18.6

Strategy:

According to equation 18.2, the number of ways of
arranging N particles in X cells is W = X^{N}.
Use this equation to calculate the number of arrangements and equation
18.1 to calculate the entropy.

Setup:

For
the setup in Figure 18.2, X = 2
when the barrier is in place and X
= 4 when it is absent. To simplify the calculations, use the power
property of natural log to write:
.

Solution:

a.

With barrier: N =
10; W = 2^{10} =
1024;
S
= (N×k)(ln W) = (10)(1.38 × 10^{–23} J/K)ln(2) = 9.57 × 10^{–23} J/K.
Without barrier: N
= 10; W = 4^{10} =
1.48 × 10^{6}; S = 1.91 × 10^{–22} J/K.

b.

With barrier: N =
50; W = 2^{50} =
1.13 × 10^{15}; S = 4.78 × 10^{–22} J/K.
Without barrier: N = 10; W = 4^{50}
= 1.27 × 10^{30}; S
= 9.57 × 10^{–22} J/K.

c.

With barrier: N =
100; W = 2^{100} = 1.26 × 10^{30}; S = 9.57 × 10^{–22}
J/K.
Without barrier: N = 100; W = 4^{100}
= 1.61 × 10^{60}; S
= 1.91 × 10^{–21} J/K.




18.7

Strategy:

According
to equation 18.2, the number of ways of arranging N particles in X
cells is W = X^{N}. Use this equation
to calculate the number of arrangements and equation 18.1 to calculate
the entropy.

Setup:

For the setup in the figure, X = 4 when the barrier is in place and X = 8 when it is absent.

Solution:

a.

With barrier: N =
2; W = 4^{2} =
16; Without barrier: N = 2; W = 8^{2}
= 64.

b.

From part (a), we know
that 16 of the 64
arrangements have both particles in the left side of the container.
Similarly, there are 16 ways
for the particles to be found on the righthand side. The number of arrangements with one
particle per side is 64 – 16 – 16 = 32.
Both particles on one side: S
= k ln W^{N} = (1.38 × 10^{–23}
J/K)ln(32) = 3.83 × 10^{–23}
J/K; particles on opposite
sides: S = (1.38 × 10^{–23} J/K)ln(32) = 4.78 × 10^{–23}. The most probable state is the one
with the larger entropy; that is, the state in which the particles are
on opposite sides.




18.12

Strategy:

Equation
18.4 gives the entropy change for the isothermal expansion of an ideal
gas. Substitute the given values into the equation and compute DS.

Solution:




18.13

Strategy:

Equation
18.4 gives the entropy change for the isothermal expansion of an ideal
gas. Substitute the given values into the equation and compute DS.

Solution:




18.14

a.

Li(l); The liquid form of any
substance always has greater entropy.

b.

At first glance there may seem to be no apparent
difference between the two substances that might affect the entropy (the
molecular formulas are identical).
However, the first has the OH
structural feature which allows it to participate in hydrogen bonding
with other molecules. This results
in fewer possible arrangements of molecules in the liquid state. The standard entropy of CH_{3}OCH_{3} is
larger.

c.

Both are monatomic
species. However, the Xe atom has a greater molar mass
than Ar. Xenon has the higher
standard entropy.

d.

Due to the extra oxygen atom, carbon dioxide molecules
have a more complex molecular structure than do carbon monoxide
molecules. Specifically, carbon
dioxide has vibrational and rotational modes that carbon monoxide does
not. So, carbon dioxide gas
has the higher standard entropy (see Appendix 2). There is also a mass effect. For gas
phase molecules of similar complexity, increasing molar mass tends to
increase the molar entropy.

e.

O_{3}
has a greater molecular complexity than O_{2} and thus has the
higher standard entropy. Ozone’s molecules are also more massive (see
(d) above).

f.

Because of its greater molecular complexity and greater mass (see
answer (d)), one mole of N_{2}O_{4} has a larger standard
entropy than one mole of NO_{2}. Compare values in Appendix 2.
Think About
It:

Use the
data in Appendix 2 to compare the standard entropy of one mole of N_{2}O_{4}
with that of two moles of NO_{2}. In this situation the number of atoms
is the same for both. Which is
higher and why?




18.15

In order of increasing entropy per mole at 25°C:
(c)
< (d) <
(e) < (a)
< (b)
a.

Ne(g): a monatomic gas
of higher molar mass than H_{2}. (For gas phase molecules,
increasing molar mass tends to increase the molar entropy. While Ne(g) is less complex structurally
than H_{2}(g), its much larger molar mass more than
offsets the complexity difference.)

b.

SO_{2}(g): a
polyatomic gas of higher complexity and higher molar mass than Ne(g) (see the explanation for
(a) above).

c.

Na(s): highly ordered, crystalline
material.

d.

NaCl(s): highly ordered crystalline
material, but with more particles per mole than Na(s).

e.

H_{2}: a
diatomic gas, hence of higher entropy than a solid.



18.16

Using Equation 18.5 of the text to calculate :
a.


b.


c.

DS= 4S°(CO_{2})
+ 6S°(H_{2}O(l)) 
[2S°(C_{2}H_{6})
+ 7S°(O_{2})]
DS= (4)(213.6 J/K×mol) + (6)(69.9 J/K×mol)  [(2)(229.5 J/K×mol) + (7)(205.0 J/K×mol)]
= 620.2
J/K×mol



18.17

Strategy:

To calculate the standard entropy change of a
reaction, we look up the standard entropies of reactants and products in
Appendix 2 of the text and apply Equation 18.7. As in the calculation of enthalpy of
reaction, the stoichiometric coefficients have no units, so is
expressed in units of J/K×mol.

Solution:

The
standard entropy change for a reaction can be calculated using the
following equation.
a.

=
(1)(33.3 J/K×mol) + (1)(188.7 J/K×mol)  [(1)(131.0 J/K×mol) + (1)(43.5 J/K×mol)]
= 47.5 J/K×mol

b.

= (1)(50.99 J/K×mol) + (3)(41.6 J/K×mol)  [(2)(28.3 J/K×mol) + (3)(43.9 J/K×mol)]
= 12.5 J/K×mol

c.

=
(1)(213.6 J/K×mol) + (2)(69.9 J/K×mol)  [(1)(186.2 J/K×mol) + (2)(205.0 J/K×mol)]
= 242.8 J/K×mol

Why was the
entropy value for water different in parts (a) and (c)?



18.20

Strategy:

Assume all reactants and
products are in their standard states. The entropy change in the surroundings is related to the
enthalpy change of the system by Equation 18.7. For each reaction in
Exercise 18.16, use Appendix 2 to calculate DH_{sys} ( = DH_{rxn}) (Section 5.3) and
then use Equation 18.7 to calculate DS_{surr}. Finally, use
Equation 18.8 to compute DS_{univ}. If DS_{univ} is positive, then
according to the second law of thermodynamics, the reaction is
spontaneous. The values for DS_{sys} are found in the solution for problem 18.16.

Solution:

a.

spontaneous

b.

not spontaneous

c.

spontaneous




18.21

Strategy:

Assume all reactants and
products are in their standard states. The entropy change in the
surroundings is related to the enthalpy change of the system by Equation
18.7. For each reaction in Exercise 18.17, use Appendix 2 to calculate DH_{sys} ( = DH_{rxn}) (Section 5.3) and
then use Equation 18.7 to calculate DS_{surr}. Finally, use
Equation 18.8 to compute DS_{univ}. If DS_{univ} is positive, then
according to the second law of thermodynamics, the reaction is
spontaneous. The values for DS_{sys} are found in the solution for problem 18.16.

Solution:

a.

spontaneous

b.

spontaneous

c.

spontaneous




18.22

Strategy:

Assume all reactants and
products are in their standard states. According to Equations 18.7, the
entropy change in the surroundings for an isothermal process is :
.
Also, Equation 18.8 states that the entropy change of the
universe is:
.
Use Appendix 2 to calculate DS_{sys} ( = DS_{rxn}) and DH_{sys} ( = DH_{rxn}). Substitute these
results into the above equations and determine the sign of DS_{univ}. If DS_{univ} is positive, then
according to the second law of thermodynamics, the reaction is
spontaneous.

Solution:

a.

not spontaneous

b.

spontaneous

c.

spontaneous

d.

To calculate DH_{rxn} for N_{2}
® 2N, use the bond
enthalpy from Table 8.6.
Calculate DS_{surr} and DS_{univ}:
not
spontaneous


Think About It:

We could have assumed that the
reaction in (d) occurred at constant volume instead of constant pressure.
Would this have changed the conclusion about the spontaneity of the
reaction?



18.23

Strategy:

Assume all reactants and
products are in their standard states. According to Equations 18.7, the
entropy change in the surroundings for an isothermal process is :
.
Also, Equation 18.8 states that the entropy change of the
universe is:
.
Use Appendix 2 to calculate DS_{sys} ( = DS_{rxn}) and DH_{sys} ( = DH_{rxn}). Substitute these
results into the above equations and determine the sign of DS_{univ}. If DS_{univ} is positive, then
according to the second law of thermodynamics, the reaction is
spontaneous.

Solution:

a.

spontaneous

b.

not spontaneous

c.

To calculate DH_{rxn} for H_{2}
® 2H, use the bond
enthalpy from Table 8.6.
Calculate DS_{surr} and DS_{univ}:
not spontaneous

d.

spontaneous


Think About It:

We could have assumed that the
reaction in (c) occurred at constant volume instead of constant pressure.
Would this have changed the conclusion about the spontaneity of the
reaction?



18.24

Strategy:

We assume the denaturation is a singlestep equilibrium
process. Therefore, because DG is zero at equilibrium, we can
use Equation 18.10, substituting 0 for DG and solving for DS, to determine the entropy change
associated with the process.

Setup:

The melting temperature of the protein is 63 + 273.15
= 336.2 K.

Solution:

The unfolding of the
protein increases the volume it occupies and increases its entropy. Therefore, we estimate the entropy of
denaturation to be positive.
Calculating the entropy
gives,
DS_{denaturation} = 1.52 kJ/K×mol
or 1.52 ´
10^{3} J/K×mol
16 J/K×mol
per amino acid



18.29

Strategy:

Use Equation 18.10 to solve for DG.

Setup:

From problem 18.24, DS = 1.52 kJ/K×mol
and DH = 510 kJ/mol.
T = (20 +
273.15) = 293.2 K.

Solution:

DG
= DH 
TDS
= 510 kJ/mol 
(293.2 K)(1.52 kJ/K×mol) =
64 kJ/mol



18.30

Using Equation 18.12 of the text
to solve for the change in standard free energy,
a.


b.


c.

= (4)(394.4
kJ/mol) + (2)(237.2
kJ/mol) 
(2)(209.2 kJ/mol)  (5)(0) = 2470
kJ/mol



18.31

Strategy:

To calculate the standard freeenergy change of a
reaction, we look up the standard free energies of formation of reactants
and products in Appendix 2 of the text and apply Equation 18.13. Note that all the stoichiometric
coefficients have no units so is
expressed in units of kJ/mol. The
standard free energy of formation of any element in its stable allotropic
form at 1 atm and 25°C is zero.

Solution:

The standard free energy change for a reaction can be
calculated using the following equation.



18.32

Reaction A: First apply Equation
18.10 of the text to compute the free energy change at 25°C
(298 K)
DG
= DH  TDS
= 10,500 J/mol  (298 K)(30 J/K×mol)
= 1560 J/mol
The +1560 J/mol shows the reaction is not spontaneous at 298 K. The DG will change sign (i.e., the reaction will become spontaneous)
above the temperature at which DG = 0.
0 = DH 
TDS
Solving for T gives
Reaction B: Calculate DG.
DG = DH 
TDS = 1800 J/mol  (298 K)(113
J/K×mol) = 35,500
J/mol
The free energy change is positive, which shows that the reaction is
not spontaneous at 298 K. Since both
terms are positive, there is no temperature at which their sum is
negative. The reaction is not spontaneous at any temperature.


18.33

a.

Calculate DG from DH and DS.
DG
= DH 
TDS
= 126,000
J/mol 
(298 K)(84 J/K×mol) =
151,000
J/mol
The free energy change is
negative so the reaction is spontaneous at 298 K. Since DH is negative and DS is positive, the reaction is
spontaneous at all temperatures.

b.

Calculate DG.
DG
= DH 
TDS
= 11,700
J/mol 
(298 K)(105
J/K×mol) = +19,600 J
The free energy change is
positive at 298 K which means the reaction is not spontaneous at that
temperature. The positive sign of DG results from the large negative
value of DS.
At lower temperatures, the TDS term will be smaller thus
allowing the free energy change to be negative.
DG will equal zero when DH = TDS.
Rearranging,
At temperatures below 111 K, DG will be negative and the
reaction will be spontaneous.



18.34

DS_{fus} = = 48.8 J/K×mol
DS_{vap} = = 74.0 J/K×mol


18.35

Strategy:

Equation 18.7 from the text relates the entropy change
of the surroundings to the enthalpy change of the system and the
temperature at which a phase change occurs.
DS_{surr} =
We know
that DS_{sys} = DS_{surr}, so we can
rewrite Equation 18.7 as
DS_{sys} =

Solution:

DS_{fus} = = 99.9 J/K×mol
DS_{vap} = = 93.6 J/K×mol

Think About It:

Remember that Celsius temperatures must be converted
to Kelvin.



18.36

Using Equation 18.12 from the text,
DG = 2DG(C_{2}H_{5}OH) + 2DG(CO_{2})  DG( C_{6}H_{12}O_{6})
DG =
(2)(174.18
kJ/mol) + (2)(394.4
kJ/mol) 
(910.56
kJ/mol) = 226.6 kJ/mol


18.37

Using Equation 18.2 from the text,
DG = DG(NO)  [1/2 DG(O_{2}) + DG( NO)]
= (110.5
kJ/mol) 
[(0 kJ/mol) + (34.6
kJ/mol) = 75.9 kJ/mol
75.9 kJ of
Gibbs free energy are released.


18.41

Strategy:

According to Equation 18.14 of the text, the
equilibrium constant for the reaction is related to the standard free
energy change; that is, DG°
= RT ln K. Since we are given DG°
in the problem, we can solve for the equilibrium constant. What temperature unit should be used?

Solution:

Solving
Equation 18.14 for K gives



18.42

Since we are
given the equilibrium constant in the problem, we can solve for DG° using Equation
18.14.
DG° = RTln K
Substitute K_{w}, R, and T into the
above equation to calculate the standard free energy change, DG°. The temperature at which K_{w} = 1.0 ´
10^{14}
is 25°C
= 298 K.
DG° = RTln K_{w}
DG° =
(8.314
J/mol×K)(298
K) ln (1.0 ´
10^{14}) = 8.0 ´
10^{4} J/mol = 8.0 ´
10^{1} kJ/mol


18.43

K_{sp} =
[Fe^{2+}][OH^{}]^{2} =
1.6 ´
10^{14}
DG° = RTln K_{sp} = (8.314 J/K×mol)(298
K)ln (1.6 ´
10^{14}) = 7.9 ´
10^{4} J/mol = 79
kJ/mol


18.44

Use standard
free energies of formation from Appendix 2 to find the standard free energy
difference.
We can calculate K_{P}
using the following equation. We
carry additional significant figures in the calculation to minimize
rounding errors when calculating K_{P}.
DG° = RTln K_{P}
4.572
´
10^{5} J/mol = (8.314 J/mol×K)(298
K) ln K_{P}
184.54 =
ln K_{P}
Taking the antiln of both sides,
e^{184.54} = K_{P}_{}
K_{P} =
7.2 ´
10^{81}


18.45

a.

We first find the
standard free energy change of the reaction.
= (1)(269.6
kJ/mol) + (1)(0)  (1)(305.0
kJ/mol) = 35.4 kJ/mol
We can calculate K_{P} by rearranging
Equation 18.14 of the text.

b.

We are
finding the free energy difference between the reactants and the products
at their nonequilibrium values.
The result tells us the direction of and the potential for further
chemical change. We use the given
nonequilibrium pressures to compute Q_{P}.
The value of DG
(notice that this is not the standard free energy difference) can be
found using Equation 18.13 of the text and the result from part (a).
DG =
DG°
+ RTln Q = (35.4 ´
10^{3} J/mol) + (8.314 J/K×mol)(298
K)ln (37) = 44.6
kJ/mol
Think About
It:

Which way
is the direction of spontaneous change for this system? What would be the value of DG if the given data were equilibrium pressures? What would be the value of Q_{P} in that case?




18.46

a.

The equilibrium constant is related to the standard free
energy change by the following equation.
DG° =
RTln K
Substitute
K_{P}, R, and T into the above equation to the standard free energy change,
DG°.
DG° =
RTln K_{P}
DG° =
(8.314
J/mol×K)(2000
K) ln (4.40) = 2.464
´
10^{4} J/mol =
24.6
kJ/mol

b.

Under nonstandardstate
conditions, DG is related to the reaction
quotient Q by the following
equation.
DG
= DG°
+ RTln Q_{P}
We are using Q_{P} in the equation
because this is a gasphase reaction.
Step 1: DG°
was calculated in part (a). We
must calculate Q_{P}. We carry additional significant figures
in this calculation to minimize rounding errors.
Step 2: Substitute DG°
= 2.46
´
10^{4} J/mol and Q_{P} into the following
equation to calculate DG.
DG
= DG°
+ RTln Q_{P}
DG
= 2.464
´
10^{4} J/mol + (8.314 J/mol×K)(2000
K) ln (4.062)
DG
= (2.464
´
10^{4} J/mol) + (2.331 ´
10^{4} J/mol)
DG = 1.33
´
10^{3} J/mol =
1.33
kJ/mol



18.47

The expression of K_{P} is:
Thus you can
predict the equilibrium pressure directly from the value of the equilibrium
constant. The only task at hand is
computing the values of K_{P}
using Equations 18.10 and 18.14 of the text.
a.

At 25°C, DG° = DH°

TDS° =
(177.8 ´
10^{3} J/mol)  (298 K)(160.5 J/K×mol)
=
130.0 ´
10^{3} J/mol

b.

At 800°C, DG° =
DH°

TDS° =
(177.8 ´
10^{3} J/mol)  (1073 K)(160.5 J/K×mol)
=
5.58 ´
10^{3} J/mol
Think About
It:

What assumptions are made in the second calculation?




18.48

We use
the given K_{P} to find
the standard free energy change.
DG° = RTln K
DG° = (8.314
J/K×mol)(298
K) ln (5.62 ´
10^{35}) = –2.04 ´ 10^{5}
J/mol = 204 kJ/mol
The standard
free energy of formation of one mole of COCl_{2} can now be found
using the standard free energy of reaction calculated above and the
standard free energies of formation of CO(g) and Cl_{2}(g).


18.49

The equilibrium constant
expression is:
We are
actually finding the equilibrium vapor pressure of water (compare to
Problem 18.47). We use Equation
18.14 of the text.
or 23.6
mmHg
Think About It:

The positive value of DG°
indicates that reactants are favored at equilibrium at 25°C. Is that what you would expect?



18.50

The standard free energy change is given by:
You can look up the standard free energy of formation
values in Appendix 2 of the text.
Thus, the
formation of graphite from diamond is favored
under standardstate conditions at 25°C. However,
the rate of the diamond to graphite conversion is very slow (due to a high
activation energy) so that it will take millions of years before the
process is complete.


18.53

C_{6}H_{12}O_{6}
+ 6O_{2} ® 6CO_{2} + 6H_{2}O DG° = 2880
kJ/mol
ADP
+ H_{3}PO_{4} ® ATP + H_{2}O DG° =
+31 kJ/mol
Maximum number of ATP
molecules synthesized:


18.54

The equation for the coupled reaction is:
glucose + ATP ® glucose 6phosphate
+ ADP
DG° =
13.4 kJ/mol  31 kJ/mol = 18 kJ/mol
As an estimate:
K =
1 ´ 10^{3}


18.55

Strategy:

Melting is an endothermic
process, DH_{fus} > 0. Also,
a liquid generally has a higher entropy than the solid at the same
temperature, so DS_{fus} > 0. To
determine the sign of DG_{fus}, use the fact that melting is spontaneous (DG_{fus} < 0) for
temperatures above the freezing point and is not spontaneous (DG_{fus} > 0) for
temperatures below the freezing point.

Solution:

a.

The temperature is above the freezing point, so
melting is spontaneous and DG_{fus} < 0.

b.

The temperature is at the freezing point, so the
solid and the liquid are in equilibrium and DG_{fus} = 0.

c.

The temperature is below the freezing point, so melting
is not spontaneous and DG_{fus} > 0.




18.56

In each part of this problem we can use the following
equation to calculate DG.
DG
= DG°
+ RTln Q
or
DG
= DG°
+ RTln( [H^{+}][OH^{}])
a.

In this case, the given
concentrations are equilibrium concentrations at 25°C. Since the reaction is at equilibrium, DG = 0. This is advantageous, because it allows
us to calculate DG°. Also recall that at equilibrium, Q = K. We can write:
DG° =
RTln K_{w}
DG° =
(8.314
J/K×mol)(298
K) ln (1.0 ´
10^{14}) =
8.0 ´
10^{4} J/mol

b.

DG
= DG°
+ RTln Q = DG°
+ RTln ([H^{+}][OH^{}])
DG
= (8.0 ´
10^{4} J/mol) + (8.314 J/K×mol)(298
K) ln [(1.0 ´
10^{3})(1.0
´
10^{4})] =
4.0 ´
10^{4} J/mol

c.

DG
= DG°
+ RTln Q = DG°
+ RTln ([H^{+}][OH^{}])
DG
= (8.0 ´
10^{4} J/mol) + (8.314 J/K×mol)(298
K) ln [(1.0 ´
10^{12})(2.0
´
10^{8})] =
3.2
´
10^{4} J/mol

d.

DG
= DG°
+ RTln Q = DG°
+ RTln [H^{+}][OH^{}]
DG
= (8.0 ´
10^{4} J/mol) + (8.314 J/K×mol)(298
K) ln [(3.5)(4.8 ´ 10^{4})] =
6.4 ´
10^{4} J/mol



18.57

Only U and H are associated with the first law alone.


18.58

One possible explanation is simply that
no reaction is possible. In other words, there is an unfavorable free
energy difference between products and reactants (DG > 0).
A second possibility is that the
potential for spontaneous change exists (DG < 0), but that the reaction is
extremely slow (very large activation energy).
A remote third choice is that the
student accidentally prepared a mixture in which the components were
already at their equilibrium concentrations.
Think About It:

Which of the above situations would be altered by the
addition of a catalyst?



18.59

Setting DG equal to zero and solving Equation
18.10for T gives
T = 42°C


18.60

For a reaction to be spontaneous, DG must be negative. If DS is negative, as it is in this case, then the reaction must be
exothermic. When water freezes, it
gives off heat (exothermic).
Consequently, the entropy of the surroundings increases and DS_{univ} > 0.


18.61

If the
process is spontaneous as well as
endothermic, the signs of DG and DH must be negative and positive, respectively. Since DG = DH  TDS, the sign of DS must be positive (DS
> 0) for DG to be negative.


18.62

The equation
is: BaCO_{3}(s)
BaO(s) + CO_{2}(g)
DG° =
(1)(528.4
kJ/mol) + (1)(394.4
kJ/mol) 
(1)(1138.9
kJ/mol) = 216.1 kJ/mol
DG° = RTln K_{P}


18.63

a.

Using the relationship:
benzene DS_{vap} =
87.8 J/K×mol
hexane DS_{vap} =
90.1 J/K×mol
mercury DS_{vap} =
93.7 J/K×mol
toluene DS_{vap} =
91.8 J/K×mol
Trouton’s rule
is a statement about DS. In most
substances, the molecules are in constant and random motion in both the
liquid and gas phases, so DS = 90 J/K×mol.

b.

Using the data in Table
12.6 of the text, we find:
ethanol DS_{vap} =
111.9 J/K×mol
water DS_{vap} =
109.4 J/K×mol
In ethanol and
water, there are fewer possible arrangements of the molecules due to the
network of Hbonds, so DS is
greater.



18.64

Evidence shows
that HF, which is strongly hydrogenbonded in the liquid phase, is still
considerably hydrogenbonded in the vapor state such that its DS_{vap} is smaller than most other substances.


18.65

a.

2CO + 2NO ® 2CO_{2} + N_{2}

b.

The oxidizing
agent is NO; the reducing agent is CO.

c.

DG° =
(2)(394.4
kJ/mol) + (0) 
(2)(137.3
kJ/mol) 
(2)(86.7 kJ/mol) = 687.6
kJ/mol
DG° =
RTln K_{P}_{}
K_{P} =
3 ´
10^{120}

d.

Q_{P}
= 1.2 ´
10^{14}
Since Q_{P} << K_{P}, the reaction will
proceed to the right.

e.

DH° =
(2)(393.5
kJ/mol) + (0) 
(2)(110.5
kJ/mol) 
(2)(90.4 kJ/mol) = 746.8
kJ/mol
Since DH°
is negative, raising the temperature will decrease K_{P} (Le Chatelier’s principle), thereby increasing
the amount of reactants and decreasing the amount of products. No, the formation of N_{2}
and CO_{2} is not favored by raising the temperature.



18.66

a.

At two different temperatures T_{1}
and T_{2},
(1)
(2)
Rearranging Equations (1) and (2),
(3)
(4)
Subtracting equation (3) from equation (4) gives,

b.

Using the equation that
we just derived, we can calculate the equilibrium constant at 65°C.
K_{1} =
4.63 ´
10^{3} T_{1} =
298 K
K_{2} =
? T_{2} =
338 K
Taking the antiln of both sides of the equation,
K_{2} =
0.074
Think About
It:

K_{2}
> K_{1}, as we
would predict for a positive DH°. Recall that an increase in
temperature will shift the equilibrium towards the endothermic
reaction; that is, the decomposition of N_{2}O_{4}.




18.67

The equilibrium reaction is:
AgCl(s)
Ag^{+}(aq) + Cl^{}(aq)
K_{sp} =
[Ag^{+}][Cl^{}] =
1.6 ´
10^{10}
We can
calculate the standard enthalpy of reaction from the standard enthalpies of
formation in Appendix 2 of the text.
DH° =
(1)(105.9 kJ/mol) + (1)(167.2 kJ/mol) 
(1)(127.0
kJ/mol) = 65.7 kJ/mol
From Problem 18.66(a):
K_{1} =
1.6 ´
10^{10} T_{1} = 298 K
K_{2} =
? T_{2} =
333 K
K_{2} =
2.6 ´
10^{9}
The increase in K
indicates that the solubility increases with temperature.


18.68

At absolute
zero. A substance can never have a
negative entropy.


18.69

Assuming that
both DH° and DS° are temperature
independent, we can calculate both DH°
and DS°.
DH° =
(1)(110.5
kJ/mol) + (1)(0)]  [(1)(241.8 kJ/mol) +
(1)(0)]
DH° = 131.3 kJ/mol
DS° = S°(CO) + S°(H_{2}) 
[S°(H_{2}O) + S°(C)]
DS° =
[(1)(197.9 J/K×mol) + (1)(131.0 J/K×mol)]

[(1)(188.7 J/K×mol)
+ (1)(5.69 J/K×mol)]
DS° = 134.5 J/K×mol
It is obvious
from the given conditions that the reaction must take place at a fairly
high temperature (in order to have redhot coke). Setting DG°
= 0
0 = DH° 
TDS°
The temperature must be greater than 703°C
for the reaction to be spontaneous.


18.70

a.

We know that HCl is a strong
acid and HF is a weak acid. Thus, the equilibrium constant will be less
than 1 (K < 1).

b.

The number of particles on each
side of the equation is the same, so DS°
»
0. Therefore DH°
will dominate.

c.

HCl is a weaker bond than HF (see Table 9.4 of the text),
therefore DH°
> 0.



18.71

Begin by
writing the equation for the process described in the problem. Hydrogen ions must be transported from a
region where their concentration is 10^{7.4} (4 ´
10^{8}
M) to a region where their
concentration is 10^{1} (0.1 M).
H^{+}(aq,
4 ´
10^{8}
M) H^{+}(aq, 0.1 M)
DG°
for this process is 0. We use
Equation 18.13 to calculate DG. (T = 37°C + 273 = 310 K.)
DG
= DG° + RTlnQ = (8.314 ´ 10^{3}
kJ/mol)(310 K)= 38 kJ/mol
Therefore, the Gibbs free
energy required for the secretion of 1 mole of H^{+} ions from the
blood plasma to the stomach is 38 kJ.


18.72

For a
reaction to be spontaneous at constant temperature and pressure, DG must be negative. The process of crystallization results in
fewer possible arrangements of ions, so DS < 0. We also
know that
DG
= DH 
TDS
Since DG must be negative, and since the
entropy term will be positive (TDS, where DS is negative), then DH must be negative (DH < 0). The reaction will be exothermic.


18.73

For the reaction: CaCO_{3}(s)
CaO(s) + CO_{2}(g)
Using the equation from Problem 18.66:
Substituting,
Solving,
DH° =
1.74 ´
10^{5} J/mol =
174 kJ/mol


18.74

For the reaction to be spontaneous, DG must be negative.
DG
= DH 
TDS
Given that DH = 19 kJ/mol = 19,000 J/mol, then
DG
= 19,000 J/mol 
(273 K + 72 K)(DS)
Solving the equation with the value of DG = 0
0 =
19,000 J/mol  (273 K + 72 K)(DS)
DS = 55 J/K×mol
This value of
DS which we solved for is the value
needed to produce a DG value of zero. Any
value of DS that is larger than 55 L/mol×K
will result in a spontaneous reaction.


18.75


18.76

The second law states that the entropy
of the universe must increase in a spontaneous process. But the entropy of the universe is the
sum of two terms: the entropy of the
system plus the entropy of the surroundings. One of the entropies can decrease, but
not both. In this case, the decrease
in system entropy is offset by an increase in the entropy of the
surroundings. The reaction in
question is exothermic, and the heat released increases the entropy of the
surroundings.


18.77

At the
temperature of the normal boiling point the free energy difference between
the liquid and gaseous forms of mercury (or any other substances) is zero,
i.e. the two phases are in equilibrium.
We can therefore use Equation 18.10 of the text to find this
temperature. For the equilibrium,
Hg(l)
Hg(g)
DG
= DH 
TDS = 0
DS
= S°[Hg(g)]  S°[Hg(l)]
= 174.7 J/K×mol

77.4 J/K×mol =
97.3 J/K×mol
T_{bp}
= 625 K
Think About It:

What assumptions are
made? Notice that the given
enthalpies and entropies are at standard conditions, namely 25°C
and 1.00 atm pressure. In
performing this calculation we
assume that DH°
and DS°
do not depend on temperature.
The actual normal boiling point of mercury is 356.58°C. Is the assumption of the temperature
independence of these quantities reasonable?



18.78

Setting DG equal to zero, because the system
is at equilibrium, and rearranging Equation 18.10, we write
The problem asks for the change in entropy for the vaporization of
0.50 moles of ethanol. The DS calculated above is for 1 mole of
ethanol.
DS for 0.50 mol
= (112 J/mol×K)(0.50
mol) = 56
J/K


18.79

No. A negative DG° tells us that a reaction has the
potential to happen, but gives no indication of the rate.


18.80

For the given
reaction we can calculate the standard free energy change from the standard
free energies of formation. Then, we can calculate the equilibrium
constant, K_{P}, from the
standard free energy change.
DG° =
(1)(587.4
kJ/mol) 
[(4)(137.3
kJ/mol) + (1)(0)] = 38.2 kJ/mol = 3.82
´
10^{4} J/mol
Substitute DG°,
R, and T (in K) into the following equation to solve for K_{P}.
DG° = RTln K_{P}
K_{P} =
4.5 ´
10^{5}


18.81

a.

DG° =
106.4
kJ/mol
K_{P} = 4
´
10^{18}

b.

DG° =
53.2
kJ/mol
K_{P} =
2 ´
10^{9}
The K_{P}
in (a) is the square of the K_{P}
in (b). Both DG°
and K_{P} depend on the
number of moles of reactants and products specified in the balanced equation.



18.82

We carry
additional significant figures throughout this calculation to minimize
rounding errors. The equilibrium
constant is related to the standard free energy change by the following
equation:
DG° = RTln K_{P}
2.12
´
10^{5} J/mol = (8.314 J/mol×K)(298
K) ln K_{P}
K_{P} =
6.894 ´
10^{38}
We can write the equilibrium constant expression for the
reaction.
This pressure is far too small to measure.


18.83

Strategy:

We assume the denaturation is a singlestep
equilibrium process. Therefore,
because DG is zero at equilibrium, we can
use Equation 18.10, substituting 0 for DG and solving for T, to determine the temperature at
which the denaturation becomes spontaneous.

Setup:

DH = 512 kJ/mol and DS = 1.60 kJ/K ∙ mol.

Solution:

DH and DS are both positive, indicating
this is an endothermic process with high entropy.
320 K =
47°C



18.84

Both (a) and
(b) apply to a reaction with a negative DG°
value. Statement (c) is not always true. An endothermic reaction that has a
positive DS° (increase in entropy)
will have a negative DG°
value at high temperatures.


18.85

a.

If DG°
for the reaction is 173.4 kJ/mol,
then,

b.

DG° = RTln K_{P}
173.4
´
10^{3} J/mol = (8.314
J/K×mol)(298
K)ln K_{P}
K_{P} =
4 ´
10^{31}

c.

DH°
for the reaction is 2 ´ (NO) = (2)(90.4 kJ/mol) =
180.8 kJ/mol
Using the equation in Problem
18.66:
K_{2} = 3
´
10^{6}

d.

Lightning supplies the energy necessary to drive this reaction,
converting the two most abundant gases in the atmosphere into NO(g). The NO gas dissolves in the rain, which
carries it into the soil where it is converted into nitrate and nitrite
by bacterial action. This “fixed”
nitrogen is a necessary nutrient for plants.



18.86

We write the
two equations as follows. The
standard free energy change for the overall reaction will be the sum of the
two steps.
CuO(s)
Cu(s) + O_{2}(g) DG° =
127.2 kJ/mol
C(graphite) + O_{2}(g) CO(g) DG° = 137.3
kJ/mol
CuO + C(graphite) Cu(s)
+ CO(g) DG° = 10.1
kJ/mol
We can now calculate the equilibrium constant from the
standard free energy change, DG°.
ln
K
= 1.81
K =
6.1


18.87

As discussed
in Chapter 18 of the text for the decomposition of calcium carbonate, a
reaction favors the formation of products at equilibrium when
DG° = DH° 
TDS° <
0
If we can
calculate DH° and DS°, we can solve for the
temperature at which decomposition begins to favor products. We use data in Appendix 2 of the text to
solve for DH° and DS°.
DH° = 601.8
kJ/mol + (393.5
kJ/mol) 
(1112.9
kJ/mol) = 117.6 kJ/mol
DS° = S°[MgO(s)] + S°[CO_{2}(g)]  S°[MgCO_{3}(s)]
DS° = 26.78 J/K×mol + 213.6 J/K×mol  65.69 J/K×mol
= 174.7 J/K×mol
For the reaction to begin to favor products,
DH°

TDS° <
0
or
T
> 673.2 K


18.88

According to Table 18.4, a reaction for which DH and DS are both negative is spontaneous at low temperatures, (d).


18.89

a.

DG° =
(1)(0) + (1)(84.9 kJ/mol) 
(1)(0) 
(2)(0)
DG° = 84.9
kJ/mol
DG° =
RTln K
84.9
´
10^{3} J/mol = (8.314
J/mol×K)(298
K) ln K
K =
7.6 ´
10^{14}

b.

DG° =
64.98 kJ/mol
DG° =
RTln K
64.98 ´
10^{3} J/mol = (8.314
J/mol×K)(298
K) ln K
K =
4.1 ´
10^{12}
The activity
series is correct. The very large
value of K for reaction (a)
indicates that products are
highly favored; whereas, the very small value of K for reaction (b) indicates that reactants are highly favored.



18.90

2NO + O_{2} 2NO_{2}
DG° =
(2)(51.8 kJ/mol)  (2)(86.7 kJ/mol) 
0 =
69.8
kJ/mol
DG° = RTln K
69.8
´
10^{3} J/mol = (8.314 J/mol×K)(298
K)ln K
K
= 1.7 ´ 10^{12} M^{1}
k_{r} = 4.2 ´
10^{3} M^{1}s^{1}


18.91

a.

It is the reverse of a disproportionation redox reaction.

b.

DG° =
(2)(228.6
kJ/mol) 
(2)(33.0
kJ/mol) 
(1)(300.4
kJ/mol)
DG° =
90.8
kJ/mol
90.8
´
10^{3} J/mol = (8.314
J/mol×K)(298
K) ln K
K =
8.2 ´
10^{15}
Because of the large value of K, this method is feasible for removing SO_{2}.

c.

DH° =
(2)(241.8
kJ/mol) + (3)(0)  (2)(20.15
kJ/mol) 
(1)(296.4
kJ/mol)
DH° =
146.9
kJ/mol
DS° =
(2)(188.7 J/K×mol) + (3)(31.88 J/K×mol)

(2)(205.64 J/K×mol)

(1)(248.5 J/K×mol)
DS° =
186.7
J/K×mol
DG° =
DH°

TDS°
Due to the negative entropy
change, DS°,
the free energy change, DG°,
will become positive at higher temperatures. Therefore, the reaction will be less
effective at high temperatures.



18.92

(1) Measure K and use DG°
= RT
(2) Measure DH°
and DS°
and use DG°
= DH°
 TDS°


18.93

2O_{3} 3O_{2}
DG° = 326.8
kJ/mol
326.8 ´
10^{3} J/mol = (8.314 J/mol×K)(243
K) ln K_{P}
K_{P} =
1.8 ´
10^{70}
Due to the large magnitude
of K, you would expect this
reaction to be spontaneous in the forward direction. However, this reaction has a large activation energy, so the
rate of reaction is extremely slow.


18.94

DS_{denaturation} = 1.2 kJ/K×mol or 1.2 ´
10^{3} J/K×mol
9.8 J/K×mol per amino acid


18.95

First convert to moles of
ice.
For a phase transition:
DS_{sys} =91.1 J/K
DS_{surr} =91.1
J/K
DS_{univ}
= DS_{sys} + DS_{surr} =
0
The system is at equilibrium.
Think About It:

We ignored the
freezing point depression of the salt water. How would inclusion of this
effect change our conclusion?



18.96

Heating the ore alone is not a
feasible process. Looking at the
coupled process:
Cu_{2}S ® 2Cu + S
DG° =
86.1 kJ/mol
S + O_{2} ® SO_{2}
DG° = 300.4
kJ/mol
Cu_{2}S + O_{2} ® 2Cu + SO_{2} DG° = 214.3 kJ/mol
Since DG°
is a large negative quantity, the coupled reaction is feasible for
extracting sulfur.


18.97

Since we are dealing with the same ion (K^{+}),
Equation 18.13 of the text can be written as:
DG
= DG°
+ RTln Q
DG = 8.5
´ 10^{3} J/mol = 8.5 kJ/mol


18.98

First, we
need to calculate DH°
and DS° for the reaction in
order to calculate DG°.
DH° = 41.2
kJ/mol DS° = 42.0
J/K×mol
Next, we
calculate DG° at 300°C
or 573 K, assuming that DH°
and DS° are temperature
independent.
DG° = DH° 
TDS°
DG° = 41.2
´
10^{3} J/mol  (573 K)(42.0
J/K×mol)
DG° = 1.71
´
10^{4} J/mol
Having solved
for DG°, we can calculate K_{P}.
DG° = RTln K_{P}
1.71 ´
10^{4} J/mol = (8.314 J/K×mol)(573
K) ln K_{P}
ln K_{P} =
3.59
K_{P} =
36
Due to the
negative entropy change calculated above, we expect that DG° will become positive
at some temperature higher than 300°C. We need to find the temperature at which DG° becomes zero. This is the temperature at which
reactants and products are equally favored (K_{P} = 1).
DG° = DH° 
TDS°
0
= DH°

TDS°
T = 981 K = 708°C
This
calculation shows that at 708°C, DG°
= 0 and the equilibrium constant KP=
1. Above 708°C, DG° is positive and K_{P} will be smaller than 1,
meaning that reactants will be favored over products. Note that the temperature 708°C
is only an estimate, as we have assumed that both DH°
and DS° are independent of
temperature.
Using a more efficient catalyst will not increase K_{P} at a given temperature,
because the catalyst will speed up both the forward and reverse reactions.
The value of K_{P}
will stay the same.


18.99

a.

DG°
for CH_{3}COOH:
DG° =
(8.314
J/mol×K)(298
K) ln (1.8 ´
10^{5})
DG° =
2.7 ´
10^{4} J/mol = 27
kJ/mol
DG°
for CH_{2}ClCOOH:
DG° =
(8.314
J/mol×K)(298
K) ln (1.4 ´
10^{3})
DG° =
1.6 ´
10^{4} J/mol = 16
kJ/mol

b.

The TDS°
term determines the value of DG°. The
system’s entropy change
dominates.

c.

The breaking and making of specific OH
bonds. Other contributions include
solvent separation and ion solvation.

d.

The CH_{3}COO^{}
ion, which is smaller than CH_{2}ClCOO^{},
can participate in hydration to a greater extent, leading to solutions
with fewer possible arrangements.



18.100

butane ® isobutane
DG° =
(1)(18.0
kJ/mol) 
(1)(15.9
kJ/mol)
DG° = 2.1
kJ/mol
For a mixture
at equilibrium at 25°C:
DG° = RTln K_{P}
2.1 ´
10^{3} J/mol = (8.314 J/mol×K)(298
K) ln K_{P}
K_{P} =
2.3
This shows
that there are 2.3 times as many moles of isobutane as moles of
butane. Or, we can say for every one
mole of butane, there are 2.3 moles of isobutane.
By
difference, the mole % of butane is 30%.
Yes,
this result supports the notion that straightchain hydrocarbons like
butane are less stable than branchedchain hydrocarbons like isobutane.


18.101

We can calculate K_{P} from DG°.
DG° = (1)(394.4 kJ/mol) + (0)  (1)(137.3 kJ/mol)  (1)(255.2 kJ/mol)
DG° = 1.9 kJ/mol
1.9 ´ 10^{3} J/mol = (8.314 J/mol×K)(1173 K) ln KP
K_{P}
= 1.2
Now, from K_{P}, we can calculate the
mole fractions of CO and CO2.
We assumed that DG° calculated from values was
temperature independent. The values in Appendix 2 of the text are measured at
25°C, but the temperature of the reaction is
900°C.


18.102

DG° = RTln K
and,
DG = DG° + RTln Q
Substituting,
DG = RTln K +
RTln Q
DG = RT(ln
Q  ln K)
If Q > K, DG > 0, and
the net reaction will proceed from right to left (see Section 15.4 of the
text).
If Q < K, DG < 0, and
the net reaction will proceed from left to right.
If Q = K, DG = 0. The system is at equilibrium.


18.103

For a phase transition, DG = 0. We write:
DG = DH 
TDS
0 = DH  TDS
Substituting DH and the temperature, (78°
+ 273°)K
= 195 K, gives
This value of DS_{sub} is for the sublimation of 1 mole of CO_{2}. We convert to the DS value for the sublimation of 84.8 g of CO_{2}.


18.104

Equation 18.10 represents the
standard freeenergy change for a reaction, and not for a particular
compound like CO_{2}. The
correct form is:
DG° = DH°
 TDS°
For a given reaction, DG°
and DH°
would need to be calculated from standard formation values (graphite,
oxygen, and carbon dioxide) first, before plugging into the equation. Also, DS°
would need to be calculated from standard entropy values.
C(graphite)
+ O_{2}(g) ® CO_{2}(g)


18.105

We can calculate DS_{sys} from standard entropy values in Appendix 2 of
the text. We can calculate DS_{surr} from the DH_{sys} value given in the
problem. Finally, we can calculate DS_{univ} from the DS_{sys} and DS_{surr} values.
DS_{sys} =
(2)(69.9 J/K×mol)  [(2)(131.0 J/K×mol)
+ (1)(205.0 J/K×mol)] = 327 J/K×mol
DS_{univ} = DS_{sys} + DS_{surr} =
(327
+ 1918) J/K×mol = 1591 J/K×mol


18.106

The equation representing the process described in the problem can
be written as
glucose(12.3
mM) glucose(0.12 mM)
The temperature is (37°C + 273) = 310 K. DG°
for this process is zero. We use
Equation 18.13 to calculate the free energy change.
DG = DG° + RTlnQ = (8.314 ´ 10^{3}
kJ/K×mol)(310
K)= 11.9 kJ/mol
_{ }
The overall free energy change for the transport of three moles of
glucose into the cell is 3 mol ´ 11.9 kJ/mol = 35.8 kJ.


18.107

q, and w are not state functions. Recall that state functions represent
properties that are determined by the state of the system, regardless of
how that condition is achieved. Heat
and work are not state functions because they are not properties of the
system. They manifest themselves
only during a process (during a change).
Thus their values depend on the path of the process and vary
accordingly.


18.108

(d) will not lead to an
increase in entropy of the system.
The gas is returned to its original state. The entropy of the system does not
change.


18.109

Since the adsorption is spontaneous, DG must be negative. When hydrogen bonds to the surface of the
catalyst, there is a decrease in the number of possible arrangements of
atoms in the system, DS must be negative. Since there is a decrease in entropy, the
adsorption must be exothermic for the process to be spontaneous, DH must be negative.


18.110

a.

An ice cube
melting in a glass of water at 20°C. The value of DG
for this process is negative so it must be spontaneous.

b.

A
"perpetual motion" machine.
In one version, a model has a flywheel which, once started up,
drives a generator which drives a motor which keeps the flywheel running
at a constant speed and also lifts a weight.

c.

A perfect air conditioner; it extracts heat energy from the room
and warms the outside air without using any energy to do so. (Note: this process does not violate
the first law of thermodynamics.)

d.

Same example as (a).

e.

The conversion of NO_{2}(g) to N_{2}O_{4}(g) in a closed flask at constant temperature containing NO_{2}(g) and N_{2}O_{4}(g) at equilibrium. (Note that the conversion of N_{2}O_{4}(g) to NO_{2}(g) is also an equilibrium process
in this example.)



18.111

a.

Each CO molecule has two possible orientations in the
crystal,
CO or OC
If there is no preferred
orientation, then for one molecule there are two, or 2^{1}, choices of orientation.
Two molecules have four or 2^{2} choices, and for 1 mole
of CO there are choices.
From Equation 18.1 of the text:
S = 5.76
J/K×mol

b.

The fact that
the actual residual entropy is 4.2 J/K×mol
means that the orientation is not totally random.



18.112

We can
calculate DG° at 872 K from the
equilibrium constant, K_{1}.
DG°
=
6.25 × 10^{4} J/mol
= 62.5 kJ/mol
We
use the equation derived in Problem 18.66 to calculate DH°.
DH° = 157.8 kJ/mol
Now that both DG° and DH° are known, we can
calculate DS° at 872 K.
DG° = DH° 
TDS°
62.5
× 10^{3} J/mol
= (157.8 × 10^{3}
J/mol)  (872 K)DS°
DS°
= 109 J/K×mol


18.113

We use data
in Appendix 2 of the text to calculate DH°
and DS°.
DH° =
82.93 kJ/mol  49.04 kJ/mol = 33.89 kJ/mol
DS° = S°[C_{6}H_{6}(g)]  S°[C_{6}H_{6}(l)]
DS°
= 269.2 J/K×mol  172.8 J/K×mol
= 96.4 J/K×mol
We can now calculate DG° at 298 K.
DG° = DH°  TDS°
DG°
= 5.2 kJ/mol
DH° is positive because this is an endothermic process. We also expect DS° to be positive because this is a
liquid ® vapor phase change. DG° is positive because we are at a temperature that is below the
boiling point of benzene (80.1°C).


18.114

The reaction mixtures are at equilibrium if K = Q. Use Equation 18.14 to
calculate K.
DG° = RTln K
−3400
J/mol = (8.314 J/mol×K)(298
K) ln K
1.372 =
ln K
Taking the antiln of both sides,
e^{1.372} = K_{}
K =
3.9 ≈ 4
To find Q of
each reaction mixture, take the concentration of products over
concentration of reactants raised to the appropriate power.
For the reaction, A_{2}(g) + B_{2}(g)
2AB(g)
Q =
The partial pressure is equal to the number of molecules
times 0.10 atm. Therefore, the
partial pressure for 3 molecules of AB is 3 × 0.10 atm = 0.30 atm.
Q_{i} =
Q_{ii} =
Q_{iii} =
a.

Since K = Q for reaction (iii), then reaction (iii) is at equilibrium.

b.

Use Equation 18.13 to
calculate DG.
Reaction i has a
negative DG value.

c.

Reactions ii and iii have positive DG values.



18.115

Strategy:

Use Equation 18.14 to solve for the equilibrium
constant, K.

Setup:

R = 8.314 ×
10^{−3} kJ/K ∙ mol and T = (20 + 273) = 293 K.
From problem 18.25, DG = 64 kJ/mol.

Solution:

Solving
Equation 18.14 for K gives


