# Silberberg7e solution manual ch 20

CHAPTER 20 THERMODYNAMICS:
ENTROPY, FREE ENERGY, AND THE
DIRECTION OF CHEMICAL REACTIONS
FOLLOW–UP PROBLEMS
20.1A

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than
that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For
substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the
products is greater than that of the reactants, S is positive.
Solution:
a) PCl5(g). For substances with the same type of atoms and in the same physical state, entropy increases with
increasing number of atoms per molecule because more types of molecular motion are available.
b) BaCl2(s). Entropy increases with increasing atomic size. The Ba2+ ion and Cl– ion are larger than the Ca2+ ion
and F– ion, respectively.
c) Br2(g). Entropy increases from solid  liquid  gas.

20.1B

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than
that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For

substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the
products is greater than that of the reactants, S is positive.
Solution:
a) LiBr(aq). For substances with the same number of atoms and in the same physical state, entropy increases with
increasing atomic size. LiBr has the lower molar mass of the two substances and, therefore, the lower entropy.
b) Quartz. Quartz has a crystalline structure and the particles have less freedom (and lower entropy) in that
structure than in glass, an amorphous solid.
c) Cyclohexane. Ethylcyclobutane has a side chain, which has more freedom of motion than the atoms in the
ring. In cyclohexane, there are no side chains. Freedom of motion is restricted by the ring structure.

20.2A

Plan: Predict the sign of Srxn
by comparing the randomness of the products with the randomness of the reactants.

Calculate Srxn
using Appendix B values and the relationship Srxn
= m S products
– n Sreactants
.

Solution:
a) 4NO (g)  N2O(g) + N2O3(g)

The Srxn
is predicted to decrease ( Srxn
< 0) because four moles of random, gaseous product are transformed
into two moles of random, gaseous product (the change in gas moles is –2).

Srxn
= m S products
– n Sreactants

Srxn
= [(1 mol N2O3)(Sº of N2O3) + (1 mol N2O)(Sº of N2O)]
– [(4 mol NO)(Sº of NO)]

Srxn
= [(1 mol N2O3)(314.7 J/mol•K) + (1 mol N2O)(219.7 J/mol•K)]
 [(4 mol NO)(210.65 J/mol•K)]
= – 308.2 J/K

Srxn
< 0 as predicted.

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20-1

b) CH3OH(g)  CO(g) + 2H2(g)

is predicted to be greater than zero.
The change in gaseous moles is +2, so the sign of Srxn

Srxn
= [(1 mol CO)(Sº of CO) + (2 mol H2)(Sº of H2)]
 [(1 mol CH3OH)(Sº of CH3OH)]

Srxn
= [(1 mol CO)(197.5 J/mol•K) + (2 mol H2)(130.6 J/mol•K)]
 [(1 mol CH3OH)(238 J/mol•K)]
= 220.7 = 221 J/K

Srxn
> 0 as predicted.

20.2B

by comparing the randomness of the products with the randomness of the reactants.
Plan: Predict the sign of Srxn

Calculate Srxn
using Appendix B values and the relationship Srxn
= m S products
– n Sreactants
.

Solution:
a) 2NaOH(s) + CO2(g)  Na2CO3(s) + H2O(l)

The Srxn
is predicted to decrease ( Srxn
< 0) because the more random, gaseous reactant is transformed into a
more ordered, liquid product.

Srxn
= m S products
– n Sreactants

Srxn
= [(1 mol Na2CO3)(Sº of Na2CO3) + (1 mol H2O)(Sº of H2O)]
– [(2 mol NaOH)(Sº of NaOH) + (1 mol CO2)(Sº of CO2)]

Srxn
= [(1 mol Na2CO3)(139 J/mol•K) + (1 mol H2O)(69.940 J/mol•K)]
 [(2 mol NaOH)(64.454 J/mol•K) + (1 mol CO2)(213.7 J/mol•K)]
= –133.668 = – 134 J/K

Srxn
< 0 as predicted.
b) 2Fe(s) + 3H2O(g)  Fe2O3(s) + 3H2(g)

The change in gaseous moles is zero, so the sign of Srxn
is difficult to predict. Iron(III) oxide has greater
entropy than Fe because it is more complex, but this is offset by the greater molecular complexity of H2O versus
H2.

Srxn
= [(1 mol Fe2O3)(Sº of Fe2O3) + (3 mol H2)(Sº of H2)]
 [(2 mol Fe)(Sº of Fe) + (3 mol H2O)(Sº of H2O)]

Srxn
= [(1 mol Fe2O3)(87.400 J/mol•K) + (3 mol H2)(130.6 J/mol•K)]
 [(2 mol Fe)(27.3 J/mol•K) + (3 mol H2O)(188.72 J/mol•K)]
= –141.56 = –141.6 J/K

The negative Srxn
shows that the greater entropy of H2O versus H2 does outweigh the greater entropy of Fe2O3
versus Fe.

20.3A

using Appendix B. Determine
Plan: Write the balanced equation for the reaction and calculate the Srxn

the Ssurr by first finding H rxn
. Add Ssurr to Srxn
to verify that Suniv is positive.
Solution:
P4(s) + 6Cl2(g)  4PCl3(g)

Srxn
= [(4 mol PCl3)(Sº of PCl3)] – [(1 mol P4)(Sº of P4) + (6 mol Cl2)(Sº of Cl2)]

Srxn
= [(4 mol PCl3)(312 J/mol•K)]
– [(1 mol P4)(41.1 J/mol•K) + (6 mol Cl2) (223.0J/mol•K)]

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20-2

Srxn
= –131J/K (The entropy change is expected to be negative because the change in gas moles is negative.)

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(4 mol PCl3)( H f of PCl3)]

– [(1 mol P4)( H f of P4) + (6 mol Cl2)( H f of Cl2)]

H rxn
= [(4 mol PCl3)(–287 kJ/mol)]
– [(1 mol P4)(0 kJ/mol) + (6 mol Cl2)(0 kJ/mol)]

H rxn
= –1148 kJ

Ssurr = 

Hrxn
1148 kJ
= 
= 3.8523 kJ/K(103 J/1 kJ) = 3850 J/K
298 K
T

 S univ = Srxn
+ Ssurr = (–131 J/K) + (3850 J/K) = 3719 J/K

Because S univ is positive, the reaction is spontaneous at 298 K.
20.3B

using Appendix B. Determine
Plan: Write the balanced equation for the reaction and calculate the Srxn

the Ssurr by first finding H rxn
. Add Ssurr to Srxn
to verify that Suniv is positive.
Solution:
2FeO(s) + 1/2O2(g)  Fe2O3(s)

Srxn
= [(1 mol Fe2O3)(Sº of Fe2O3)] – [(2 mol FeO)(Sº of FeO) + (1/2 mol O2)(Sº of O2)]

Srxn
= [(1 mol Fe2O3)(87.400 J/mol•K)]
– [(2 mol FeO)(60.75 J/mol•K) + (1/2 mol O2) (205.0J/mol•K)]

Srxn
= –136.6 J/K (entropy change is expected to be negative because gaseous reactant
is converted to solid product).

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(1 mol Fe2O3)( H f of Fe2O3)]

– [(2 mol FeO)( H f of FeO) + (1/2 mol O2)( H f of O2)]

H rxn
= [(1 mol Fe2O3)(–825.5 kJ/mol)]
– [(2 mol FeO)(–272.0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

H rxn
= –281.5 kJ

Ssurr = 

Hrxn
281.5 kJ
= 
= 0.94463 kJ/K(103 J/1 kJ) = 944.63 J/K
T
298 K

 S univ = Srxn
+ Ssurr = (–136.6 J/K) + (944.63 J/K) = 808.03 = 808 J/K

Because S univ is positive, the reaction is spontaneous at 298 K.
This process is also known as rusting. Common sense tells us that rusting occurs spontaneously. Although the
entropy change of the system is negative, the increase in entropy of the surroundings is large enough to offset

Srxn
.

20.4A

Plan: Calculate the H rxn
using H f values from Appendix B. Calculate Srxn
from tabulated Sº values and then

use the relationship Grxn
= H rxn
– T Srxn
.
Solution:

H rxn
= m H f (products) – n H f (reactants)

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20-3

H rxn
= [(2 mol NOCl)( H f of NOCl)] – [(2 mol NO)( H f of NO) + (1 mol Cl2)( H f of Cl2)]

H rxn
= [(2 mol NOCl)(51.71 kJ/mol)] – [(2 mol NO)(90.29 kJ/mol) + (1 mol Cl2)(0 kJ/mol)]

H rxn
= –77.16 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(2 mol NOCl)(Sº of NOCl)] – [(2 mol NO)(Sº of NO) + (1 mol Cl2)(Sº of Cl2)]

Srxn
= [(2 mol NOCl)(261.6 J/mol•K)] – [(2 mol NO)(210.65 J/mol•K) + (1 mol Cl2)(223.0 J/mol•K)]

Srxn
= –121.1 J/K

Grxn
= H rxn
– T Srxn
= –77.16 kJ – [(298 K)(–121.1 J/K)(1 kJ/103 J)] = –41.0722 = –41.1 kJ

20.4B

Plan: Calculate the H rxn
using H f values from Appendix B. Calculate Srxn
from tabulated Sº values and then

use the relationship Grxn
= H rxn
– T Srxn
.
Solution:

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(2 mol NO2)( H f of NO2)] – [(2 mol NO)( H f of NO) + (1 mol O2)( H f of O2)]

H rxn
= [(2 mol NO2)(33.2 kJ/mol)] – [(2 mol NO)(90.29 kJ/mol) + (1 mol O2)(0 kJ/mol)]

H rxn
= –114.18 = –114.2 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(2 mol NO2)(Sº of NO2)] – [(2 mol NO)(Sº of NO) + (1 mol O2)(Sº of O2)]

Srxn
= [(2 mol NO2)(239.9 J/mol•K)]
– [(2 mol NO)(210.65 J/mol•K) + (1 mol O2)(205.0 J/mol•K)]

Srxn
= –146.5 J/K

Grxn
= H rxn
– T Srxn
= –114.2 kJ – [(298 K)(–146.5 J/K)(1 kJ/103 J)] = –70.543 = –70.5 kJ

20.5A

using the relationship
Plan: Use Gf values from Appendix B to calculate Grxn

Grxn
= m Gf (products) – n Gf (reactants) .

Solution:

a) Grxn
= [(2 mol NOCl)( Gf of NOCl)] – [(2 mol NO)( Gf of NO) + (1 mol Cl2)( Gf of Cl2)]

Grxn
= [(2 mol NOCl)(66.07 kJ/mol)] – [(2 mol NO)(86.60 kJ/mol) + (1 mol Cl2)(0 kJ/mol)]

Grxn
= –41.06 kJ

b) Grxn
= [(2 mol Fe)( Gf of Fe) + (3 mol H2O)( Gf of H2O)] –

[(3 mol H2)( Gf of H2) + (1 mol Fe2O3)( Gf of Fe2O3)]

Grxn
= [(2 mol Fe)(0 kJ/mol) + (3 mol H2O)(–228.60 kJ/mol)] –
[(3 mol H2)(0 kJ/mol) + (1 mol Fe2O3)( –743.6 kJ/mol)]

Grxn
= 57.8 kJ

20.5B

Plan: Use Gf values from Appendix B to calculate Grxn
using the relationship

Grxn
= m Gf (products) – n Gf (reactants) .

Solution:
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20-4

= [(2 mol NO2)( Gf of NO2)] – [(2 mol NO)( Gf of NO) + (1 mol O2)( Gf of O2)]
a) Grxn

Grxn
= [(2 mol NO2)(51 kJ/mol)] – [(2 mol NO)(86.60 kJ/mol) + (1 mol O2)(0 kJ/mol)]

Grxn
= –71.2 = –71 kJ

b) Grxn
= [(2 mol CO)( Gf of CO)] – [(2 mol C)( Gf of C) + (1 mol O2)( Gf of O2)]

Grxn
= [(2 mol CO)(–137.2 kJ/mol)] – [(2 mol C)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)]

Grxn
= –274.4 kJ

20.6A

Plan: Predict the sign of Srxn
by comparing the randomness of the products with the randomness of the

reactants. Use the relationship Grxn
= H rxn
– T Srxn
Solution:
a) The reaction is X2Y2(g) → X2(g) + Y2(g). Since there are more moles of gaseous product than there are of
gaseous reactant, entropy increases and S > 0.
b) The reaction is only spontaneous above 325ºC or in other words, at high temperatures. In the relationship

Grxn
= H rxn
– T Srxn
, when S > 0 so that – TS° is < 0, G° will only be negative at high T if H° > 0.

20.6B

Plan: Predict the sign of Srxn
by comparing the randomness of the products with the randomness of the

reactants. Use the relationship Grxn
= H rxn
– T Srxn
Solution:
a) A solid forms a gas and a liquid, so S > 0. A crystalline array breaks down, so H > 0.
b) For the reaction to occur spontaneously (G < 0), – TS° must be greater than H, which would occur only at
higher T.

20.7A

Plan: Use the equation G° = H° – TS° to determine if the reaction is spontaneous (G° < 0). Then examine
the same equation to determine the effect of raising the temperature on the spontaneity of the reaction.
Solution:

a) Grxn
= H rxn
– T Srxn
= –192.7 kJ – [(298 K)(–308.2 J/K)(1 kJ/103 J)] = –100.8564 = –100.9 kJ
Because G < 0, the reaction is spontaneous at 298 K.
b) As temperature increases, – TS° becomes more positive, so the reaction becomes less spontaneous at higher
temperatures.

c) Grxn
= H rxn
– T Srxn
= –192.7 kJ – [(773 K)(–308.2 J/K)(1 kJ/103 J)] = 45.5386 = 45.5 kJ

20.7B

Plan: Examine the equation G° = H° – TS° and determine which combination of enthalpy and entropy will
describe the given reaction.
Solution:
Two choices can already be eliminated:
1) When H > 0 (endothermic reaction) and S < 0 (entropy decreases), the reaction is always
nonspontaneous, regardless of temperature, so this combination does not describe the reaction.
2) When H < 0 (exothermic reaction) and S > 0 (entropy increases), the reaction is always spontaneous,
regardless of temperature, so this combination does not describe the reaction.
Two combinations remain: 3) H° > 0 and S° > 0, or 4) H° < 0 and S° < 0. If the reaction becomes
spontaneous at –40°C, this means that G° becomes negative at lower temperatures. Case 3) becomes
spontaneous at higher temperatures, when the –TS° term is larger than the positive enthalpy term. By process of
elimination, Case 4) describes the reaction. At a lower temperature, the negative H° becomes larger than the
positive (–TS°) value, so G° becomes negative.

20.8A

Plan: To find the temperature at which the reaction becomes spontaneous, use

Grxn
= 0 = H rxn
– T Srxn
and solve for temperature.

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20-5

Solution:
The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs
when ΔG is 0.

Grxn
= H rxn
– T Srxn
0 = –192.7x103 J – (T)(–308.2 J/K)
T = 625.2 K – 273.15 = 352.0°C

20.8B

Plan: To find the temperature at which the reaction becomes spontaneous, use

Grxn
= 0 = H rxn
– T Srxn
and solve for temperature. H rxn
can be calculated from the individual H f values
of the reactants and products by using the relationship

H rxn
= m H f (products) – n H f (reactants) . Srxn
can be calculated from the individual S  values of the

reactants and products by using the relationship Srxn
= m S products
– n Sreactants
.

Solution:
CaO(s) + CO2(g) → CaCO3(s)

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(1 mol CaCO3)( H f of CaCO3)]

– [(1 mol CaO)( H f of CaO) + (1 mol CO2)( H f of CO2)]

H rxn
= [(1 mol CaCO3)( –1206.9 kJ/mol)]
– [(1 mol CaO)(–635.1 kJ/mol) + (1 mol CO2)( –393.5 kJ/mol)]

H rxn
= –178.3 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(1 mol CaCO3)(Sº of CaCO3)] – [(1 mol CaO)(Sº of CaO) + (1 mol CO2)(Sº of CO2)]

Srxn
= [(1mol CaCO3)(92.9 J/mol•K)]
– [(1 mol CaO)(38.2 J/mol•K) + (1 mol CO2)(213.7 J/mol•K)]

Srxn
= –159.0 J/K = – 0.159 kJ/K

Grxn
= 0 = H rxn
– T Srxn

H rxn
= T Srxn

H 

–178.3 kJ
= 1121.384 = 1121 K
–0.1590
kJ/K
S
The reaction becomes spontaneous at temperatures < 1121 K.

T =

20.9A

=

Plan: First find G°, then calculate K from G° = –RT ln K. Calculate G  using Gf values in the relationship

Grxn
= m Gf (products) – n Gf (reactants) .

Solution:
2C(graphite) + O2(g) 2CO(g)

Grxn
= m Gf (products) – n Gf (reactants)

Grxn
= [(2 mol CO)(–137.2 kJ/mol)] – [(2 mol C)(0 kJ/mol) + (1 mol O2)( 0 kJ/mol)] = –274.4 kJ

ln K = 

274.4 kJ/mol
G 
 1000 J 
= 
= 110.7536 = 111
8.314
J/mol•K
298
K
RT


  1 kJ 

K = e111 = 1.6095x1048 = 1.6x1048
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20-6

20.9B

Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K.
Solution:
G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (2.22x10–15) = 8.35964x104 J = 83.6 kJ/mol

20.10A Plan: Write the equilibrium expression for the reaction and calculate Qc for each scene. Remember that each
particle represents 0.10 mol and that the volume is 1.0 L. A reaction that is proceeding to the right will have G°
< 0 and a reaction that is proceeding to the left will have G° > 0. A reaction at equilibrium has G° = 0.
Solution:
a) A(g) + 3B(g)  AB3(g)
b) Qc =
Mixture 1: Qc =

=

Mixture 2: Qc =

=

Mixture 3: Qc =

=

.
.

.
.

.

.
.

.

.

= 0.98
=8
= 250

Mixture 2 is at equilibrium since Qc = Kc
c) Qc < Kc for Mixture 1 and the reaction is proceeding right to reach equilibrium; thus G° < 0. Mixture 2 is at
equilibrium and G° = 0. Mixture 3 proceeds to the left to reach equilibrium since Qc > Kc and G° > 0. The
ranking for most positive to most negative is 3 > 2 > 1.
20.10B Plan: Write the equilibrium expression for the reaction and calculate Qc for each scene. A reaction that
is proceeding to the right will have G° < 0 and a reaction that is proceeding to the left will have G° > 0. A
reaction at equilibrium has G° = 0.
Solution:
a) X2(g) + 2Y2(g)  2XY2(g)
[XY2 ]2
Qc =
[X 2 ][Y2 ]2
Mixture 1: Qc =
Mixture 2: Qc =
Mixture 3: Qc =

[XY2 ]2
2

[X 2 ][Y2 ]
[XY2 ]2

[X 2 ][Y2 ]2
[XY2 ]2

=
=
=

[5]2
[2][1]2
[4]2
[2][2]2

= 12.5
=2

[2]2

= 0.25
[X 2 ][Y2 ]2
[4][2]2
Mixture 2 is at equilibrium since Qc = Kc
b) Qc > Kc for Mixture 1 and the reaction is proceeding left to reach equilibrium; thus G° > 0. Mixture 2 is at
equilibrium and G° = 0. Mixture 3 proceeds to the right to reach equilibrium since Qc < Kc and G° < 0. The
ranking for most negative to most positive is 3 < 2 < 1.
c) Any reaction mixture moves spontaneously towards equilibrium so both changes have a negativeG°.
20.11A Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K. The free energy of the
reaction under non-standard state conditions is calculated using G = G° + RT ln Q.
Solution:
33.5 kJ/mol
G 
 1000 J 
a) ln K = 
= 
= 13.5213
RT
8.314 J/mol•K  298 K   1 kJ 
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20-7

K = e13.5213 = 7.4512x105 = 7.45x105
b) Q =

=

.
.

.

= 3.0

G = G° + RT ln Q = (–33.5 kJ/mol) + (8.314 J/mol•K) (1 kJ/1000 J) (298 K) ln (3.0)
= –30.7781 = –30.8 kJ/mol
20.11B Plan: Write a balanced equation for the dissociation of hypobromous acid in water. The free energy of the
reaction at standard state is calculated using G° = –RT ln K. The free energy of the reaction under non-standard
state conditions is calculated using G = G° + RT ln Q.
Solution:
HBrO(aq) + H2O(l)  BrO–(aq) + H3O+(aq)
a) G° = –RT ln K = – (8.314 J/mol•K)(298)ln (2.3x10–9) = 4.927979x104 = 4.9x104 J/mol = 49 kJ/mol
 H3O   BrO 
6.0x104   0.10

b) Q =
=
 HBrO
0.20
G = G° + RT ln Q = (4.927979x104 J/mol) + (8.314 J/mol•K) (298 K) ln

6.0x104   0.10

0.20
 

= 2.9182399x104 = 2.9x104 J/mol = 29 kJ/mol
The value of Ka is very small, so it makes sense that G° is a positive number. The natural log of a negative
exponent gives a negative number (ln 3.0 x 10–4), so the value of G decreases with concentrations lower than the
standard state 1 M values.
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B20.1

Plan: Convert mass of glucose (1 g) to moles and use the ratio between moles of glucose and moles of ATP to
find the moles and then molecules of ATP formed. Do the same calculation with tristearin.
Solution:
a) Molecules of ATP/g glucose =
 1 mol glucose   36 mol ATP   6.022x1023 molecules ATP 
1 g glucose  


 
1 mol ATP
 180.16 g glucos e   1 mol glucose  

= 1.20333x1023 = 1.203x1023 molecules ATP/g glucose
b) Molecules of ATP/g tristearin =
 1 mol tristearin   458 mol ATP   6.022x1023 molecules ATP 
1 g tristearin  


 
1 mol ATP
 897.50 g tristearin   1 mol tristearin  

= 3.073065x1023 = 3.073x1023 molecules ATP/g tristearin

B20.2

Plan: Add the two reactions to obtain the overall process; the values of the two reactions are
then added to obtain for the overall reaction.
Solution:
creatine phosphate  creatine + phosphate
G° = – 43.1 kJ/mol
ADP + phosphate  ATP
G° = +30.5 kJ/mol
creatine phosphate + ADP  creatine + ATP
G = –43.1 kJ/mol + 30.5 kJ/mol = –12.6 kJ/mol

END–OF–CHAPTER PROBLEMS
20.1

Spontaneous processes proceed without outside intervention. The fact that a process is spontaneous does not mean
that it will occur instantaneously (in an instant) or even at an observable rate. The rusting of iron is an example of

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20-8

20.2

a process that is spontaneous but very slow. The ignition of gasoline is an example of a process that is not
spontaneous but very fast.
A spontaneous process occurs by itself (possibly requiring an initial input of energy), whereas a nonspontaneous
process requires a continuous supply of energy to make it happen. It is possible to cause a nonspontaneous
process to occur, but the process stops once the energy source is removed. A reaction that is found to be
nonspontaneous under one set of conditions may be spontaneous under a different set of conditions (different
temperature, different concentrations).

20.3

a) The energy of the universe is constant.
b) Energy cannot be created or destroyed.
c) Esystem = –Esurroundings
The first law is concerned with balancing energy for a process but says nothing about whether the process can, in
fact, occur.

20.4

Entropy is related to the freedom of movement of the particles. A system with greater freedom of movement has
higher entropy.
a) and b) Probability is so remote as to be virtually impossible. Both would require the simultaneous, coordinated
movement of a large number of independent particles, so are very unlikely.

20.5

Vaporization is the change of a liquid substance to a gas so Svaporization = Sgas – Sliquid. Fusion is the change of a
solid substance into a liquid so Sfusion = Sliquid – Ssolid. Vaporization involves a greater change in volume than
fusion. Thus, the transition from liquid to gas involves a greater entropy change than the transition from solid to
liquid.

20.6

In an exothermic process, the system releases heat to its surroundings. The entropy of the surroundings increases
because the temperature of the surroundings increases (Ssurr > 0). In an endothermic process, the system absorbs
heat from the surroundings and the surroundings become cooler. Thus, the entropy of the surroundings decreases
(Ssurr < 0). A chemical cold pack for injuries is an example of a spontaneous, endothermic chemical reaction as
is the melting of ice cream at room temperature.

20.7

a) According to the third law the entropy is zero.
b) Entropy will increase with temperature.
c) The third law states that the entropy of a pure, perfectly crystalline element or compound may be taken as zero
at zero Kelvin. Since the standard state temperature is 25°C and entropy increases with temperature, S° must be
greater than zero for an element in its standard state.
d) Since entropy values have a reference point (0 entropy at 0 K), actual entropy values can be determined, not
just entropy changes.

20.8

Plan: A spontaneous process is one that occurs by itself without a continuous input of energy.
Solution:
a) Spontaneous, evaporation occurs because a few of the liquid molecules have enough energy to break away
from the intermolecular forces of the other liquid molecules and move spontaneously into the gas phase.
b) Spontaneous, a lion spontaneously chases an antelope without added force. This assumes that the lion has not
just eaten.
c) Spontaneous, an unstable substance decays spontaneously to a more stable substance.

20.9

a) The movement of Earth about the Sun is spontaneous.
b) The movement of a boulder against gravity is nonspontaneous.
c) The reaction of an active metal (sodium) with an active nonmetal (chlorine) is spontaneous.

20.10

Plan: A spontaneous process is one that occurs by itself without a continuous input of energy.
Solution:
a) Spontaneous, with a small amount of energy input, methane will continue to burn without additional energy
(the reaction itself provides the necessary energy) until it is used up.

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20-9

20.11

b) Spontaneous, the dissolved sugar molecules have more states they can occupy than the crystalline sugar, so
the reaction proceeds in the direction of dissolution.
c) Not spontaneous, a cooked egg will not become raw again, no matter how long it sits or how many times it is
mixed.
a) If a satellite slows sufficiently, it will fall to Earth’s surface through a spontaneous process.
b) Water is a very stable compound; its decomposition at 298 K and 1 atm is not spontaneous.
c) The increase in prices tends to be spontaneous.

20.12

Plan: Particles with more freedom of motion have higher entropy. Therefore, Sgas > Sliquid > Ssolid. If the products
of the process have more entropy than the reactants, Ssys is positive. If the products of the process have less
entropy than the reactants, Ssys is negative.
Solution:
a) Ssys positive, melting is the change in state from solid to liquid. The solid state of a particular substance
always has lower entropy than the same substance in the liquid state. Entropy increases during melting.
b) Ssys negative, the entropy of most salt solutions is greater than the entropy of the solvent and solute
separately, so entropy decreases as a salt precipitates.
c) Ssys negative, dew forms by the condensation of water vapor to liquid. Entropy of a substance in the gaseous
state is greater than its entropy in the liquid state. Entropy decreases during condensation.

20.13

a) Ssys positive

20.14

Plan: Particles with more freedom of motion have higher entropy. Therefore, Sgas > Sliquid > Ssolid. If the products
of the process have more entropy than the reactants, Ssys is positive. If the products of the process have less
entropy than the reactants, Ssys is negative.
Solution:
a) Ssys positive, the process described is liquid alcohol becoming gaseous alcohol. The gas molecules have
greater entropy than the liquid molecules.
b) Ssys positive, the process described is a change from solid to gas, an increase in possible energy states for the
system.
c) Ssys positive, the perfume molecules have more possible locations in the larger volume of the room than inside
the bottle. A system that has more possible arrangements has greater entropy.

20.15

a) Ssys negative

20.16

Plan: Ssys is the entropy of the products – the entropy of the reactants. Use the fact that Sgas > Sliquid > Ssolid; also,
the greater the number of particles of a particular phase of matter, the higher the entropy.
Solution:
a) Ssys negative, reaction involves a gaseous reactant and no gaseous products, so entropy decreases. The
number of particles also decreases, indicating a decrease in entropy.
b) Ssys negative, gaseous reactants form solid product and number of particles decreases, so entropy decreases.
c) Ssys positive, when a solid salt dissolves in water, entropy generally increases since the entropy of the
aqueous mixture has higher entropy than the solid.

20.17

a) Ssys negative

20.18

Plan: Ssys is the entropy of the products – the entropy of the reactants. Use the fact that Sgas > Sliquid > Ssolid; also,
the greater the number of particles of a particular phase of matter, the higher the entropy.
Solution:
a) Ssys positive, the reaction produces gaseous CO2 molecules that have greater entropy than the physical states
of the reactants.
b) Ssys negative, the reaction produces a net decrease in the number of gaseous molecules, so the system’s
entropy decreases.
c) Ssys positive, the reaction produces a gas from a solid.

b) Ssys positive

b) Ssys negative

b) Ssys negative

c) Ssys negative

c) Ssys negative

c) Ssys negative

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20-10

b) Ssys positive

c) Ssys negative

20.19

a) Ssys negative

20.20

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than
that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For
substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the
products is greater than that of the reactants, S is positive.
Solution:
a) Ssys positive, decreasing the pressure increases the volume available to the gas molecules so entropy of the
system increases.
b) Ssys negative, gaseous nitrogen molecules have greater entropy (more possible states) than dissolved
nitrogen molecules.
c) Ssys positive, dissolved oxygen molecules have lower entropy than gaseous oxygen molecules.

20.21

a) Ssys negative

20.22

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than
that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For
substances in the same phase, entropy increases with atomic size and molecular complexity.
Solution:
a) Butane has the greater molar entropy because it has two additional CH bonds that can vibrate and has greater
rotational freedom around its bond. The presence of the double bond in 2-butene restricts rotation.
b) Xe(g) has the greater molar entropy because entropy increases with atomic size.
c) CH4(g) has the greater molar entropy because gases in general have greater entropy than liquids.

20.23

a) N2O4(g); it has greater molecular complexity.
b) CH3OCH3(l); hydrogen bonding in CH3CH2OH would increase order.
c) HBr(g); it has greater mass.

20.24

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than
that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For
substances in the same phase, entropy increases with atomic size and molecular complexity.
Solution:
a) Ethanol, C2H5OH(l), is a more complex molecule than methanol, CH3OH, and has the greater molar entropy.
b) When a salt dissolves, there is an increase in the number of possible states for the ions. Thus, KClO3(aq) has
the greater molar entropy.
c) K(s) has greater molar entropy because K(s) has greater mass than Na(s).

20.25

a) P4(g); it has greater molecular complexity.
b) HNO3(aq); because S(solution) > S(pure).
c) CuSO4 · 5H2O; it has greater molecular complexity.

20.26

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than
that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For
substances in the same phase, entropy increases with atomic size and molecular complexity.
Solution:
a) Diamond < graphite < charcoal. Diamond has an ordered, three-dimensional crystalline shape, followed by
graphite with an ordered two-dimensional structure, followed by the amorphous (disordered) structure of
charcoal.
b) Ice < liquid water < water vapor. Entropy increases as a substance changes from solid to liquid to gas.
c) O atoms < O2 < O3. Entropy increases with molecular complexity because there are more modes of movement
(e.g., bond vibration) available to the complex molecules.

20.27

a) Ribose < glucose < sucrose; entropy increases with molecular complexity.
b) CaCO3(s) < (CaO(s) + CO2(g)) < (Ca(s) + C(s) + 3/2O2(g)); entropy increases with moles of gas particles.

b) Ssys positive

c) Ssys negative

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20-11

c) SF4(g) < SF6(g) < S2F10(g); entropy increases with molecular complexity.
20.28

Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than
that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For
substances in the same phase, entropy increases with atomic size and molecular complexity.
Solution:
a) ClO4–(aq) > ClO3–(aq) > ClO2–(aq). The decreasing order of molar entropy follows the order of decreasing
molecular complexity.
b) NO2(g) > NO(g) > N2(g). N2 has lower molar entropy than NO because N2 consists of two of the same atoms
while NO consists of two different atoms. NO2 has greater molar entropy than NO because NO2 consists of three
atoms while NO consists of only two.
c) Fe3O4(s) > Fe2O3(s) > Al2O3(s). Fe3O4 has greater molar entropy than Fe2O3 because Fe3O4 is more complex
and more massive. Fe2O3 and Al2O3 contain the same number of atoms but Fe2O3 has greater molar entropy
because iron atoms are more massive than aluminum atoms.

20.29

a) Ba(s) > Ca(s) > Mg(s); entropy decreases with lower mass.
b) C6H14 > C6H12 > C6H6; entropy decreases with lower molecular complexity and lower molecular flexibility.
c) PF2Cl3(g) > PF5(g) > PF3(g); entropy decreases with lower molecular complexity.

20.30

a) X2(g) + 3Y2(g) → 2XY3(g)
b) ΔS < 0 since there are fewer moles of gas in the products than in the reactants.
c) XY3 is the most complex molecule and thus will have the highest molar entropy.

20.31

A system at equilibrium does not spontaneously produce more products or more reactants. For either reaction
direction, the entropy change of the system is exactly offset by the entropy change of the surroundings. Therefore,
for system at equilibrium, Suniv = Ssys + Ssurr = 0. However, for a system moving to equilibrium, Suniv > 0,
because the second law states that for any spontaneous process, the entropy of the universe increases.

20.32

Plan: Since entropy is a state function, the entropy changes can be found by summing the entropies of the
products and subtracting the sum of the entropies of the reactants.
Solution:

Srxn
= [(2 mol HClO)(Sº of HClO)] – [(1 mol H2O)(Sº of H2O) + (1 mol Cl2O)(Sº of Cl2O)]

Rearranging this expression to solve for Sº of Cl2O gives: Sº of Cl2O = 2(Sº of HClO) – Sº of H2O – Srxn

20.33

= m S products
– n Sreactants
.
Plan: To calculate the standard entropy change, use the relationship Srxn

To predict the sign of entropy recall that in general Sgas > Sliquid > Ssolid, and entropy increases as the number of
particles of a particular phase of matter increases, and with increasing atomic size and molecular complexity.
Solution:
a) Prediction: S° negative because number of moles of (n) gas decreases.
S° = [(1 mol N2O)(S° of N2O) + (1 mol NO2)(S° of NO2)] – [(3 mol NO)(S° of NO)]
S° = [(1 mol)(219.7 J/mol•K) + (1 mol)(239.9 J/mol•K)] – [(3 mol)(210.65 J/mol•K)]
S° = –172.35 = –172.4 J/K
b) Prediction: Sign difficult to predict because n = 0, but possibly S° positive because water vapor has greater
complexity than H2 gas.
S° = [(2 mol Fe)(S° of Fe) + (3 mol H2O)(S° of H2O)] – [(3 mol H2)(S° of H2) + (1 mol Fe2O3)(S° of Fe2O3)]
S° = [(2 mol)(27.3 J/mol•K) + (3 mol)(188.72 J/mol•K)] – [(3 mol)(130.6 J/mol•K) + (1 mol)(87.400 J/mol•K)]
S° = 141.56 = 141.6 J/K
c) Prediction: S° negative because a gaseous reactant forms a solid product and also because the number of
moles of gas (n) decreases.
S° = [(1 mol P4O10)(S° of P4O10)] – [(1 mol P4)(S°of P4) + (5 mol O2)(S° of O)]
S° = [(1 mol)(229 J/mol•K)] – [(1 mol)(41.1 J/mol•K) + (5 mol)(205.0 J/mol•K)]
S° = –837.1 = –837 J/K
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20-12

20.34

a) 3NO2(g) + H2O(l)  2HNO3(l) + NO(g) S° negative
S° = [(2 mol HNO3)(S° of HNO3) + (1 mol NO)(S° of NO)]
– [(3 mol NO2)(S° of NO2) +( 1 mol H2O)(S° of H2O)]
S° = [(2 mol)(155.6 J/K•mol) + (1 mol)(210.65 J/K•mol)]
– [(3 mol)(239.9 J/K•mol) + (1 mol)(69.940 J/K•mol)]
S° = –267.79 = –267.8 J/K
S° negative
b) N2(g) + 3F2(g)  2NF3(g)
S° = [(2 mol NF3)(S° of NF3] – [(1 mol N2)(S° of N2) + (3 mol F2)(S° of F2)]
S° = [(2 mol)(260.6 J/K•mol)] – [(1 mol)(191.5 J/K•mol) + (3 mol)(202.7 J/K•mol)]
S° = –278.4 J/K
S° positive
c) C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)
S° = [(6 mol CO2)(S° of CO2) + (6 mol H2O)(S° of H2O)]
– [(1 mol C6H12O6)(S° of C6H12O6) + (6 mol O2)(S° of O2)]
S° = [(6 mol)(213.7 J/K•mol) + (6 mol H2O)(188.72 J/K•mol)]
– [(1 mol)(212.1 J/K•mol) + (6 mol)(205.0 J/K•mol)]
S° = 972.42 = 972.4 J/K

20.35

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship

Srxn
= m S products
– n Sreactants
. To predict the sign of entropy recall that in general Sgas > Sliquid > Ssolid,

entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with
increasing atomic size and molecular complexity.
Solution:
The balanced combustion reaction is:
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
S° = [(4 mol CO2)(S° of CO2) + (6 mol H2O)(S° of H2O)] – [(2 mol C2H6)(S° of C2H6) + (7 mol O2)(S° of O2)]
S° = [(4 mol)(213.7 J/mol•K) + (6 mol)(188.72 J/mol•K)] – [(2 mol)(229.5 J/mol•K) + (7 mol)(205.0 J/mol•K)]
S° = 93.12 = 93.1 J/K
The entropy value is not per mole of C2H6 but per two moles. Divide the calculated value by two to obtain entropy
per mole of C2H6.
Yes, the positive sign of S° is expected because there is a net increase in the number of gas molecules from
nine moles as reactants to ten moles as products.
20.36

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
S° = [(1 mol CO2)(S° of CO2) + (2 mol H2O)(S° of H2O)] – [(1 mol CH4)(S° of CH4) + (2 mol O2)(S° of O2)]
S° = [(1 mol)(213.7 J/K•mol) + (2 mol)(69.940 J/K•mol)] – [(1 mol)(186.1 J/K•mol) + (2 mol)(205.0 J/K•mol)]
S° = –242.52 = –242.5 J/K
Yes, a decrease in the number of moles of gas should result in a negative S° value.

20.37

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship

Srxn
= m S products
– n Sreactants
. To predict the sign of entropy recall that in general Sgas > Sliquid > Ssolid,

entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with
increasing atomic size and molecular complexity.
Solution:
The balanced chemical equation for the described reaction is:
2NO(g) + 5H2(g)  2NH3(g) + 2H2O(g)
Because the number of moles of gas decreases, i.e., n = 4 – 7 = –3, the entropy is expected to decrease.
S° = [(2 mol NH3)(S° of NH3) + (2 mol H2O)(S° of H2O)] – [(2 mol NO)(S° of NO) + (5 mol H2)(S° of H2)]
S° = [(2 mol)(193 J/mol•K) + (2 mol)(188.72 J/mol•K)] – [(2 mol)(210.65 J/mol•K) + (5 mol)(130.6 J/mol•K)]
S° = –310.86 = –311 J/K
Yes, the calculated entropy matches the predicted decrease.

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20-13

20.38

4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)
S° = [(4 mol NO2)(S° of NO2) + (6 mol H2O)(S° of H2O)] – [(4 mol NH3)(S° of NH3) + (7 mol O2)(S° of O2)]
S° = [(4 mol)(239.9 J/K•mol) + (6 mol)(188.72 J/K•mol)] – [(4 mol)(193 J/K•mol) + (7 mol)(205.0 J/K•mol)]
S° = –115.08 = –115 J/K
Yes, a loss of one mole of a gas should result in a small negative S° value.

20.39

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship

Srxn
= m S products
– n Sreactants
.

To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship

H rxn
= m H f (products) – n H f (reactants)

Use ΔHo of the reaction to calculate ΔS of the surroundings.

Hrxn
T
Add ΔS of the surroundings and ΔSo of the reaction to calculate ΔS of the universe. If ΔS of the universe is
greater than zero, the reaction is spontaneous at the given temperature.

 S univ = Srxn
+ Ssurr
Solution:
a) The reaction for forming Cu2O from copper metal and oxygen gas is:
2Cu(s) + 1/2O2(g)  Cu2O(s)
S° = [(1 mol Cu2O)(S° of Cu2O)] – [(2 mol Cu)(S° of Cu) + (1/2 mol O2)(S° of O2)]
S° = [(1 mol)(93.1 J/mol•K)] – [(2 mol)(33.1 J/mol•K) + (1/2 mol)(205.0 J/mol•K)]
S° = –75.6 J/K

H rxn
= m H f (products) – n H f (reactants)
b)

Ssurr = 

H rxn
= [(1 mol Cu2O)( H f of Cu2O)]

– [(2 mol Cu)( H f of Cu) + (1/2 mol O2)( H f of O2)]

H rxn
= [(1 mol Cu2O)(–168.6 kJ/mol)]
– [(2 mol Cu)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

H rxn
= –168.6 kJ

168.6 kJ
Hrxn
=
= 0.56577 kJ/K(103 J/1 kJ) = 565.77 J/K
298 K
T

 S univ = Srxn
+ Ssurr = (–75.6 J/K) + (565.77 J/K) = 490.17 = 490. J/K

Ssurr = 

Because S univ is positive, the reaction is spontaneous at 298 K.
20.40

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship

Srxn
= m S products
– n Sreactants
.

To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship

H rxn
= m H f (products) – n H f (reactants)

Use ΔHo of the reaction to calculate ΔS of the surroundings.

Hrxn
T
Add ΔS of the surroundings and ΔSo of the reaction to calculate ΔS of the universe. If ΔS of the universe is
greater than zero, the reaction is spontaneous at the given temperature.

Ssurr = 

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20-14

 S univ = Srxn
+ Ssurr

Solution:
a) One mole of hydrogen iodide is formed from its elements in their standard states according to the following
equation:
1/2H2(g) + 1/2I2(s)  HI(g)
S° = [(1 mol HI)(S° of HI)] – [(1/2 mol H2)(S° of H2) + (1/2 mol I2)(S° of I2)]
S° = [(1 mol)(206.33 J/K•mol)] – [(1/2 mol)(130.6 J/K•mol) + (1/2 mol)(116.14 J/K•mol)]
S°= 82.96 = 83.0 J/K
b)

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(1 mol HI)( H f of HI)]

– [(1/2 mol H2)( H f of H2) + (1/2 mol I2)( H f of I2)]

H rxn
= [(1 mol HI)(25.9 kJ/mol)]
– [(1/2 mol H2)(0 kJ/mol) + (1/2 mol I2)(0 kJ/mol)]

H rxn
= 25.9 kJ

25.9 kJ
Hrxn
=
= – 0.0869128 kJ/K(103 J/1 kJ) = –86.9128 J/K
298 K
T

 S univ = Srxn
+ Ssurr = (83.0 J/K) + (–86.9128 J/K) = –3.9128 = –3.9 J/K

Ssurr = 

Because S univ is negative, the reaction is not spontaneous at 298 K.
20.41

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship

Srxn
= m S products
– n Sreactants
.

To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship

H rxn
= m H f (products) – n H f (reactants)

Use ΔHo of the reaction to calculate ΔS of the surroundings.

Hrxn
T
Add ΔS of the surroundings and ΔSo of the reaction to calculate ΔS of the universe. If ΔS of the universe is
greater than zero, the reaction is spontaneous at the given temperature.

 S univ = Srxn
+ Ssurr
Solution:
a) One mole of methanol is formed from its elements in their standard states according to the following equation:
C(g) + 2H2(g) + 1/2O2(g)  CH3OH(l)
S° = [(1 mol CH3OH)(S° of CH3OH)] – [(1 mol C)(S° of C) + (2 mol H2)(S° of H2) + (1/2 mol O2)(S° of O2)]
S° = [(1 mol)(127 J/mol•K)] – [(1 mol)(5.686 J/mol•K) + (2 mol)(130.6 J/mol•K) + (1/2 mol)(205.0 J/mol•K)]
S° = –242.386 = –242 J/K

H rxn
= m H f (products) – n H f (reactants)
b)

Ssurr = 

H rxn
= [(1 mol CH3OH)( H f of CH3OH)]

– [(1 mol C)( H f of C) + (2 mol H2)( H f of H2) + (1/2 mol O2)( H f of O2)]

H rxn
= [(1 mol CH3OH)(–238.6 kJ/mol)]
– [(1 mol C)(0 kJ/mol) + (2 mol H2)( 0 kJ/mol) + (1/2 mol O2)( 0 kJ/mol)]

H rxn
= –238.6 kJ
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20-15

Ssurr = 

Hrxn
=
T

238.6 kJ
298 K

= 0.80067 kJ/K(103 J/1 kJ) = 800.67 J/K

 S univ = Srxn
+ Ssurr = (–242 J/K) + (800.67 J/K) = 558.67 = 559 J/K

Because S univ is positive, the reaction is spontaneous at 298 K.
20.42

Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship

Srxn
= m S products
– n Sreactants
.

To calculate ΔSo of the universe in part b, first calculate ΔHo of the reaction using the relationship

H rxn
= m H f (products) – n H f (reactants)

Use ΔHo of the reaction to calculate ΔS of the surroundings.

Hrxn
T
Add ΔS of the surroundings and ΔSo of the reaction to calculate ΔS of the universe. If ΔS of the universe is
greater than zero, the reaction is spontaneous at the given temperature.

 S univ = Srxn
+ Ssurr
Solution:
a) One mole of dinitrogen oxide is formed from its elements in their standard states according to the following
equation:
N2(g) + 1/2O2(g)  N2O(g)
S° = [(1 mol N2O)(S° of N2O] – [(1 mol N2)(S° of N2) + (1/2 mol O2)(S° of O2)]
S° = [(1 mol)(219.7 J/K•mol)] – [(1 mol)(191.5 J/K•mol) + (1/2 mol)(205.0 J/K•mol)]
S° = –214.775 = –74.3 J/K

H rxn
= m H f (products) – n H f (reactants)
b)

Ssurr = 

H rxn
= [(1 mol N2O)( H f of N2O)]

– [(1 mol N2)( H f of N2) + (1/2 mol O2)( H f of O2)]

H rxn
= [(1 mol N2O)(82.05 kJ/mol)]
– [(1 mol N2)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

H rxn
= 82.05 kJ

82.05 kJ
Hrxn
=
= 0.27534 kJ/K(103 J/1 kJ) = 275.34 J/K
298 K
T

 S univ = Srxn
+ Ssurr = (–74.3 J/K) + ( 275.34J/K) = –349.64 = –340.6 J/K

Ssurr = 

Because S univ is negative, the reaction is not spontaneous at 298 K.
20.43

SO2(g) + Ca(OH)2(s)  CaSO3(s) + H2O(l)
S° = [(1 mol CaSO3)(S° of CaSO3) + (1 mol H2O)(S° of H2O)]
– [(1 mol SO2)(S° of SO2) + (1 mol Ca(OH)2)(S° of Ca(OH)2)]
S° = [1 mol)(101.4 J/K•mol) + (1 mol)(69.940 J/K•mol)] – [(1 mol)(248.1 J/K•mol) + (1 mol)(83.39 J/K•mol)]
S° = –160.15 = –160.2 J/K

20.44

Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship

Srxn
= m S products
– n Sreactants
.

Solution:
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20-16

Complete combustion of a hydrocarbon includes oxygen as a reactant and carbon dioxide and water as the
products.
C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(g)
S° = [(2 mol CO2)(S° of CO2) + (1 mol H2O)(S° of H2O)] – [(1 mol C2H2)(S° of C2H2) + (5/2 mol O2)(S° of O2)]
S° = [(2 mol)(213.7 J/mol•K) + (1 mol)(188.72 J/mol•K)]
– [(1 mol)(200.85 J/mol•K) + (5/2 mol)(205.0 J/mol•K)]
S° = –97.23 = –97.2 J/K
20.45

Reaction spontaneity may now be predicted from the value of only one variable (Gsys) rather than two
(Ssys and Ssurr).

20.46

A spontaneous process has Suniv > 0. Since the Kelvin temperature is always positive, Gsys must be negative
(Gsys < 0) for a spontaneous process.

20.47

a) G = H – TS. Since TS > H for an endothermic reaction to be spontaneous, the reaction is more likely
to be spontaneous at higher temperatures.
b) The change depicted is the phase change of a solid converting to a gas (sublimation).
1. Energy must be absorbed to overcome intermolecular forces to convert a substance in the solid phase
to the gas phase. This is an endothermic process and H is positive.
2. Since gases have higher entropy values than solids, the process results in an increase in entropy and S is
positive.
3. This is an endothermic process so the surroundings lose energy to the system. Ssurr is negative.
4. G = H – TS. Both H and S are positive. At low temperature, the H term will predominate and
G will be positive; at high temperatures, the TS term will predominate and G will be negative.

20.48

Plan: Examine the provided diagrams. Determine whether bonds are being formed or broken in order to determine
the sign of ΔH. Determine the relative number of particles before and after the reaction in order to determine the
sign of ΔS.
Solution:
a) The sign of ΔH is negative. Bonds are being formed, so energy is released and ΔH is negative.
b) The sign of ΔS is negative. There are fewer particles in the system after the reaction, so entropy decreases.

Hrxn
Since ΔHrxn is negative, ΔSsurr will be positive.
T
d) Because both ΔH and ΔS are negative, this reaction will become more spontaneous (ΔG will become more
negative) as temperature decreases.

c) The sign of ΔSsurr is positive. Ssurr = 

20.49

H° is positive and S° is positive. The reaction is endothermic (H° > 0) and requires a lot of heat from
its surroundings to be spontaneous. The removal of heat from the surroundings results in S° < 0. The only way
an endothermic reaction can proceed spontaneously is if S° > 0, effectively offsetting the decrease in
surroundings entropy. In summary, the values of H° and S° are both positive for this reaction. Melting is an
example.

20.50

For a given substance, the entropy changes greatly from one phase to another, e.g., from liquid to gas. However,
the entropy changes little within a phase. As long as the substance does not change phase, the value of S° is
relatively unaffected by temperature.

20.51

Plan: G° can be calculated with the relationship m Gf (products) – n Gf (reactants) .
Solution:
a) G° = [(2 mol MgO)( Gf of MgO)] – [(2 mol Mg)( Gf of Mg) + (1 mol O2)( Gf of O2)]
Both Mg(s) and O2(g) are the standard-state forms of their respective elements, so their Gf values are zero.
G° = [(2 mol)(–569.0 kJ/mol)] – [(2 mol)(0) + (1 mol)(0)] = –1138.0 kJ

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20-17

b)G° = [(2 mol CO2)( Gf of CO2) + (4 mol H2O)( Gf of H2O)]
– [(2 mol CH3OH)( Gf of CH3OH) + (3 mol O2)( Gf of O2)]
 G° = [(2 mol)(–394.4 kJ/mol) + (4 mol)(–228.60 kJ/mol)]  [(2 mol)(–161.9 kJ/mol) + (3 mol)(0)]
G° = –1379.4 kJ
c) G° = [(1 mol BaCO3)( Gf of BaCO3)] – [(1 mol BaO)( Gf of BaO) + (1 mol CO2)( Gf of CO2)]
G° = [(1 mol)(–1139 kJ/mol)] – [(1 mol)(–520.4 kJ/mol) + (1 mol)(–394.4 kJ/mol)]
G° = –224.2 = –224 Kj
20.52

a) H2(g) + I2(s)  2HI(g)
G° = [(2 mol HI)( Gf of HI] – [(1 mol H2)( Gf of H2) + (1 mol I2)( Gf of I2)]
G° = [(2 mol)(1.3 kJ/mol)] – [(1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)]
G° = 2.6 kJ
b) MnO2(s) + 2CO(g)  Mn(s) + 2CO2(g)
G° = [(1 mol Mn)( Gf of Mn) + (2 mol CO2)( Gf of CO2)]
– [(1 mol MnO2)( Gf of MnO2) + (2 mol CO)( Gf of CO)]
G° = [(1 mol)(0 kJ/mol) + (2 mol)(–394.4 kJ/mol)]
– [(1 mol)(–466.1 kJ/mol) + (2 mol)(–137.2 kJ/mol)]
G° = –48.3 kJ
c) NH4Cl(s)  NH3(g) + HCl(g)
G° = [(1 mol NH3)( Gf of NH3) + (1 mol HCl)( Gf of HCl)] – [(1 mol NH4Cl)( Gf of NH4Cl)]
G° = [(1 mol)(–16 kJ/mol) + (1 mol)(–95.30 kJ/mol)] – [1 mol)(–203.0 kJ/mol)]
G° = 91.7 = 92 kJ

20.53

Plan: H rxn
can be calculated from the individual H f values of the reactants and products by using the

relationship H rxn
= m H f (products) – n H f (reactants) . Srxn
can be calculated from the individual S  values

of the reactants and products by using the relationship Srxn
= m S products
– n Sreactants
. Once H rxn
and

Srxn
are known, G° can be calculated with the relationship Grxn
= H rxn
– T Srxn
. Srxn
values in J/K

must be converted to units of kJ/K to match the units of H rxn
.
Solution:

a) H rxn
= [(2 mol MgO)( H f of MgO)] – [(2 mol Mg)( H f of Mg) + (1 mol O2)( H f of O2)]

H rxn
= [(2 mol)(–601.2 kJ/mol)] –[(2 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)]

H rxn
= –1202.4 kJ

Srxn
= [(2 mol MgO)( S  of MgO)] – [(2 mol Mg)( S  of Mg) + (1 mol O2)( S  of O2)]

Srxn
= [(2 mol)(26.9 J/mol•K)] – [(2 mol)(32.69 J/mol•K) + (1 mol)(205.0 J/mol•K)]

Srxn
= –216.58 J/K

Grxn
= H rxn
– T Srxn
= –1202.4 kJ – [(298 K)(–216.58 J/K)(1 kJ/103 J)] = –1137.859 = –1138 kJ

b) H rxn
= [(2 mol CO2)( H f of CO2) + (4 mol H2O)( H f of H2O)]

– [(2 mol CH3OH)( H f of CH3OH) + (3 mol O2)( H f of O2)]

H rxn
= [(2 mol)(–393.5 kJ/mol) + (4 mol)(–241.826 kJ/mol)] – [(2 mol)(–201.2 kJ/mol) + (3 mol)(0 kJ/mol)]

H rxn
= –1351.904 kJ

Srxn
= [(2 mol CO2)( S  of CO2) + (4 mol H2O)( S  of H2O)]
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20-18

– [(2 mol CH3OH)( S  of CH3OH) + (3 mol O2)( S  of O2)]

Srxn
= [(2 mol)(213.7 J/mol•K) + (4 mol)(188.72 J/mol•K)]
– [(2 mol)(238 J/mol•K) + (3 mol)(205.0 J/mol•K)] = 91.28 J/K

Grxn
= H rxn
– T Srxn
= –1351.904 kJ – [(298 K)(91.28 J/K)(1 kJ/103 J)] = –1379.105 = –1379 kJ

c) H rxn
= [(1 mol BaCO3)( H f of BaCO3)] – [(1 mol BaO)( H f of BaO) + (1 mol CO2)( H f of CO2)]

H rxn
= [(1 mol)(–1219 kJ/mol)] – [(1 mol)(–548.1 kJ/mol) + (1 mol)(–393.5 kJ/mol)]

H rxn
= –277.4 kJ

Srxn
= [(1 mol BaCO3)( S  of BaCO3)] – [(1 mol BaO)( S  of BaO) + (1 mol CO2)( S  of CO2)]

Srxn
= [(1 mol)(112 J/mol•K)] – [(1 mol)(72.07 J/mol•K) + (1 mol)(213.7 J/mol•K)]

Srxn
= –173.77 J/K

Grxn
= H rxn
– T Srxn
= –277.4 kJ – [(298 K)(–173.77 J/K)(1 kJ/103 J)] = –225.6265 = –226 kJ

20.54

a) H rxn
= [(2 mol HI)(25.9 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1 mol I2)(0 kJ/mol)]

H rxn
= 51.8 kJ

Srxn
= [(2 mol HI)(206.33 J/K•mol)] – [(1 mol H2)(130.6 J/K•mol) + (1 mol I2)(116.14 J/K•mol)]

Srxn
= 165.92 J/K

Grxn
= H rxn
– T Srxn
= 51.8 kJ – [(298 K)(165.92 J/K)(1 kJ/103 J)] = 2.3558 = 2.4 kJ

b) H rxn
= [(1 mol Mn)(0 kJ/mol) + (2 mol CO2)(–393.5 kJ/mol)]
– [(1 mol MnO2)(–520.9 kJ/mol) + (2 mol CO)(–110.5 kJ/mol)]

H rxn
= –45.1 kJ

Srxn
= [(1 mol Mn)(31.8 J/K•mol) + (2 mol CO2)(213.7 J/K•mol)]
– [(1 mol MnO2)(53.1 J/K•mol) + (2 mol CO)(197.5 J/K•mol)]

Srxn

= 11.1 J/K

Grxn
= H rxn
– T Srxn
= –45.1 kJ – [(298 K (11.1 J/K)(1 kJ/103 J)] = –48.4078 = –48.4 kJ

c) H rxn
= [(1 mol NH3)(–45.9 kJ/mol) + (1 mol HCl)(–92.3 kJ/mol)] – [(1 mol NH4Cl)(–314.4 kJ/mol)]

H rxn
= 176.2 kJ

Srxn
= [(1 mol NH3)(193 J/K•mol) + (1 mol HCl)(186.79 J/K•mol)] – [(1 mol NH4Cl)(94.6 J/K•mol)]

Srxn
= 285.19 J/K

Grxn
= H rxn
– T Srxn
= 176.2 kJ – [(298 K)(285.19 J/K)(1 kJ/103 J)] = 91.213 = 91.2 kJ

20.55

can be calculated with the relationship m Gf (products) – n Gf (reactants) . Alternatively, Grxn
Plan: Grxn

can be calculated with the relationship Grxn
= H rxn
– T Srxn
. Entropy decreases (is negative) when there
are fewer moles of gaseous products than there are of gaseous reactants.
Solution:
a) Entropy decreases ( ΔS o negative) because the number of moles of gas decreases from reactants (1 1/2 mol) to

products (1 mole). The oxidation (combustion) of CO requires initial energy input to start the reaction, but then
releases energy (exothermic, ΔH o negative) which is typical of all combustion reactions.

b) Method 1: Calculate Grxn
from Gf values of products and reactants.
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

20-19

Grxn
= m Gf (products) – n Gf (reactants)

Grxn
= [(1 mol CO2)( Gf of CO2)] – [(1 mol CO)( Gf of CO) + (1/2 mol)( Gf of O2)]

Grxn
= [(1 mol)(–394.4 kJ/mol)] – [(1 mol)(–137.2 kJ/mol) + (1/2 mol)(0 kJ/mol)] = –257.2 kJ

from H rxn
and Srxn
at 298 K (the degree superscript indicates a reaction at
Method 2: Calculate Grxn
standard state, given in the Appendix at 25°C).

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(1 mol CO2)( H f of CO2)] – [(1 mol CO)( H f of CO) + (1/2 mol)( H f of O2)]

H rxn
= [(1 mol)(–393.5 kJ/mol)] – [(1 mol)(–110.5 kJ/mol) + (1/2 mol)(0 kJ/mol)] = –283.0 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(1 mol CO2)( S  of CO2)] – [(1 mol CO)( S  of CO) + (1/2 mol)( S  of O2)]

Srxn
= [(1mol)(213.7 J/mol•K)] – [(1mol)(197.5 J/mol•K) + (1/2 mol)(205.0 J/mol•K)]

Srxn
= –86.3 J/K

Grxn
= H rxn
– T Srxn
= (–283.0 kJ) – [(298 K)(–86.3 J/K)(1 kJ/103 J)] = –257.2826 = –257.3 kJ

20.56

C4H10(g) + 13/2O2(g)  4CO2(g) + 5H2O(g)
a) An increase in the number of moles of gas should result in a positive S° value. The combustion of C4H10(g)
will result in a release of energy or a negative H° value.

b) H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(4 mol CO2)(–393.5 kJ/mol) + (5 mol H2O)(–241.826 kJ/mol)]
– [(1 mol C4H10)(–126 kJ/mol) + (13/2 mol O2)(0 kJ/mol)]

H rxn
= –2657.13 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(4 mol CO2)(213.7 J/K•mol) + (5 mol H2O)(188.72 J/K•mol)]
– [(1 mol C4H10)(310 J/K•mol) + (13/2 mol O2)(205.0 J/K•mol)]

Srxn
= 155.9 J/K

Grxn
= H rxn
– T Srxn
= –2657.13 kJ – [(298 K)(155.9 J/K)(1 kJ/103 J)] = –2703.588 = –2704 kJ

Grxn
= m Gf (products) – n Gf (reactants)

Grxn
= [(4 mol CO2)(–394.4 kJ/mol) + (5 mol H2O)(–228.60 kJ/mol)]
– [(1 mol C4H10)(–16.7 kJ/mol) + (13/2 mol O2)(0 kJ/mol)]

Grxn
= –2703.9 kJ

20.57

Plan: Use the relationship Grxn
= H rxn
– T Srxn
to find Srxn
, knowing H rxn
and Grxn
. This

relationship is also used to find Grxn
at a different temperature.
Solution:
Reaction is Xe(g) + 3F2(g)  XeF6(g)

a) Grxn
= H rxn
– T Srxn

S° =

402 kJ/mol   280. kJ/mol 
H  G
=
= –0.40939597 = –0.409 kJ/mol•K
298 K
T

b) Grxn
= H rxn
– T Srxn
= (–402 kJ/mol) – [(500. K)(–0.40939597 kJ/mol•K)] = –197.302 = –197 kJ/mol
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20-20

20.58

a) S° =

 220. kJ/mol    206 kJ/mol 
H  G
=
= –0.046979865 = –0.047 kJ/mol•K
T
298 K

= H rxn
– T Srxn
= –220. kJ/mol – (450. K)(– 0.046979865 kJ/K•mol) = –198.859 = –199 kJ/mol
b) Grxn

20.59

Plan: H rxn
can be calculated from the individual H f values of the reactants and products by using the

relationship H rxn
= m H f (products) – n H f (reactants) . Srxn
can be calculated from the individual S  values

of the reactants and products by using the relationship Srxn
= m S products
– n Sreactants
. Once H rxn
and

Srxn
are known, G° can be calculated with the relationship Grxn
= H rxn
– T Srxn
. Srxn
values in J/K

must be converted to units of kJ/K to match the units of H rxn
. The temperature at which a reaction becomes
spontaneous can be calculated by setting ΔG to zero in the free energy equation (assuming the reaction is at
equilibrium) and solving for T.
Solution:

a) H rxn
= [(1 mol CO)( H f of CO) + (2 mol H2)( H f of H2)] – [(1 mol CH3OH)( H f of CH3OH)]

H rxn
= [(1 mol)(–110.5 kJ/mol) + (2 mol)(0 kJ/mol)] – [(1 mol)(–201.2 kJ/mol)]

H rxn
= 90.7 kJ

Srxn
= [(1 mol CO)( S  of CO) + (2 mol H2)( S  of H2)] – [(1 mol CH3OH)( S  of CH3OH)]

Srxn
= [(1 mol)(197.5 J/mol•K) + (2 mol)(130.6 J/mol•K)] – [(1 mol)(238 J/mol•K)]

Srxn
= 220.7 = 221 J/K

b) Grxn
= H rxn
– T Srxn
T1 = 28 + 273 = 301 K G° = 90.7 kJ – [(301 K)(220.7 J/K)(1 kJ/103 J)] = 24.2693 = 24.3 kJ
T2 = 128 + 273 = 401 K G° = 90.7 kJ – [(401 K)(220.7 J/K)(1 kJ/103 J)] = 2.1993 = 2.2 kJ
T3 = 228 + 273 = 501 K G° = 90.7 kJ – [(501 K)(220.7 J/K)(1 kJ/103 J)] = –19.8707 = –19.9 kJ
c) For the substances in their standard states, the reaction is nonspontaneous at 28°C, near equilibrium at 128°C,

and spontaneous at 228°C. Reactions with positive values of H rxn
and Srxn
become spontaneous at high
temperatures.
d) The reaction will become spontaneous when ΔG changes from being positive to being negative. This point
occurs when ΔG is 0.

Grxn
= H rxn
– T Srxn
3
0 = 90.7 x 10 J – (T)(221 J/K)
T = 410. K
At temperatures above 410. K, this reaction is spontaneous. (Because both ΔH and ΔS are positive, the
reaction becomes spontaneous above this temperature.)

20.60

Plan: H rxn
can be calculated from the individual H f values of the reactants and products by using the

relationship H rxn
= m H f (products) – n H f (reactants) . Srxn
can be calculated from the individual S  values

of the reactants and products by using the relationship Srxn
= m S products
– n Sreactants
. Once H rxn
and

Srxn
are known, G° can be calculated with the relationship Grxn
= H rxn
– T Srxn
. Srxn
values in J/K

must be converted to units of kJ/K to match the units of H rxn
. The temperature at which a reaction becomes
spontaneous can be calculated by setting ΔG to zero in the free energy equation (assuming the reaction is at
equilibrium) and solving for T.
Solution:
a) N2(g) + O2(g)  2NO(g)

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20-21

H rxn
= [(2 mol NO)(90.29 kJ/mol)] – [(1 mol N2)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)]

H rxn
= 180.58 kJ

Srxn
= [(2 mol NO)(210.65 J/K•mol)] – [(1 mol N2)(191.5 J/K•mol) + (1 mol O2)(205.0 J/K•mol)]

Srxn
= 24.8 J/K
b) G°373 = H° – ((273 + 100.)K) (S°)
= 180.58 kJ – [(373 K)(24.8 J/K)(1 kJ/103 J)]
= 171.3296 = 171.33 kJ
G°2833 = H° – ((273 + 2560.)K) (S°)
= 180.58 kJ – [(2833 K)(24.8 J/K)(1 kJ/103 J)]
= 110.3216 = 110.3 kJ
G°3813 = H° – ((273 + 3540.)K) (S°)
= 180.58 kJ – [(3813 K)(24.8 J/K)(1 kJ/103 J)]
= 86.0176 = 86.0 kJ
c) The values of G became smaller at higher temperatures. The reaction is not spontaneous at any of these
temperatures; however, the reaction becomes less nonspontaneous as the temperature increases.
d) The reaction will become spontaneous when ΔG changes from being positive to being negative. This point
occurs when ΔG is 0.

Grxn
= H rxn
– T Srxn
3
0 = 180.58 x 10 J – (T)(24.8 J/K)
T = 7280 K
At temperatures above 7280 K, this reaction is spontaneous. (Because both ΔH and ΔS are positive, the
reaction becomes spontaneous above this temperature.)

20.61

Plan: H rxn
can be calculated from the individual H f values of the reactants and products by using the

relationship H rxn
= m H f (products) – n H f (reactants) . Srxn
can be calculated from the individual S  values

of the reactants and products by using the relationship Srxn
= m S products
– n Sreactants
. Once H rxn
and

Srxn
are known, G° can be calculated with the relationship Grxn
= H rxn
– T Srxn
. Srxn
values in J/K

must be converted to units of kJ/K to match the units of H rxn
. To find the temperature at which the reaction

becomes spontaneous, use Grxn
= 0 = H rxn
– T Srxn
and solve for temperature.
Solution:
a) The reaction for this process is H2(g) + 1/2O2(g)  H2O(g). The coefficients are written this way (instead of
2H2(g) + O2(g)  2H2O(g)) because the problem specifies thermodynamic values “per (1) mol H2,” not per
2 mol H2.

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(1 mol H2O)( H f of H2O)] – [(1 mol H2)( H f of H2) + (1/2 mol O2)( H f of O2)]

H rxn
= [(1 mol H2O)(–241.826 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1/2 mol O2)(0 kJ/mol)]

H rxn
= –241.826 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(1 mol H2O)( S  of H2O)] – [(1 mol H2)( S  of H2) + (1/2 mol O2)( S  of O2)]

Srxn
= [(1 mol)(188.72 J/mol•K)] – [(1 mol)(130.6 J/mol•K) + (1/2 mol)(205.0 J/mol•K)]

Srxn
= –44.38 = –44.4 J/K = –0.0444 kJ/K

Grxn
= H rxn
– T Srxn

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20-22

Grxn
= –241.826 kJ – [(298 K)( –0.0444 kJ/K)]

Grxn
= –228.6 kJ
b) Because H < 0 and S < 0, the reaction will become nonspontaneous at higher temperatures because the
positive (–TS) term becomes larger than the negative H term.

c) The reaction becomes spontaneous below the temperature where Grxn
=0

Grxn
= 0 = H rxn
– T Srxn

H rxn
= T Srxn

T =
20.62

H 
S

=

–241.826 kJ
= 5446.53 = 5.45x103 K
–0.0444 kJ/K

C6H12O6(s)  2C2H5OH(l) + 2CO2(g)

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(2 mol C2H5OH)(–277.63 kJ/mol) + (2 mol CO2)(–393.5 kJ/mol)] – [1 mol C6H12O6)(–1273.3 kJ/mol)]

H rxn
= –68.96 kJ = –69.0 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(2 mol C2H5OH)(161 J/K•mol) + (2 mol CO2)(213.7 J/K•mol)] – [(1 mol C6H12O6)(212.1 K•mol)]

Srxn
= 537.3 J/K = 537 J/K

Grxn
= H rxn
– T Srxn

Grxn
= –68.96 kJ – [(298 K)( 0.5373 kJ/K)]

Grxn
= –229.0754 kJ/mol = –229.1 kJ/mol
No, a reaction with a negative value for H and a positive value for S is spontaneous at all temperatures.

20.63

a) An equilibrium constant that is much less than one indicates that very little product is made to reach
equilibrium. The reaction, thus, is not spontaneous in the forward direction and G° is a relatively large positive
value.
b) A large negative G° indicates that the reaction is quite spontaneous and goes almost to completion. At
equilibrium, much more product is present than reactant so K > 1. Q depends on initial conditions, not equilibrium
conditions, so its value cannot be predicted from G°.

20.64

For a spontaneous process, G is the maximum useful work obtainable from the system. In reality, the actual
amount of useful work is less due to energy lost as heat. If the process is run in a slower or more controlled
fashion, the actual amount of available work approaches G.

20.65

a) Point x represents the difference between Greactants and Gproducts or G°, the standard free energy change
for the reaction.
b) Scene A corresponds to Point 1 on the graph. This point corresponds to the pure substances, not a mixture.
c) Scene C corresponds to Point 2 on the graph. Point 2 represents equilibrium; for this reaction, products
dominate at equilibrium (the minimum in the curve is close to the XY side of the graph).

20.66

The standard free energy change, G°, occurs when all components of the system are in their standard states.
Standard state is defined as 1 atm for gases, 1 M for solutes, and pure solids and liquids. Standard state does not
specify a temperature because standard state can occur at any temperature. G° = G when all concentrations
equal 1 M and all partial pressures equal 1 atm. This occurs because the value of Q = 1 and ln Q = 0 in the
equation G = G° + RT ln Q.

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20-23

20.67

Plan: For each reaction, first find G°, then calculate K from G° = –RT ln K. Calculate G  using Gf values

in the relationship Grxn
= m Gf (products) – n Gf (reactants) .

Solution:
a) MgCO3(s) Mg2+(aq) + CO32–(aq)

Grxn
= m Gf (products) – n Gf (reactants)

Grxn
= [(1 mol Mg2+)(–456.01 kJ/mol) + (1 mol CO32–)(–528.10 kJ/mol)]
– [(1 mol MgCO3)(–1028 kJ/mol)] = 43.89 kJ

  103 J 
G 
43.89 kJ /mol
ln K =
= 
 
 = –17.7149

 RT
   8.314 J /mol•K  298 K    1 kJ 
K = e–17.7149 = 2.0254274x10–8 = 2.0x10–8
b) H2(g) + O2(g)  H2O2(l)

Grxn
= [(1 mol H2O2)(–120.4 kJ/mol)] – [(1 mol H2)(0 kJ/mol) + (1 mol O2)(0 kJ/mol)] = –120.4 kJ/mol

  103 J 
G 
120.4 kJ /mol
= 
 
 = 48.59596

 RT
   8.314 J /mol•K  298 K    1 kJ 
K = e48.59596 = 1.2733777x1021 = 1.27x1021

ln K =

20.68

Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K.
Solution:
a) G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (6.57 x 10173) = –9.91598 x105 J/mol = –9.92x105 J/mol
b) G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (4.46 x 10-15) = 8.18680 x 104 J/mol = 8.19x104 J/mol

20.69

Plan: For each reaction, first find G°, then calculate K from G° = –RT ln K. Calculate G  using Gf values

in the relationship Grxn
= m Gf (products) – n Gf (reactants) .

Solution:

a) Grxn
= [(1 mol NaCN)( Gf of NaCN) + (1 mol H2O)( Gf of H2O)]

– [(1 mol HCN)( Gf of HCN) + (1 mol NaOH)( Gf of NaOH)]
NaCN(aq) and NaOH(aq) are not listed in Appendix B.
Converting the equation to net ionic form will simplify the problem:
HCN(aq) + OH–(aq)  CN–(aq) + H2O(l)

Grxn
= [(1 mol CN–)( Gf of CN–) + (1 mol H2O)( Gf of H2O)]

– [(1 mol HCN)( Gf of HCN) + (1 mol OH–)( Gf of OH–)]

Grxn
= [(1 mol)(166 kJ/mol) + (1 mol)(–237.192 kJ/mol)]
– [(1 mol)(112 kJ/mol) + (1 mol)(–157.30 kJ/mol)]

Grxn
= –25.892 kJ

  103 J 
G 
25.892 kJ /mol
= 
= 10.45055
   8.314 J /mol•K  298 K    1 kJ 
 RT


K = e10.45055 = 3.4563x104 = 3.46x104
b) SrSO4(s)  Sr2+(aq) + SO42–(aq)

ln K =

Grxn
= [(1 mol Sr2+)(–557.3 kJ/mol) + (1 mol SO42–)(–741.99 kJ/mol)]
– [(1 mol SrSO4)(–1334 kJ/mol)] = 34.71 kJ

  103 J 
G
34.71 kJ/mol
ln K =
= 
 
 = –14.00968

 RT
   8.314 J /mol•K  298 K    1 kJ 
K = e–14.00968 = 8.23518x10–7 = 8.2x10–7
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20-24

20.70

Plan: The equilibrium constant, K, is related to G° through the equation G° = –RT ln K.
Solution:
a) G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (1.58 x 107) = –4.10670 x104 J/mol = –4.11x104 J/mol
b) G° = –RT ln K = –(8.314 J/mol•K)(298 K) ln (3.25 x 1037) = –2.13999 x 105 J/mol = –2.14x105 J/mol

20.71

Plan: At the normal boiling point, defined as the temperature at which the vapor pressure of the liquid equals 1
atm, the phase change from liquid to gas is at equilibrium. For a system at equilibrium, the change in Gibbs free
energy is zero. Since the gas is at 1 atm and the liquid assumed to be pure, the system is at standard state and

= 0 = H rxn
– T Srxn
. H rxn
can be
G° = 0. The temperature at which this occurs can be found from Grxn

calculated from the individual H f values of the reactants and products by using the relationship

H rxn
= m H f (products) – n H f (reactants) . Srxn
can be calculated from the individual S  values of the

= m S products
– n Sreactants
.
reactants and products by using the relationship Srxn

Solution:
Br2(l) Br2(g)

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(1 mol Br2)( H f of Br2(g))] – [(1 mol Br2)( H f of Br2(l))]

H rxn
= [(1 mol)(30.91 kJ/mol)] – [(1 mol)(0 kJ/mol)] = 30.91 kJ

Srxn
= m S products
– n Sreactants

Srxn
= [(1 mol Br2)( S  of Br2(g))] – [(1 mol Br2)( S  of Br2(l))]

Srxn
= [(1 mol)(245.38 J/K•mol)] – [(1 mol)(152.23 J/K•mol)] = 93.15 J/K = 0.09315 kJ/K

Grxn
= 0 = H rxn
– T Srxn

H rxn
= T Srxn

T =
20.72

H 
S

=

30.91 kJ
= 331.830 = 331.8 K
0.09315 kJ/K

S(rhombic)  S(monoclinic)

H rxn
= [1 mol S(monoclinic)(0.30 kJ/mol)] – [1 mol S(rhombic)(0 kJ/mol)] = 0.30 kJ

Srxn
= [1 mol S(monoclinic)(32.6 J/K•mol)] – [1 mol S(rhombic)(31.9 J/K•mol)] = 0.7 J/K = 0.0007 kJ/K

Grxn
= 0 = H rxn
– T Srxn

H rxn
= T Srxn

T =

20.73

H 
S

=

30.91 kJ
= 428.571 = 4x102 K
0.09315 kJ/K

Plan: Write the balanced equation. First find G°, then calculate K from G° = –RT ln K. Calculate G  using

Gf values in the relationship Grxn
= m Gf (products) – n Gf (reactants) .

Solution:
The solubility reaction for Ag2S is
Ag2S(s) + H2O(l)  2Ag+(aq) + HS–(aq) + OH–(aq)

Grxn
= m Gf (products) – n Gf (reactants)

Grxn
= [(2 mol Ag+)( Gf of Ag+) + (1 mol HS–)( Gf of HS–) + (1 mol OH–)( Gf of OH–]
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20-25

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