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CHAPTER SIXTEEN
SPONTANEITY, ENTROPY, AND FREE ENERGY
For Review
1.
a. A spontaneous process is one that occurs without any outside intervention.
b. Entropy is a measure of disorder or randomness.
c. Positional probability is a type of probability that depends on the number of arrangements
in space that yield a particular state.
d. The system is the portion of the universe in which we are interested.
e. The surroundings are everything else in the universe besides the system.
f.
The universe is everything; universe = system + surroundings.
2.
Second law of thermodynamics: in any spontaneous process, there is always an increase in
the entropy of the universe. Suniv = Ssys + Ssurr; When both Ssys and Ssurr are positive,
Suniv must be positive (so process is spontaneous). Suniv is always negative (so process is
nonspontaneous) when both Ssys and Ssurr are negative. When the signs of Ssys are
opposite of each other [(Ssys (+), Ssurr () or vice versa], the process may or may not be
spontaneous.
3.
Ssurr is primarily determined by heat flow. This heat flow into or out of the surroundings
comes from the heat flow out of or into the system. In an exothermic process (H < 0), heat
flows into the surroundings from the system. The heat flow into the surroundings increases
the random motions in the surroundings and increases the entropy of the surroundings (Ssurr
> 0). This is a favorable driving force for spontaneity. In an endothermic reaction (H > 0),
heat is transferred from the surroundings into the system. This heat flow out of the
surroundings decreases the random motions in the surroundings and decreases the entropy of
the surroundings (Ssurr < 0). This is unfavorable. The magnitude of Ssurr also depends on
the temperature. The relationship is inverse; at low temperatures, a specific amount of heat
exchange makes a larger percent change in the surroundings than the same amount of heat
flow at a higher temperature. The negative sign in the Ssurr = H/T equation is necessary to
get the signs correct. For an exothermic reaction where H is negative, this increases Ssurr so
the negative sign converts the negative H value into a positive quantity. For an endothermic
process where H is positive, the sign of Ssurr is negative and the negative sign converts the
positive H value into a negative quantity.
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4.
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633
Suniv = G/T (at constant T and P); When G is negative (Suniv > 0), the process is
spontaneous. When G is positive (Suniv < 0), the process in nonspontaneous (the reverse
process is spontaneous). When G = 0, the process is at equilibrium.
G = H  TS; See Table 16.5 for the four possible sign conventions and the temperature
dependence for these sign combinations. When the signs for H and S are both the same,
then temperature determines if the process is spontaneous. When H is positive (unfavorable)
and S is positive (favorable), high temperatures are needed so the favorable S term
dominates making the process spontaneous (G < 0). When H is negative (favorable) and
S is negative (unfavorable), low temperatures are needed so the favorable H term
dominates making the process spontaneous (G < 0). Note that if G is positive for a
process, then the reverse process has a negative G value and is spontaneous.
At the phase change temperature (melting point or boiling point), two phases are in
equilibrium with each other so G = 0. The G = H  TS equation reduces to H = TS at
the phase change temperature. For the s → l phase change, G is negative above the freezing
point because the process is spontaneous (as we know). For the l → g phase change, the sign
of G is positive below the boiling point as the process is nonspontaneous (as we know).
5.
Third law of thermodynamics: the entropy of a perfect crystal at 0 K is zero. Standard entropy
values (S) represent the increase in entropy that occurs when a substance is heated from 0 K
to 298 K at 1 atm pressure. The equation to calculate S for a reaction using the standard
entropy values is: ΔSorxn = ΣnpSoproducts  ΣnrSoreactants
This equation works because entropy is a state function of the system (it is not pathwaydependent). Because of this, one can manipulate chemical reactions with known Svalues
to determine S for a different reaction. The entropy change for a different reaction equals
the sum of the entropy changes for the reactions added together that yield the different
reaction. This is utilizing Hess’s law. The superscript indicates conditions where T = 25C
and P = 1 atm.
To predict signs for gas phase reactions, you need to realize that the gaseous state represents a
hugely more disordered state as compared to the solid and liquid states. Gases dominate sign
predictions for reactions. Those reactions that show an increase in the number of moles of gas
as reactants are converted to products have an increase in disorder which translates into a
positive S value. S values are negative when there is a decrease in the moles of gas as
reactants are converted into products. When the moles of gaseous reactants and products are
equal, S is usually difficult to predict for chemical reactions. However, predicting signs for
phase changes can be done by realizing the solid state is the most ordered phase (lowest S
values), the liquid state is a slightly more disordered phase than the solid state, with the
o
gaseous state being the most disordered phase by a large margin (Ssolid
 Soliquid  Sogas ) .
Another process involving condensed phases whose sign is also easy to predict (usually) is
the dissolution of a solute in a solvent. Here, the mixed up solution state is usually the more
disordered state as compared to the solute and solvent separately.
6.
Standard free energy change: the change in free energy that will occur for one unit of reaction
if the reactants in their standard states are converted to products in their standard state. The
standard free energy of formation (G of ) of a substance is the change in free energy that
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accompanies the formation of 1 mole of that substance from its constituent elements with all
reactants and products in their standard states. The equation that manipulates G of values to
determining ΔG oreaction is: ΔG° = Σnp Δnof (products)  Σnr ΔGof (reactants) .
Because ΔG is a state function (path independent), chemical reactions with known ΔG
values can be manipulated to determine ΔG for a different reaction. ΔG for the different
reaction is the sum of ΔG for all the steps (reactions) added together to get the different
reaction. This is Hess’s law.
Another way to determine ΔG° for a reaction is to utilize the ΔG° = ΔH°  TΔS° equation.
Here, you need to know ΔH°, ΔS°, and the temperature, then you can use the above equation
to calculate ΔG°.
Of the functions ΔG, ΔH, and ΔS, ΔG has the greatest dependence on temperature. The
temperature is usually assumed to be 25C. However, if other temperatures are used in a
reaction, we can estimate ΔG° at that different temperature by assuming ΔH° and ΔS° are
temperature independent (which is not always the best assumption). We calculate ΔH and
ΔS° values for a reaction using Appendix 4 data, then use the different temperature in the
ΔG° = ΔH  TS° equation to determine (estimate) ΔG° at that different temperature.
7.
No; When using G of values in Appendix 4, we have specified a temperature of 25°C.
Further, if gases or solutions are involved, we have specified partial pressures of 1 atm and
solute concentrations of 1 molar. At other temperatures and compositions, the reaction may
not be spontaneous. A negative ΔG° value means the reaction is spontaneous under standard
conditions.
The free energy and pressure relationship is G = G° + RT ln (P). The RT ln P term corrects
for nonstandard pressures (or concentrations if solutes are involved in the reaction). The
standard pressure for a gas is 1 atm and the standard concentration for solutes is 1 M. The
equation to calculate ΔG for a reaction at nonstandard conditions is: ΔG = ΔG° + RT ln Q
where Q is the reaction quotient determined at the nonstandard pressures and/or
concentrations of the gases and/or solutes in the reaction. The reaction quotient has the exact
same form as the equilibrium constant K. The difference is that the partial pressures or
concentrations used may or may not be the equilibrium concentrations.
All reactions want to minimize their free energy. This is the driving force for any process. As
long as ΔG is a negative, the process occurs. The equilibrium position represents the lowest
total free energy available to any particular reaction system. Once equilibrium is reached, the
system cannot minimize its free energy anymore. Converting from reactants to products or
products to reactants will increase the total free energy of the system which reactions do not
want to do.
8.
At equilibrium, ΔG = 0 and Q = K (the reaction quotient equals the equilibrium constant
value). From the ΔG° = RT ln K equation, when a reaction has K < 1, the ln K term is
negative, giving a positive ΔG° value. When K > 1, the ln K term is positive so ΔG° is negative. When ΔG° = 0, this tell us that K for the process is equal to one (K = 1) because ln 1 =
0.
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635
The sign of ΔG (positive or negative) tells us which reaction is spontaneous (the forward or
reverse reaction). If ΔG < 0, then the forward reaction is spontaneous and if ΔG > 0, then the
reverse reaction is spontaneous. If ΔG = 0, then the reaction is at equilibrium (neither the
forward or reverse reactions are spontaneous). ΔG° gives the equilibrium position by
determining K for a reaction utilizing the equation ΔG° = RT ln K. ΔG° can only be used to
predict spontaneity when all reactants and products are present at standard pressures of 1 atm
and/or standard concentrations of 1 M.
9.
A negative ΔG value does not guarantee that a reaction will occur. It does say that it can
occur (is spontaneous), but whether it will occur also depends on how fast the reaction is
(depends on the kinetics). A process with a negative ΔG may not occur because it is too slow.
The example used in the text is the conversion of diamonds into graphite. Thermodynamics
says the reaction can occur (ΔG > 0), but the reaction is so slow that it doesn’t occur.
The rate of a reaction is directly related to temperature. As temperature increases, the rate of a
reaction increases. Spontaneity, however, does not necessarily have a direct relationship to
temperature. The temperature dependence of spontaneity depends on the signs of ΔH and ΔS
(see Table 16.5 of the text). For example, when ΔH and ΔS are both negative, the reaction
becomes more favorable thermodynamically (ΔG becomes more negative) with decreasing
temperature. This is just the opposite of the kinetics dependence on temperature. Other sign
combinations of ΔH and ΔS have different spontaneity temperature dependence.
10.
wmax = ΔG; When ΔG is negative, the magnitude of ΔG is equal to the maximum possible
useful work obtainable from the process (at constant T and P). When ΔG is positive, the
magnitude of ΔG is equal to the minimum amount of work that must be expended to make
the process spontaneous. Due to waste energy (heat) in any real process, the amount of useful
work obtainable from a spontaneous process is always less than wmax, and for a nonspontaneous reaction, an amount of work greater than wmax must be applied to make the
process spontaneous.
Reversible process: a cyclic process carried out by a hypothetical pathway, which leaves the
universe the same as it was before. No real process is reversible.
Questions
7.
Living organisms need an external source of energy to carry out these processes. Green plants
use the energy from sunlight to produce glucose from carbon dioxide and water by
photosynthesis. In the human body, the energy released from the metabolism of glucose helps
drive the synthesis of proteins. For all processes combined, ΔSuniv must be greater than zero
(2nd law).
8.
Dispersion increases the entropy of the universe because the more widely something is
dispersed, the greater the disorder. We must do work to overcome this disorder. In terms of
the 2nd law, it would be more advantageous to prevent contamination of the environment
rather than to clean it up later. As a substance disperses, we have a much larger area that must
be decontaminated.
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9.
It appears that the sum of the two processes has no net change. This is not so. By the second
law of thermodynamics, ΔSuniv must have increased even though it looks as if we have gone
through a cyclic process.
10.
The introduction of mistakes is an effect of entropy. The purpose of redundant information is
to provide a control to check the "correctness" of the transmitted information.
11.
As a process occurs, Suniv will increase; Suniv cannot decrease. Time, like Suniv, only goes
in one direction.
12.
This reaction is kinetically slow but thermodynamically favorable (ΔG < 0). Thermodynamics only tells us if a reaction can occur. To answer the question will it occur, one also needs to
consider the kinetics (speed of reaction). The ultraviolet light provides the activation energy
for this slow reaction to occur.
13.
Possible arrangements for one molecule:
1 way
1 way
Both are equally probable.
Possible arrangements for two molecules:
1 way
2 ways
most probable
1 way
Possible arrangement for three molecules:
1 way
3 ways
3 ways
1 way
equally most probable
14.
Ssurr = H/T; Heat flow (H) into or out of the system dictates Ssurr. If heat flows into the
surroundings, the random motions of the surroundings increase and the entropy of the surroundings increase. The opposite is true when heat flows from the surroundings into the
system (an endothermic reaction). Although the driving force described here really results
from the change in entropy of the surroundings, it is often described in terms of energy.
Nature tends to seek the lowest possible energy.
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15.
16.
17.
SPONTANEITY, ENTROPY, AND FREE ENERGY
637
Note that these substances are not in the solid state, but are in the aqueous state; water
molecules are also present. There is an apparent increase in ordering when these ions are
placed in water as compared to the separated state. The hydrating water molecules must be in
a highly ordered arrangement when surrounding these anions.
G = RTlnK = H  TS; HX(aq) ⇌ H+(aq) + X(aq) Ka reaction; The value of
Ka for HF is less than one while the other hydrogen halide acids have K a > 1. In terms of
G, HF must have a positive G orxn value while the other H-X acids have Grxn < 0. The
reason for the sign change in the Ka value between HF versus HCl, HBr, and HI is entropy.
S for the dissociation of HF is very large and negative. There is a high degree of ordering
that occurs as the water molecules associate (hydrogen bond) with the small F ions. The
entropy of hydration strongly opposes HF dissociating in water, so much so that it
overwhelms the favorable hydration energy making HF a weak acid.
One can determine S and H for the reaction using the standard entropies and standard
enthalpies of formation in Appendix 4, then use the equation G = H  TS. One can
also use the standard free energies of formation in Appendix 4. And finally, one can use
Hess’s law to calculate G. Here, reactions having known G values are manipulated to
determine G for a different reaction.
For temperatures other than 25C, G is estimated using the G = H  TS equation.
The assumptions made are that the H and S values determined from Appendix 4 data
are temperature independent. We use the same H and S values as determined when T =
25C, then plug in the new temperature in Kelvin into the equation to estimate G at the
new temperature.
18.
The sign of G tells us if a reaction is spontaneous or not at whatever concentrations are
present (at constant T and P). The magnitude of G equals wmax. When G < 0, the
magnitude tells us how much work, in theory, could be harnessed from the reaction. When
G > 0, the magnitude tells us the minimum amount of work that must be supplied to make
the reaction occur. G gives us the same information only when the concentrations for all
reactants and products are at standard conditions (1 atm for gases, 1 M for solute). These
conditions rarely occur.
G = RTlnK; From this equation, one can calculate K for a reaction if G is known at
that temperature. Therefore, G gives the equilibrium position for a reaction. To determine
K at a temperature other than 25C, one needs to know G at that temperature. We assume
H and S are temperature independent and use the equation G = H - TS to
estimate G at the different temperature. For K = 1, we want G = 0, which occurs when
H = TS. Again, assume H and S are temperature independent, then solve for T
(=H/S). At this temperature, K = 1 because G = 0. This only works for reactions
where the signs of H and S are the same (either both positive or both negative). When
the signs are opposite, K will always be greater than one (when H is negative and S is
positive) or K will always be less than one (when H is positive and S is negative).
When the signs of H and S are opposite, K can never equal one.
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Exercises
Spontaneity, Entropy, and the Second Law of Thermodynamics: Free Energy
19.
a, b and c; From our own experiences, salt water, colored water and rust form without any
outside intervention. A bedroom, however, spontaneously gets cluttered. It takes an outside
energy source to clean a bedroom.
20.
c and d; It takes an outside energy source to build a house and to launch and keep a satellite
in orbit.
21.
We draw all of the possible arrangements of the two particles in the three levels.
2 kJ
1 kJ
0 kJ
x
Total E =
xx
x
x
x
0 kJ
1 kJ
2 kJ
x
x
xx
xx
__
2 kJ
3 kJ
4 kJ
The most likely total energy is 2 kJ.
22.
2 kJ
AB
1 kJ
AB
B
B
A
A
B
A
A
B
B
A_
A
B_
0 kJ
AB
_
ET =
0 kJ 2 kJ 4 kJ 1 kJ 1 kJ 2 kJ 2 kJ 3 kJ 3 kJ
The most likely total energy is 2 kJ.
23.
a. H2 at 100°C and 0.5 atm; Higher temperature and lower pressure means greater volume
and hence, greater positional probability.
b. N2 at STP has the greater volume.
c. H2O(l) is more positional probability than H2O(s).
24.
Of the three phases (solid, liquid, and gas), solids are most ordered and gases are most
disordered. Thus, a , b, and f (melting, sublimation, and boiling) involve an increase in the
entropy of the system since going from a solid to a liquid or a solid to a gas or a liquid to a
gas increases disorder. For freezing (process c), a substance goes from the more disordered
liquid state to the more ordered solid state, hence, entropy decreases. Process d (mixing)
involves an increase in disorder (entropy) while separation increases order (decreases the
entropy of the system). So of all the processes, a, b, d, and f result in an increase in the
entropy of the system.
25.
a. To boil a liquid requires heat. Hence, this is an endothermic process. All endothermic
processes decrease the entropy of the surroundings (ΔSsurr is negative).
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639
b. This is an exothermic process. Heat is released when gas molecules slow down enough to
form the solid. In exothermic processes, the entropy of the surroundings increases (ΔSsurr
is positive).
26.
27.
a. ΔSsurr =
 (2221 kJ )
 ΔH
= 7.45 kJ/K = 7.45 × 103 J/K

T
298 K
b. ΔSsurr =
 ΔH
 112 kJ
= 0.376 kJ/K = 376 J/K

T
298 K
ΔG = ΔH  TΔS; When ΔG is negative, then the process will be spontaneous.
a. ΔG = ΔH  TΔS = 25 × 103 J  (300. K)(5.0 J/K) = 24,000 J, Not spontaneous
b. ΔG = 25,000 J  (300. K)(100. J/K) = 5000 J, Spontaneous
c. Without calculating ΔG, we know this reaction will be spontaneous at all temperatures.
ΔH is negative and ΔS is positive (-TΔS < 0). ΔG will always be less than zero with
these sign combinations for ΔH and ΔS.
d. ΔG = 1.0 × 104 J  (200. K)(40. J/K) = 2000 J, Spontaneous
28.
ΔG = ΔH  TΔS; A process is spontaneous when ΔG < 0. For the following, assume ΔH and
ΔS are temperature independent.
a. When ΔH and ΔS are both negative, ΔG will be negative below a certain temperature
where the favorable ΔH term dominates. When ΔG = 0, then ΔH = TΔS. Solving for this
temperature:
T=
ΔH
 18,000 J
= 3.0 × 102 K

ΔS
 60. J / K
At T < 3.0 × 102 K, this process will be spontaneous (ΔG < 0).
b. When ΔH and ΔS are both positive, ΔG will be negative above a certain temperature
where the favorable ΔS term dominates.
T=
ΔH 18,000 J
= 3.0 × 102 K

ΔS
60. J / K
At T > 3.0 × 102 K, this process will be spontaneous (ΔG < 0).
c. When ΔH is positive and ΔS is negative, this process can never be spontaneous at any
temperature because ΔG can never be negative.
d. When ΔH is negative and ΔS is positive, this process is spontaneous at all temperatures
because ΔG will always be negative.
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At the boiling point, ΔG = 0 so ΔH = TΔS.
ΔS =
ΔH
27.5 kJ / mol
= 8.93 × 10 2 kJ/Kmol = 89.3 J/Kmol

T
(273  35) K
ΔH
58.51  10 3 J / mol

= 629.7 K
ΔS
92.92 J / K  mol
30.
At the boiling point, ΔG = 0 so ΔH = TΔS. T =
31.
a. NH3(s) → NH3(l); ΔG = ΔH  TΔS = 5650 J/mol  200. K (28.9 J/Kmol)
ΔG = 5650 J/mol  5780 J/mol = T = =130 J/mol
Yes, NH3 will melt because ΔG < 0 at this temperature.
b. At the melting point, ΔG = 0 so T =
32.
ΔH
5650 J / mol
= 196 K.

ΔS
28.9 J / K  mol
C2H5OH(l) → C2H5OH(g); At the boiling point, G = 0 and Suniv = 0. For the vaporization
process, S is a positive value while H is a negative value. To calculate Ssys, we will
determine Ssurr from H and the temperature, then Ssys = Ssurr for a system at equilibrium.
Ssurr =
 ΔH
38.7  10 3 J / mol

= 110. J/Kmol
T
351 K
Ssys = Ssurr = (110.) = 110. J/K mol
Chemical Reactions: Entropy Changes and Free Energy
33.
a. Decrease in disorder; ΔS°()
c. Decrease in disorder (Δn < 0); ΔS°()
b. Increase in disorder; ΔS°(+)
d. Increase in disorder (Δn > 0); ΔS°(+)
For c and d, concentrate on the gaseous products and reactants. When there are more gaseous
product molecules than gaseous reactant molecules (Δn > 0), then ΔS° will be positive
(disorder increases). When Δn is negative, then ΔS° is negative (disorder decreases or order
increases).
34.
35.
a. Decrease in disorder (Δn < 0); ΔS°()
b. Decrease in disorder (Δn < 0); ΔS°()
c. Increase in disorder; ΔS°(+)
d. Increase in disorder; ΔS°(+)
a. Cgraphite(s); Diamond has a more ordered structure than graphite.
b. C2H5OH(g); The gaseous state is more disordered than the liquid state.
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SPONTANEITY, ENTROPY, AND FREE ENERGY
641
c. CO2(g); The gaseous state is more disordered than the solid state.
a. He (10 K); S = 0 at 0 K
b. N2O; More complicated molecule
c. H2O(l): The liquid state is more disordered than the solid state.
37.
a. 2 H2S(g) + SO2(g) → 3 Srhombic(s) + 2 H2O(g); Because there are more molecules of
reactant gases as compared to product molecules of gas (Δn = 2 - 3 < 0), ΔS° will be
negative.
ΔS° =
 n pSoproducts   n rSoreactants
ΔS° = [3 mol Srhombic(s) (32 J/Kmol) + 2 mol H2O(g) (189 J/Kmol)]
 [2 mol H2S(g) (206 J/Kmol) + 1 mol SO2(g) (248 J/Kmol)]
ΔS° = 474 J/K  660. J/K = 186 J/K
b. 2 SO3(g) → 2 SO2(g) + O2(g); Because Δn of gases is positive (Δn = 3  2), ΔS° will be
positive.
ΔS = 2 mol(248 J/Kmol) + 1 mol(205 J/Kmol)  [2 mol(257 J/Kmol)] = 187 J/K
c. Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g); Because Δn of gases = 0 (Δn = 3  3), we
can’t easily predict if ΔS° will be positive or negative.
ΔS = 2 mol(27 J/Kmol) + 3 mol(189 J/Kmol) 
[1 mol(90. J/Kmol) + 3 mol (141 J/Kmol)]
= 138
J/K
38.
a. H2(g) + 1/2 O2(g) → H2O(l); Since Δn of gases is negative, then ΔS° will be negative.
ΔS° = 1 mol H2O(l) (70. J/Kmol) 
[1 mol H2(g) (131 J/Kmol) + 1/2 mol O2(g) (205
J/Kmol)]
ΔS° = 70. J/K  234 J/K = 164 J/K
b. 2 CH3OH(g) + 3 O2(g) → 2 CO2(g) + 4 H2O(g); Because Δn of gases is positive, ΔS°
will be positive.
[2 mol (214 J/Kmol) + 4 mol (189 J/Kmol)] –
[2 mol (240. J/Kmol) + 3 mol (205 J/Kmol)] = 89 J/K
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c. HCl(g) → H+(aq) + Cl(aq); The gaseous state dominates predictions of ΔS°. Here the
gaseous state is more disordered than the ions in solution so ΔS° will be negative.
39.
ΔS° = 1 mol H+(0) + 1 mol Cl(57 J/Kmol)  1 mol HCl(187 J/Kmol) = 130. J/K
C2H2(g) + 4 F2(g) → 2 CF4(g) + H2(g); ΔS° = 2 SoCF4  SoH2  [SoC2H2  4SoF2 ]
358 J/K = (2 mol) SoCF4 + 131 J/K  [201 J/K + 4(203 J/K)], SoCF4 = 262 J/Kmol
40.
144 J/K = (2 mol) SoAlBr3  [2(28 J/K) + 3(152 J/K)], SoAlBr3 = 184 J/Kmol
41.
a. Srhombic → Smonoclinic; This phase transition is spontaneous (ΔG < 0) at temperatures above
95°C. ΔG = ΔH  TΔS; For ΔG to be negative only above a certain temperature, then ΔH
is positive and ΔS is positive (see Table 16.5 of text).
b. Because ΔS is positive, Srhombic is the more ordered crystalline structure.
42.
Enthalpy is not favorable, so ΔS must provide the driving force for the change. Thus, ΔS is
positive. There is an increase in disorder, so the original enzyme has the more ordered
structure.
43.
a. When a bond is formed, energy is released so ΔH is negative. There are more reactant
molecules of gas than product molecules of gas (Δn < 0), so ΔS will be negative.
b. ΔG = ΔH  TΔS; For this reaction to be spontaneous (ΔG < 0), the favorable enthalpy
term must dominate. The reaction will be spontaneous at low temperatures where the ΔH
term dominates.
44.
Because there are more product gas molecules than reactant gas molecules (Δn > 0), ΔS will
be positive. From the signs of ΔH and ΔS, this reaction is spontaneous at all temperatures. It
will cost money to heat the reaction mixture. Because there is no thermodynamic reason to do
this, the purpose of the elevated temperature must be to increase the rate of the reaction, i.e.,
kinetic reasons.
45
a.
CH4(g)
+ 2 O2(g) → CO2(g ) + 2 H2O(g)
_________________________________________________________
ΔH of
75 kJ/mol
0
393.5
242
ΔG of
51 kJ/mol
0
394
229 Data from Appendix 4
S°
186 J/Kmol
205
214
189
__________________________________________________________
ΔH° =
 n pΔHof, products   n r ΔHof, reactants;
ΔS° =
 n pSoproducts   n rSoreactants
ΔH° = 2 mol(242 kJ/mol) + 1 mol(393.5 kJ/mol)  [1 mol(75 kJ/mol)] = 803 kJ
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
ΔS° = 2 mol(189 J/K∙mol) + 1 mol(214 J/Kmol)
 [1 mol(186 J/Kmol) + 2 mol(205 J/Kmol)] = 4 J/K
There are two ways to get ΔG°. We can use ΔG° = ΔH°  TΔS° (be careful of units):
ΔG° = ΔH°  TΔS° = -803 × 103 J  (298 K)( 4 J/K) = 8.018 × 105 J = 802 kJ
or we can use ΔG of values where ΔG° =
 n pΔGof , products   n r ΔGof, reactants :
ΔG° = 2 mol(229 kJ/mol) + 1 mol(394 kJ/mol)  [1 mol(51 kJ/mol)]
ΔG° = 801 kJ (Answers are the same within round off error.)
b.
6 CO2(g) + 6 H2O(l) → C6H12O6(s) + 6 O2(g)
______________________________________________________
ΔH of
393.5 kJ/mol
286
1275
0
S°
214 J/Kmol
70.
212
205
______________________________________________________
ΔH° = 1275  [6(286) + 6(393.5)] = 2802 kJ
ΔS° = 6(205) + 212  [6(214) + 6(70.)] = 262 J/K
ΔG° = 2802 kJ  (298 K)( 0.262 kJ/K) = 2880. kJ
c.
P4O10(s) + 6 H2O(l) → 4 H3PO4(s)
_____________________________________________
ΔH of
2984
286
1279
(kJ/mol)
S°
229
70.
110.
(J/Kmol)
_____________________________________________
ΔH° = 4 mol(1279 kJ/mol)  [1 mol(2984 kJ/mol) + 6 mol(286 kJ/mol)] = 416 kJ
ΔS° = 4(110.)  [229 + 6(70.)] = 209 J/K
ΔG° = ΔH°  TΔS° = 416 kJ  (298 K)( 0.209 kJ/K) = 354 kJ
d.
HCl(g) + NH3(g) → NH4Cl(s)
___________________________________________
ΔH of
92
46
314
(kJ/mol)
643
644
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
S° (J/Kmol) 187
193
96
____________________________________________
ΔH° = 314  [92  46] = 176 kJ; ΔS° = 96  [187 + 193] = 284 J/K
46.
ΔG° = ΔH°  TΔS° = 176 kJ  (298 K)( 0.284 kJ/K) = 91 kJ
a. ΔH° = 2(46 kJ) = 92 kJ; ΔS° = 2(193 J/K)  [3(131 J/K) + 192 J/K] = 199 J/K
ΔG° = ΔH°  TΔS° = 92 kJ  298 K(0.199 kJ/K) = 33 kJ
b. ΔG° is negative, so this reaction is spontaneous at standard conditions.
c. ΔG° = 0 when T =
 92 kJ
ΔH o

= 460 K
o
 0.199 kJ / K
ΔS
At T < 460 K and standard pressures (1 atm), the favorable ΔH° term dominates and the
reaction is spontaneous (ΔG° < 0).
47.
ΔG° = 58.03 kJ  (298 K)( 0.1766 kJ/K) = 5.40 kJ
ΔG° = 0 = ΔH°  TΔS°, T =
ΔH o
 58.03 kJ

= 328.6 K
o
 0.1766 kJ / K
ΔS
ΔG° is negative below 328.6 K where the favorable ΔH° term dominates.
48.
H2O(l) → H2O(g); ΔG° = 0 at the boiling point of water at 1 atm and 100.C.
ΔH° = TΔS°, ΔS° =
ΔH o
40.6  10 3 J / mol

= 109 J/Kmol
T
373 K
At 90.C: ΔG° = ΔH°  TΔS° = 40.6 kJ/mol – (363 K)(0.109 kJ/Kmol) = 1.0 kJ/mol
As expected, ΔG° > 0 at temperatures below the boiling point of water at 1 atm (process is
nonspontaneous).
At 110.C: ΔG° = ΔH°  TΔS° = 40.6 kJ/mol – (383 K)(0.109 J/Kmol) = 1.1 kJ/mol
When ΔG° < 0, the boiling of water is spontaneous at 1 atm and T > 100.C (as expected).
49.
CH4(g) + CO2(g) → CH3CO2H(l)
ΔH° = 484  [75 + (393.5)] = 16 kJ; ΔS° = 160  [186 + 214] = 240. J/K
ΔG° = ΔH°  TΔS° = 16 kJ  (298 K)( 0.240 kJ/K) = 56 kJ
This reaction is spontaneous only at temperatures below T = ΔH°/ΔS° = 67 K (where the
favorable ΔH° term will dominate giving a negative ΔG° value). This is not practical.
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
645
Substances will be in condensed phases and rates will be very slow at this extremely low
temperature.
CH3OH(g) + CO(g) → CH3CO2H(l)
ΔH° = -484  [110.5 + (201)] = 173 kJ; ΔS° = 160  [198 + 240.] = 278 J/K
ΔG° = 173 kJ  (298 K)( 0.278 kJ/K) = 90. kJ
This reaction also has a favorable enthalpy and an unfavorable entropy term. This reaction is
spontaneous at temperatures below T = ΔH°/ΔS° = 622 K. The reaction of CH3OH and CO
will be preferred. It is spontaneous at high enough temperatures that the rates of reaction
should be reasonable.
50.
C2H4(g) + H2O(g) → CH3CH2OH(l)
ΔH° = 278  (52  242) = 88 kJ; ΔS° = 161  (219 + 189) = 247 J/K
When ΔG° = 0, ΔH° = TΔS°, T =
ΔH o
 88  10 3 J

= 360 K
o
 247 J / K
ΔS
Because the signs of ΔH° and ΔS° are both negative, this reaction will be spontaneous at
temperatures below 360 K (where the favorable ΔH° term will dominate).
C2H6(g) + H2O(g) → CH3CH2OH(l) + H2(g)
ΔH° = 278  (84.7  242) = 49 kJ; ΔS° = 131 + 161  (229.5 + 189) = 127 J/K
This reaction can never be spontaneous because of the signs of ΔH° and ΔS°.
Thus the reaction C2H4(g) + H2O(g) → C2H5OH(l) would be preferred.
CH4(g) → 2 H2(g) + C(s)
51.
2 H2(g) + O2(g) → 2 H2O(l)
ΔG° = (51 kJ)
ΔG° = 2(237 kJ)
C(s) + O2(g) → CO2(g)
ΔG° = 394 kJ
______________________________________________________________
CH4(g) + 2 O2(g) → 2 H2O(l) + CO2(g)
52.
ΔG° = 817 kJ
6 C(s) + 6 O2(g) → 6 CO2(g)
ΔG° = 6(394 kJ)
3 H2(g) + 3/2 O2(g) → 3 H2O(l)
ΔG° = 3(237 kJ)
6 CO2(g) + 3 H2O(l) → C6H6(l) + 15/2 O2(g)
ΔG° = 1/2 (6399 kJ)
______________________________________________________________
6 C(s) + 3 H2(g) → C6H6(l)
53.
ΔG° =
 n pΔGof , products   n r ΔGof, reactants ,
ΔG° = 125 kJ
374 kJ = 1105 kJ  ΔG of , SF4
646
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
ΔG of , SF4 = 731 kJ/mol
54.
5490. kJ = 8(394 kJ) + 10(237 kJ)  2 ΔG of , C4H10 , ΔG of , C4H10 = 16 kJ/mol
55.
ΔG° =
 n pΔGof , products   n r ΔGof, reactants
ΔG° = [57.37 kJ + (68.85 kJ) + 3(95.30 kJ)] – [3(0) + 2(50.72 kJ)] = 310.68 kJ
For a temperature change from 25°C to ~20°C (room temperature), the magnitude of ΔG°
will not change much. Therefore, ΔG° will be a negative value at room temperature (~20°C),
so the reaction will be spontaneous.
56.
a. ΔG° = 2(270. kJ)  2(502 kJ) = 464 kJ
b. Because ΔG° is positive, this reaction is not spontaneous at standard conditions at
298 K.
c. ΔG° = ΔH°  TΔS°, ΔH° = ΔG° + TΔS° = 464 kJ + 298 K(0.179 kJ/K) = 517 kJ
We need to solve for the temperature when ΔG° = 0:
ΔH o
517 kJ

ΔG° = 0 = ΔH°  TΔS°, T =
= 2890 K
o
0.179 kJ / K
ΔS
This reaction will be spontaneous at standard conditions (ΔG° < 0) when T > 2890 K.
Here the favorable entropy term will dominate.
Free Energy: Pressure Dependence and Equilibrium
57.
ΔG = ΔG° + RT ln Q; For this reaction: ΔG = ΔG° + RT ln
PNO2  PO 2
PNO  PO3
ΔG° = 1 mol(52 kJ/mol) + 1 mol(0)  [1 mol(87 kJ/mol) + 1 mol(163 kJ/mol)] = 198 kJ
ΔG = 198 kJ +
8.3145 J / K  mol
(1.00  10 7 ) (1.00  10 3 )
(298 K ) ln
1000 J / kJ
(1.00  10 6 ) (1.00  10 6 )
ΔG = 198 kJ + 9.69 kJ = 188 kJ
58.
ΔG° = 3(0) + 2(229)  [2(-34) + 1(300.)] = 90. kJ
ΔG = ΔG° + RT ln
PH2 O
2
PH2 S  PSO 2
2
= 90. kJ +


(8.3145 ) (298)
(0.030 ) 2
kJ ln

4
1000
 (1.0  10 ) (0.010 ) 
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
647
ΔG = 90. kJ + 39.7 kJ = 50. kJ
59.
ΔG = ΔG° + RT ln Q = ΔG° + RT ln
PN 2O 4
2
PNO
2
ΔG° = 1 mol(98 kJ/mol)  2 mol(52 kJ/mol) = 6 kJ
a. These are standard conditions so ΔG = ΔG° because Q = 1 and ln Q = 0. Because ΔG° is
negative, the forward reaction is spontaneous. The reaction shifts right to reach equilibrium.
b. ΔG = 6 × 103 J + 8.3145 J/Kmol (298 K) ln
0.50
(0.21) 2
ΔG = 6 × 103 J + 6.0 × 103 J = 0
Because ΔG = 0, this reaction is at equilibrium (no shift).
c. ΔG = 6 × 103 J + 8.3145 J/Kmol (298 K) ln
1.6
(0.29) 2
ΔG = 6 × 103 J + 7.3 × 103 J = 1.3 × 103 J = 1 × 103 J
Because ΔG is positive, the reverse reaction is spontaneous, and the reaction shifts to the
left to reach equilibrium.
60.
a. ΔG = ΔG° + RT ln
ΔG = 34 kJ +
2
PNH
3
PN 2  PH2 2
; ΔG° = 2 ΔG of , NH3 = 2(17) = 34 kJ
(8.3145 J / K  mol ) (298 K )
(50.) 2
ln
1000 J / kJ
(200 .) (200 .)3
ΔG = 34 kJ  33 kJ = 67 kJ
b. ΔG = 34 kJ
(8.3145 J / K  mol ) (298 K )
(200 .) 2
ln
1000 J / kJ
(200 .) (600 .)3
ΔG = 34 kJ  34.4 kJ = 68 kJ
61.
At 25.0°C: ΔG° = ΔH° − TΔS° = −58.03 × 103 J/mol − (298.2 K)(−176.6 J/Kmol)
= −5.37 × 103 J/mol
ΔG° = −RT ln K, ln K =
K = e2.166 = 8.72


 ΔG o
 (5.37 10 3 J / mol )
 = 2.166
 exp 

RT
 (8.3145 J / K  mol ) (298 .2 K ) 
648
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
At 100.0°C: ΔG° = −58.03 × 103 J/mol − (373.2 K)(−176.6 J/Kmol) = 7.88 × 103 J/mol
ln K =
62.
 (7.88  10 3 J / mol )
= −2.540, K = e-2.540 = 0.0789
(8.3145 J / K  mol ) (373 .2 K )
Note: When determining exponents, we will round off after the calculation is complete. This
helps eliminate excessive round off error.
a. ΔH° = 2 mol(−92 kJ/mol) − [1 mol(0) + 1 mol(0)] = −184 kJ
ΔS° = 2 mol(187 J/Kmol) − [1 mol(131 J/Kmol) + 1 mol(223 J/Kmol)] = 20. J/K
ΔG° = ΔH° − TΔS° = −184 × 103 J − 298 K (20. J/K) = −1.90 × 105 J = −190. kJ
ΔG° = −RT ln K, ln K =
 ΔG o
 (1.90  10 5 J )

= 76.683
RT
8.3145 J / K  mol (298 K )
K = e76.683 = 2.01 × 1033
b. These are standard conditions so ΔG = ΔG° = −190. kJ. When ΔG is negative, the
forward reaction is spontaneous so the reaction shifts right to reach equilibrium.
63.
When reactions are added together, the equilibrium constants are multiplied together to
determine the K value for the final reaction.
H2(g) +O2(g) ⇌ H2O2(g)
K = 2.3 × 106
H2O(g) ⇌ H2(g) + 1/2O2(g)
K = (1.8 × 1037)1/2
_______________________________________________________________________
H2O(g) + 1/2O2(g) ⇌ H2O2(g)
K = 2.3 × 106(1.8 × 1037) 1/2 = 5.4 × 10 13
ΔG° = −RT ln K = −8.3145 J/Kmol (600. K) ln(5.4 × 10 13 ) = 1.4 × 105 J/mol = 140 kJ/mol
64.
ΔH of (kJ/mol)
a.
NH3(g)
O2(g)
NO(g)
H2O(g)
NO2(g)
HNO3(l)
H2O(l)
−46
0
90.
−242
34
−174
−286
S° (J/Kmol)
193
205
211
189
240.
156
70.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
ΔH° = 6(−242) + 4(90.) − [4(−46)] = −908 kJ
ΔS° = 4(211) + 6(189) − [4(193) + 5(205)] = 181 J/K
ΔG° = −908 kJ − 298 K (0.181 kJ/K) = −962 kJ
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
ΔG° = − RT ln K, ln K =
649


 ΔG o
 (962  10 3 J)
 = 388
 

RT
8
.
3145
J
/
K

mol

298
K


ln K = 2.303 log K, log K = 168, K = 10168 (an extremely large number)
2 NO(g) + O2(g) → 2 NO2(g)
ΔH° = 2(34) − [2(90.)] = −112 kJ; ΔS° = 2(240.) − [2(211) + (205)] = −147 J/K
ΔG° = −112 kJ − (298 K)( −0.147 kJ/K) = −68 kJ
K = exp


 ΔG o
 (68,000 J)
 = e27.44 = 8.3 × 1011
 exp 
RT
8
.
3145
J
/
K

mol
(
298
K
)


Note: When determining exponents, we will round off after the calculation is complete.
3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)
ΔH° = 2(−174) + (90.) − [3(34) + (−286)] = −74 kJ
ΔS° = 2(156) + (211) − [3(240.) + (70.)] = −267 J/K
ΔG° = −74 kJ − (298 K)( −0.267 kJ/K) = 6 kJ
K = exp


 ΔG o
 6000 J
 = e 2.4 = 9 × 10 2
 exp 
RT
 8.3145 J / K  mol (298 K) 
b. ΔG° = −RT ln K; T = 825°C = (825 + 273) K = 1098 K; We must determine ΔG°
at 1098 K.
o
= ΔH° − TΔS° = −908 kJ − (1098 K)(0.181 kJ/K) = −1107 kJ
ΔG1098
K = exp


 ΔG o
 (1.107  10 6 J)
 = e121.258 = 4.589 × 1052
 exp 

RT
8
.
3145
J
/
K

mol
(
1098
K
)


c. There is no thermodynamic reason for the elevated temperature because ΔH° is negative
and ΔS° is positive. Thus, the purpose for the high temperature must be to increase the
rate of the reaction.
65.
K=
2
PNF
3
PN 2  PF32

(0.48) 2
= 4.4 × 104
0.021(0.063)3
o
= −RT ln K = −8.3145 J/Kmol (800. K) ln (4.4 × 104) = −7.1 × 104 J/mol =
ΔG 800
−71 kJ/mol
66.
2 SO2(g) + O2(g) → 2 SO3(g); ΔG° = 2(−371 kJ) − [2(−300. kJ)] = −142 kJ
650
CHAPTER 16
ΔG° = −RT ln K, ln K =
SPONTANEITY, ENTROPY, AND FREE ENERGY
 ΔG o
 (142 .000 J )

= 57.311
RT
8.3145 J / K  mol (298 K )
K = e57.311 = 7.76 × 1024
K = 7.76 × 1024 =
2
PSO
3
2
PSO
 PO2
2

(2.0) 2
, PSO 2 = 1.0 × 10 12 atm
2
PSO

(
0
.
50
)
2
From the negative value of ΔG°, this reaction is spontaneous at standard conditions. There
are more molecules of reactant gases than product gases, so ΔS° will be negative
(unfavorable). Therefore, this reaction must be exothermic (ΔH° < 0). When ΔH° and ΔS°
are both negative, the reaction will be spontaneous at relatively low temperatures where the
favorable ΔH° term dominates.
67.
 ΔH o  1  ΔSo
is in the form of a straight line equation
 
R T
R
(y = mx + b). A graph of ln K vs. 1/T will yield a straight line with slope = m = ΔH°/R and
a y-intercept = b = ΔS°/R.
The equation ln K =
From the plot:
slope =
Δy
0  40.
= −1.3 × 104 K

Δx
3.0  10 3 K 1  0
−1.3 × 104 K = −ΔH°/R, ΔH° = 1.3 × 104 K × 8.3145 J/Kmol = 1.1 × 105 J/mol
y-intercept = 40. = ΔS°/R, ΔS° = 40. × 8.3145 J/Kmol = 330 J/Kmol
As seen here, when ΔH° is positive, the slope of the ln K vs. 1/T plot is negative. When ΔH°
is negative as in an exothermic process, then the slope of the ln K vs. 1/T plot will be positive
(slope = −ΔH°/R).
68.
The ln K vs. 1/T plot gives a straight line with slope = H/R and y-intercept = S/R.
1.352 × 104 K = H/R, H = (8.3145 J/ Kmol) (1.352 × 104 K)
H = 1.124 × 105 J/mol = 112.4 kJ/mol
14.51 = S/R, S = (14.51)(8.3145 J/ Kmol) = 120.6 J / Kmol
Note that the signs for ΔH° and ΔS° make sense. When a bond forms, ΔH° < 0 and ΔS° < 0.
Additional Exercises
69.
From Appendix 4, S = 198 J/Kmol for CO(g) and S = 27 J/Kmol for Fe(s).
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
651
Let S ol = Sfor Fe(CO)5(l) and S og = S for Fe(CO)5(g).
S = 677 J/K = 1 mol(S ol ) – [1 mol (27 J/Kmol) + 5 mol(198 J/ Kmol]
S ol = 340. J/Kmol
S = 107 J/K = 1 mol (S og ) – 1 mol (340. J/Kmol)
S og = S for Fe(CO)5(g) = 447 J/Kmol
70.
When an ionic solid dissolves, one would expect the disorder of the system to increase, so
ΔSsys is positive. Because temperature increased as the solid dissolved, this is an exothermic
process and ΔSsurr is positive (ΔSsurr = ΔH/T). Because the solid did dissolve, the dissolving
process is spontaneous, so ΔSuniv is positive.
71.
ΔS will be negative because 2 mol of gaseous reactants form 1 mol of gaseous product. For
ΔG to be negative, ΔH must be negative (exothermic). For exothermic reactions, K decreases
as T increases. Therefore, the ratio of the partial pressure of PCl5 to the partial pressure of
PCl3 will decrease when T is raised.
72.
At boiling point, ΔG = 0 so ΔS =
ΔH vap
T
; For methane: ΔS =
8.20  10 3 J / mol
112 K
= 73.2
J/molK
For hexane: ΔS =
Vmet =
28 .9  10 3 J / mol
= 84.5 J/molK
342 K
nRT
nRT
1.00 mol (0.08206 ) (112 K)
=
= 9.19 L; Vhex =
= R(342 K) = 28.1 L
P
P
1.00 atm
Hexane has the larger molar volume at the boiling point so hexane should have the larger
entropy. As the volume of a gas increases, positional disorder increases.
73.
solid I → solid II; Equilibrium occurs when ΔG = 0.
ΔG = ΔH  TΔS, ΔH = TΔS, T = ΔH/ΔS =
74.
 743 .1 J / mol
= 43.7 K = 229.5°C
 17.0 J / K  mol
a. ΔG° = RT ln K = (8.3145 J/K mol)(298 K) ln 0.090 = 6.0 × 103 J/mol = 6.0 kJ/mol
b. H‒O‒H + Cl‒O‒Cl → 2 H‒O‒Cl
On each side of the reaction there are 2 H‒O bonds and 2 O‒Cl bonds. Both sides have
the same number and type of bonds. Thus, ΔH ≈ ΔH° ≈ 0.
c. ΔG° = ΔH° TΔS°, ΔS° =
ΔH o  ΔG o
0  6.0  10 3 J

= 20. J/K
T
298 K
652
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
d. For H2O(g), ΔH of = 242 kJ/mol and S° = 189 J/Kmol
ΔH° = 0 = 2 ΔH of , HOCl  [1 mol(242 kJ/mol) + 1 mol(80.3 J/K/mol)], ΔHof , HOCl
= 81 kJ/mol
20. J/K = 2 SoHOCl  [1 mol(189 J/Kmol) + 1 mol(266.1 J/K∙mol)], SoHOCl
= 218 J/Kmol
o
e. Assuming ΔH° and ΔS° are T independent: Δ G 500
= 0  (500. K)(20. J/K) = 1.0 × 104 J
  ΔG o 
  1.0  10 4 
2.41
  exp 

ΔG° = RT ln K, K = exp
= 0.090

 (8.3145 ) (500.)  = e
RT




f.
ΔG = ΔG° + RT ln
2
PHOCl
; From part a, ΔG° = 6.0 kJ/mol.
PH 2O  PCl 2O
We should express all P’s in atm. However, we perform the pressure conversion the same
number of times in the numerator and denominator, so the factors of 760 torr/atm will all
cancel. Thus, we can use the pressures in units of torr.
ΔG =
75.
6.0 kJ / mol  (8.3145 J / K  mol ) (298 K )  (0.10) 2 
 = 6.0  20. = 14 kJ/mol
ln 

1000 J / kJ
(
18
)
(
2
.
0
)


HgbO2
→ Hgb + O2
ΔG° =  (70 kJ)
Hgb + CO → HgbCO
ΔG° = 80 kJ
____________________________________________
HgbO2 + CO → HgbCO + O2
ΔG° = 10 kJ
  ΔG o 


 (10  10 3 J)
  exp 
 = 60
ΔG° = RT ln K, K = exp 



 RT 
 8.3145 J / K  mol (298 K ) 
76.
Ba(NO3)2(s) ⇌ Ba2+(aq) + 2 NO3(aq) K = Ksp; ΔG° = 561 + 2(109)  (797) = 18 kJ
ΔG° = RT ln Ksp, ln Ksp =
 ΔG o
 18,000 J

= 7.26, Ksp = e 7.26
RT
8.3145 J / K  mol (298 K)
= 7.0 × 10 4
77.
HF(aq)
⇌
H+(aq) + F(aq); ΔG = ΔG° + RT ln
[H  ][F ]
[HF]
ΔG° = RT ln K =  (8.3145 J/Kmol) (298 K) ln (7.2 × 10 4 ) = 1.8 × 104 J/mol
a. The concentrations are all at standard conditions so ΔG = ΔG = 1.8 × 104 J/mol
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
653
(Q = 1.0 and ln Q = 0). Because ΔG° is positive, the reaction shifts left to
reach equilibrium.
b. ΔG = 1.8 × 104 J/mol + (8.3145 J/Kmol) (298 K) ln
ΔG = 1.8 × 104 J/mol  1.8 × 104 J/mol = 0
( 2.7  10 2 ) 2
0.98
ΔG = 0, so the reaction is at equilibrium (no shift).
c. ΔG = 1.8 × 104 J/mol + 8.3145 (298 K) ln
(1.0  10 5 ) 2
= 1.1 × 104 J/mol; shifts right
1.0  10 5
7.2  10 4 (0.27 )
= 1.8 × 104  1.8 × 104 = 0;
0.27
at equilibrium
3
1.0  10 (0.67 )
e. ΔG = 1.8 × 104 + 8.3145 (298) ln
= 2 × 103 J/mol; shifts left
0.52
 [K  ] 
K+ (blood) ⇌ K+ (muscle) ΔG° = 0; ΔG = RT ln   m  ; ΔG = wmax
 [ K ]b 
8.3145 J
 0.15 
ΔG =
(310. K) ln 
 , ΔG = 8.8 × 103 J/mol = 8.8 kJ/mol
0
.
0050
K mol


d. ΔG = 1.8 × 104 + 8.3145 (298) ln
78.
At least 8.8 kJ of work must be applied.
Other ions will have to be transported in order to maintain electroneutrality. Either anions
must be transported into the cells, or cations (Na+) in the cell must be transported to the
blood. The latter is what happens: [Na+] in blood is greater than [Na+] in cells as a result of
this pumping.
79.


 (30,500 J)
 = 2.22 × 105
a. ΔG° = RT ln K, K = exp(ΔG°/RT) = exp 
 8.3145 J / K  mol  298 K 
b. C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
ΔG° = 6 mol(394 kJ/mol) + 6 mol(-237 kJ/mol)  1 mol(911 kJ/mol) = 2875 kJ
2875 kJ
1 mol ATP 94.3 mol ATP
; 94.3 molecules ATP/molecule glucose


mol glucose
30.5 kJ
mol glucose
This is an overstatement. The assumption that all of the free energy goes into this reaction
is false. Actually only 38 moles of ATP are produced by metabolism of one mole of
glucose.
c. From Exercise 17.78, ΔG = 8.8 kJ in order to transport 1.0 mol K+ from the blood to the
muscle cells.
654
CHAPTER 16
8.8 kJ ×
80.
SPONTANEITY, ENTROPY, AND FREE ENERGY
1 mol ATP
= 0.29 mol ATP
30.5 kJ
a. ΔG° = RT ln K
ln K =
b.
 ΔG o
 14,000 J

= 5.65, K = e 5.65 = 3.5 × 10 3
RT
(8.3145 J / K  mol ) (298 K)
Glutamic acid + NH3 → Glutamine + H2O
ΔG° = 14 kJ
ATP + H2O → ADP + H2PO4
ΔG° = 30.5 kJ
_______________________________________________________________________
Glutamic acid + ATP + NH3 → Glutamine + ADP + H2PO4 ΔG° = 14  30.5
= 17 kJ
o
 ΔG
 (17,000 J)

ln K =
= 6.86, K = e6.86 = 9.5 × 102
RT
8.3145 J / K  mol (298 K)
81.
ΔS is more favorable for reaction two than for reaction one, resulting in K2 > K1. In reaction
one, seven particles in solution are forming one particle. In reaction two, four particles form
one particle which results in a smaller decrease in disorder than for reaction one.
82.
A graph of ln K vs. 1/T will yield a straight line with slope equal to -ΔH°/R and y-intercept
equal to ΔS°/R.
Temp (°C)
0
25
35
40.
50.
T(K)
273
298
308
313
323
1000/T (K-1)
3.66
3.36
3.25
3.19
3.10
Kw
ln Kw
1.14 × 10 15
1.00 × 10 14
2.09 × 10 14
2.92 × 10 14
5.47 × 10 14
34.408
32.236
31.499
31.165
30.537
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
655
1
The straight line equation (from a calculator) is: ln K = 6.91 × 103    9.09
T
o
 ΔH
Slope = 6.91 × 103 K =
, ΔH° =  (6.91 × 103 K × 8.3145 J/Kmol)
R
= 5.75 × 104 J/mol
y-intercept = 9.09 =
83.
ΔS o
, ΔS° = 9.09 × 8.3145 J/Kmol = 75.6 J/Kmol
R
ΔG° = RT ln K; When K = 1.00, ΔG° = 0 since ln 1.00 = 0. ΔG° = 0 = ΔH°  TΔS°
ΔH° = 3(242 kJ)  [-826 kJ] = 100. kJ; ΔS° = [2(27 J/K) + 3(189 J/K)] 
[90. J/K + 3(131 J/K)] = 138 J/K
ΔH° = TΔS°, T =
ΔH o
100 . kJ

= 725 K
o
0.138 kJ / K
ΔS
Challenge Problems
84.
The liquid water will evaporate at first and eventually an equilibrium will be reached
(physical equilibrium).
 Because evaporation is an endothermic process, ΔH is positive.
 Because H2O(g) is more disordered (greater positional probability), ΔS is positive.
 The water will become cooler (the higher energy water molecules leave), thus ΔTwater will
be negative.
 The vessel is insulated (q = 0), so ΔSsurr = 0.
 Because the process occurs, it is spontaneous, so ΔSuniv is positive.
85.
3 O2(g)
ln K =
⇌
2 O3(g); ΔH° = 2(143 kJ) = 286 kJ; ΔG° = 2(163 kJ) = 326 kJ
 326  10 3 J
 ΔG o

= 131.573, K = e 131.573 = 7.22 × 10 58
RT
(8.3145 J / K  mol ) (298 K )
We need the value of K at 230. K. From Section 16.8 of the text:
 ΔG o
ΔSo

ln K =
RT
R
For two sets of K and T:
656
CHAPTER 16
ln K1 =
SPONTANEITY, ENTROPY, AND FREE ENERGY
 ΔH o  1  ΔS
 ΔH o  1  ΔSo
  
  
; ln K2 =
R  T1 
R
R  T2 
R
Subtracting the first expression from the second:
ln K2  ln K1 =
K
ΔH o  1
1 
ΔH o  1
1 
   or ln 2 
  
R  T1 T2 
K1
R  T1 T2 
Let K2 = 7.22 × 10 58 , T2 = 298 K; K1 = K230, T1 = 230. K; ΔH° = 286 × 103 J
ln
7.22  10 58
286  10 3  1
1 



 = 34.13
K 230
8.3145  230 . 298 
7.22  10 58
= e34.13 = 6.6 × 1014, K230 = 1.1 × 10 72
K 230
K230 = 1.1 × 10 72 =
PO23
PO3 3

PO23
(1.0  10
3
atm)
3
, PO3 = 3.3 × 10 41 atm
The volume occupied by one molecule of ozone is:
V =
nRT
P
=
(1 / 6.022  10 23 mol ) (0.08206 L atm / mol  K ) (230 . K )
= 9.5 × 1017 L
 41
(3.3  10 atm)
Equilibrium is probably not maintained under these conditions. When only two ozone
molecules are in a volume of 9.5 × 1017 L, the reaction is not at equilibrium. Under these
conditions, Q > K and the reaction shifts left. But with only 2 ozone molecules in this huge
volume, it is extremely unlikely that they will collide with each other. At these conditions, the
concentration of ozone is not large enough to maintain equilibrium.
86.
Arrangement I and V:
S = k ln W; W = 1; S = k ln 1 = 0
Arrangement II and IV: W = 4; S = k ln 4 = 1.38 × 10 23 J/K ln 4, S = 1.91 × 10 23 J/K
Arrangement III:
87.
W = 6; S = k ln 6 = 2.47 × 10 23 J/K
a. From the plot, the activation energy of the reverse reaction is Ea + (ΔG°) = Ea  ΔG°
(ΔG° is a negative number as drawn in the diagram).
  (E a  ΔG o )  k f
  Ea 
,
kf = A exp 
 and kr = A exp 
 k 
RT
 RT 
r


  Ea 
A exp 

 RT 
  (E a  ΔG o ) 

A exp 

RT


CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
If the A factors are equal:
657
E
  ΔG o 
(E  ΔG o ) 
kf
 = exp 

 exp  a  a
 RT 

kr
RT


 RT

  ΔG o 
k
 ; Because K and f are both equal to the
From ΔG° = RT ln K, K = exp 

kr
 RT 
same expression, K =
kf
.
kr
b. A catalyst will lower the activation energy for both the forward and reverse reaction (but
not change ΔG°). Therefore, a catalyst must increase the rate of both the forward and
reverse reactions.
88.
At equilibrium:
PH 2
 1.10  1013 molecules
  0.08206 L atm 


 (298 K )
23

mol K
6.022  10 molecules / mol  
nRT




V
1.00 L
PH 2 = 4.47 × 10 10 atm
The pressure of H2 decreased from 1.00 atm to 4.47 × 10 10 atm. Essentially all of the H2 and
Br2 has reacted. Therefore, PHBr = 2.00 atm because there is a 2:1 mole ratio between HBr and
H2 in the balanced equation. Because we began with equal moles of H2 and Br2, we will have
equal moles of H2 and Br2 at equilibrium. Therefore, PH 2  PBr2 = 4.47 × 10 10 atm.
2
PHBr
(2.00) 2

= 2.00 × 1019; Assumptions good.
10 2
PH 2  PBr2
(4.47  10 )
K=
ΔG° = RT ln K =  (8.3145 J/Kmol)(298 K) ln (2.00 × 1019) = 1.10 × 105 J/mol
ΔS° =
89.
ΔH o  ΔG o
 103,800 J / mol  (1.0  10 5 J / mol )

= 20 J/Kmol
T
298 K
a. ΔG° = G oB  G oA = 11,718  8996 = 2722 J
  ΔG o 


 2722 J
 = exp 
 = 0.333
K = exp 


 (8.3145 J / K  mol) (298 K) 
 RT 
b. When Q = 1.00 > K, the reaction shifts left. Let x = atm of B(g) which reacts to reach
equilibrium.
A(g)
Initial
Equil.
1.00 atm
1.00 + x
⇌
B(g)
1.00 atm
1.00  x
658
CHAPTER 16
K=
SPONTANEITY, ENTROPY, AND FREE ENERGY
PB 1.00  x
= 0.333, 1.00  x = 0.333 + 0.333 x, x = 0.50 atm

PA 1.00  x
PB = 1.00  0.50 = 0.50 atm; PA = 1.00 + 0.50 = 1.50 atm
c. ΔG = ΔG° + RT ln Q = ΔG° + RT ln (PB/PA)
ΔG = 2722 J + (8.3145)(298) ln (0.50/1.50) = 2722 J  2722 J = 0 (carrying extra
sig. figs.)
90.
 ΔH o
ΔSo

. For K at two temperatures T1 and T2, the
RT
R
K
ΔH o  1
1 


equation can be manipulated to give (see Exercise 16.77): ln 2 

K1
R  T1
T2 
 3.25  10 2 
 1
ΔH o
1 



ln 


8.84
348 K 

 8.3145 J / K  mol  298 K
From Exercise 16.67, ln K =
5.61 = (5.8 ×10-5 mol/J) (ΔH°), ΔH° = 9.7 × 104 J/mol
For K = 8.84 at T = 25°C:
ln 8.84 =
 (9.7  10 4 J / mol )
ΔSo
ΔSo

,
= 37
(8.3145 J / K  mol ) (298 K )
8.3145 J / K  mol 8.3145
ΔS° = 310 J/Kmol
We get the same value for ΔS° using K = 3.25 × 10 2 at T = 348 K data.
ΔG° = RT ln K; When K = 1.00, then ΔG° = 0 since ln 1.00 = 0. Assuming ΔH° and ΔS° do
not depend on temperature:
ΔG° = 0 = ΔH°  TΔS°, ΔH° = TΔS°, T =
91.
ΔH o
 9.7  10 4 J / mol

= 310 K
 310 J / K  mol
ΔSo
K = PCO 2 ; To insure Ag2CO3 from decomposing, PCO 2 should be greater than K.
ΔH o
ΔSo

. For two conditions of K and T, the equation is:
RT
R
K
ΔH o  1
1 


ln 2 

K1
R  T1
T2 
From Exercise 16.67, ln K =
Let T1 = 25°C = 298 K, K1 = 6.23 × 10 3 torr; T2 = 110.°C = 383 K, K2 = ?
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
659
 1
1 



 298 K 383 K 
ln
K2
79.14  10 3 J / mol

8.3145 J / K  mol
6.23  10 3 torr
ln
K2
K2
= e7.1 = 1.2 × 103, K2 = 7.5 torr
 7.1,
3
6.23  10
6.23  10 3
To prevent decomposition of Ag2CO3, the partial pressure of CO2 should be greater than 7.5
torr.
92.
L
From the problem, χ CL 6H6  χ CCl
= 0.500. We need the pure vapor pressures (Po) in order to
4
calculate the vapor pressure of the solution. Using the thermodynamic data:
C6H6(l)
⇌
C6H6(g) K = PC6H6  PCo6H6 at 25°C
ΔG orxn  ΔG of , C6H6 (g)  ΔG of , C6H6 (l) = 129.66 kJ/mol  124.50 kJ/mol = 5.16 kJ/mol
ΔG° = RT ln K, ln K =
 5.16  10 3 J / mol
 ΔG o
= exp
= 2.08
RT
(8.3145 J / K  mol ) (298 K )
K = PCo6H6 = e 2.08 = 0.125 atm
For CCl4: ΔG orxn  ΔG of , CCl 4 (g)  ΔG of , CCl 4 (l) = 60.59 kJ/mol  (65.21 kJ/mol)
= 4.62 kJ/mol
  ΔG o 


 4620 J / mol
o


K = PCCl
=
exp
4
 RT  = exp  8.3145 J / K  mol  298 K  = 0.155 atm




o
= 0.500 (0.155 atm)
PC6H6  χ CL6H6 PCo6H6 = 0.500 (0.125 atm) = 0.0625 atm; PCCl
4
= 0.0775 atm
χ CV6 H 6 =
PC 6 H 6
Ptot
=
0.0625
0.0625 atm
=
= 0.446
0.0625 atm  0.0775 atm 0.1400
V
= 1.000  0.446 = 0.554
χ CCl
4
93.
Use the thermodynamic data to calculate the boiling point of the solvent.
At boiling point: ΔG = 0 = ΔH  TΔS, T =
ΔH
33.90  10 3 J / mol

= 353.3 K
ΔS
95.95 J / K  mol
660
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
2.1
= 0.84 mol/kg
2.5
ΔT = Kbm, (355.4 K 353.3 K) = 2.5 K kg/mol (m), m =
0.879 g
1 kg
= 0.132 kg

mL
1000 g
mass solvent = 150. mL ×
0.84 mol solute 142 g
= 15.7 g = 16 g solute

kg solvent
mol
mass solute = 0.132 kg solvent ×
94.
ΔSsurr = ΔH/T = qP/T
q = heat loss by hot water = moles × molar heat capacity × ΔT
q = 1.00 × 103 g H2O ×
ΔSsurr =
1 mol H 2 O
75.4 J
× (298.2 – 363.2) = 2.72 × 105 J

18.02 g
K  mol
 ( 2.72 10 5 J )
= 912 J/K
298 .2 K
95.
HX
Initial
Equil.
⇌
0.10 M
0.10 – x
H+
+ X
~0
x
Ka =
[H  ][X  ]
[HX]
0
x
From problem, x = [H+] = 10 5.83 = 1.5 × 10 6 ; Ka =
(1.5  10 6 ) 2
= 2.3  10 11
0.10  1.5  10 6
ΔG = RT ln K = 8.3145 J/Kmol (298 K) ln(2.3 × 10 11 ) = 6.1 × 104 J/mol = 61 kJ/mol
96.
NaCl(s)
⇌
Na+(aq) + Cl(aq)
K = Ksp = [Na+][Cl]
ΔG = [(262 kJ) + (131 kJ)] – (384 kJ) = 9 kJ = 9000 J


 (9000 J)
ΔG = = RT ln Ksp, Ksp = exp 
 = 38 = 40
 8.3145 J / K  mol  298 K 
NaCl(s)
⇌
Initial s = solubility (mol/L)
Equil.
Na+(aq) + Cl(aq)
0
s
0
s
Ksp = 40 = s(s), s = (40)1/2 = 6.3 = 6 M = [Cl]
Ksp = 40
CHAPTER 16
SPONTANEITY, ENTROPY, AND FREE ENERGY
661
Integrative Problems
97.
Because the partial pressure of C(g) decreased, the net change that occurs for this reaction to
reach equilibrium is for products to convert to reactants.
A(g)
+
2 B(g)
⇌
C(g)
Initial
0.100 atm
0.100 atm
0.100 atm
Change
+x
+2x

x
Equil.
0.100 + x
0.100 + 2x
0.100  x
From the problem, PC = 0.040 atm = 0.100 – x, x = 0.060 atm
The equilibrium partial pressures are: PA = 0.100 + x = 0.100 + 0.060 = 0.160 atm,
PB = 0.100 + 2((0.60) = 0.220 atm, and PC = 0.040 atm
K=
0.040
= 5.2
0.160 (0.220 ) 2
G = RT ln K = 8.3145 J/Kmol (298 K) ln 5.2 = 4.1 × 103 J/mol = 4.1 kJ/mol
98.
G = H  TS = 28.0 × 103 J – 298 K(175 J/K) = 24,200 J
G = RT ln K, ln K =
 ΔG o
 24,000 J

= 9.767
RT
8.3145 J / K  mol  298 K
9.767
K= e
= 5.73 × 10 5
B
Initial
Change
Equil.
+
0.125 M
x
0.125  x
Kb = 5.73 × 10 5 =
H2O
⇌
BH+
0
+x
x
+
OH
K = Kb = 5.73 × 10 5
~0
+x
x
[BH  ][OH  ]
x2
x2
=

, x = [OH] = 2.68 × 10 3 M
0.125  x
0.125
[B]
pH = log(2.68 × 10 3 ) = 2.572; pOH = 14.000 – 2.572 = 11.428; Assumptions good
Marathon Problem
99.
a. ΔS° will be negative because there is a decrease in the number of moles of gas.
b. Because ΔS° is negative, ΔH° must be negative for the reaction to be spontaneous at
some temperatures. Therefore, ΔSsurr is positive.
662
CHAPTER 16
c. Ni(s) + 4 CO(g)
⇌
SPONTANEITY, ENTROPY, AND FREE ENERGY
Ni(CO)4(g)
ΔH° = 607  [4(110.5)] = 165 kJ; ΔS° = 417  [4(198) + (30.)] = 405 J/K
d. ΔG° = 0 = ΔH°  TΔS°, T =
ΔH o
 165  10 3 J

= 407 K or 134°C
 405 J / K
ΔS o
e. T = 50.°C + 273 = 323 K
o
ΔG 323
= 165 kJ  (323 K)(0.405 kJ/K) = 34 kJ
ln K =
f.
 ΔG o
 (34,000 J )

= 12.66, K = e12.66 = 3.1 × 105
RT
8.3145 J / K  mol (323 K )
T = 227°C + 273 = 500. K
o
ΔG 500
= 165 kJ  (500. K)( 0.405 kJ/K) = 38 kJ
ln K =
 38,000 J
= 9.14, K = e 9.14 = 1.1 × 10 4
(8.3145 J / K  mol )(500. K)
g. The temperature change causes the value of the equilibrium constant to change from a
large value favoring formation of Ni(CO)4 to a small value favoring the decomposition of
Ni(CO)4 into pure Ni and CO. This is exactly what is wanted in order to purify a nickel
sample.
h. Ni(CO)4(l) ⇌ Ni(CO)4(g)
K = PNi( CO ) 4
At 42°C (the boiling point): ΔG° = 0 = ΔH°  TΔS°
ΔS° =
ΔH o
29.0  10 3 J

= 92.1 J/K
T
315 K
o
At 152°C: ΔG 152
= ΔH°  TΔS° = 29.0 × 103 J  425 K (92.1 J/K) = 10,100 J
ΔG° = RT ln K, ln K =
 (10,100 J)
= 2.858, Kp = e2.858 = 17.4
8.3145 J / K  mol(425 K)
A maximum pressure of 17.4 atm can be attained before Ni(CO)4(g) will liquify.