Download 18.3 Standard Entropies and the Third Law of

CHAPTER 18
Thermodynamics and Equilibrium
CHAPTER TERMS AND DEFINITIONS
Numbers in parentheses after definitions give the text section in which the terms are explained. Starred
terms are italicized in the text. Where a term does not fall directly under a text section heading,
additional information is given for you to locate it.
thermodynamics* study of the relationship between heat and other forms of energy involved in a
chemical or physical process (chapter introduction)
internal energy (U)
(18.1)
sum of the kinetic and potential energies of the particles making up a system
state function
property of a system, such as temperature and pressure, that depends only on its
present condition and that is a defining factor of that condition (18.1)
work (w) energy exchange that results when a force F moves an object through a distance d;
w = F  d (18.1)
first law of thermodynamics
the change in internal energy of a system ΔU equals the heat absorbed
or lost by the system ±q plus the work done on or by the system ±w; ΔU=q+ w (18.1)
enthalpy (H)
thermodynamic quantity defined by the equation H = U + PV (18.1)
spontaneous process
physical or chemical change that occurs by itself (18.2, introductory section)
nonspontaneous* describes a physical or chemical change that occurs only with the aid of an
outside force or agent (18.2, introductory section)
entropy (S) thermodynamic quantity that is a measure of how dispersed the energy of a system is
among the different possible ways that system can contain energy (18.2)
second law of thermodynamics
the total entropy of a system and its surroundings always increases
for a spontaneous process; ΔS > q/T (18.2)
statistical thermodynamics*
rigorous method of defining the entropy of a system in terms of a sum
over the energy levels available to the system (18.2)
reversible*
describes a process that can go in one direction or the other (18.2, marginal note)
third law of thermodynamics
(18.3)
standard (absolute) entropy (S)
free energy (G)
a substance that is perfectly crystalline at 0 K has an entropy of zero
entropy value for the standard state of a species (18.3)
thermodynamic quantity defined by the equation G = H – TS (18.4)
standard states* conditions under which to report the properties of matter: 1 atm pressure for pure
substances, 1 M concentration for ions in solution (18.4)
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
standard free energy of formation ( G f )
393
free-energy change when one mole of substance is
formed from its elements in their stablest states at one atmosphere of pressure and at a specified
temperature (usually 25C) (18.4)
thermodynamic equilibrium constant (K)
equilibrium constant in which the concentrations of
gases are expressed in partial pressures in atmospheres, whereas the concentrations of solutes in liquid
solutions are expressed in molarities (18.6)
CHAPTER DIAGNOSTIC TEST
1.
Determine whether each of the following statements is true or false. If the statement is false,
change it so that it is true.
a.
A spontaneous process is a physical or chemical change that occurs without warning.
True/False: ________________________________________________________________
__________________________________________________________________________
b.
Entropy is a measure of disorder in a physical or chemical system. True/False: __________
__________________________________________________________________________
c.
Entropy increases going from solid to liquid to gas. True/False: _______________________
__________________________________________________________________________
d.
A reaction is spontaneous as written if ΔG is a large positive number. True/False:
__________________________________________________________________________
__________________________________________________________________________
2.
Calculate the change in entropy ΔS for the following precipitation reaction:
Ag+(aq) + Cl–(aq)  AgCl(s)
Use text Table 18.1 for standard entropy values.
3.
4.
Predict by inspection the sign of ΔS for each of the following reactions, and explain your
answers.
a.
H2(g) + C2H6(g)
CH4(g)
b.
2SO2(g) + O2(g)
2SO3(g)
c.
2NO2(g)
d.
Pb(NO3)2(aq) + Na2SO4(aq)  PbSO4(s) + 2NaNO3(aq)
2NO(g) + O2(g)
The heat of vaporization of liquid bromine, Br2(l), at 25C is 30.7 kJ/mol:
Br2(l)
Br2(g)
If liquid bromine at 25C has an entropy of 152.2 J/(K ∙ mol), what is the entropy of one mole of
the vapor in equilibrium with the liquid at that temperature?
5.
Calculate ΔG for the following precipitation reaction at 25C:
2Ag+(aq) + S2–(aq)  Ag2S(s)
Use values of H f and S from text Appendix C.
Copyright © Houghton Mifflin Company. All rights reserved.
394
6.
Chapter 18: Thermodynamics and Equilibrium
Calculate the free-energy change ΔG for the following reaction using values of G f from text
Appendix C:
2Li(s) + 2H2O(l)  2Li+(aq) + 2OH–(aq) + H2(g)
7.
Give the expression for K for the following reactions.
a.
2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
b.
NH4OH(aq)
c.
2H2O2(l)
NH3(g) + H2O(l)
2H2O(l) + O2(g)
8.
Determine whether each of the reactions in Problem 3 is spontaneous or nonspontaneous as
written. Use values in text Appendix C and give reasons for your answers.
9.
Calculate Ka for the ionization of formic acid, HCOOH, in aqueous solution. Use values of G f
in text Appendix C.
10. Calculate the value of the thermodynamic equilibrium constant at 648 K for the following
reaction:
N2(g) + 3H2(g)
2NH3(g)
ΔG for this reaction at the given temperature is 43.2 kJ.
ANSWERS TO CHAPTER DIAGNOSTIC TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
2.
a.
False. A spontaneous process is a physical or chemical change that can occur by itself,
requiring no continuing outside agency to make it happen. (18.2, introductory section)
b.
True. (18.2)
c.
True. (18.2)
d.
False. A reaction is spontaneous as written if ΔG is a large negative number. (18.4)
ΔS= –32.9 J/K (18.3, PS Sk. 3)
3.
a.
ΔS  0; there is no change in the number of gas molecules.
b.
ΔS < 0; there are three molecules of gas on the left but only two on the right.
c.
ΔS > 0; there are two molecules of gas on the left and three on the right.
d.
ΔS < 0; a solid is formed on the right. (18.3, PS Sk. 2)
4.
ΔS = 255 J/(K ∙ mol) (18.2, PS Sk. 1)
5.
ΔG = –278.3 kJ (18.4, PS Sk. 4)
6.
ΔG = –427.8 kJ (18.4, PS Sk. 5)
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
395
7.
a.
K = PCO2 PH2O
b.
K=
c.
Water is not a solvent and must be included in the expression:
K=
PNH3
[ NH 4 OH]
[H 2 O] 2 PO2
[H 2 O 2 ]2
(18.6, PS Sk. 7)
8.
9.
a.
Spontaneous; ΔG < 0
b.
Spontaneous; ΔG < 0
c.
Nonspontaneous; ΔG > 0
d.
Spontaneous; ΔG < 0 (18.4, PS Sk. 6)
Ka = K = 2.1  10–4 (18.6, PS Sk. 8)
10. K = 3.3  10–4 (18.6, PS Sk. 8)
SUMMARY OF CHAPTER TOPICS
18.1 First Law of Thermodynamics; Enthalpy
Learning Objectives

Define internal energy, state function, work, and first law of thermodynamics.

Explain why the work done by the system as a result of expansion or contraction during a
chemical reaction is –PV.

Relate the change of internal energy U and heat of reaction q.

Define enthalpy H.

Show how heat of reaction at constant pressure qp equals the change of enthalpy H.
Exercise 18.1
A gas is enclosed in a system similar to that shown in text Figure 18.2. More weights are added to the
piston, giving a total mass of 2.20 kg. As a result, the gas is compressed, and the weights are lowered
0.250 m. At the same time, 1.50 J of heat evolves from (leaves) the system. What is the change in
internal energy of the system ΔU? The force of gravity on a mass m is mg, where g is the constant
acceleration of gravity (g = 9.80 m/s2).
Wanted: ΔU, in joules
Given:
Mass of weights = 2.20 kg, piston dropped 0.250 m, heat evolved (q) = –1.50 J; F =
mg, g = 9.80 m/s2.
Known: q = –1.50 J, as heat is evolved; ΔU = q + w; w = F  d, the sign of w is + because the
work is done on the system; 1 J = 1 kg ∙ m2/s2.
Copyright © Houghton Mifflin Company. All rights reserved.
396
Chapter 18: Thermodynamics and Equilibrium
Solution: Substitute algebraically, then with numbers, into
ΔU = q + w = q + (F  d) = q + (mg  d)
= 1.50 J + 2.20 kg  9.80 m/s2  0.250 m 
J
kg • m 2 /s 2
ΔU = 3.89 J
Exercise 18.2
Consider the combustion (burning) of methane, CH4, in oxygen:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
The heat of reaction at 25C and 1.00 atm is –890.2 kJ. What is the change in volume when 1.00 mol
CH4 reacts with 2.00 mol O2? (You can ignore the volume of liquid water, which is insignificant
compared with volumes of gases.) What is w for this change? Calculate ΔU for the change indicated by
the chemical equation.
Wanted: ΔV, in L; w, in kJ; ΔU, in kJ
Given:
equation; t = 25C; P = 1 atm; q = –890.2 kJ; 1 mol CH4; 2.00 mol O2.
Known: PΔV = ΔnRT; w = PΔV and is + because work is done on the system (the volume of the
system decreases); ΔU = q + w, 1 L = 10–3 m3; 1 Pa ∙ m3 = 1 J; 1 atm = 1.01  105 Pa..
Solution: Since there are 3 mol of gas on the left of the equation and 1 mol on the right, Δn = 2
mol, and
ΔV =
2 mol  0.0821 L • atm  298 K
nRT
=
= 48.9 L
1.00 atm (K • mol)
P
w = +PΔV = 1.01  105 Pa  48.9 L 
1J
10 3 m 3
=
L
1 Pa • m 3
= 4.94  103 J = 4.94 kJ
ΔU = q + w = –890.2 kJ + 4.94 kJ = –885.3 kJ
18.2 Entropy and the Second Law of Thermodynamics
Learning Objectives

Define spontaneous process.

Define entropy.

Relate entropy to disorder in a molecular system (energy dispersal).

State the second law of thermodynamics in terms of system plus surroundings.

State the second law of thermodynamics in terms of the system only.

Calculate the entropy change for a phase transition. (Example 18.1)

Describe how H – TS functions as a criterion of a spontaneous reaction.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
397
Problem-Solving Skill
1.
Calculating the entropy change for a phase transition. Given the heat of phase transition and
the temperature of the transition, calculate the entropy change ΔS of the system (Example 18.1).
Stop a minute and think of your room at home or in the dorm. If you are like most of us, you envision
clothes hanging over a chair, books scattered, maybe a sock or two half hidden under the bed. Such a
room gives a vivid example of the second law of thermodynamics—that entropy increases in a
spontaneous process. It takes work to hang up the clothes, to arrange the books neatly, and to put the
socks in the hamper.
It is a good idea to memorize the expression for entropy, S = q/T, and to remember that the entropy
change in a system during a spontaneous change is ΔS > q/T (i.e.,nonreversible).
Exercise 18.3
Liquid ethanol, C2H5OH(l), at 25C has an entropy of 161J/(K∙mol). If the heat of vaporization ΔHvap
at 25C is 42.3 kJ/mol, what is the entropy of the vapor in equilibrium with the liquid at 25C?
Wanted: entropy of the vapor, in J/(K ∙ mol)
Given:
S for liquid = 161 J/(K ∙ mol); ΔHvap = 42.3 kJ/mol; T = 25C.
Known: The entropy of the vapor = entropy of the liquid plus ΔS of vaporization; qp= ΔH;
ΔSvap = ΔHvap/T; 25C = 298 K.
Solution: The calculation is
ΔSvap =
42 .3  10 3 J
= 1.420  102 J/(K ∙ mol)
298 K • mol
Svap = (161 + 142) J/(K ∙ mol) = 303 J/(K ∙ mol)
18.3 Standard Entropies and the Third Law of Thermodynamics
Learning Objectives

State the third law of thermodynamics.

Define standard entropy (absolute entropy).

State the situations in which the entropy usually increases.

Predict the sign of the entropy change of a reaction. (Example 18.2)

Express the standard change of entropy of a reaction in terms of standard entropies of
products and reactants.

Calculate S for a reaction. (Example 18.3)
Problem-Solving Skills
2.
Predicting the sign of the entropy change of a reaction. Predict the sign of ΔS for a reaction to
which the rules listed in the text can be clearly applied (Example 18.2).
3.
Calculating ΔS for a reaction. Given the standard entropies of reactants and products, calculate
the entropy of reaction ΔS (Example 18.3).
Copyright © Houghton Mifflin Company. All rights reserved.
398
Chapter 18: Thermodynamics and Equilibrium
It is important to note that the zero point of entropy is at 0 K and that we talk of absolute entropies S of
substances rather than of entropies of formation. Recall that substances have an enthalpy of formation
H f and that the zero point for these values is the element in its most stable state at standard
conditions.
Exercise 18.4
Predict the sign of ΔS for each of the following reactions.
a.
CaCO3(s)  CaO(s) + CO2(g)
b.
CS2(l)  CS2(g)
c.
2Hg(l) + O2(g)  2HgO(s)
d.
2Na2O2(s) + 2H2O(l)  4NaOH(aq) + O2(g)
Known: Entropy is a measure of disorder; solids are more ordered than liquids, which are more
ordered than gases.
Solution:
a.
ΔS > 0; a molecule of gas is formed.
b.
ΔS > 0; a molecule of gas is formed.
c.
ΔS < 0; the solid product is more ordered than the reactants.
d.
ΔS > 0; a molecule of gas is formed.
Exercise 18.5
Calculate the change of entropy ΔS for the reaction given in Example 18.2(a). The standard entropy of
glucose, C6H12O6(s), is 212 J/(K ∙ mol). See text Table 18.1 for other values.
Wanted:
ΔS in J/K
Given:
S for C6H12O6 = 212 J/(K ∙ mol)
Known:
reaction from Example 18.2(a); text Table 18.1 for S values
Solution: The calculation is
C6H12O6(s)  2C2H5OH(l) + 2CO2(g)
S
212
2(161)
2(213.7) J/(K ∙ mol)
ΔS = [2(161) + 2(213.7) – 212] J/K
ΔS = 537 J/K
It is important to recall that in Section 14.2 you learned that the equilibrium-constant expression and its
value depend on the coefficients used to write the reaction. It is customary to write the coefficients
using the smallest ratio of whole numbers. But the reaction is still correct as long as you use any
numbers in that ratio.
Thus the values for the entropy of reaction calculated in Example 18.3 and in Exercise18.5 both
depend on the coefficients used to write the individual reactions. The entropies would be different if
different coefficients were used.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
399
This is also true of the enthalpies of reaction ΔH calculated in Chapter 6. Sometimes the term mole of
reaction is used to describe this idea. This use of the word mole does not designate 6.02  1023 particles.
It merely means “the reaction as written” or “the quantities specified in the reaction as written.” In
Example 18.3, where the reaction is written as
2NH3(g) + CO2(g)
NH2CONH2(aq) + H2O(l)
“one mole of reaction” means “2  6.02  1023 molecules of NH3 and 6.02  1023 molecules of CO2
react to form 6.02  1023 molecules of urea and 6.02  2023 molecules of water.” In Exercise 18.5, “one
mole of reaction” means “6.02  1023 molecules of glucose yield 26.02 1023 molecules of ethyl
alcohol and 2  6.02  1023 molecules of CO2.” So the units for the entropies of reaction could be
written J/(K ∙ mol), mol meaning “mole of reaction.” We will refer to this concept again in the next
section and in the following chapter.
18.4 Free Energy and Spontaneity
Learning Objectives

Define free energy G.

Define the standard free-energy change.

Calculate G from H and S. (Example 18.4)

Define the standard free energy of formation G.

Calculate G from standard free energies of formation. (Example 18.5)

State the rules for using G as a criterion for spontaneity.

Interpret the sign of G. (Example 18.6)
Problem-Solving Skills
4.
Calculating ΔG from ΔH and ΔS. Given enthalpies of formation and standard entropies of
reactants and products, calculate the standard free-energy change ΔG for a reaction (Example
18.4).
5.
Calculate ΔG from standard free energies of formation. Given the free energies of formation
of reactants and products, calculate the standard free-energy change ΔG for a reaction (Example
18.5).
6.
Interpreting the sign of ΔG. Use the standard free-energy change to determine the spontaneity
of a reaction (Example 18.6).
Note that the zero point of free energy of formation G f is the element in its most stable state at 1 atm
pressure and usually 25C, as with the enthalpy of formation. It is a good idea to memorize the
expression for the free-energy change of a reaction: ΔG = ΔH – TΔS. You must know that ΔG for a
spontaneous reaction is negative.
Exercise 18.6
Calculate ΔG for the following reaction at 25C. Use data given in text Tables 6.2 and 18.1.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Known:
ΔG = ΔH – TΔS; H f values are in text Table 6.2; S values are in text Table 18.1.
Copyright © Houghton Mifflin Company. All rights reserved.
400
Chapter 18: Thermodynamics and Equilibrium
Solution: Calculations follow:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H f
S°
–74.9
2(0)
–393.5
186.1
2(205.0) 213.7
2(–241.8) kJ/mol
2(188.7) J/(K ∙ mol)
ΔH = [–393.5 + 2(–241.8) – (–74.9)] kJ
= –802.2 kJ
ΔS = {[213.7 + 2(188.7)] – [186.1 + 2(205.0)]} J/K
= –5.0 J/K
ΔG = ΔH – TΔS
ΔG = 802.2 kJ  {298 K 
5.0 J
10 3 kJ

}
J
K
ΔG = –800.7 kJ
Let us again refer to the term mole of reaction. The free-energy change also could be thought of as per
“mole of reaction” for the reaction with coefficients specified. This means that the units for the ΔH of
the preceding reaction could be calculated as
ΔG =
1 mol CO 2
2 mol H 2 O( g )
393 .5 kJ
241 .8 kJ

+

mol (of rxn )
mol (of rxn )
mol CO 2
mol H 2 O( g )

1 mol CH 4
74.9 kJ

mol (of rxn)
mol CH 4
= 802.2 kJ / mol (of rxn)
We could do the same for the calculation of ΔS for the reaction.
In another text you might see ΔH or ΔG written as kJ/mol, where mol means “mole of reaction.”
Exercise 18.7
Calculate ΔG at 25C for the following reaction using values of G f .
CaCO3(s)  CaO(s) + CO2(g)
Known:
G f values given in text Table 18.2
Solution: The calculation is
CaCO3(s)  CaO(s) + CO2(g)
G f
–1128.8
–603.5
–394.4 kJ/mol
ΔG = [–603.5 + (–394.4) – (–1128.8)] kJ
ΔG = 130.9 kJ
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
401
Exercise 18.8
Which of the following reactions are spontaneous in the direction written? See text Table 18.2 for data.
a.
C (graphite) + 2H2(g)  CH4(g)
b.
2H2(g) + O2(g) 2H2O(l)
c.
4HCN(g) + 5O2(g) 2H2O(l) + 4CO2(g) + 2N2(g)
d.
Ag+(aq) + I–(aq)  AgI(s)
Wanted: spontaneity of reaction
Given:
G f values in text Table 18.2
Known:
If ΔG > 0, reaction is nonspontaneous; if ΔG < 0, reaction is spontaneous.
Solution: All the reactions are spontaneous, as the following calculations show:
C (graphite) + 2H2(g)  CH4(g)
a.
G f
0
2(0)
–50.8 kJ/mol
ΔG = –50.8 kJ
2H2(g) + O2(g)  2H2O(l)
b.
G f
2(0)
2(0)
2(–237.1) kJ/mol
ΔG = [2(–237.1)] kJ
ΔG = –474.2 kJ
4HCN(g) + 5O2(g)  2H2O(l) + 4CO2(g) + 2N2(g)
c.
G f
4(124.7)
5(0)
2(–237.1) 4(–394.4)
2(0) kJ/mol
ΔG = [2(–237.1) + 4(–394.4) – 4(124.7)] kJ
ΔG = –2551 kJ
Ag+(aq) + I–(aq)  AgI(s)
d.
G f
77.12
–51.59
–66.19 kJ/mol
ΔG = [–66.19 – (–51.59 + 77.12)] kJ
ΔG = –91.72 kJ
18.5 Interpretation of Free Energy
Learning Objectives

Relate the free-energy change to maximum useful work.

Describe how the free energy changes during a chemical reaction.
Copyright © Houghton Mifflin Company. All rights reserved.
402
Chapter 18: Thermodynamics and Equilibrium
18.6 Relating ΔG° to the Equilibrium Constant
Learning Objectives

Define the thermodynamic equilibrium constant K.

Write the expression for a thermodynamic equilibrium constant. (Example 18.7)

Indicate how the free-energy change of a reaction and the reaction quotient are related.

Relate the standard free-energy change to the thermodynamic equilibrium constant.

Calculate K from the standard free-energy change (molecular equation). (Example 18.8)

Calculate K from the standard free-energy change (net ionic equation). (Example 18.9)
Problem-Solving Skills
7.
Writing the expression for a thermodynamic equilibrium constant. For any balanced chemical
equation, write the expression for the thermodynamic equilibrium constant (Example18.7).
8.
Calculating K from the standard free-energy change. Given the standard free-energy change
for a reaction, calculate the thermodynamic equilibrium constant (Examples 18.8 and 18.9).
Memorize the relationship between ΔG and the thermodynamic equilibrium constant: ΔG = RT ln K.
Remember also that K = Kc for reactions of solutes in liquid solution, K = Kp when the reaction involves
only gases, and K includes both pressures and concentrations in heterogeneous systems.
You may well ask why R, the gas constant, shows up here when many reactions do not involve gases.
The presence of R in the expression for ΔG is due to the intimate relationship between R and how the
mole is defined. Recall from your study of the ideal gas law (text Section 5.3) that the amount of gas in
a given volume (22.41 L) at a specified temperature (273.15 K) and pressure (1 atm) was called a mole.
Also recall that R is the constant ratio of these standard conditions, PV/T. Thus R defines the mole. In
this expression for ΔG, the word mole in the units of R, J/(K ∙ mol), means “mole of reaction,”
explained previously in this chapter.
Exercise 18.9
Give the expression for K for each of the following reactions.
a.
CaCO3(s)
CaO(s) + CO2(g)
Pb (aq) + 2I–(aq)
2+
b.
PbI2(s)
c.
H+(aq) + HCO3–(aq)
H2O(l) + CO2(g)
Solution: The calculations are
a.
K = PCO 2 = Kp
b.
K = [Pb2+][I–]2 = Ksp
c.
K=
PCO
2


[ H ][HCO 3 ]
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
403
Exercise 18.10
Use data from text Table 18.2 to obtain the equilibrium constant Kp at 25C for the reaction
CaCO3(s)
CaO(s) + CO2(g)
Note that values of G f are needed for CaCO3 and CaO, even though the substances do not appear in
Kp = PCO 2 .
Wanted: Kp (from ΔG)
Given:
reaction; text Table 18.2 ( G f )
Known: ΔG = RT ln K; K for this reaction = Kp; R = 8.31 J/(K ∙ mol), so in order to have the
units work out properly, we must report ΔG as kJ/mol; T = 298 K.
Solution: Calculations follow:
CaCO3(s)  CaO(s) + CO2(g)
G f
–1128.8
–603.5
–394.4 kJ/mol
ΔG = [–603.5 + (–394.4) – (–1128.8)] kJ
ΔG = 130.9 kJ/mol
Rearrange the expression for ΔG to solve for Kp:
ln Kp =
 G o
 130 .0  10 3 J  K • mol
=
= 52.50
RT
mol  8.31 J  298 K
Kp = 2  1023
Exercise 18.11
Calculate the solubility product constant for Mg(OH)2 at 25C. The G f values (in kJ/mol) are as
follows: Mg2+(aq), –454.8; OH–(aq), –157.3; Mg(OH)2(s), –833.7.
Known:
ΔG = –RT ln Ksp; R (see Exercise 18.10) and T given above
Solution: Calculations follow:
Mg(OH)2(s)
G f
–933.9
Mg2+(aq) + 2OH–(aq)
–456.0
2(–157.3) kJ/mol
ΔG = [–454.8 + 2(–157.3) – (–833.7)] kJ
ΔG = 64.3 kJ/mol
 64.3  10 J  K • mol
 G o
ln Ksp =
=
= 25.97
RT
mol  8.31 J  298 K
3
K = 5  1012
Copyright © Houghton Mifflin Company. All rights reserved.
404
Chapter 18: Thermodynamics and Equilibrium
18.7 Change of Free Energy with Temperature
Learning Objectives

Describe how G at a given temperature (GT) is approximately related to H and S at
that temperature.

Describe how the spontaneity or nonspontaneity of a reaction is related to each of the four
possible combinations of signs of H and S.

Calculate G and K at various temperatures. (Example 18.10)
Problem-Solving Skill
9.
Calculating ΔG and K at various temperatures. Given ΔH and ΔS at 25C, calculate ΔG
and K for a reaction at a temperature other than 25C (Example 18.10).
Exercise 18.12
The thermodynamic equilibrium constant for the vaporization of water,
H2O(l)
H2O(g)
is Kp = PH 2O . Use thermodynamic data to calculate the vapor pressure of water at 45C. Compare your
answer with the value given in text Table 5.6.
Wanted:
PH 2O at 45C
Known:
PH 2O = Kp; ΔG = –RT ln Kp; need values from text Tables 6.2 and 18.1;
R= 8.31 J/(K ∙ mol); T = 318 K
Solution: Calculations follow:
H2O(l)  H2O(g)
H f
–285.8
S
–241.8 kJ/mol
188.7 J/(K ∙ mol)
69.9
ΔH = [–241.8 – (–285.8)] kJ = 44.0 kJ
ΔS = (188.7 – 69.95) J/K = 118.75 J/K
ΔG = ΔH – TΔS
ΔG = 44.0 kJ 
118.75 J
10 3 J
 318 K 
= 6.237  103 J
K
kJ
which we will express as 6.237  103 J/mol.
ln Kp =
 G o
6.237  103 J  K • mol
=
= 2.360
RT
mol  8.31 J  318 K
Kp = 9.4  102 atm
To compare with the value in text Table 5.6, convert to mmHg:
9.5  102 atm 
760 mmHg
= 72 mmHg
1 atm
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
405
The text table value is 71.9 mmHg. The calculated value is the text table value to two significant
figures.
Exercise 18.13
To what temperature must magnesium carbonate be heated to decompose it to MgO and CO2 at 1 atm?
Is this higher or lower than the temperature required to decompose CaCO3? Values of H f (in kJ/mol)
are as follows: MgO(s),601.2; MgCO3(s), –1111.7. Values of S(in J/K) are as follows: MgO(s),26.9;
MgCO3(s), 65.9. Data for CO2 are given in text Tables 6.2 and 18.1.
Wanted: T at which MgCO3 will decompose; compare with T for CaCO3
Given:
H f values; S values; see text Tables 6.2 and 18.1.
Known: ΔG = 0 at that T; ΔGo = ΔHo – TΔSo; assume ΔH and ΔS are constant, so we can use
their values at 25C for the calculation.
Solution: If ΔG = 0, then ΔH = TΔS, and T = ΔHo/ΔSo. Calculations of ΔH and ΔS follow:
MgCO3(s)  MgO(s) + CO2(g)
H f
ΔS
–1111.7
–601.2
65.9
26.9
–393.5 kJ/mol
213.7 J/(K ∙ mol)
ΔH = [–601.2 + (–393.5) – (–1111.7)] kJ
ΔH = 117.0 kJ
ΔS = [(26.9 + 213.7) – 65.9] J/K
ΔS = 174.7 J/K
To solve for temperature:
T=
3
117.0  10 J
174.7 J/K
T = 6.70  102 K; T = 398C
This temperature is considerably lower than the 848C decomposition temperature of CaCO3
given in the text immediately preceding this exercise.
ADDITIONAL PROBLEMS
1.
As a sample of gas cools at 1 atm pressure, it loses 54 J of heat, and the volume changes by 0.200
L. What are the values of q, w, and ΔU for this process?
2.
Gray tin, which exhibits the diamond structure, is transformed above 13.2C to white tin, which
has a metallic nature. The S values at 25C are 44.8 and 51.5 J/(K ∙ mol), respectively. Predict
ΔH for the phase transition.
3.
The ΔH value for the transition of PCl3(l) to PCl3(g) is 32.7 kJ. The observed boiling point is
75C. Predict the entropy change on boiling 2 mol of the liquid.
Copyright © Houghton Mifflin Company. All rights reserved.
406
4.
5.
Chapter 18: Thermodynamics and Equilibrium
Will the entropy change for each of the following processes probably be positive or negative?
Explain why.
a.
 C (graphite)
C (diamond) 
b.
 NH4Cl(s)
NH3(g) + HCl(g) 
c.
water
 Na+(aq) + Cl–(aq)
NaCl(s) 
d.
 CaCO3(s)
CaO(s) + CO2(g) 
The enthalpy of formation of ICl(g) is 17.8 kJ/mol at 25C. Calculate the absolute entropy change
for the reaction
1
1
 ICl(g)
I2(s) + Cl2(g) 
2
2
S for I2(s) = 116.1 J/(K ∙ mol); S for Cl2(g) = 223.0 J/(K ∙ mol); S for ICl(g) = 247.4 J/(K ∙ mol).
6.
a.
Using the following values for H f and S:
H2O(g): H f = –241.8 kJ/mol; S = 188.7 J/(K ∙ mol)
H2O(l): H f = –285.8 kJ/mol; S = 69.9 J/(K ∙ mol)
calculate ΔH and ΔS for the process
H2O(l)  H2O(g) at 1 atm and 25C
b.
Use your calculated values to determine ΔG for this process.
c.
Is this process spontaneous at 25C?
d.
If not, how could you make it spontaneous?
7.
Calculate ΔG for the complete combustion of 1 mol methanol, CH3OH(l), to CO2(g) and H2O(g).
The G f values are –166.2, –394.4, and –228.6 kJ/mol, respectively.
8.
Use the following information to calculate ΔG at 325.37 K for the reaction given. What is
important about this temperature?
2NO2(g)
N2O4(g)
H f
33.18
9.16 kJ/mol
S°
239.9
304.2 J/(K ∙ mol)
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
9.
407
The reaction
 CaO(s) + CO2(g)
CaCO3(s) 
is nonspontaneous at 25C and 1 atm. This does not surprise us when we know that marble, which
is essentially CaCO3 with small amounts of impurities present, is an excellent building stone. The
white marble Parthenon on the Acropolis in Athens has stood for 2500 years and has not
decomposed to CaO and CO2. Limestone (CaCO3 with more impurities than occur in white
marble) is used extensively for making quicklime, CaO, an important ingredient in building
plaster. The limestone is mixed with fuel that burns and heats the stone, decomposing it according
to the foregoing equation. How hot must we heat the limestone to make quicklime? Use the
following values for H f :
CaCO3(s), –1206 kJ/mol; CaO(s), –635.5 kJ/mol; CO2(g), –393.5 kJ/mol
and for S:
CaCO3(s), 92.9 J/(K ∙ mol); CaO(s), 38.21 J/(K ∙ mol); CO2(g), 213.7 J/(K ∙ mol)
10. For each of the following systems, calculate the thermodynamic equilibrium constant at 25C and
1 atm. From this value of the constant, predict whether the reaction will produce only negligible
amounts of the products, will yield moderate amounts of products, or will go essentially to
completion toward the right.
a.
 2HCl(g) + I2(g)
2HI(g) + Cl2(g) 
ΔG = –174.6 kJ
b.
 N2O4(g)
2NO2(g) 
ΔG = –4.78 kJ
c.
 2NO2(g)
N2(g) + 2O2(g) 
ΔG = 103 kJ
ANSWERS TO ADDITIONAL PROBLEMS
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
q = –54 J, w = +20.2 J, ΔU = –34 J (18.1)
2.
ΔH ≈ 1.9 kJ/mol (18.2, PS Sk. 1)
3.
ΔS ≈ 188 J/K (18.1, 18.2, PS Sk. 1)
4.
5.
a.
ΔS should be positive because graphite is softer and the disorder greater than in diamond.
b.
ΔS will be negative; the disorder in NH4Cl(s) crystals is much less than the disorder in
NH3(g) and HCl(g). Furthermore, the reaction proceeds from 2 mol of a gas to 1 mol of a
solid—with a decrease in the number of individual molecular units.
c.
ΔS will be positive because NaCl in solution is dissociated into 2 mol of ions per mole of
NaCl dissolved, and these ions, with their freedom of motion, are more disordered than the
highly ordered Na+ and Cl– ions in crystalline NaCl.
d.
ΔS will be negative because a gas with large entropy becomes incorporated into a solid with
much less entropy, and the reaction proceeds with 2 mol of reactants forming 1 mol of
product. (18.3, PS Sk. 2)
ΔS = 77.8 J/K (18.3, PS Sk. 3)
Copyright © Houghton Mifflin Company. All rights reserved.
408
Chapter 18: Thermodynamics and Equilibrium
6.
a.
ΔH = 44.0 kJ (6.8, 18.1)
ΔS = 118.8 J/K (18.3, PS Sk. 3)
b.
ΔG = 8.60 kJ (18.4, PS Sk. 4)
c.
Because ΔG is positive, the process is not spontaneous at 25C. (18.4, PS Sk. 6)
d.
The process can be made spontaneous by increasing the temperature until –TΔS exceeds ΔH,
giving a negative value of ΔG. We know that water boils at 100C. At this temperature,
H2O(l) and H2O(g) are in equilibrium, and ΔG for the vaporization process must be zero. For
the process to become spontaneous, the temperature must be raised an infinitesimal amount
above 100C. (18.7, PS Sk. 9)
7.
ΔG = –685.4 kJ (18.4, PS Sk. 5)
8.
ΔG = ~ 0. This is the temperature at which the stoichiometry of the balanced equation represents
the relative amounts of the substances at equilibrium. (18.7, PS Sk. 9)
9.
The temperature must be kept above 1100 K. (18.2, 18.3, 18.4, 18.7, PS Sk. 3, 4, 6, 9)
10.
a.
K = Kp = 4  1030; the reaction goes essentially to completion.
b.
K = Kp = 6.89; only moderate amounts of products form.
c.
K = Kp = 9  10–19; negligible amounts of products form. (18.6, PS Sk. 8)
CHAPTER POST-TEST
1.
The enthalpy change when liquid dimethyl sulfide, (CH3)2S(l), vaporizes at 25C is 28.0 kJ/mol.
What is the entropy change when one mole of vapor in equilibrium with the liquid condenses to
liquid at 25C if the entropy of this vapor at 25C is 285.7J/(K∙ mol)?
2.
Predict by inspection the sign of ΔS for each of the following reactions, and explain your
answers.
3.
a.
2NOCl(g)
2NO(g) + Cl2(g)
b.
 ZnCl2(aq) + H2(g)
Zn(s) + 2HCl(aq) 
c.
N2(g) + 3H2(g)
d.
 HCOO–(aq) + H2O(l)
HCOOH(aq) + OH–(aq) 
2NH3(g)
Calculate the change in entropy ΔS for the following neutralization reaction:
 CaSO4(s) + 2H2O(l)
Ca2+(aq) + 2OH–(aq) + H2SO4(aq) 
Use text Appendix C for standard entropy values.
4.
Calculate the ΔG for the following reaction at 25C:
2NO2(g)
2NO(g) + O2(g)
Is the reaction spontaneous as written? Use values of H f and S from text AppendixC.
Copyright © Houghton Mifflin Company. All rights reserved.
Chapter 18: Thermodynamics and Equilibrium
5.
409
Determine whether each of the following statements is true or false. If the statement is false,
change it so that it is true.
a.
A flow of entropy, a transfer of disorder, always follows a temperature change.
True/False: ________________________________________________________________
__________________________________________________________________________
b.
The second law of thermodynamics states that the total entropy of a system and
surroundings always increases for a spontaneous process. True/False:
__________________________________________________________________________
__________________________________________________________________________
c.
The free-energy change of a reacting system is the amount of energy available to do work.
True/False: ________________________________________________________________
__________________________________________________________________________
d.
The zero point for free energy is the same as that for entropy. True/False:
__________________________________________________________________________
__________________________________________________________________________
6.
Calculate the free-energy change ΔG for the following reaction using values of G f from text
Appendix C:
 Mg2+(aq) + 2Cl–(aq) + H2(g)
Mg(s) + 2HCl(aq) 
7.
Determine whether each of the reactions in Problem 2 is spontaneous or nonspontaneous as
written. Give reasons for your answers. [Use text Appendix C. G f (NOCl) = 66.36 kJ/mol.]
8.
The work that a reacting system open to the atmosphere can do is PΔV work, where the
atmosphere is pushed away. Using the ideal gas law (text Section 5.3), express this work as a
more useful term, and calculate the maximum amount of work that could be done in the following
reaction at 25C:
 16CO2(g) + 18H2O(g)
2C8H18(l) + 25O2(g) 
9.
Use data from text Table 18.2 to calculate the thermodynamic equilibrium constant for the
reaction
2H2O2(l)  2H2O(l) + O2(g)
at 25C. G f for H2O2(l) = –120.4 kJ/mol.
10. To what temperature must potassium chlorate be heated to decompose it to potassium chloride
and oxygen gas? Use values of H f and S from text Appendix C. Values for KClO3 are
H f = –391.20 kJ/mol, S = 142.97 J/K ∙ mol.
Copyright © Houghton Mifflin Company. All rights reserved.
410
Chapter 18: Thermodynamics and Equilibrium
ANSWERS TO CHAPTER POST-TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
ΔS = –94.0 J/K (18.2, PS Sk. 1)
2.
a.
ΔS > 0; there are two molecules of gas on the left and three on the right.
b.
ΔS > 0; there is one molecule of gas on the right.
c.
ΔS < 0; there are four molecules of gas on the left and only two on the right.
d.
ΔS  0. One would expect the molecular HCOOH to be associated with water molecules and
the small OH– to be tightly surrounded by water molecules. In the products, the COOH– ion
would not have quite the association with water the OH– does, but water associates with
itself. (The calculated value is 8.08 J/K.) (18.3, PS Sk. 2)
3.
ΔS = 305 J/K (18.3, PS Sk. 3)
4.
ΔG = 70.5 kJ; reaction is nonspontaneous. (18.4, PS Sk. 4)
5.
6.
a.
True. (18.2)
b.
True. (18.2)
c.
True. (18.5)
d.
False. The zero point for free energy is the element in its standard state at 1 atm pressure and
a specified temperature (usually 25C). The zero point for entropy is a perfect crystalline
substance at 0 K. (18.3, 18.4)
ΔG = –456.01 kJ (18.4, PS Sk. 5)
7.
a.
Nonspontaneous; ΔG > 0
b.
Spontaneous; ΔG < 0
c.
Spontaneous; ΔG < 0
d.
Spontaneous; ΔG < 0 (18.4, PS Sk. 5, 6)
8.
wmax = ΔnRT = 22.3 kJ (18.1, 18.5)
9.
K = Kp = 9  1040 (18.6, PS Sk. 8)
10. The reaction is spontaneous at any temperature. (18.7, PS Sk. 9)
Copyright © Houghton Mifflin Company. All rights reserved.