CHAPTER 14
Thermodynamics:
Spontaneous Processes,
Entropy, and Free Energy
CHEMISTRY
THIRD EDITION
Gilbert | Kirss | Foster | Davies
© 2012 by W. W. Norton & Company
Chapter Outline
 14.1 Spontaneous Processes and Entropy
» What Does “Spontaneous” Mean?
» Statistical Entropy and Microstates
» A Mathematical View of Entropy Microstates
 14.2 Thermodynamic Entropy
 14.3 Absolute Entropy, the Third Law
of Thermodynamics, and Structure
 14.4 Calculating Entropy Changes
 14.5 Free Energy and Free-Energy Change
 14.6 Driving the Human Engine: Coupled Reactions
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Spontaneous Processes
 Spontaneous process:
• One that proceeds in a given direction without
outside intervention.
 Nonspontaneous process:
• Occurs only as long as energy is added to the
system.
 Spontaneity depends on dispersion of
energy that occurs during a process.
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Spontaneity and Enthalphy (ΔH)
 Although many spontaneous processes are
exothermic (e.g., combustion), not true for all
spontaneous reactions:
 Endothermic (ΔH > 0), but reaction is
spontaneous.
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Thermodynamics: Entropy
 Second Law of Thermodynamics:
• The total entropy of the universe increases
in any spontaneous process.
 Entropy (S):
• A measure of the distribution of energy
in a system at a specific temperature.
• Energy distribution affected by molecular
motion, volume.
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Entropy and Microstates
 The motion of molecules is quantized:
• Different molecular states related to molecular
motion are separated by specific energies.
 Energy state or energy level:
• An allowed value of energy.
 Microstate:
• A unique distribution of particles among
energy levels.
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Types of Molecular Motion
 Three types of motion:
• Translational—movement
through space.
• Rotational—spinning motion
around axis  to bond.
• Vibrational—movement of
atoms toward/away from each
other.
 As temperature increases, the
amount of motion increases.
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Quantized Energy States
Quantized vibrational
states of O2 molecule.
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Quantized rotational
states.
Microstates: Energy Distribution
 The number of times
a molecule occupies
an accessible energy
level follows a
Boltzmann distribution:
• Number of energy
states increases as
volume increases.
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Microstates: Energy
Distribution (cont.)
 Increasing volume
“stretches” the energy
distribution:
• Accessible energy
levels move closer
together.
• Number of accessible
energy states
increases.
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Statistical Entropy
 Boltzmann Equation:
S = k ln W
• S = entropy
• W = # of microstates.
• k = Boltzmann constant (1.38 × 10-23 J/K)
= R/NA
 This equation indicates that entropy
increases as the number of microstates
increases.
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Chapter Outline
 14.1 Spontaneous Processes and Entropy
 14.2 Thermodynamic Entropy
» Isothermal Processes
» Entropy Changes for Some Common Processes
 14.3 Absolute Entropy, the Third Law
of Thermodynamics, and Structure
 14.4 Calculating Entropy Changes
 14.5 Free Energy and Free-Energy Change
 14.6 Driving the Human Engine: Coupled Reactions
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Thermodynamic Entropy
 Isothermal Process:
• A process that takes place at constant T.
 Reversible Process:
• A process that can be run in the reverse
direction with no net heat flow into or out of
the system.
 For an isothermal process: ΔSsys= qrev/T
• qrev = flow of heat for reversible process.
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Entropy of the Universe
 3rd Law: “Entropy of the universe increases
for a spontaneous process.”
• ΔSuniv = ΔSsys + ΔSsurr
 Consider an ice cube melting at 0.0°C (273 K)
on a counter top at room temperature (293 K).
• Heat flows into ice cube: ΔSsys = qrev/273
• Heat flows out of counter top: ΔSsurr = −qrev/293
• ΔSuniv= (qrev/273) + (−qrev/293) > 0 spontaneous!
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Entropy of Cold Packs
 ΔHsoln = positive, (heat
lost by surroundings →
ΔSsurr = negative).
 ΔSsys = positive, due
to increased freedom of
movement of dissolved
ions.
 Net: ΔSuniverse > 0
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Spontaneity and Entropy
 If ΔSsys and ΔSsurr > 0, then ΔSuniv > 0.
 If ΔSsys and ΔSsurr < 0, then ΔSuniv < 0.
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Spontaneity and Entropy (cont.)
 If ΔSsys < 0 but ΔSsurr > 0:
• if │ΔSsys│< │ ΔSsurr │, then ΔSuniv > 0.
• if │ΔSsys│> │ ΔSsurr │, then ΔSuniv < 0.
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Spontaneity and Entropy (cont.)
 If ΔSsys > 0 but ΔSsurr < 0:
• if │ΔSsys│> │ ΔSsurr │, then ΔSuniv > 0.
• if │ΔSsys│< │ ΔSsurr │, then ΔSuniv < 0.
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Chapter Outline
 14.1 Spontaneous Processes and Entropy
 14.2 Thermodynamic Entropy
 14.3 Absolute Entropy, the Third Law
of Thermodynamics, and Structure
» Entropy and Temperature.
» Standard Entropies: The 3rd Law
of Thermodynamics
 14.4 Calculating Entropy Changes
 14.5 Free Energy and Free-Energy Change
 14.6 Driving the Human Engine: Coupled Reactions
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Entropy and Temperature
 Entropy increases as temperature
increases.
• Increases in kinetic energy increase
the number of accessible microstates.
 Decreasing temperature decreases
entropy.
 At what temperature does all molecular
motion cease, and entropy equal zero?
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Third Law of Thermodynamics
 Third Law of Thermodynamics:
• The entropy of a perfect crystal is zero at absolute zero.
 Absolute Entropy:
• The entropy of a substance at some temperature
above 0 Kelvin.
• Calculated from measurement of molar heat capacities
as a function of temperature.
 Standard Molar Entropy (S°):
•
The absolute entropy of 1 mole of a substance in its
standard state at 298 K and 1 bar of pressure.
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Standard States
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Selected Standard Molar
Entropy Values
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Trends in Entropies
Ssolid < Sliquid < Sgas
Example:
H2O at 298 K
H2O(l) = 69.9 J/(mol∙K)
H2O(g) =188.8 J/(mol∙K)
ΔSvap = 109 J/(mol∙K)
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Other Trends
 Entropy increases as the complexity
of molecular structure increases.
• More bonds, more opportunities for internal
motion (more microstates).
S° (J/(mol∙K):
186
230
270
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310
Entropy Differences in Allotropes
Rigid network of covalent More flexibility and range
bonds; less range of
of motion in planar rings
motion = less entropy.
= more entropy.
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Factors Affecting Entropy
 Temperature:
• Entropy increases as T increases.
 Volume:
• Entropy increases as volume increases.
 Number of particles:
• Entropy increases as the number of particles
increases.
 Can use these observations to predict entropy
changes associated with reactions.
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Chapter Outline
 14.1 Spontaneous Process and Entropy
 14.2 Thermodynamic Entropy
 14.3 Absolute Entropy, the Third Law
of Thermodynamics, and Structure
 14.4 Calculating Entropy Changes
» Entropy Changes for Reactions: ΔSsystem
» Entropy Changes for Surroundings: ΔHsystem
 14.5 Free Energy and Free-Energy Change
 14.6 Driving the Human Engine: Coupled Reactions
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Calculating Entropy Changes
 Entropy change for the system (ΔSrxn°):
ΔSrxn° = ΣnproductsSproducts° − ΣnreactantsSreactants°
• Where n = coefficients of the products/reactants
in the balanced equation.
 Entropy change for the universe:
• Heat gained/lost by system affects entropy
of the surroundings.
• ΔSsurr° = −qsys / T
(qsys = ΔHrxn°)
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Practice: Calculating ΔS°
Using values from Appendix 4, calculate the
standard entropy change for the reaction:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
•
•
•
•
Collect and Organize:
Analyze:
Solve:
Think about It:
© 2012 by W. W. Norton & Company
Chapter Outline
 14.1 Spontaneous Processes and Entropy
 14.2 Thermodynamic Entropy
 14.3 Absolute Entropy, the Third Law
of Thermodynamics, and Structure
 14.4 Calculating Entropy Changes
 14.5 Free Energy and Free-Energy Change
» The Meaning of Free Energy
» Calculating Free-Energy Changes
» Temperature and Spontaneity
 14.6 Driving the Human Engine: Coupled Reactions
© 2012 by W. W. Norton & Company
Free Energy
 Free energy (G):
• The energy available to do useful work.
 Free energy change (ΔG):
• The change in free energy of a process
occurring at constant temperature and
pressure.
• For spontaneous processes, ΔG < 0.
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ΔG and Entropy of the Universe
 For a spontaneous process:
• Suniverse = Ssys + Ssurroundings > 0
 Change in S of system = ΔSsys
 Change in S of surroundings = −ΔHsys/T
 Change in S of universe:
• ΔSuniverse = ΔSsys − ΔHsys/T > 0
• Rearrange:
» −TΔSuniverse = ΔHsys − TΔSsys (ΔG = −TΔSuniverse)
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Driving Forces for Spontaneous
Chemical Processes
• The formation of low energy products:
• i.e., exothermic processes; ΔHrxn < 0
• The formation of products that have
greater entropy than the reactants:
• ΔSrxn > 0
• Free energy (G) relates enthalpy, entropy,
and temperature for a process:
• G = H − TS or ΔG = ΔH − TΔS
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Effects of ΔH, ΔS, and T on ΔG
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Free Energy
 Free Energy = energy available to do
useful work (w).
• Change in internal energy of system can be
used to perform work: ΔE = q + w
• Some energy lost as heat to the
surroundings (q).
• Some energy lost due to entropy (−TΔS).
 Efficiency = (work done)/(energy
consumed).
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Calculating Free-Energy Changes
 Standard Free Energy of Formation (ΔGf°):
• Change in free energy associated with formation
of 1 mole of a compound from its constituent
elements.
• Standard free energy of formation of most stable
form of element in standard state = 0.
 Free-energy change for a reaction:
• ΔGrxn° = Σ(Δnprod.ΔGf, prod.°) − Σ(Δnreact.ΔGf, react.°)
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Practice: Free-Energy Change
Calculate the free-energy change for the
following reaction using the ΔGf° values in
the appendix:
C12H22O11(s) + 12O2 (g) → 12CO2 (g) + 11H2O (l)
•
•
•
•
Collect and Organize:
Analyze:
Solve:
Think about It:
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Temperature and Spontaneity
Consider:
H2O(s) → H2O(l) at 273 K
ΔH = positive.
ΔS = positive.
ΔG = ΔH − TΔS
= negative above a
certain temperature, when
TΔS > ΔH.
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Practice: Transition Temperature
Given the data below, calculate the
temperature at which the reaction becomes
spontaneous: N2(g) + 3H2(g) ⇌ 2NH3(g)
(ΔH° = −92 kJ/mol and ΔS° = −199J/mol)
•
•
•
•
Collect and Organize:
Analyze:
Solve:
Think about It:
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Chapter Outline
 14.1 Spontaneous Processes and Entropy
 14.2 Thermodynamic Entropy
 14.3 Absolute Entropy, the Third Law
of Thermodynamics, and Structure
 14.4 Calculating Entropy Changes
 14.5 Free Energy and Free-Energy Change
 14.6 Driving the Human Engine: Coupled Reactions
» Combining Spontaneous and Non-Spontaneous
Processes
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ChemTour: Dissolution
of Ammonium Chloride
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this ChemTour
This ChemTour identifies the intermolecular forces that contribute to
the overall dissolution of ammonium nitrate in water. The enthalpy
changes that accompany each step in the process are evaluated.
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ChemTour: Gibbs Free Energy
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this ChemTour
Students learn to calculate the maximum potential energy available to do work in a system. An
interactive “Gibbs free energy calculator” allows students to manipulate variables of entropy,
enthalpy, and temperature to explore the effect on ΔG of a reaction. It includes Practice
Exercises.
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Sample Exercise 14.1
Predict whether ΔSsys is positive or negative for each of
these spontaneous isothermal processes:
 Collect and Organize: We are given four reactions and
are to predict whether the entropy change for the system
is positive or negative. We can compare these
processes to those we have studied in detail so far: the
melting of ice and the dissolution of ammonium nitrate in
water. All four reactions are spontaneous, so we know
that ΔSuniv > 0.
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Sample Exercise 14.1 (cont.)
 Analyze:
a. One mole of liquid water molecules becomes one
mole of water vapor molecules, which increases
their freedom of motion.
b. Two moles of gas form one mole of solid, which
decreases the freedom of motion of NH3 and HCl.
c. Two moles of ions in solution form one mole of
solid, which decreases the freedom of motion of the
ions.
d. One mole of a solid dissolves, forming one mole
of molecules dispersed in an aqueous solution,
which increases the molecules’ freedom of motion.
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Sample Exercise 14.1 (cont.)
 Solve:
a. ΔSsys > 0 because molecules of a gas have greater
average kinetic energy and more accessible microstates.
b. ΔSsys< 0 because formation of a solid causes a decrease in
the freedom of motion of the particles and results in a more
ordered (less random) arrangement of the particles than in the
gas phase. The solid has fewer microstates.
c. ΔSsys < 0 because formation of a solid causes a decrease
in the freedom of motion of the particles and results in a more
ordered (less random) arrangement of the particles than in a
solution. The solid has fewer microstates.
d. ΔSsys > 0 because particles in a solution are less ordered
than in the solid, and have access to more microstates.
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Sample Exercise 14.1 (cont.)
 Think about It: Entropy generally increases
when solids melt and when liquids vaporize
because of the increased freedom of motion
of the particles that make up these
substances. Similarly, when solids dissolve,
entropy usually increases, but when a gas
dissolves in a liquid, it loses freedom of
motion and undergoes a decrease in entropy.
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Sample Exercise 14.2
Predict whether each reaction results in an increase or
decrease in the entropy of the system. Assume the
reactants and products are at the same temperature and
pressure.
a. CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(ℓ)
b. NH3(g) + BF3(g) → NH3BF3(s)
 Collect and Organize: We are given two reactions and
are to predict whether they result in an increase or a
decrease in the entropy of the system. The reaction in
each case is the system.
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Sample Exercise 14.2 (cont.)
 Analyze: Entropy increases if temperature,
volume, or the number of independently
moving particles increases. The reactants
and products are at the same temperature,
so temperature is not a factor. We must
evaluate each reaction in terms of volume
and number of particles.
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Sample Exercise 14.2 (cont.)
 Solve:
a. A total of 3 moles of reactants in the solid or liquid
phase (condensed phases) yields 2 moles of products in
the liquid phase (a condensed phase) and, more
significantly from the point of view of entropy, 1 mole of
gaseous product. Recall that molecules in the gas phase
(Figure 14.6c) have many more microstates available
and have large entropies, ensuring that this reaction has
a positive ΔSsys.
b. There are 2 moles of gas in the reactants but no
gaseous products. Fewer microstates are available to
the product (a solid) than to the reactants (gases).
Therefore this reaction results in a loss in entropy by the
system (ΔSsys < 0).
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Sample Exercise 14.2 (cont.)
 Think about It: In Section 14.1 we remarked
that entropy changes are reflected in
changes in the randomness of the system.
When a chemical reaction produces a gas—
as in part a—the gas represents a more
random arrangement (more microstates). The
opposite situation exists in part b, where the
degree of randomness decreases when
gases react to form a solid.
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Sample Exercise 14.3
Calculate ΔS°rxn for the dissolution of ammonium nitrate
(Figure 14.13), given the following standard molar entropy
values:
 Collect and Organize: Entropy changes associated with
a chemical reaction depend on the entropies of the
reactants and products. We are also given the standard
molar entropies for the species involved in the reaction.
The reaction takes place under standard conditions.
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Sample Exercise 14.3 (cont.)
 Analyze: We can calculate ΔS°rxn using the given S°
values and Equation 14.4. Dissolving an ionic solid in
water increases the randomness of the system, so we
predict that ΔS°rxn will be positive.
 Solve:
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Sample Exercise 14.3 (cont.)
 Think about It: As predicted, the value of
ΔS°rxn is positive, so the entropy of the
system increases. Because the number of
independently moving particles increases, we
expect the entropy of the system to increase,
so our answer is logical.
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Sample Exercise 14.4
Consider the reaction of nitrogen gas and hydrogen
gas (Figure 14.15) at 298 K to make ammonia at the
same temperature:
N2(g) + 3 H2(g) → 2 NH3(g)
a. Before doing any calculations, predict the sign of
ΔS°sys.
b. Use data from Table 14.2 to calculate ΔS°sys.
c. Use data from Appendix 4 to calculate ΔSsurr.
d. Is the reaction, as written, spontaneous at 298 K
and 1 bar pressure?
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Sample Exercise 14.4 (cont.)
 Collect and Organize: We are to predict the
sign and calculate the value of the standard
molar entropy change for a reaction. We are
also to calculate the entropy change for the
surroundings and to determine whether the
reaction is spontaneous under the stated
conditions. We are to use the tabulated data
for standard molar entropies found in Table
14.2 and Appendix 4.
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Sample Exercise 14.4 (cont.)
 Analyze:
a. From the balanced equation, we can determine the sign of the
change in the number of moles of gaseous reactants and products,
and from that determine the sign of ΔS°sys.
b. We need to look up S° values for the reactants and products and
use Equation 14.4 to calculate ΔS°rxn, which equals ΔS°sys.
c. To calculate ΔSsurr, we need to know how much heat the reaction
gives off to the surroundings. We can calculate that from the heat of
formation (ΔH°f) values in Appendix 4.
d. We can use Equation 14.3 to calculate ΔSuniv and predict
spontaneity based on its sign: the reaction is spontaneous if ΔSuniv >
0. It is difficult to predict whether this reaction is spontaneous when
reactants and products are in their standard states without knowing
the magnitudes of ΔHsurr and ΔSsurr.
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Sample Exercise 14.4 (cont.)
 Solve:
a. According to the balanced chemical equation, four moles of
gaseous reactants are converted into two moles of gaseous
products. We predict that this decrease in the number of
moles of gases means ΔS°rxn < 0.
b. Using data from Table 14.2 in Equation 14.4 gives
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Sample Exercise 14.4 (cont.)
The entropy change for the reaction is negative, as predicted
in part a.
c. If we consider the reaction to be reversible, we can
calculate the entropy change of the surroundings from the
enthalpy change of the reaction:
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Sample Exercise 14.4 (cont.)
Because we were given the temperature (298 K), we now
have all we need to calculate ΔSsurr. Combining Equations
14.2 and 14.5 (ΔHrxn = qrev), and changing 92.2 kJ to joules so
that the units match, from the perspective of the surroundings,
we have
The surroundings experience an increase in entropy.
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Sample Exercise 14.4 (cont.)
The entropy change of the universe is positive, so the reaction
is spontaneous as written.
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Sample Exercise 14.4 (cont.)
 Think about It: At the start of this exercise,
we were unable to make a prediction about
the spontaneity of the reaction because we
didn’t have an intuitive feel for the
magnitudes of ΔHsurr and ΔSrxn. As long as
the formation of ammonia is exothermic
(ΔHrxn < 0), ΔHsurr and ΔSsurr will be
positive. Since ΔSuniv > 0, ΔSsurr > 0, and
ΔSsys < 0, the reaction is an example of the
situation in Figure 14.9(b).
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Sample Exercise 14.5
The standard free energies of formation for three structural
isomers of the alkane containing eight carbons are shown in
Figure 14.19. All three isomers burn in air by the combustion
reaction
2 C8H18(ℓ) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
Predict whether the ΔG°rxn values for the combustion
reactions of these isomers are all the same or different.
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Sample Exercise 14.5 (cont.)
 Collect and Organize: Structural isomers have
the same molecular formula but different
arrangements of bonds. A combustion reaction
converts all of the carbon in a hydrocarbon to
CO2 and all of the hydrogen to water. We are to
predict whether the ΔG° values for the
combustion of a set of structural isomers are the
same or different.
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Sample Exercise 14.5 (cont.)
 Analyze: The standard free energy of
combustion can be calculated using Equation
14.10; however, there is actually no need to do
the calculation. Because the products of the
reaction are the same for all three isomers,
ΔG°f,products is the same in each combustion
reaction. We predict that the values are related
to the values for ΔG°f for the three isomers.
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Sample Exercise 14.5 (cont.)
 Solve: Let’s use Equation 14.10 to express ΔG° for the
combustion of all three isomers:
Notice that the term for the products is the same in all three
equations. The difference in ΔG°rxn values are determined by
the differences in ΔG°f between the isomers in the reactant
term. Since ΔG°f is different for each isomer, ΔG°rxn will also
be different.
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Sample Exercise 14.5 (cont.)
 Think about It: As predicted, the value of
ΔG°rxn depends on the values for ΔG°f when
we compare structural isomers in a
combustion reaction.
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Sample Exercise 14.6
Calculate ΔG° for the dissolution of 1 mole of
ammonium nitrate in water (total volume = 1 liter) at
298 K, given ΔH°= 28.1 kJ/mol and ΔS° = 108.7
J/(mol · K).
 Collect and Organize: We are to find the standard
free-energy change for NH4NO3(s) dissolving in
water. We are given the change in enthalpy and the
change in entropy for the process under standard
conditions (1 M solution, 1 bar, and 298 K).
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Sample Exercise 14.6 (cont.)
 Analyze: We can use Equation 14.8 for this calculation.
Since the dissolution of ammonium nitrate (a chemical
cold pack) is spontaneous at 298 K (room temperature),
we predict that ΔG° will be negative.
 Solve: Substituting the values of ΔH° and ΔS° for ΔH
and ΔS in Equation 14.8 and converting 108.7 J/(mol ·
K) to kJ allows us to calculate ΔG°:
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Sample Exercise 14.6 (cont.)
 Think about It: As predicted, ΔG° is
negative, which is consistent with a
spontaneous process. A chemical cold pack
is only useful if the dissolution of ammonium
nitrate is spontaneous at ambient
temperature. This sample exercise illustrates
a situation where a positive value for ΔH
(endothermic process) is more than offset by
the increase in entropy, a positive value for
ΔS.
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Sample Exercise 14.7
A certain chemical reaction is spontaneous at
low temperatures but not at high temperatures.
Use Equation 14.8 to determine the sign of the
enthalpy and entropy changes for this reaction.
 Collect and Organize: We are to determine the
signs of ΔH (enthalpy change) and ΔS (entropy
change) based on the change in spontaneity of
a reaction as temperature changes. This means
we need to think about how the signs of ΔH and
ΔS determine how ΔG varies with temperature.
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Sample Exercise 14.7 (cont.)
 Analyze: There are three possible
combinations for ΔH and ΔS that lead to
spontaneous reactions: ΔH < 0 and ΔS > 0;
ΔH > 0 and ΔS > 0; ΔH < 0 and ΔS < 0. Only
one of these combinations of ΔH and ΔS will
satisfy the condition that ΔG > 0 at higher
temperature and ΔG < 0 at lower
temperature.
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Sample Exercise 14.7 (cont.)
 Solve: The importance of ΔS increases with
increasing temperature because the product T ΔS
appears in Equation 14.8.The reaction is
nonspontaneous at higher temperatures, where the
magnitude of T ΔS is more likely to be larger than
the magnitude of ΔH. The reaction is spontaneous at
low temperatures, however, where the impact of a
negative ΔS value is more than offset by a decrease
in enthalpy, a change that favors the reaction. The
reaction must have negative ΔS and negative ΔH
values.
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Sample Exercise 14.7 (cont.)
 Think about It: Table 14.4 confirms our
prediction that a process that is spontaneous
only at low temperatures is one in which
there is a decrease in both entropy and
enthalpy.
© 2012 by W. W. Norton & Company
Sample Exercise 14.8
The body would rapidly run out of ATP if there were not some
process for regenerating it from ADP, and that process is the
hydrolysis of 1,3-diphosphoglycerate4- (1,3-DPG4-) to 3phosphoglycerate3- (3-PG3-) (Figure 14.26):
ADP3- + 1,3-diphosphoglycerate4- → 3-phosphoglycerate3- + ATP4This hydrolysis is spontaneous. Calculate its ΔG° value from these
values:
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Sample Exercise 14.8 (cont.)
 Collect and Organize: We can calculate ΔG° for a
reaction that is the sum of two reactions. If the reactions
in equations (1) and (2) add up to the overall reaction,
then overall ΔG° is the sum of the ΔG° values for the
individual reactions.
 Analyze: First, we add the reactions described by
equations (1) and (2). Assuming that the overall reaction
between ADP and 1,3-diphosphoglycerate4- is the sum
of the reactions describing the hydrolysis of 1,3diphosphoglycerate4- and the phosphorylation of ADP,
we know that the sum of ΔG°1 and ΔG°2 will be less than
zero because the overall reaction is spontaneous.
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Sample Exercise 14.8 (cont.)
 Solve: Summing the reactions in equations (1) and (2)
confirms that they equal the overall reaction:
We sum the ΔG° values for steps 1 and 2 to determine
ΔG° for the overall reaction:
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Sample Exercise 14.8 (cont.)
 Think about It: The hydrolysis of 1,3diphosphoglycerate4- provides more than
sufficient energy for the conversion of ADP
into ATP.
© 2012 by W. W. Norton & Company
Clicker Question: Entropy
of Four Atoms in Two Boxes
Shown to the left are three
possible configurations (A, B,
and C) for placing 4 atoms in
two boxes. Which of the
following processes is
accompanied by the largest
increase in entropy, ΔS?
A) A → B
B) B → C
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C) C → A
Clicker Question: Entropy
of Four Atoms in Two Boxes
Consider the following arguments for each answer and
vote:
A. The entropy change from state A, which has 1 microstate, to
state B, which has 4 microstates, is the greatest.
B. State C is the most probable equilibrium state, so the B → C
transition has the largest entropy.
C. The entropy of the C → A transition is equal to the sum of the
entropies for the A → B and B → C transitions.
© 2012 by W. W. Norton & Company
Clicker Question: Isothermal
Expansion of an Ideal Gas
An ideal gas in a sealed piston is
allowed to expand isothermally
and reversibly against an external
pressure of 1.0 atm. What can
be said of the change in the
entropy of the surroundings,
ΔSsurr, for this process?
A) ΔSsurr > 0
B) ΔSsurr = 0
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C) ΔSsurr < 0
Clicker Question: Isothermal
Expansion of an Ideal Gas
Consider the following arguments for each answer and:
A. The gas is doing work, thereby increasing the entropy of the
surroundings.
B. For a reversible expansion, entropy is constant, so
ΔSsys = ΔSsurr = ΔSuniv = 0.
C. The expansion is isothermal, ΔSsys > 0, and reversible,
ΔSuniv = 0. Therefore, ΔSsurr < 0.
© 2012 by W. W. Norton & Company
Clicker Question:
Gas Expansion into a Vacuum
An ideal gas is expanded
into a vacuum so that q = 0.
Which of the following
statements is true for this
process?
A) ΔEsys < 0
B) ΔGsys < 0
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C) ΔSsys < 0
Clicker Question:
Gas Expansion into a Vacuum
Consider the following arguments for each answer
and vote again:
A. During the expansion, the gas performs work, so the
energy decreases.
B. The expansion of a gas into a vacuum is a
spontaneous process, so ΔGsys is negative.
C. Although the gas volume increases, the temperature
decreases dramatically, thereby reducing the entropy.
© 2012 by W. W. Norton & Company
Clicker Question: Formation
of CH2Cl2 from CH4 and CCl4
Consider the following possible gas phase reaction:
Which of the following is probably true for this reaction?
A) ΔH > 0
B) ΔS > 0
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C) ΔG > 0
Clicker Question: Formation
of CH2Cl2 from CH4 and CCl4
Consider the following arguments for each answer and vote
again:
A. It is energetically more favorable to have all of the same type
of bonds in a molecule than it is to mix and match.
B. There are more microstates for the arrangements of H and
Cl atoms on 2 CH2Cl2 molecules than on a CH4 and a CCl4
molecule.
C. Free energy is required to accommodate formation of the
electric dipole moment on CH2Cl2.
© 2012 by W. W. Norton & Company
Clicker Question: ΔG° for
Condensation of Water at 25° C
What can be said of ΔG° for the
condensation of water vapor,
H2O(g) → H2O(l),
at 25°C if the partial pressure of
H2O(g) is 1.0 atm?
A) ΔG° > 0
B) ΔG° = 0
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C) ΔG° < 0
Clicker Question: ΔG° for
Condensation of Water at 25° C
Consider the following arguments for each answer and vote again:
A. The vaporization of water is spontaneous when the partial pressure
of H2O(g) is 1.0 atm at 25°C. Therefore, ΔG° > 0 for the
condensation of water vapor.
B. At 25°C, condensation will occur spontaneously only when the
partial pressure of H2O(g) rises above the equilibrium partial
pressure of 1.0 atm.
C. The equilibrium partial pressure of H2O(g) is less than 1.0 atm at
25°C, so water vapor will condense spontaneously.
© 2012 by W. W. Norton & Company
Clicker Question:
ΔG° of Vaporization of Ethanol
To the left is a plot of vapor
pressure versus temperature
for the vaporization of ethanol.
C2H5OH(l) → C2H5OH(g).
At which temperature is
ΔG° = 0 for the vaporization of
ethanol at 1.0 atm?
A) > 100°C
B) 100°C
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C) < 100°C
Clicker Question:
ΔG° of Vaporization of Ethanol
Consider the following arguments for each answer and vote
again:
A. Ethanol has a higher molecular mass than water and so
requires more heat for vaporization.
B. The temperature at which ΔG° = 0 for vaporization is 100°C
for all liquids.
C. For ethanol, ΔG° = 0 when the vapor pressure equals 1.0
atm, which occurs at a temperature lower than 100°C.
© 2012 by W. W. Norton & Company
Clicker Question: ΔG° Versus T
for the Sublimation of I2(s)
Which of the following plots
shows the correct relationship
between ΔG° (y-axis) and
temperature (x-axis) for the
sublimation of solid iodine to
iodine vapor at 1.0 atm?
A)
B)
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C)
Clicker Question: ΔG° Versus T
for the Sublimation of I2(s)
Consider the following arguments for each answer and vote again:
A. Once the temperature becomes high enough that the equilibrium
partial pressure of I2(g) is greater than 1.0 atm, the reaction will be
spontaneous.
B. This process becomes less spontaneous at higher temperatures
because more iodine must be vaporized.
C. As the temperature increases and the system approaches
equilibrium, ΔG° will decrease. When the system moves past
equilibrium, ΔG° will begin to increase.
© 2012 by W. W. Norton & Company
Clicker Question:
Spontaneity of Ozone Formation
The formation of ozone, O3(g), from
molecular oxygen is an endothermic
process, with ΔH° = 85 J/mole.
3O2(g) ⇌ 2O3(g)
At what temperatures will the reaction
proceed spontaneously if PO2 = PO3 = 1.0
atm?
A) High temperatures B) Low temperatures C) No temperatures
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Clicker Question:
Spontaneity of Ozone Formation
Consider the following arguments for each answer and vote
again:
A. Because the formation of ozone is an endothermic reaction,
it will be spontaneous only at high temperatures.
B. The formation of ozone will occur only at low temperatures,
where the O2(g) molecules will begin to condense and form
O3(g) molecules.
C. The reaction is endothermic and ΔS° < 0, so at no
temperature can this reaction be spontaneous at 1.0 atm.
© 2012 by W. W. Norton & Company