
Tabulated half-cell potentials provide a powerful way to determine the equilibrium constant in an electrochemical cell using Equation (11.36). To determine K, the
overall reaction must be separated into the oxidation and reduction half-reactions,
and the number of electrons transferred must be determined. For example, consider
the reaction
2 MnO4-1aq2 + 6 H +1aq2 + 5 HOOCCOOH1aq2 ∆
2 Mn2+1aq2 + 8 H2O1l) + 10 CO21g2
(11.37)
The reduction and oxidation half-reactions are
MnO4-1aq2 + 8 H +1aq2 + 5e- S Mn2+1aq2 + 4 H2O1l2
HOOCCOOH1aq2 S 2 H +1aq2 + 2 CO21g2 + 2e-
E ~ = +1.51 V (11.38)
E ~ = +0.49 V
(11.39)
To eliminate the electrons in the overall equation, the first equation must be multiplied
by 2, and the second equation must be multiplied by 5; E ~ is unchanged by doing so.
However, n is affected by the multipliers. In this case, n = 10. Therefore,
∆ rG ~ = -nFE ~ = -10 * 96485 C mol-1 * 11.51 V + 0.49 V2
= -1.93 * 102 kJ mol-1
ln K = -
∆ rG ~
1.93 * 103 kJ mol-1
=
= 778
RT
8.314 J K-1 mol-1 * 298.15 K
(11.40)
As this result shows, the equilibrium corresponds to essentially complete conversion of reactants to products. Example Problem 11.5 shows a calculation of K for
the Daniell cell and Example Problem 11.6 shows a calculation for Ksp of a weakly
soluble salt.
EXAMPLE PROBLEM 11.5
For the Daniell cell E ~ = 1.10 V. Calculate K for the reaction at 298.15 K
Zn1s2 + Cu2+1aq2 ∆ Zn2+1aq2 + Cu1s2.
Solution
ln K =
nF ~
2 * 96485 C mol-1 * 1.10 V
E =
RT
8.314 J K-1 mol-1 * 298.15 K
= 85.63
K = 1.55 * 1037
Note that the equilibrium constant calculated in Example Problem 11.5 is so large
that it could not have been measured by determining the activities of aZn2+ and aCu2+
by spectroscopic methods. This would require a measurement technique that is accurate over more than 30 orders of magnitude in the activity. By contrast, the equilibrium
constant in an electrochemical cell can be determined with high accuracy using only a
voltmeter.
A further example of the use of electrochemical measurements to determine equilibrium constants is the solubility constant for a weakly soluble salt. If the overall
reaction corresponding to dissolution can be generated by combining half-cell potentials, then the solubility constant can be calculated from the potentials. For example,
the following half-cell reactions can be combined to calculate the solubility product
of AgBr.
317
Concept
The equilibrium constant for a
reaction in an electrochemical cell
can be determined from the standard
potential.
318
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
AgBr1s2 + e- S Ag1s2 + Br -1aq2 E ~ = 0.07133 V and
Ag1s2 S Ag + 1aq2 + e- E ~ = -0.7996 V
AgBr1s2 ∆ Ag +1aq2 + Br -1aq2 E ~ = -0.7283 V
ln Ksp =
1 * 96485 C mol-1 * 1 -0.7283 V)
nF ~
E =
= -28.35
RT
8.314 J K-1 mol-1 * 298.15 K
The value of the solubility constant is Ksp = 4.88 * 10-13.
EXAMPLE PROBLEM 11.6
A concentration cell consists of two half-cells that are identical except for the activities
of the redox components. Consider two half-cells based on the Ag +1aq2 + e- S Ag1s2
reaction. The left half-cell contains AgNO3 at unit activity, and the right half-cell initially
had the same concentration of AgNO3, but just enough NaCl1aq2 has been added to precipitate the Ag +1aq2 as AgCl. Write an equation for the overall cell reaction. If the emf of
this cell is 0.29 V, what is Ksp for AgCl?
Solution
The overall reaction is Ag +1aq, a = 1) ∆ Ag +1aq, a2 with a concurrent transport
of Cl- in the opposite direction through the salt bridge between the half-cells. In the
right half-cell we have the equilibrium AgCl1s2 ∆ Ag +1aq, a {) + Cl-1aq, a {2,
so that aAg+ aCl- = a2{ = Ksp.
Because the half-cell reactions are the same, E ~ = 0 and
E = = log10 Ksp = -
a{
0.05916 V
0.05916
log10
= log10 2Ksp
n
1
1
0.05916 V
log10 Ksp
2
2 * 10.29 V2
2E
= = -9.804
0.05916
0.05916 V
Ksp = 1.57 * 10-10
11.8 DETERMINATION OF E ° AND
ACTIVITY COEFFICIENTS USING
AN ELECTROCHEMICAL CELL
Concept
The mean ionic activity coefficient for
a half-cell reaction can be determined
from the concentration dependence
of the cell potential.
The main challenge in determining standard potentials lies in knowing the value of
the activity coefficient g{ for a given solute concentration. The best strategy is to
carry out measurements of the cell potential at low concentrations, where g{ S 1,
rather than near unit activity, where g{ differs appreciably from 1. Consider an electrochemical cell consisting of the Ag + >Ag and standard hydrogen electrode half-cells
at 298 K. The cell reaction is Ag +1aq2 + 1>2 H21g2 ∆ Ag1s2 + H +1aq2 and
Q = 1aAg+)-1. Because the activities of H21g2 and H +1aq2 are 1, they do not appear in
Q. Assume that the Ag + arises from the dissociation of AgNO3. Recall that the activity
of an individual ion cannot be measured directly. It must be calculated from the mean2
sured activity a { and the definition an{ = an+
+ a - . In this case, a { = aAg + aNO-3 and
a { = aAg+ = aNO-3 . Similarly, g{ = gAg+ = gNO-3 and mAg+ = mNO-3 = m { = m,
and E is given by
~
+
E = E Ag
>Ag +
RT
RT
RT
~
+
ln aAg+ = E Ag
ln1m>m ~ 2 +
ln g{
>Ag +
F
F
F
(11.41)
319
11.9 Cell Nomenclature and Types of Electrochemical Cells
The left-hand side of this equation can be calculated from measurements and plotted
as a function of 21m>m ~ 2. The results will resemble the graph shown in Figure 11.7.
An extrapolation of the line that best fits the data to m = 0 gives E ~ as the intercept
with the vertical axis. Once E ~ has been determined, Equation (11.41) can be used to
calculate g{.
Electrochemical cells provide a powerful method of determining activity coefficients because cell potentials can be measured more accurately and more easily than
colligative properties such as freezing point depression or boiling point elevation. Note
that although the Debye–Hückel limiting law was used to determine E ~ , it is not necessary to use the limiting law to calculate activity coefficients once E ~ is known.
OF ELECTROCHEMICAL CELLS
It is useful to use an abbreviated notation to describe an electrochemical cell. This notation includes all species involved in the cell reaction and phase boundaries within the
cell, which are represented by a vertical line. As will be seen later in this section, the
metal electrodes appear at the ends of this notation, the half-cell in which oxidation
occurs is written on the left, and the electrode is called the anode. The half-cell in
which reduction occurs is written on the right, and the electrode is called the cathode.
In addition to the electrochemical reactions at the anode and cathode, we need to
briefly discuss an additional small contribution to the cell potential that arises from the
differing diffusion rates of large and small ions in an electrical field. As an electrochemical reaction proceeds, ions that diffuse rapidly across a liquid–liquid junction,
such as H +, will travel farther than ions that diffuse slowly, such as Cl-, in a given
time. At steady state, a dipole layer is built up across this junction, and the rates of ion
transfer through this dipole layer become equal. This kinetic effect will give rise to a
small junction potential between two liquids of different composition or concentration. Such a junction potential is largely eliminated by a salt bridge. An interface for
which the junction potential has been eliminated is indicated by a pair of vertical lines.
The separation of different phases that are in contact and allow electron transfer is
shown by a solid vertical line. A single dashed line is used to indicate a liquid–liquid
interface across which charge transfer can occur.
For example, the abbreviated notation for the Daniell cell containing a salt bridge is
2+
and a cell made up of the Zn>Zn
described by
(11.43)
half-cell and the standard hydrogen electrode is
Zn1s2 ZnSO41aq2 H +1aq2 H21g2 Pt1s2
+
(11.44)
2+
E
m/m
Figure 11.7
Application of the Debye–Hückel law
in the determination of E ~ and Gt. The
value of E ~ and the activity coefficient
can be measured by plotting the left-hand
side of Equation (11.42) against the
square root of the molality.
11.9 CELL NOMENCLATURE AND TYPES
Zn1s2 ZnSO41aq2 } CuSO41aq2 Cu1s2
E20.05916 V log10 m/m
(11.42)
[
~
~
+
= E Ag
>Ag - 0.0301121m>m 2
[
~
~
+
E - 0.05916 log10 1m>m ~ 2 = E Ag
>Ag - 0.05916 * 0.509221m>m 2
[
At low concentrations, the Debye–Hückel limiting law is valid and log g{ =
-0.50922m { >m ~ at 298 K as discussed in Section 10.4. Using this relation, we can
rewrite Equation (11.41) in the form
The overall reaction in this cell is Zn1s2 + 2 H 1aq2 ∆ Zn 1aq2 + H21g2. In
general, in such a cell the solutions are physically separated by a porous membrane
to prevent mixing of the solutions. In this case, the junction potential has not been
eliminated.
The half-cell and overall reactions can be determined from the abbreviated notation
in the following way. An electron is transferred from the electrode at the far left of
the abbreviated notation to the electrode on the far right through the external circuit.
The number of electrons is then adjusted to fit the half-cell reaction. This procedure is
illustrated in Example Problem 11.7.
320
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
EXAMPLE PROBLEM 11.7
Determine the half-cell reactions and the overall reaction for the cell designated
Ag1s2 AgCl1s2 Cl-1aq, a { = 0.00102 } Fe2+1aq, a { = 0.502
Fe3+1aq, a { = 0.102 Pt1s2
Solution
The anode and cathode reactions are
Ag1s2 + Cl-1aq2 S AgCl1s2 + eThe overall reaction is
Fe3+1aq2 + e- S Fe2+1aq2
Ag1s2 + Cl-1aq2 + Fe3+1aq2 ∆ AgCl1s2 + Fe2+1aq2
Only after the cell potential is calculated is it clear whether the reaction or the reverse
reaction is spontaneous.
Although we have already discussed several specific half-cells, we will now formally list the most common types of half-cells. One type of electrode is the standard
hydrogen electrode, which involves the equilibrium between a gas and an ion. A second
such electrode is the Cl2 > Cl- electrode. The reduction reaction for this half-cell is
Cl21g2 + 2e- S 2 Cl-1aq2 E ~ = +1.36 V
(11.45)
Zn2+1aq2 + 2e- S Zn1s2 E ~ = -0.76 V
(11.46)
AgCl1s2 + e- S Ag1s2 + Cl-1aq2 E ~ = +0.22 V
(11.47)
Hg2 Cl21s2 + 2e- S 2 Hg1l2 + 2 Cl-1aq2 E ~ = +0.27 V
(11.48)
Fe3+1aq2 + e- S Fe2+1aq2 E ~ = 0.771 V
(11.49)
Another type of half-cell that is frequently encountered in chemistry and industrial laboratories involves a metal and a metal ion in solution. Both half-cells in the Daniell cell
fall into this category.
A number of half-cells consist of a metal, an insoluble salt containing the metal, and an
aqueous solution containing the anion of the salt. Two examples of this type of half-cell
are the Ag–AgCl half-cell for which the reduction reaction is
and the calomel (mercurous chloride) electrode, which is frequently used as a reference
electrode in electrochemical cells:
In a further type of half-cell, both species are present in solution, and the electrode is
an inert conductor such as Pt, which allows an electrical connection to be made to the
solution. For example, in the Fe3+ >Fe2+ half-cell, the reduction reaction is
11.10 THE ELECTROCHEMICAL SERIES
Concept
In a redox couple formed from two
entries in the electrochemical series,
the species appearing higher in the
list will be reduced and the species
appearing lower in the list will be
oxidized.
Tables 11.1 and 11.2 (see Appendix A, Data Tables) list the reduction potentials of
commonly encountered half-cells. The emf of a cell constructed from two of these halfcells with standard reduction potentials E 1~ and E 2~ is given by
~
E cell
= E 1~ - E 2~
(11.50)
The potential E cell will be positive and, therefore, ∆G 6 0, if the reduction potential
for reaction 1 is more positive than that of reaction 2. Therefore, the relative strength of
a species as an oxidizing agent follows the order of the numerical value of its reduction
potential in Table 11.2. The electrochemical series shown in Table 11.3 is obtained if
~
11.11 Thermodynamics of Batteries and Fuel Cells
the oxidation of neutral metals to their most common oxidation state is considered. For
example, the entry for gold in Table 11.3 refers to the reduction reaction
Au3+1aq2 + 3e- S Au1s2 E ~ = 1.498 V
In a redox couple formed from two entries in the list shown in Table 11.3, the species
appearing higher in the list will be reduced, and the species appearing lower in the list will
be oxidized in the spontaneous reaction. For example, the table predicts that the spontaneous reaction in the copper–zinc couple is Zn1s2 + Cu2+1aq2 S Zn2+1aq2 + Cu1s2
and not the reverse reaction. An example of determining which element of a redox couple
is reduced is shown in Example Problem 11.8.
321
TABLE 11.3
The Electrochemical Series
Most Strongly Reducing (The metal
is least easily oxidized and has the
most positive reduction potential.)
Gold
Platinum
Palladium
Silver
Rhodium
EXAMPLE PROBLEM 11.8
Copper
MnO4-
2+
For the reduction of the permanganate ion
to Mn in an acidic solution,
E ~ = +1.51 V. The reduction reactions and standard potentials for Zn2+, Ag +, and
Au+ are given here:
Zn2+1aq2 + 2e- S Zn1s2
Nickel
Au+1aq2 + e- S Au1s2 E ~ = 1.692 V
Iron
Zinc
ion?
Solution
Assuming reduction of the permanganate ion and oxidation of the metal, the cell
potentials are
Chromium
Vanadium
Manganese
Magnesium
Zn: 1.51 V + 0.761 V = 2.27 V 7 0
Sodium
Ag: 1.51 V - 0.7996 V = 0.710 V 7 0
Calcium
Potassium
Au: 1.51 V - 1.692 V = -0.18 V 6 0
If E ~ 7 0, ∆ rG ~ 6 0. On the basis of the sign of the cell potential, we conclude
that only Zn and Ag will be oxidized by the MnO4- ion.
AND FUEL CELLS
Two types of electrochemical cells are designed to maximize the ratio of output power
to the cell weight or volume. Batteries contain the reactants needed to support the overall electrochemical reaction, whereas fuel cells are designed to accept a continuous
flow of reactants from the surroundings. Batteries that cannot be recharged are called
primary batteries, whereas rechargeable batteries are called secondary batteries.
It is useful to compare the relative amount of work that can be produced through an
electrochemical reaction with the work that a heat engine could produce using the same
overall reaction. The maximum electrical work is given by
T∆ rS
b
∆ rH
(11.51)
whereas the maximum work available through a reversible heat engine operating
between Th and Tc is
wthermal = qhot e = - ∆ rH a
Th - Tc
b
Th
Rubidium
Cesium
Lithium
Least Strongly Reducing (The metal
is most easily oxidized and has the
most negative reduction potential.)
11.11 THERMODYNAMICS OF BATTERIES
welectrical = - ∆ rG = - ∆ rH a1 -
Zero reduction potential
by convention
Tin
Ag +1aq2 + e- S Ag1s2 E ~ = 0.7996 V
Which of these metals will be oxidized by the
Hydrogen
Lead
E ~ = -0.7618 V
MnO4-
Mercury
(11.52)
where e is the efficiency of a reversible heat engine (see Section 5.2). To compare
the maximum thermal and electrical work, we use the overall reaction for the familiar
322
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Concept
Significantly more work is produced
if a reaction is carried out in an
electrochemical cell than in a heat
engine.
lead–acid battery used in cars. For this reaction, ∆ rG ~ = -376.97 kJ mol-1, ∆ rH ~ =
-227.58 kJ mol-1, and ∆ rS ~ = 501.1 J K-1 mol-1. Assuming Th = 600. K and
Tc = 300. K and that the battery operates at 300. K, then
welectrical
= 3.31
(11.53)
wthermal
This calculation shows that much more work can be produced in the electrochemical
reaction than in the thermal reaction. This comparison does not even take into account
that the lead–acid battery can be recharged, whereas the thermal reaction can only be
run once.
11.12 ELECTROCHEMISTRY OF COMMONLY
USED BATTERIES
The lead–acid battery was invented in 1859 and is still widely used in automobiles.
Because the power required to start an automobile engine is on the order of a kilowatt,
the current capacity of such a battery must be on the order of a hundred amperes.
Additionally, such a battery must be capable of 500 to 1500 recharging cycles from a
deep discharge. In recharging batteries, the reaction product in the form of a solid must be
converted back to the reactant, also in the form of a solid. Because the solids in general
have a different crystal structure and density, the conversion induces mechanical stress
in the anode and cathode, which ultimately leads to a partial disintegration of these electrodes. This is the main factor that limits the number of charge–discharge cycles that a
battery can tolerate.
The electrodes in the lead–acid battery consist of Pb powder and finely divided PbO
and PbSO4 held on a Pb frame. The electrodes are supported in a container containing
concentrated H2SO4. In use, a battery is discharged as the reactants and products approach
their equilibrium concentrations. Some batteries, including the lead–acid battery, can be
recharged by applying an external voltage to convert products back to reactants. In that
case, the roles of anode and cathode are reversed. To avoid confusion, we write half-cell
and overall reactions for the discharge mode. In the lead–acid battery, the cell reactions at
the cathode and anode are
PbO21s2 + 4 H +1aq2 + SO24 - 1aq2 + 2e- S PbSO41s2 + 2 H2O1l2 E ~ = 1.685 V
(11.54)
Pb1s2 + SO24 - 1aq2 S PbSO41s2 + 2e-
E ~ = -0.356 V
(11.55)
respectively, and the overall reaction is
PbO21s2 + Pb1s2 + 2 H2 SO41aq2 ∆ 2 PbSO41s2 + 2 H2O1l2 E ~ = 2.04 V
(11.56)
The arrows in Equations 11.54 and 11.55 would point in the opposite direction, and the
sign of the emfs would be reversed for the charging mode.
Six such cells connected in series are required for a battery that provides a nominal potential of 12 V. The lead–acid battery is very efficient in that more than 90% of
the electrical charge used to charge the battery is available in the discharge part of the
cycle. This means that side reactions such as the electrolysis of water play a minimal
role in charging the battery. However, only about 50% of the lead in the battery is
converted between PbO2 and PbSO4. Because Pb has a large atomic mass, this limited
convertibility decreases the power per unit mass figure of merit for the battery. Parasitic side reactions also lead to a self-discharge of the cell without current flowing in
the external circuit. For the lead–acid battery, the available energy is diminished by
approximately 0.5% per day through self-discharge.
As batteries have become more common in portable devices such as cell phones
and laptop computers, energy density is a major criterion in choosing the most suitable
11.12 Electrochemistry of Commonly Used Batteries
Figure 11.8
Smaller
Energy density for a number of different battery types. Battery types are
classified according to their specific energy
density per unit volume and per unit mass.
Ni-MH is a shorthand version of nickel–
metal hydride. Wh is an abbreviation for
watt hours.
800
Energy density/volume (Wh/L)
323
700
600
Li-ion
500
AA Alkaline
400
Ni-MH
Li-metal
300
200
Ni-Cd
100
Lead-Acid
0
Lighter
0
500
300
400
100
200
Energy density/mass (Wh/kg)
battery chemistry for a specific application. Figure 11.8 shows a comparison of different battery types. The lead–acid battery has the lowest specific energy in terms of either
volume or mass. Next we discuss the chemistry of three commonly used rechargeable
batteries: the alkaline, nickel–metal hydride, and lithium-ion batteries.
The individual elements of the alkaline cell are shown in Figure 11.9. The anode in
this cell is powdered zinc, and the cathode is in the form of a MnO2 paste mixed with
powdered carbon to impart conductivity. KOH is used as the electrolyte. The anode
and cathode reactions are
Anode: Zn1s2 + 2 OH -1aq2 S ZnO1s2 + H2O1l2 + 2e-
E ~ = 1.1 V
Concept
Energy density per volume and per
mass are useful criteria for evaluating
different batteries.
(11.57)
-
Cathode: 2 MnO21s2 + H2O1l2 + 2e S Mn2O31s2 + 2 OH 1aq2 E ~ = -0.76 V
(11.58)
Positive cover:
plated steel
Can: steel
Metalized
plastic film label
Electrolyte:
potassium
hydroxide/
water
Anode:
powdered zinc
Cathode:
manganese
dioxide,
carbon
Current collector:
brass pin
Separator:
nonwoven
fabric
Seal: nylon
Inner cell cover:
steel
Negative cover:
plated steel
Metal
washer
Metal spur
Figure 11.9
Schematic diagram of an alkaline cell
battery.
324
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Nickel–metal hydride batteries are currently used in hybrid vehicles that rely on dc
motors to drive the vehicle in city traffic and use a gasoline engine for higher-speed
driving. The Toyota Prius uses 28 modules of six cells, each with a nominal voltage of
1.2 V and a total voltage of 201.6 V to power the vehicle. The capacity of the battery
pack is ∼1100 Wh. The anode and cathode reactions are
Anode: MH1s2 + OH -1aq2 S M + H2O1l2 + e-
E ~ = 0.83 V
(11.59)
Cathode: NiOOH1s2 + H2O1l2 + e- S Ni1OH221s2 + OH -1aq2
The electrolyte is KOH1aq2, and the overall reaction is
E ~ = 0.52 V
(11.60)
MH1s2 + NiOOH1s2 ∆ M1s2 + Ni1OH221s2 E ~ = 1.35 V
(11.61)
where M designates an alloy that can contain V, Ti, Zr, Ni, Cr, Co, and Fe.
Lithium-ion batteries find applications as diverse as cell phones, where a highenergy density per unit volume is required, and electric vehicles, where a high-energy
density per unit mass is required. The electrodes in lithium-ion batteries contain Li+
ions, and the cell voltage reflects the difference in the binding strength of Li+ in the
two materials. The structure of a cylindrical lithium-ion battery is shown schematically
in Figure 11.10. The two electrodes are separated by an electrolyte–saturated polymer
membrane through which the Li+ ions move in the internal circuit. The electrolyte is
a lithium salt dissolved in an organic solvent—for example, 1M LiPF6 in a mixture of
ethylene carbonate and diethyl carbonate. Aqueous electrolytes would limit the cell
voltage to 1.2 V because at larger potentials, water is reduced or oxidized. Figure 11.11
shows a number of materials that can be used as electrodes. Materials that fall outside
of the band gap of the electrolyte are unsuitable because their use initiates reduction
or oxidation of the solvent. It would appear that carbon is unsuitable, but the formation of a thin solid-electrolyte interface layer stabilizes carbon with respect to solvent
reactions, and it is the most widely used anode electrode. Using a carbon anode and
LiCoO21s2 as the cathode allows a cell potential of 3.7 V to be achieved. As Figure
11.11 shows, greater potentials are possible, but batteries with long life cycles using
materials other than LiCoO21s2 have not yet been developed.
Cathode
cover
Cathode lead
Separator
Gasket
Figure 11.10
Schematic structure of a cylindrical
lithium-ion battery. The anode and cathode material are formed of thin sheets to
optimize the transport kinetics of Li+ ions.
Anode
container
Cathode
Anode lead
Anode
11.12 Electrochemistry of Commonly Used Batteries
Figure 11.11
Eg of electrolyte
Cell potential of a lithium-ion battery
plotted against the energy density per
unit mass for a number of electrode
materials. The dashed lines indicate the
voltage range in which 1M LiPF6 in a
1:1 mixture of ethylene carbonate and
diethyl carbonate is stable with respect to
reduction or oxidation. The vertical axis
shows the maximum energy that can be
withdrawn from the battery in units of
milliamp hours.
400
Graphite
LiTi2(PS4)3
Capacity (mAh/g)
300
LiTiS2
200
TiS2
VS2
Li3V2(PO4)3
LixMn0.5Ni0.5O2
LiVS2
LiTi2(PO4)3
Li4Ti5O12
100
325
LixNi0.5Mn1.5O4
Li11xMn2O4
LiFePO4
LixCoO2
Source: Adapted from Goodenough J. B., and
Kim Y. Chemistry of Materials 22 (2010): 587.
LiCoPO4
Li12xMn2O4
0
1
0
3
2
Potential (V) vs. Li1/Li0
4
5
Rechargeable lithium batteries have the following half-cell reactions while discharging the battery:
positive electrode: Li1 - x CoO21s2 + xLi+1solution2 + xe- S LiCoO21s2 (11.62)
negative electrode: CLix S C1s2 + xLi+1solution2 + xe-
(11.63)
Li1 - xCoO21s2 + CLix ∆ LiCoO21s2 + C1graphite2 E ~ ∼ 3.7 V
(11.64)
Concept
Lithium ion batteries are used in
many types of devices as dissimilar as
cell phones and electric vehicles.
The right arrows indicate the discharge directions. In these equations, x is a small positive number. The overall cell reaction is
and the fully charged battery has a cell potential of ∼3.7 V. The structures of LiCoO21s2
and CLix are shown schematically in Figure 11.12. CLix designates Li atoms intercalated
between sheets of graphite; it is not a stoichiometric compound. In a lithium-ion battery,
the lithium ions are transported to and from the cathode or anode, with the transition
metal, cobalt (Co), in LixCoO2 being oxidized from Co3+ to Co4+ during charging and
reduced from Co4+ to Co3+ during discharge.
Positive electrode
Negative electrode
Charge
Li1
Co
O
Li
Figure 11.12
Li1
Discharge
LiCoO2
Specialty carbon
Illustration of atomic scale processes
in a lithium battery. The cell voltage
is generated by moving lithium atoms
between a lattice site in LiCoO2 and an
intercalation position between sheets of
graphite.
326
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
11.13 FUEL CELLS
Concept
A fuel cell can be thought of as a
battery in which the reactants are
continually regenerated.
The primary advantage of fuel cells relative to batteries is that they can be continually
refueled and do not require a downtime for recharging. A number of different technologies are used to operate fuel cells. Of these, most are still in the research and development stage; only phosphoric acid fuel cells are available as off-the-shelf technology.
Figure 11.13 shows a number of different types of fuel cells. All use O2 or air as
the oxidant, and the fuel can be either pure H2 or H2 produced on board by reforming
hydrocarbons, methanol, or glucose. For the fuel cells shown in Figure 11.13, the
operating temperature and available power output are shown in Figure 11.14.
The remainder of this section will concentrate on the most widely understood fuel
cell technology, the proton exchange membrane fuel cell (PEMFC). Proton exchange
membrane fuel cells that use H21g2 and O21g2 as reactants were originally employed
in the NASA Gemini space flights of the 1960s. The principles underlying this technology have not changed in the intervening years. However, advances in technology have
significantly increased the power generated per unit weight and per unit area of the electrodes. A schematic drawing of a single PEMFC and the relevant half-cell reactions is
shown in Figure 11.15. The anode and cathode have channels that transport H2 and O2 to
the catalyst and membrane. An intermediate diffusion layer ensures uniform delivery of
reactants and separation of gases from the water formed in the reaction. The membrane
is coated with a catalyst (usually Pt or Pt alloys) that facilitates the H + formation and
the reaction between O2 and H +. Individual units can be arranged back to back to form
a stack so that the potential across the stack is a multiple of the individual cell potential.
The heart of this fuel cell is the proton exchange membrane, which functions as a solid
electrolyte. This membrane facilitates the passage of H + from the anode to the cathode in
the internal circuit, and it also prevents electrons and negative ions from moving in the
opposite direction. The membrane must be thin 1∼10-100 mm2 to allow rapid charge
transport at reasonably high current densities and must be unreactive under the potentials
present in the cell. The most widely used membranes are polymeric forms of perfluorosulfonic acids. These membranes are quite conductive if fully hydrated. The membrane
structure consists of spherical cavities, or inverted micelles ∼4 nm in diameter, connected
by cylindrical channels ∼1 nm in diameter. The interior of the cavities and channels is
lined with SO3- groups. The -CF2 – backbone of the polymer provides rigidity, makes the
membrane unreactive, and increases the acidity of the terminal SO3- groups. The positively
Types of fuel cells
CO2
H1
A E C
e2
A E C
e2
A E C
e2
H2O
H2
H1
H2O
H2O
O2
O2
PEM
PEM
CH3OH
Direct methanol fuel cell
Polymer electrolyte fuel cell
Proton exchange membrane fuel cells (PEMFC)
A E C
e2
H2
H1
H2PO4
C6H10O6
C6H12O6
O2
Phosphoric acid fuel cell
CO2
H2O
CO322
Na2
HPO4
H2O
O2
Enzymatic fuel cell
A E C
e2
H2O
H1
A E C
e2
A E C
e2
O2
CO2
H2
X2CO3 (X5Li,Na,K)
CO
Molten carbonate fuel cell
CO2
H2O
O22
O2
H2
YSZ
CO
Solid oxide fuel cell
H2O
H2
OH2
O2
H2O
KOH
Alkaline fuel cell
Legend of
components:
A 5 anode
E 5 electrolyte
C 5 cathode
PEM 5 proton exchange
membrane
YSZ 5 yttria-stabilized
zirconia
Figure 11.13
Schematic diagram of the most common types of fuel cells. All use a combination of O2 with
either H2 or organic molecules.
11.13 Fuel Cells
Implantable
devices
(mW)
Portable
devices
(1–50 W)
Auxillary
power unit
(1–10 kW)
Traction
(<300 kW)
Combined heat &
power generation
(kW–MW)
Power plants
(MW–GW)
Typical
operating
temperature (°C)
EFC
AFC
DMFC
PEFC
PAFC
MCFC
SOFC
35
60
80
<160
200
620
<1000
Figure 11.14
The optimal temperature of operation and application potential of different types of fuels
cells. Fuel cell type abbreviations are as follows: EFC, enzymatic fuel cell; AFC, alkaline fuel
cell; DMFC, direct methanol fuel cell; PEFC, polymer electrolyte fuel cell; PAFC, phosphoric
acid fuel cell; MCFC, molten carbonate fuel cell; and SOFC, solid oxide fuel cell. For more
details, see citation to this author in Further Reading.
Source: Adapted from Sundermacher, K. Industrial Engineering Chemical Research 49 (2010): 10159.
charged H + can migrate through this network under the influence of the electrical potential
gradient across the membrane that arises through the half-cell reactions. By contrast, negatively charged species in a cavity cannot migrate through the network because they are
repelled by the negatively charged SO3- groups that are covalently bonded to long polymer
chains in the narrow channels that connect adjacent cavities.
The electrochemistry of the two half-cell reactions shown in Figure 11.15 is well
understood. The main challenges in manufacturing a fuel cell based on the overall
reaction H2 + 1>2 O2 S H2O lie in maintaining a high cell potential at the necessary
current output. A number of strategies have been pursued to improve the performance
of proton exchange membrane fuel cells. Optimizing the interface between the gas diffusion element and the membrane has been the most important area of progress. The
use of porous and finely divided materials allows the electrode surface area to be maximized per unit volume of the fuel cell. Careful integration of the catalyst and membrane
surface also allows the amount of the expensive Pt-based catalysts to be minimized.
A future goal is to develop operating conditions that do not require highly pure H2.
Currently, high purity is required because trace amounts 1∼50 ppm2 of CO poison the
Pt catalyst surface and render it inactive for H2 dissociation. Fuel cells that operate at
temperatures above 100°C are much less sensitive to poisoning by CO. However, operation at these temperatures requires proton exchange membranes that use ionic liquids
other than water to maintain a high conductivity.
327
328
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Porous gas
diffusion layer
Figure 11.15
Schematic diagram of a proton exchange
membrane fuel cell. The half-cell reactions
are shown below the figure. The channels
in the anode and cathode facilitate the
supply of O2 and H2 to the cell and carry
away the H2O reaction product. The gas
diffusion layer ensures that the reactants
are uniformly distributed over the membrane surface.
Flow field
Catalyst layer
Membrane
H2O
e2
e2
H1
Anode
Cathode
H2
H2
O2
2H1 1 2e2
1/
2O2
1 2H1 1 2e2
H2O
The goals of much current research on fuel cells are to develop inexpensive catalysts that are not based on noble metals and that use fuels other than pure hydrogen.
Because gaseous H2 has a low-energy density per unit volume, it is not an economically
useful fuel unless it is produced at the site of the fuel cell using renewable techniques.
Reforming fuels, such as methane to produce hydrogen or direct methanol, oxidation
are the most promising avenues to pursue. For a more detailed discussion of the different types of fuel cells, see the article by Sundmacher (2010) cited in Further Reading.
SU P P LE ME NTAL
SE CTION
11.14 ELECTROCHEMISTRY AT THE
ATOMIC SCALE
The half-cell reaction Cu2+1aq2 + 2e- S Cu1s2 describes a macroscopic scale process.
What is known about this reaction at an atomic scale? In this section, we will briefly
investigate this topic. For a more detailed discussion, see the article by Kolb (2002)
cited in Further Reading.
To investigate atomic-scale processes, we will first describe the structure of the
interface between the solution and a metal electrode in an electrochemical cell (see
Figure 11.16a), which has the structure of a capacitor and is known as the electrical
double layer. In Figure 11.16b, the positively charged plate of the capacitor is the metal
electrode, and the negative plate consists of negative ions that are surrounded by their
solvation shells. The two types of negative ions at the interface are distinguished by
the forces that hold them in this region. Specifically bound ions are those that form a
chemical bond with one or more of the metal atoms at the surface of the electrode.
Examples of specifically bound ions are Cl- and Br -. It is advantageous for these ions to
form bonds with the surface because the water molecules in their solvation shell are not
as strongly bound as positive ions such as Na+ and K+. The part of the solvation shell
directed toward the electrode atoms, to which the ion is bonded, is missing. The plane
that passes through the center of the specifically adsorbed ions is known as the inner
Helmholtz plane. The second type of ion that occurs close to the positively charged
electrode consists of fully solvated ions, which are called nonspecifically adsorbed ions.
The plane that passes through the center of the nonspecifically adsorbed ions is known
as the outer Helmholtz plane. Outside the electrical double layer, ions are fully solvated.
11.14 Electrochemistry at the Atomic Scale
E
329
Figure 11.16
Outer Helmholtz plane (OHP)
Schematic visualization of the processes
at the interface between an aqueous
solution and a metal electrode in an
electrochemical cell. (a) A beaker with an
electrochemical cell. The circled region is
greatly magnified in part (b). (b) The electrical double layer, showing specifically
and nonspecifically adsorbed ions and
water molecules and the inner and outer
Helmholtz planes. The blue solid line
passes through the center of a nonspecifically adsorbed ion, and the dashed line
passes through the center of a specifically
adsorbed ion.
Inner Helmholtz plane (IHP)
Source: Adapted from Kolb, D. M. Surface
Science 500 (2002): 722.
(a)
1
2
Specifically
adsorbed
ion
1
Water
molecule
1
Metal
Electrolyte
Normal water
structure
2
Fully
solvated ion
1
(b)
The falloff in the electrical potential between the metal electrode and the electrolyte solution occurs in a very small distance of approximately 3 nm, resulting in an
electrical field in the electrical double layer as large as 3 * 107 V cm-1. Because the
charge that can be accommodated at the surface of the electrode by the specifically and
nonspecifically adsorbed ions is on the order of 0.1–0.2 electron per atom of metal at
the surface, the electrical double layer has a very large capacitance of 20-50 mC cm-2.
This property makes the metal–electrolyte interface useful in supercapacitors, which
can provide energy storage at a very high-energy density. For a typical difference in
potential between the electrode and the solution of ∼1 V, the energy stored in the
interfacial capacitance can be as large as 150 kJ kg -1. Because this density is so large,
electrochemical capacitors can be used to provide backup electrical energy in the event
of a power failure in electronic devices. All important aspects of an electrochemical
reaction occur in this region immediately adjacent to the electrode surface.
Having described the structure of the solution adjacent to the electrode surface, what
can we now say about the structure of the electrode surface at an atomic scale? Much of
what is known about the structure of the electrode surface at an atomic scale has been
obtained using an instrument called a scanning tunneling microscope (STM). In an
STM, a metal tip is positioned within 0.5–2 nm of a conducting surface, and at these distances, electrons can tunnel between the tip and surface. The tunneling current decreases
330
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Figure 11.17
Schematic illustration of a scanning
tunneling microscope (STM). The
piezoelectric elements labeled x and y are
used to scan the tip parallel to the surface,
and the z piezoelectric element is used to
vary the tip–surface distance. The inset
shows a greatly magnified image of the
region between the end of the tip and the
surface. The dashed line is a cut through a
contour map of the surface resulting from
the measurement.
Piezoelectric
ceramics
0.5 – 2 nm
z
y
x
3107
Tunneling
tip
Concept
Electrochemical reactions on metal
electrodes can be studies at the
atomic level using the scanning
tunneling microscope.
exponentially as the tip–surface distance increases. Using piezoelectric elements, we can
vary the tip height as the tip is scanned over the surface to keep the tunneling current at
a constant value, typically 1 nA. The voltage applied to the piezoelectric element normal
to the surface is a direct measure of the height of the surface at a point parallel to the
surface. These voltage values are used to construct a topographical map of the surface. A
schematic picture of the essential elements of an STM is shown in Figure 11.17.
The structure of an electrode surface in an electrochemical environment depends
on the possible reactions that the electrode can undergo in the solution and the applied
potential. For most metals, it is possible by careful preparation to obtain an electrode
that is virtually identical to what is expected from terminating the crystal structure of
the electrode material. An example of this ideal case is shown in Figure 11.18 for a
metal electrode surface. The great majority of metal atoms are in their ideal lattice
positions. A much smaller number are located at the steps that form the edges of large
terraces that compose the long-range structure of the surface. Although the step atoms
are a small minority of the total number of surface atoms, they play a significant role in
electrochemical processes, as we will next discuss.
We now turn to the influence of surface defects in electrochemical reactions.
If an Au electrode is immersed in a CuSO4 solution, and the potential is adjusted
appropriately, Cu will be deposited on the electrode, as described by the reaction
Cu2+1aq2 +2e- S Cu1s2. How does the Cu layer grow on the Au surface? The answer
Atoms in
topmost layer
Atoms in underlying
atomic layer
Figure 11.18
Details of the surface of a metal electrode. Depiction of atomic layer structure
at a metal electrode surface. The surface
consistes of wide terraces separated by
one atom high steps.
Atomic layer structure depicting individual
atoms as spheres.
11.15 Using Electrochemistry for Nanoscale Machining
Growing row of deposited copper
Cu deposits are anchored
at step edges of the
underlying gold electrode.
Scan direction
is provided by the STM image shown in Figure 11.19. The initial stage of the Cu film
is the formation of small Cu islands that are exclusively located at step edges of the
underlying Au surface. This result can be understood by realizing that the coordination
number of a Cu atom at a gold site at the step edge is larger than on a flat portion of
the surface. As more copper is deposited, the initially formed islands grow laterally and
eventually merge to form a uniform layer.
From these studies, several important conclusions can be made concerning the growth
of an electrochemically deposited metal film. The fact that island growth is observed rather
than random deposition of Cu atoms indicates that the initially adsorbed Cu atoms have a
high mobility parallel to the surface. They diffuse across a limited region of the surface until
they encounter the edge of an island. Because these edge sites have a higher coordination
number than a site on the terrace surface, they are strongly bound and diffuse no further.
Another conclusion that can be drawn from the imaging studies is that the metal film grows
one layer at a time. A second layer does not nucleate until the underlying layer is nearly
complete. Layer-by-layer growth ensures that a compact and crystalline film is formed.
As deposition of Cu on a gold electrode shows, step edges play a central role in determining the formation of electrochemically deposited layers. Although small in number, step edges or surface defects such as vacancies play a major role in a wide range of
chemical processes including catalytic cracking of crude oil and the corrosion of metals.
Electrochemical scanning tunneling microscopy has also provided new insight into
the atomic-level processes underlying the electrochemical dissolution of a metal electrode. To carry out this type of study, an STM was constructed that allowed 25 images
to be obtained per second. A comparison of images obtained at closely spaced times
allows researchers to follow the ongoing processes. For more details, see the citation to
Magnussen et al. in Further Reading. A mode of growth was observed in which individual atoms were added to single rows at the edge of a terrace, as shown in Figure 11.20
when the electrical potential is changed suddenly to favor copper deposition.
331
Au
Cu
170 nm 3 170 nm
Figure 11.19
STM image of a gold electrode with
small amounts of deposited copper.
Electrochemically deposited copper
appears as the bright, roundish, small
islands. The triangular features are Au
islands on a large flat Au terrace. The
onset of the deposition is indicated by
the horizontal arrow in the figure. At
that point, the electrode potential was
increased to a value that allows the
reaction Cu2+1aq2 + 2e- S Cu1s2 to
proceed. Lower portions of the figure
are imaged at successively longer times.
Source: Adapted from Kolb et al. Angewandte
Chemie International Edition 39 (2000): 1123.
Figure 11.20
Time-lapse STM imaging (interval of
100 ms) of electrochemically deposited
copper on a copper electrode. Note that
the uppermost layer grows rapidly and
in an ordered fashion.
Source: Adapted from Magnussen et al,
Faraday Discussions 121 (2002):43–52.
S U PPL EM EN TAL
SEC TI O N
11.15 USING ELECTROCHEMISTRY
FOR NANOSCALE MACHINING
Many current technologies require fabrication of miniature devices with dimensional
control of the micron to nanometer scale. For example, lithographic methods are used
to make the masks employed in the fabrication of integrated circuits. Until recently,
electrochemical machining techniques have not been able to provide dimensional control better than 10 mm.
332
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
Figure 11.21
Possible current paths between a tool
and an electrode immersed in an electrolyte solution. Before a reaction can
proceed, the capacitance of the electrical
double layer must be charged to its maximum value.
Source: Adapted from A. L. Trimmer et al.,
Applied Physics Letters 82, No. 19 (2003): 3327.
Electrochemical
double layer
Tool
Current
distribution
1
2
Potential
source
Substrate
Concept
Nanoscale machining of metals can
be carried out in an electrochemical
environment using short voltage
pulses.
1
Below tool
V/V0
0.8
0.6
0.4
Away from tool
0.2
2
Figure 11.22
4
6
t/RCtool
8
10
The ratio V , V0 plotted as a function
of t , RCtool for two points relative to a
metal tool in an electrolyte solution.
One point is immediately below the tool,
and the other is at a point where the
resistance along the path is greater by a
factor of 10 compared to that immediately
below the tool. The light blue bar indicates the voltage range in which the metal
dissolution reaction is initiated.
If the value of external voltage is changed in an electrochemical cell, how long
does it take for the cell to reach equilibrium with respect to the potential distribution?
As discussed in the previous section, the electrical double region can be modeled as a
large capacitor that must be charged up to the value of the new potential as the external
voltage is changed. The current must flow through the electrolyte solution, which has
an ohmic resistance, and the cell can be modeled as a series R-C circuit, with R equal to
the electrical resistance of the solution and C equal to the capacitance of the electrical
double layer.
If a metal tool is placed very close to an electrode, the resistance associated with
a current path between the tool and the electrode depends on the path, as shown in
Figure 11.21. Those paths immediately below the tool correspond to the smallest electrical resistance. It can be shown that the time-dependent voltage across the capacitor in
an R-C series circuit is given by
V1t2 = V011 - e-t>RC2
(11.65)
The rate at which V>V0 increases for a point on the surface directly below the tool
and a point well to one side of the tool is shown in Figure 11.22. Below the tool, the
potential is built up to its full value at much shorter times than outside of this region. Until
the potential is built up to the value required for an electrochemical reaction of interest,
the reaction will not proceed. Trimmer et al. have used this fact to supply the external
voltage in the form of a short pulse, rather than as a dc voltage. As a consequence, the
reaction of interest occurs directly under the tool, and only to a negligible extent outside
of this region. For example, if the dissolution reaction in which material is removed
from the work piece only takes place at potentials higher than that indicated by the horizontal bar in Figure 11.22, it will only take place under the tool for a short enough pulse.
Note that if a dc potential is applied instead of a pulse, the reaction will not be localized.
For more details, see the citation for Trimmer, Hudson, Kock, and Schuster (2003) in
Further Reading.
Examples of the patterns that have been generated using the procedures described
above by metal dissolution are shown in Figure 11.23. A resolution of ∼20 nm has
been obtained by using pulses of 200-ps duration.
Key Equations
333
Figure 11.23
Nanoscale machining using electrochemical dissolution techniques.
Images obtained by STM techniques.
The machining process performed in an
HCl solution. (a) Tip of electrochemical
machining tool. (b) Results produced by
the tool shown in (a).
Source: Adapted from Trimmer, A. L., et al.
Applied Physics Letters 82 (2003): 3327.
(a) Tip of electrochemical machining tool.
(b) Results produced by the tool shown
in (a) for different pulse lengths
(values given in ns on image).
Note improved resolution with
shorter pulse lengths.
VOCABULARY
anode
cathode
Daniell cell
electrical double layer
electrochemical cell
electrochemical potential
electrochemical series
electromotive force (emf)
Faraday constant
fuel cell
half-cell potential
junction potential
Nernst equation
proton exchange membrane fuel cell (PEMFC)
reference electrode
salt bridge
KEY EQUATIONS
Equation
Significance of Equation
∼ = m + zfF
m
Definition of electrochemical potential
11.4
Condition for equilibrium in electrochemical cell
11.7
Standard state for ions in solution
11.8
Standard state for electrons in electrode
11.9
Potential of hydrogen electrode
11.13
Relation between cell emf and Gibbs energy of reaction
11.23
Nernst equation; allows calculation of the emf for an electrochemical cell
11.26
Relation between cell and half-cell potentials
11.32
Relation between reaction entropy and temperature dependence of cell potential
11.35
Relation between standard state cell potential and equilibrium
constant
11.36
∼ =
∼
∆m
a ni mi = 0, rather than ∆m = a ni mi = 0
i
i
∼ = m 1ions in solution2
m
i
i
∼ - = - fF (electrons in metal electrode2
m
e
f H +>H2
1
1>2
mH~ + - mH~ 2
2
RT ƒH2
=
ln
aH +
F
F
∆ rG = -nF∆f = -nFE
E = E~ -
RT
ln Q
nF
~
~
~
E cell
= E red
+ E ox
∆ rS ~ = - a
E~ =
0∆ rG ~
0E ~
b = nF a
b
0T
0T P
P
RT
ln K
nF
Equation Number
334
CHAPTER 11 Electrochemical Cells, Batteries, and Fuel Cells
CONCEPTUAL PROBLEMS
Q11.1 To determine standard cell potentials, measurements are
carried out in very dilute solutions rather than at unit activity.
Why is this the case?
Q11.2 Show that if ∆ f G ~ 1H +, aq2 = 0 for all T, the potential of the standard hydrogen electrode is zero.
Q11.3 Explain why the magnitude of the maximum work
available from a battery can be greater than the magnitude of
the reaction enthalpy of the overall cell reaction.
Q11.4 How can one conclude from Figure 11.20 that Cu atoms
can diffuse rapidly over a well-ordered Cu electrode in an electrochemical cell?
Q11.5 The temperature dependence of the potential of a cell
is very small. What does this tell you about the thermodynamics of the cell reaction?
Q11.6 Why is the capacitance of an electrolytic capacitor so
large compared with conventional capacitors?
Q11.7 What is the difference in the chemical potential and
the electrochemical potential for an ion and for a neutral species in solution? Under what conditions is the electrochemical
potential equal to the chemical potential for an ion?
Q11.8 Why is it possible to achieve high-resolution electrochemical machining by applying a voltage pulse rather than a
dc voltage to the electrode being machined?
Q11.9 Why is it not necessary to know absolute half-cell
potentials to determine the emf of an electrochemical cell?
Q11.10 What is the voltage between the