Lecture PowerPoint
Chemistry
The Molecular Nature of
Matter and Change
Seventh Edition
Martin S. Silberberg
and Patricia G. Amateis
20-1 Copyright  McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.
Chapter 20
Thermodynamics: Entropy, Free Energy,
and the Direction of Chemical Reactions
20-2
Thermodynamics: Entropy, Free Energy, and the
Direction of Chemical Reactions
20.1 The Second Law of Thermodynamics: Predicting
Spontaneous Change
20.2 Calculating Entropy Change of a Reaction
20.3 Entropy, Free Energy, and Work
20.4 Free Energy, Equilibrium, and Reaction Direction
20-3
Spontaneous Change
A spontaneous change is one that occurs without a
continuous input of energy from outside the system.
All chemical processes require energy (activation energy)
to take place, but once a spontaneous process has begun,
no further input of energy is needed.
A nonspontaneous change occurs only if the
surroundings continuously supply energy to the system.
If a change is spontaneous in one direction, it will be
nonspontaneous in the reverse direction.
20-4
The First Law of Thermodynamics Does
Not Predict Spontaneous Change
Energy is conserved. It is neither created nor destroyed,
but is transferred in the form of heat and/or work.
DE = q + w
The total energy of the universe is constant:
DEsys = −DEsurr or DEsys + DEsurr = DEuniv = 0
The law of conservation of energy applies to all
changes, and does not allow us to predict the direction
of a spontaneous change.
20-5
DH Does Not Predict Spontaneous Change
A spontaneous change may be exothermic or endothermic.
Spontaneous exothermic processes include:
• freezing and condensation at low temperatures,
• combustion reactions,
• oxidation of iron and other metals.
Spontaneous endothermic processes include:
• melting and vaporization at higher temperatures,
• dissolving of most soluble salts.
The sign of DH does not by itself predict the direction
of a spontaneous change.
20-6
Figure 20.1
A spontaneous endothermic chemical reaction.
Ba(OH)2·8H2O(s) + 2NH4NO3(s) → Ba2+(aq) + 2NO3–(aq) + 2NH3(aq) + 10H2O(l)
DH°rxn = +62.3 kJ
water
This reaction occurs spontaneously when the solids are mixed. The
reaction mixture absorbs heat from the surroundings so quickly that
the beaker freezes to a wet block.
20-7
Freedom of Particle Motion
All spontaneous endothermic processes result in an
increase in the freedom of motion of the particles in the
system.
solid → liquid → gas
crystalline solid + liquid → ions in solution
less freedom of particle motion
localized energy of motion
more freedom of particle motion
dispersed energy of motion
A change in the freedom of motion of particles in a system
is a key factor affecting the direction of a spontaneous
process.
20-8
Microstates and Dispersal of Energy
• Just as the electronic energy levels within an atom are
quantized, a system of particles also has different
allowed energy states.
• Each quantized energy state for a system of particles is
called a microstate.
– At any instant, the total energy of the system is dispersed
throughout one microstate.
• At a given set of conditions, each microstate has the
same total energy as any other.
– Each microstate is therefore equally likely.
• The larger the number of possible microstates, the larger
the number of ways in which a system can disperse its
energy.
20-9
Entropy
The number of microstates (W) in a system is related to
the entropy (S) of the system.
S = k lnW
A system with fewer microstates has lower entropy.
A system with more microstates has higher entropy.
All spontaneous endothermic processes exhibit an
increase in entropy.
Entropy, like enthalpy, is a state function and is therefore
independent of the path taken between the final and initial
states.
20-10
Figure 20.2
Spontaneous expansion of a gas.
When the stopcock is opened, the gas spontaneously expands to
fill both flasks.
Increasing the volume increases the number of translational
energy levels the particles can occupy. The number of microstates
– and the entropy – increases.
20-11
Figure 20.3
The entropy increase due to the expansion of a gas.
Opening the stopcock increases the number of possible energy
levels, which are closer together on average. More distributions of
particles are possible.
20-12
Figure 20.4
Expansion of a gas and the increase in number of
microstates.
When the stopcock opens,
the number of microstates
is 2n, where n is the
number of particles.
20-13
DS for a Reversible Process
DSsys = qrev
T
A reversible process is one that occurs in such tiny
increments that the system remains at equilibrium, and
the direction of the change can be reversed by an
infinitesimal reversal of conditions.
20-14
Figure 20.5
Simulating a reversible process.
A sample of Ne gas is confined to a
volume of 1 L by the “pressure” of a
beaker of sand on the piston.
We remove one grain of sand (an
“infinitesimal” decrease in pressure),
causing the gas to expand a tiny amount.
The gas does work on its surroundings,
absorbing a tiny increment of heat, q, from
the heat reservoir.
This simulates a reversible process, since
it can be reversed by replacing the grain
of sand.
20-15
The Second Law of Thermodynamics
The sign of DS for a reaction does not, by itself, predict the
direction of a spontaneous reaction.
If we consider both the system and the surroundings, we
find that all real processes occur spontaneously in the
direction that increases the entropy of the universe.
For a process to be spontaneous, a decrease in the
entropy of the system must be offset by a larger
increase in the entropy of the surroundings.
DSuniv = DSsys + DSsurr > 0
20-16
Comparing Energy and Entropy
The total energy of the universe remains constant.
DEsys + DEsurr =DEuniv = 0
DH is often used to approximateDE.
For enthalpy there is no zero point; we can only measure
changes in enthalpy.
The total entropy of the universe increases.
DSuniv =DSsys +DSsurr > 0
For entropy there is a zero point, and we can determine
absolute entropy values.
20-17
The Third Law of Thermodynamics
A perfect crystal has zero entropy at absolute zero.
Ssys = 0 at 0 K
A “perfect” crystal has flawless alignment of all its
particles. At absolute zero, the particles have minimum
energy, so there is only one microstate.
S = k lnW = k ln 1 = 0
To find the entropy of a substance at a given temperature, we cool it
as close to 0 K as possible. We then heat in small increments,
measure q and T, and calculate DS for each increment. The sum of
these DS values gives the absolute entropy at the temperature of
interest.
20-18
Standard Molar Entropies
S° is the standard molar entropy of a substance,
measured for a substance in its standard state in
units of J/mol·K.
The conventions for defining a standard state include:
• 1 atm for gases
• 1 M for solutions, and
• the pure substance in its most stable form for
solids and liquids.
20-19
Factors Affecting Entropy
• Entropy depends on temperature.
– For any substance, S° increases as temperature increases.
• Entropy depends on the physical state of a substance.
– S° increases as the phase changes from solid to liquid to gas.
• The formation of a solution affects entropy.
• Entropy is related to atomic size and molecular
complexity.
– Remember to compare substances in the same physical state.
20-20
Figure 20.6A
Visualizing the effect of temperature on entropy.
Computer simulations show each particle in a crystal moving about its
lattice position. Adding heat increases T and the total energy, so the
particles have greater freedom of motion, and their energy is more
dispersed. S therefore increases.
20-21
Figure 20.6B
Visualizing the effect of temperature on entropy.
At any T, there is a range of occupied energy levels and therefore a
certain number of microstates. Adding heat increases the total energy
(area under the curve), so the range of occupied energy levels
becomes greater, as does the number of microstates.
20-22
Figure 20.6C
Visualizing the effect of temperature on entropy.
A system of 21 particles occupy energy levels (lines) in a box whose
height represents the total energy. When heat is added, the total
energy increases and becomes more dispersed, so S increases.
20-23
Figure 20.7
20-24
The increase in entropy during phase changes
from solid to liquid to gas.
Figure 20.8 The entropy change accompanying the dissolution
of a salt.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
pure solid
MIX
pure liquid
solution
The entropy of a salt solution is usually greater than that of the solid
and of water, but it is affected by the organization of the water molecules
around each ion.
20-25
Figure 20.9
The small increase in entropy when ethanol
dissolves in water.
Ethanol (A) and water (B) each have many H bonds between their own
molecules. In solution (C) they form H bonds to each other, so their
freedom of motion does not change significantly.
20-26
Figure 20.10
20-27
The entropy of a gas dissolved in a liquid.
Entropy and Atomic Size
S° is higher for larger atoms or molecules of the same type.
Li
Atomic radius (pm) 152
Na
K
Rb
Cs
186
227
248
265
Molar mass (g/mol)
6.941
22.99
39.10
85.47
132.9
S°(s)
29.1
51.4
64.7
69.5
85.2
HF
HCl
Molar mass (g/mol)
S°(s)
20-28
20.01
173.7
36.46
186.8
HBr
80.91
198.6
HI
127.9
206.3
Entropy and Structure
For allotropes, S° is higher in the form that allows the
atoms more freedom of motion.
S° of graphite is 5.69 J/mol·K, whereas S° of diamond is 2.44 J/mol·K.
For compounds, S° increases with chemical
complexity.
S°(s)
S°(g)
NaCl
AlCl3
72.1
167
P4O10
NO
NO2
N 2 O4
211
240
304
229
These trends only hold for substances in the same
physical state.
20-29
Figure 20.11
20-30
Entropy, vibrational motion, and molecular
complexity.
Sample Problem 20.1
Predicting Relative Entropy Values
PROBLEM: Choose the member with the higher entropy in each of the
following pairs, and justify your choice [assume constant
temperature, except in part (e)]:
(a) 1 mol of SO2(g) or 1 mol of SO3(g)
(b) 1 mol of CO2(s) or 1 mol of CO2(g)
(c) 3 mol of O2(g) or 2 mol of O3(g)
(d) 1 mol of KBr(s) or 1 mol of KBr(aq)
(e) seawater at 2°C or at 23°C
(f) 1 mol of CF4(g) or 1 mol of CCl4(g)
PLAN: In general, particles with more freedom of motion have more
microstates in which to disperse their kinetic energy, so they
have higher entropy. Raising the temperature or having more
particles increases entropy.
20-31
Sample Problem 20.1
SOLUTION:
(a) 1 mol of SO3(g). For equal numbers of moles of substances with
the same types of atoms in the same physical state, the more
atoms in the molecule the higher the entropy.
(b) 1 mol of CO2(g). For a given substance, entropy increases as the
phase changes from solid to liquid to gas.
(c) 3 mol of O2(g). The two samples contain the same number of
oxygen atoms but different numbers of molecules. Although each
O3 molecule is more complex than each O2 molecule, the greater
number of molecules dominates because there are many more
microstates possible for 3 mol of particles than for 2.
20-32
Sample Problem 20.1
(d) 1 mol of KBr(aq). The two samples have the same number of
ions, but their motion is more limited and their energy less
dispersed in the solid than in the solution. An ionic substance in
solution usually has a higher entropy than the solid.
(e) Seawater at 23°C. Entropy increases with rising temperature.
(f)
20-33
1 mol of CCl4(g). For similar compounds, entropy increases with
molar mass.
Entropy Changes in the System
The standard entropy of reaction, DS°rxn, is the
entropy change that occurs when all reactants and
products are in their standard states.
DS°rxn = SmS°products − SnS°reactants
where m and n are the amounts (mol) of products and
reactants, given by the coefficients in the balanced equation.
20-34
Predicting the Sign of DS°rxn
We can often predict the sign of DS°rxn for processes
that involve a change in the number of moles of gas.
•
DS°rxn is positive if the amount of gas increases;
– e.g., H2(g) + I2(s) → 2HI(g)
•
DS°rxn is negative if the amount of gas decreases;
– e.g., N2(g) + 3H2(g)
•
2NH3(g)
DS° is likely to be positive if a new structure forms that
has more freedom of motion.
H2
C
H2C
20-35
CH2 (g)
CH3
CH CH2 (g)
Sº > 0
Sample Problem 20.2
Calculating the Standard Entropy of
Reaction, DS°rxn
PROBLEM: Predict the sign of DS°rxn and calculate its value for the
combustion of 1 mol of propane at 25°C.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
PLAN: From the change in the number of moles of gas (6 mol yields
3 mol), the entropy should decrease (DS°rxn < 0). To find
DS°rxn, we apply Equation 20.4, using the S° values from
Appendix B.
SOLUTION:
ΔS°rxn = [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)]
– [(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)]
= [(3mol)(213.7J/K·mol) + (4 mol)(69.9J/K·mol)]
– [(1mol)(269.9 J/K·mol) + (5 mol)(205.0 J/K·mol)] = –374 J/K
20-36
Entropy Changes in the Surroundings
A decrease in the entropy of the system is outweighed
by an increase in the entropy of the surroundings.
The surroundings function as a heat source or heat sink.
In an exothermic process, the surroundings absorbs the
heat released by the system, and Ssurr increases.
qsys < 0; qsurr > 0 and DSsurr > 0
In an endothermic process, the surroundings provides the
heat absorbed by the system, and Ssurr decreases.
qsys > 0; qsurr < 0 and DSsurr < 0
20-37
Temperature at which Heat is Transferred
Since entropy depends on temperature, DS°surr is also
affected by the temperature at which heat is transferred.
For any reaction, qsys = –qsurr. The heat transferred is
specific for the reaction and is the same regardless of the
temperature of the surroundings.
The impact on the surroundings is larger when the
surroundings are at lower temperature, because there is
a greater relative change in Ssurr.
DSsurr = – qsys
T
20-38
DSsurr = – DHsys for a process at constant P
T
Sample Problem 20.3
Determining Reaction Spontaneity
PROBLEM: At 298 K, the formation of ammonia has a negative
DS°sys;
N2(g) + 3H2(g) → 2NH3(g); DS°sys = –197 J/K
Calculate DS°univ, and state whether the reaction occurs
spontaneously at this temperature.
PLAN: For the reaction to occur spontaneously, DS°univ > 0, so DS°surr
must be greater than +197 J/K. To find DS°surr we need DH°sys,
which is the same as DH°rxn. We use the standard enthalpy
values from Appendix B to calculate this value.
SOLUTION:
DH°rxn = [(2 mol)(DH°f of NH3)] – [(1 mol)(DH°f of N2) + (3 mol)(DH°f
of H2)]
= [(2 mol)(-45.9 kJ/mol)] – [(1 mol)(0 kJ/mol) + (3 mol)(0 kJ/mol)]
= –91.8 kJ
20-39
Sample Problem 20.3
DSsurr = -
DHsys
T
–91.8 kJ x 1000 J
1 kJ
=–
298 K
= 308 J/K
DS°univ = DS°sys + DS°surr = –197 J/K + 308 J/K = 111
J/K
Since DS°univ > 0, the reaction occurs spontaneously at 298 K.
Although the entropy of the system
decreases, this is outweighed by
the increase in the entropy of the
surroundings.
20-40
Figure 20.12
A whole-body calorimeter.
In this room-sized apparatus, a person exercises while respiratory gases,
energy input and output, and other physiological variables are monitored.
Biological systems also obey the laws of thermodynamics. For all
processes, however complex, the entropy of the university increases.
20-41
DS and the Equilibrium State
For any process approaching equilibrium, DS°univ > 0.
At equilibrium, there is no further net change, and DS°sys
is balanced by DS°surr.
DS°univ = DS°sys + DS°surr = 0, so DS°sys = –DS°surr
When a system reaches equilibrium, neither the forward
nor the reverse reaction is spontaneous, so there is no
net reaction in either direction.
20-42
Figure 20.13AB
Components of DS°univ for spontaneous
reactions.
exothermic
For an exothermic reaction in
which DSsys > 0, the size of DSsurr
is not important since DSsurr > 0
for all exothermic reactions. The
reaction will always be
spontaneous.
20-43
exothermic
For an exothermic reaction in
which DSsys < 0, DSsurr must be
larger than DSsys for the reaction
to be spontaneous.
Figure 20.13C Components of DS°univ for spontaneous
reactions.
For an endothermic reaction in which
DSsys > 0, DSsurr must be smaller
than DSsys for the reaction to be
spontaneous. DSsurr is always < 0 for
endothermic reactions.
20-44
Gibbs Free Energy
The Gibbs free energy (G) combines the enthalpy and
entropy of a system; G = H – TS.
The free energy change (DG) is a measure of the
spontaneity of a process and of the useful energy available
from it.
DGsys = DHsys – T DSsys
DG < 0 for a spontaneous process
DG > 0 for a nonspontaneous process
DG = 0 for a process at equilibrium
20-45
Calculating DG°
DG°rxn can be calculated using the Gibbs equation:
DG°sys = DH°sys – T DS°sys
DG°rxn can also be calculated using values for the
standard free energy of formation of the components.
DG°rxn = SmDG°products – SnDG°reactants
20-46
Calculating DG°rxn from Enthalpy and
Entropy Values
PROBLEM: Potassium chlorate, a common oxidizing agent in
fireworks and matchheads, undergoes a solid-state
disproportionation reaction when heated.
+5
+7
–1
Δ
4KClO3(s)
3KClO4(s) + KCl(s)
Sample Problem 20.4
PLAN: To solve for DG°, we need values from Appendix B. We use
DHf° to calculate DH°rxn, use S° values to calculate DS°rxn,
and then apply the Gibbs equation.
SOLUTION:
DH°rxn = SmDHf°(products) - SnDHf°(reactants)
= [(3 mol KClO4)(DHf° of KClO4) + (1 mol KCl)(DHf° of
KCl)]
‒ [(4 mol KClO3)(DHf° of KClO3)]
= [(3 mol)(-432.8 kJ/mol) + (1 mol)(–436.7 kJ/mol)
– [(4 mol)(-397.7 kJ/mol)]
= –144 kJ
20-47
Sample Problem 20.4
DS°rxn = SmS°(products) – SS°(reactants)
= [(3 mol of KClO4)(S° of KClO4) + (1 mol KCl)(S° of KCl)]
‒ [(4 mol KClO3)(S° of KClO3)]
= [(3 mol)(151.0 J/mol·K) + (1 mol)(82.6 J/mol·K)
– [(4 mol)(143.1 J/mol·K)]
= –36.8 J/K
DG°sys = DH°sys – T DS°sys = –144 kJ – (298 K)(–36.8 J/K)
= –133 kJ
20-48
1 kJ
1000 J
Sample Problem 20.5
Calculating DG°rxn from DG°f Values
PROBLEM: Use DG°f values to calculate DG°rxn for the reaction
in Sample Problem 20.4:
4KClO3(s)
Δ
3KClO4(s) + KCl(s)
PLAN: We apply Equation 20.8 and use the values from Appendix
B to calculate DG°rxn.
SOLUTION:
DG°rxn = SmDG°products - SnDG°reactants
= [(3 mol KClO4)(DGf° of KClO4) + (1 mol KCl)(DGf° of KCl)]
‒ [(4 mol KClO3)(DGf° of KClO3)]
= [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)]
‒ [(4 mol)(-296.3 kJ/mol)]
= –134 kJ
20-49
DG° and Useful Work
DG is the maximum useful work done by a system
during a spontaneous process at constant T and P.
In practice, the maximum work is never done. Free energy not used
for work is lost to the surroundings as heat.
DG is the minimum work that must be done to a system
to make a nonspontaneous process occur at constant T
and P.
A reaction at equilibrium (DGsys = 0) can no longer do any
work.
20-50
Effect of Temperature on Reaction Spontaneity
DGsys = DHsys - T DSsys
Reaction is spontaneous at
all temperatures
If DH < 0 and DS > 0
DG < 0 for all T
Reaction is nonspontaneous
at all temperatures
If DH > 0 and DS < 0
DG > 0 for all T
20-51
Effect of Temperature on Reaction Spontaneity
Reaction becomes spontaneous
as T increases
If DH > 0 and DS > 0
DG becomes more negative as T
increases.
Reaction becomes spontaneous
as T decreases
If DH < 0 and DS < 0
DG becomes more negative as T
decreases.
20-52
Table 20.1 Reaction Spontaneity and the Signs of DH, DS, and DG
DH
DS
-TDS
DG
–
+
–
–
Spontaneous at all T
+
–
+
+
Nonspontaneous at all T
+
+
–
+ or –
Spontaneous at higher T;
Description
nonspontaneous at lower T
–
–
+
+ or –
Spontaneous at lower T;
nonspontaneous at higher T
20-53
Sample Problem 20.6
Using Molecular Scenes to Determine the
Signs of DH, DS, and DG
PROBLEM: The following scenes represent a familiar phase change
for water (blue spheres).
(a) What are the signs of DH and DS for this process? Explain.
(b) Is the process spontaneous at all T, no T, low T, or high T? Explain.
PLAN: (a) From the scenes, we determine any change in the amount
of gas, which indicates the sign of DS, and any change in
the freedom of motion of the particles, which indicates
whether heat is absorbed or released.
(b) To determine reaction spontaneity we need to consider the
sign of DG at different temperatures.
20-54
Sample Problem 20.6
SOLUTION:
(a)
The scene represents the condensation of water vapor, so the
amount of gas decreases dramatically, and the separated molecules
give up energy as they come closer together.
DS < 0 and DH < 0
(b) Since DS is negative, the –TDS term is positive. In order for DG to
be < 0, the temperature must be low.
The process is spontaneous at low temperatures.
20-55
Sample Problem 20.7
Determining the Effect of Temperature on ΔG
PROBLEM: A key step in the production of sulfuric acid is the oxidation
of SO2(g) to SO3(g):
2SO2(g) + O2(g) → 2SO3(g)
At 298 K, DG = –141.6 kJ; DH = –198.4 kJ; and DS = –187.9 J/K
(a) Use the data to decide if this reaction is spontaneous at 25°C, and
predict how DG will change with increasing T.
(b) Assuming DH and DS are constant with increasing T (no phase
change occurs), is the reaction spontaneous at 900.° C?
PLAN: We note the sign of DG to see if the reaction is spontaneous
and the signs of DH and DS to see the effect of T. We can then
calculate whether or not the reaction is spontaneous at the
higher temperature.
20-56
Sample Problem 20.7
SOLUTION:
(a)
DG < 0 at 209 K (= 25°C), so the reaction is spontaneous.
With DS < 0, the term –TDS > 0 and this term will become more
positive at higher T.
DG will become less negative, and the reaction less spontaneous,
with increasing T.
(b)
DG = DH – TDS
Convert T to K: 900. + 273.15 = 1173 K
Convert S to kJ/K: –187.9 J/K = –0.1879 kJ/K
DG = –198.4 kJ – [1173 K)(–0.1879 kJ/K) = 22.0 kJ
DG > 0, so the reaction is nonspontaneous at 900.°C.
20-57
Figure 20.14 The effect of temperature on reaction spontaneity.
The sign of DG
switches at
T = DH
DS
20-58
Sample Problem 20.8
Finding the Temperature at Which a
Reaction Becomes Spontaneous
PROBLEM: At 25°C (298 K), the reduction of copper(I) oxide is
nonspontaneous (DG = 8.9 kJ). Calculate the temperature
at which the reaction becomes spontaneous.
PLAN: We need to calculate the temperature at which DG crosses
over from a positive to a negative value. We set DG equal to
zero, and solve for T, using the values for DH and DS from
the text.
SOLUTION:
DH
58.1 kJ
= 352 K
T=
=
DS
0.165 kJ/K
At any temperature above 352 K (= 79°C), the reaction becomes
spontaneous.
20-59
Chemical Connections
Figure B20.1
The coupling of a nonspontaneous reaction to
the hydrolysis of ATP.
A spontaneous reaction can be coupled to a nonspontaneous reaction
so that the spontaneous process provides the free energy required to
drive the nonspontaneous process. The coupled processes must be
physically connected.
20-60
Chemical Connections
Figure B20.2 The cycling of metabolic free energy.
20-61
Chemical Connections
Figure B20.3 ATP is a high-energy molecule.
When ATP is hydrolyzed to ADP, the decrease in charge repulsion (A)
and the increase in resonance stabilization (B) causes a large
amount of energy to be released.
20-62
DG, Equilibrium, and Reaction Direction
A reaction proceeds spontaneously to the right if
Q < K; Q < 1 so ln Q < 0 and DG < 0
K
K
A reaction proceeds spontaneously to the left if
Q > K; Q > 1 so ln Q > 0 and DG > 0
K
K
A reaction is at equilibrium if
Q = K; Q = 1 so ln Q = 0 and DG = 0
K
K
20-63
DG, Q, and K
DG = RT ln Q = RT lnQ – RT lnK
K
If Q and K are very different, DG has a very large value
(positive or negative). The reaction releases or absorbs a
large amount of free energy.
If Q and K are nearly the same, DG has a very small value
(positive or negative). The reaction releases or absorbs very
little free energy.
20-64
ΔG and the Equilibrium Constant
For standard state conditions, Q = 1 and
DG° = –RT lnK
A small change in DG° causes a large change in K, due
to their logarithmic relationship.
As DG° becomes more positive, K becomes smaller.
As DG° becomes more negative, K becomes larger.
To calculate DG for any conditions:
DG = DG° + RT lnQ
20-65
Sample Problem 20.9
Exploring the relationship Between DGº and K
PROBLEM: (a) Use Appendix B to find K at 298 K for the following reaction:
NO(g) + ½O2(g)
NO2(g)
(b) Use the equilibrium constant to calculate DGº at 298 K for the following
reaction:
2HCl(g)
H2(g) + Cl2(g); K = 3.89x10–34 at 298 K
PLAN: (a) We use DG°f values in Appendix B with Equation 20.8 to find
DG°, and then we use Equation 20.12 to solve for K.
(b) In this calculation, we do the reverse of the second calculation in part (a):
We are given K (3.89x10–34), so we use Equation 20.12 to solve for DG°.
20-66
Sample Problem 20.9
SOLUTION: (a) Solving for DGº using DG°f values from Appendix B and
Equation 20.8:
DGº = [(1 mol NO2)(DG°f of NO2)] – [(1 mol NO)(DG°f of NO) +
(½ mol O2)(DG°f of O2)]
= [(1 mol)(51 kJ/mol)] – [(1 mol)(86.60 kJ/mol) + (½ mol)(0 kJ/mol)]
= –36 kJ/mol
Solving for K using Equation 20.12:
DGº = –RT ln K; so, ln K = –DGº/RT
= – (–36 kJ/mol)(1000 J/1 kJ)/(8.314 J/molK)(298 K)
= 14.53
K = e14.53 = 2.0x106
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Sample Problem 20.9
SOLUTION: (b) Solving for DGº using Equation 20.12:
K = 3.89x10–34; so, ln K = –76.9
DGº = –RT ln K = –(8.314 J/molK)(298 K)(–76.9)
= 1.91x105 J = 191 kJ
CHECK: In (a), a negative free energy change is consistent
with a positive K. Rounding to confirm the value of DGº,
50 kJ – 90 kJ = –40 kJ, close to the calculated value.
In (b), a very negative K is consistent with a highly positive
DGº. Looking up DG°f values to check the value of DGº gives
(0 kJ + 0 kJ) – [2 x (–95.30 kJ)] = –190.6 kJ.
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Table 20.2 The Relationship Between DG° and K at 298 K
DG°
(kJ)
9x10–36
100
3x10–18
50
2x10–9
10
2x10–2
1
7x10–1
0
1
–1
1.5
–10
5x101
–50
6x108
–100
3x1017
–200
1x1035
{Essentially no forward reaction;
reverse reaction goes to completion
{Forward and reverse reactions proceed
to same extent
{Forward reaction goes to completion;
essentially no reverse reaction
REVERSE REACTION
200
Significance
FORWARD REACTION
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K
Using Molecular Scenes to Find DG for a
Reaction at Nonstandard Conditions
PROBLEM: These molecular scenes represent three mixtures in which
A2 (black) and B2 (green) are forming AB. Each molecule
represents 0.10 atm. The equation is
A2(g) + B2(g)
2AB(g) DGo = –3.4 kJ/mol
Sample Problem 20.10
(a) If mixture 1 is at equilibrium, calculate K.
(b) Which mixture has the most negative DG, and which has the most
positive?
(c) Is the reaction spontaneous at the standard state, that is,
PA2 = PB2 = PAB = 1.0 atm?
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Sample Problem 20.10
PLAN:
(a) Mixture 1 is at equilibrium, so we first write the expression for Q
and then find the partial pressure of each substance from the
number of molecules and calculate K.
(b) To find DG, we apply equation 20.13. We can calculate the value
of T using K from part (a), and substitute the partial pressure of
each substance to get Q.
(c) Since 1.0 atm is the standard state for gases, DG = DG°.
SOLUTION:
(a) A2(g) + B2(g)
(0.40)2
K=
(0.20)(0.20)
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= 4.0
2AB(g)
PAB2
Q=
PA2 x PB2
Sample Problem 20.10
(b) DG° = –RT lnK
–3.4 kJ
8.314 J T ln 4.0
=–
mol
mol·K
T=
–3.4 kJ 1000 J
mol
1 kJ
–
8.314 J
mol·K
Mixture 1 is at equilibrium, so DG = 0;
DG = DG° + RT lnQ
= –3400 J/mol + (8.314 J/mol·K)(295 K)(ln 4.0)
= –3400 J + 3400 J = 0.0 J
Mixture 2:
(0.20)2
Q=
= 0.44
(0.30)(0.30)
DG = DG° + RT lnQ
= –3400 J/mol + (8.314 J/mol·K)(295 K)(ln 0.44)
= –3400 J – 2010 J = –5.4x10-3 J
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ln 4.0
= 295 K
Sample Problem 20.10
Mixture 3:
(0.60)2
Q=
= 36
(0.10)(0.10)
DG = DG° + RT lnQ
= –3400 J/mol + (8.314 J/mol·K)(295 K)(ln 36)
= –3400 J + 8800 J = +5.4x10–3 J
Mixture 2 has the most negative DG, and mixture 3 has the
most positive DG.
(c)
Under standard conditions, DG = DG° = –3.4 kJ/mol.
Yes, the reaction is spontaneous when the components are in
their standard states.
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Sample Problem 20.11
Calculating DG at Nonstandard Conditions
PROBLEM: The oxidation of SO2(g) is too slow at 298 K to be useful
in the manufacture of sulfuric acid, so the reaction is run
at high T.
2SO2(g) + O2(g) → 2SO3(g)
(a) Calculate K at 298 K and at 973 K, (DG°298 = −141.6 kJ/mol of
reaction as written; using DH° and DS°values at 973 K,
DG°973 = −12.12 kJ/mol for reaction as written.)
(b) Two containers are filled with 0.500 atm of SO2, 0.0100 atm of
O2, and 0.100 atm of SO3; one is kept at 25°C and the other at
700.°C. In which direction, if any, will the reaction proceed to
reach equilibrium at each temperature?
(c) Calculate DG for the system in part (b) at each temperature.
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Sample Problem 20.11
PLAN:
(a) We know DG°, T, and R, so we can calculate the values of K at
each temperature.
(b) To determine if a net reaction will occur, we find Q from the given
partial pressures and compare Q to each K from part (a).
(c) These are not standard-state pressures, so we find DG at each
temperature using the equation DG = DG° + RT lnQ.
SOLUTION:
(a) DG° = −RT lnK so K = e −(DG°/RT)
−141.6x103 J/mol
= 57.2
At 298 K, −(DG°)/RT =
8.314 J/mol·K x 298 K
K at 298 K = e57.2 = 7x1024
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Sample Problem 20.11
−12.12x103 J/mol
= 1.50
At 973 K, −(DG°)/RT =
8.314 J/mol·K x 973 K
K at 973 K = e1.50 = 4.5
(b) Calculating the value of Q
Q=
2
PSO
3
2
PSO
2
(0.100)2
=
= 4.00
2
x PO
(0.500) (0.0100)
2
Since Q < K at both temperatures, the reaction will proceed
toward the products (to the right) in both cases until Q equals K.
At 298 K, the system is far from equilibrium and will proceed far to
the right. At 973 K the system is closer to equilibrium and will
proceed only slightly toward the right.
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Sample Problem 20.11
(c)
DG298 = DG° + RT ln Q
= −141.6 kJ/mol + (8.314x10–3 kJ/mol·K x 298 K x ln 4.00)
= −138.2 kJ/mol
DG973 = DG° + RT ln Q
= −12.12 kJ/mol + (8.314x10–3 kJ/mol·K x 973 K x ln 4.00)
= −0.9 kJ/mol
20-77
Figure 20.15
Free energy and the extent of reaction.
Each reaction proceeds spontaneously (green curved arrows) from
reactants or products to the equilibrium mixture, at which point DG = 0.
After that, the reaction is nonspontaneous in either direction (red curved
arrows). Free energy reaches a minimum at equilibrium.
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