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Transcript
Chapter 13
13.1
(a) A sand castle represents an ordered structure constructed by a person. Waves destroy
that structure, returning the sand grains to a disordered arrangement.
(b) Two separate liquids represent an ordered arrangement (all molecules of one kind in
one container). Upon mixing, the molecules in the liquids are distributed randomly
throughout the container, an increase in disorder.
(c) Sticks in a bundle are ordered (all aligned in the same direction). When dropped, the
sticks lose their alignment and become more disordered.
(d) Water in a puddle is relatively ordered, as it is confined to a small volume. When the
water evaporates, the molecules spread over a much larger volume and become more
disordered.
13.2
(a) "Straightening up" involves making the materials on the desk more ordered (regular
piles of related documents, for example). The secretary expends energy in the process,
which leads to increased disorder in the surroundings of the desk.
(b) Wood represents a relatively ordered structure of aligned long-chain carbohydrates,
while gaseous CO2 and H2O vapors are disordered.
(c) I2 molecules dissolved in a solvent are relatively disordered. As crystals form, the
solid I2 is more ordered than the solution, but energy that is given off during
crystallization increases the disorder in the surroundings.
(d) A torn-down engine is highly disordered. The reassembly leads to an organized
structure, but the mechanic must expend energy that increases the disorder of the
surroundings.
13.3
The molecules in a drop of ink are relatively ordered, because they occupy one particular
part of the total volume. As they move about randomly, they spread throughout the
volume of the container and become more disordered.
13.4
An egg (or any other living thing) is a highly ordered system. When an egg breaks, it
becomes disordered. To reassemble a disordered egg into its original form would require
a very large expenditure of energy (which, in turn, would lead to disorder in the
surroundings).
13.5
(a) The air molecules in a tire are relatively ordered, because they are confined to a
specific, small volume. A puncture allows gas molecules to escape from the tire and fill a
much larger volume, becoming less ordered in the process.
(b) Like air in a tire, the fragrant molecules in a perfume bottle are relatively ordered
because they are confined to a specific, small volume. When the bottle is open,
molecules escape from the confined volume into a much larger space of the room,
becoming more disordered in the process.
13.6
(a) The glass has a specific shape which indicates that it is relatively ordered. When it
breaks, the pieces lose the specific shape of the glass, so the fragments are more
disordered.
(b) A pile of raked leaves is relatively ordered because the leaves are confined to a
particular location. Scattering the leaves increases the amount of disorder by giving each
leaf a random location.
414
Chapter 13
13.7
Calculate W by determining how many ways each symbol can be placed. The first X can
be placed in any of the nine compartments, the first O in any of the eight empty
compartments, so W = (9)(8) = 72.
13.8
Calculate W by determining how many ways each marble can be placed. The first can be
placed in any of the 16 compartments. There are 15 compartments into which the second
can be placed, giving W = (16)(15) = 240 for marbles in different compartments. There
is only one way to put the second marble in the same compartment as the first, giving
W = (16)(1) = 16 for both marbles in the same compartment.
13.9
The entropy change accompanying a constant-temperature process is ∆S =
q
.
T
(a) Melting involves absorption of heat:
 1 mol 6.01 kJ 103 J 


13
.
8
g
qice = n∆Hfus =
18.02 g 
 1 mol 

 = 4.60 x 103 J;
 1 kJ 


4.60 x 103 J
= 16.8 J/K.
273.15 K
(b) The heat absorbed by the ice cube must be supplied by the pool water: qpool = –qice;
3
∆Spool = − 4.60 x 10 J = –15.3 J/K.
27.5 + 273.15 K
(c) ∆Stotal = ∆Sice + ∆Spool = (16.8 J/K) + (–15.3 J/K) = 1.5 J/K.
∆Sice =
13.10 The entropy change accompanying a constant-temperature process is ∆S =
q
.
T
(a) Sublimation involves absorption of heat:
 1 mol  25.2 kJ 103 J 


2
7.5
g
qdry ice = n∆Hsubl =
 44.01 g 
 1 mol 

= 1.575 x 104 J;
 1 kJ 


4
∆Sdry ice = 1.575 x 10 J = 80.8 J/K.
195 K
(b) The heat absorbed by the dry ice must be supplied by air in the room: qair = –qdry ice;
4
∆Sair = − 1.575 x 10 J = –52.6 J/K.
26.5 + 273.15 K
(c) ∆Stotal = ∆Sice + ∆Spool = (80.8 J/K) + (–52.6 J/K) = 28.2 J/K.
13.11 The sign of the entropy change will be the same as q.
(a) The system (water) absorbs heat to boil, so qsys is positive and ∆Ssys is positive. The
surroundings (stove) release heat to the system, so qsurr is negative and ∆Ssurr is negative.
415
Chapter 13
(b) The system (ice cubes) absorbs heat to melt, so qsys is positive and ∆Ssys is positive.
The surroundings (table top) release heat to the system, so qsurr is negative and ∆Ssurr is
negative.
(c) The system (coffee) absorbs heat, so qsys is positive and ∆Ssys is positive. The
surroundings (microwave oven) release heat to the system, so qsurr is negative and ∆Ssurr
is negative.
13.12 The sign of the entropy change will be the same as q.
(a) The system (water) releases heat to freeze, so qsys is negative and ∆Ssys is negative.
The surroundings (atmosphere) absorb the heat released by the system, so qsurr is positive
and ∆Ssurr is positive.
(b)The system (coffee) releases heat to cool, so qsys is negative and ∆Ssys is negative.
The surroundings (atmosphere) absorb the heat released by the system, so qsurr is positive
and ∆Ssurr is positive.
(c) The system (popsicle) absorbs heat to melt, so qsys is positive and ∆Ssys is positive.
The surroundings (table top) release heat to the system, so qsurr is negative and ∆Ssurr is
negative.
13.13 The entropy change accompanying a constant-temperature process is ∆S =
q
.
T
Condensation involves release of heat:
 1 mol  40.79 kJ 103 J 


1
5.5
g
qsteam = –n∆Hcondensation = –
18.02 g 
 1 mol 

 = –3.509 x 104 J;
 1 kJ 


4
∆Ssteam = − 3.509 x 10 J = –94.0 J/K.
373.15 K
Knowing that ∆Stotal must be positive, we can say without doing further calculations that
∆Ssurr > 94.0 J/K.
13.14 The entropy change accompanying a constant-temperature process is ∆S =
q
.
T
(a) Freezing involves release of heat:
 1 mol 6.01 kJ 103 J 


7
5.4
g
qwater = –n∆Hfus = –
18.02 g 
 1 mol 

= –2.515 x 104 J;
 1 kJ 


4
∆Swater = − 2.515 x 10 J = –92.1 J/K.
273.15 K
Knowing that ∆Stotal must be positive, we can say without doing further calculations that
∆Ssurr > 92.1 J/K.
13.15 The molar entropy change accompanying fusion is ∆Smolar =
416
∆H fus
.
Tfus
Chapter 13
3
1.3 kJ/mol 10 J 




(a) ∆Smolar =
 = 16 J/mol K;

 83 K  1 kJ 
3
0.84 kJ/mol 10 J 



(b) ∆Smolar = 
 = 9.3 J/mol K;
 90 K
 1 kJ 
3
7.61 kJ/mol 10 J 




(c) ∆Smolar =
 = 48.8 J/mol K;

 156 K  1 kJ 
3
 23.4 kJ/mol 10 J 



(d) ∆Smolar = 
= 1.00 x 102 J/mol K.
 234 K  1 kJ 
13.16 The molar entropy change accompanying vaporization is ∆Smolar =
∆H vap
Tvap
.
3
9.8 kJ/mol 10 J 


(a) ∆Smolar = 

 = 1.1 x 102 J/mol K;
 90 K  1 kJ 
3
15.5 kJ/mol 10 J 




(b) ∆Smolar =

 = 84.2 J/mol K;
 184 K  1 kJ 
3
31.0 kJ/mol 10 J 


(c) ∆Smolar = 

 = 87.8 J/mol K;
 353 K  1 kJ 
3
59.0 kJ/mol 10 J 




(d) ∆Smolar =

 = 93.7 J/mol K.
 630 K  1 kJ 
13.17 (a) Both are ionic solutions, but MgCl2 produces three moles of ions per mole of
substance, while NaCl produces two moles of ions per mole of substance, so MgCl2 has
the larger molar entropy;
(b) Both are ionic solids, but HgS has a higher molar mass than HgO, so HgS has the
larger molar entropy;
(c) Both are diatomic molecules, but Br2(l) is liquid while I2(s) is solid, so Br2(l) has the
larger molar entropy.
13.18 (a) Both are diatomic molecules, but Br2 is a liquid and Cl2 is a gas under standard
conditions, so Cl2 has the larger molar entropy;
(b) Both are metallic elements, but Pt has a higher molar mass than Ni, so Pt has the
larger molar entropy;
(c) Both are organic liquids, but C8H18 has more atoms per molecule than C5H12, so
C8H18 has the larger molar entropy;
(d) Both are gases, but SiF4 has a larger molar mass than CH4, so SiF4 has the larger
molar entropy.
13.19 All these substances are small gaseous molecules. Methane is a highly symmetric
molecule, so it has fewer distinguishable orientations in space. Thus it has lower entropy
417
Chapter 13
than the other two despite having five atoms per molecule. Ozone has more entropy than
O2 because it has three atoms per molecule while O2 has only two.
13.20 All these substances are relatively simple in nature and are liquids under standard
conditions. Water is the most ordered because of strong hydrogen-bonding interactions
which give it a regular structure, so it has the lowest entropy. Mercury has a lower
entropy than bromine because it is monatomic and bromine is diatomic.
13.21 Absolute entropies are tabulated in Appendix D of your textbook. To obtain the entropy
per mole of atoms, divide each value by the number of atoms in one particle:
 1 mol He 
He: S° = 126.153 J/mol K 
= 126.153 J/mol K;
1 mol atom 
 1 mol H 2 
= 65.340 J/mol K;
H2: S° = 130.680 J/mol K 
 2 mol atom 
 1 mol CH 4 
= 37.26 J/mol K;
CH4: S° = 186.3 J/mol K 
5 mol atom 
1 mol C 3H 6 
C3H6: S° = 226.9 J/mol K 
= 25.21 J/mol K.
 9 mol atom 
Entropy per mole of atoms decreases as the number of atoms in a species increases,
because tying together atoms into a molecule increases the amount of order among those
atoms.
13.22 Absolute entropies are tabulated in Appendix D of your textbook. To obtain the entropy
per mole of atoms, divide each value by the number of atoms in one particle:
 1 mol Ar 
Ar: S° = 154.843 J/mol K 
 =154.843 J/mol K;
1 mol atom 
 1 mol O 2 
= 102.576 J/mol K;
O2: S° = 205.152 J/mol K 
 2 mol atom 
 1 mol O 3 
 = 79.63 J/mol K;
O3: S° = 238.9 J/mol K 
3 mol atom 
1 mol C 2 H 6 
 = 28.65 J/mol K.
C2H6: S° = 229.2 J/mol K 
 8 mol atom 
Entropy per mole of atoms decreases as the number of atoms in a species increases,
because tying together atoms into a molecule increases the amount of order among those
atoms.
13.23 Standard entropy changes can be calculated using tabulated values of absolute entropies
(Appendix D of your textbook) and Equation 13-5:
∆Soreaction = Σcoeffp So(products) – Σcoeffr So(reactants)
(a) ∆S°reaction = 2 mol(192.8 J/mol K)
– [1mol(191.61 J/mol K) + 3 mol(130.680 J/mol K)] = –198.1 J/K;
418
Chapter 13
(b) ∆S°reaction = 2 mol(238.9 J/mol K) – 3 mol(205.152 J/mol K) = –137.7 J/K;
(c) ∆S°reaction = [1 mol(64.8 J/mol K) + 2 mol(37.99 J/mol K)]
– [1 mol(68.7 J/mol K) + 2 mol(29.9 J/mol K)] = 12.3 J/K;
(d) ∆S°reaction = [2 mol(213.8 J/mol K) + 2 mol(69.95 J/mol K)]
– [1 mol(219.3 J/mol K) + 3 mol (205.152 J/mol K)] = –267.3 J/K.
13.24 Standard entropy changes can be calculated using tabulated values of absolute entropies
(Appendix D of your textbook) and Equation 13-5:
∆Soreaction = Σcoeffp So(products) – Σcoeffr So(reactants)
(a) ∆S°reaction = 2 mol(69.95 J/mol K)
– [2 mol(130.680 J/mol K) + 1 mol(205.152 J/mol K)] = –326.61 J/K;
(b) ∆S°reaction = 1 mol(186.3 J/mol K)
– [1 mol(5.7 J/mol K) + 2 mol(130.680 J/mol K)] = –80.8 J/K;
(c) ∆S°reaction = [2 mol(213.8 J/mol K) + 3 mol(69.95 J/mol K)]
– [1 mol(160.7 J/mol K) + 3 mol (205.152 J/mol K)] = –138.7 J/K;
(d) ∆S°reaction = [1 mol(50.9 J/mol K) + 2 mol(27.3 J/mol K)]
– [1 mol(87.4 J/mol K) + 2 mol(28.3 J/mol K)] = –38.5 J/K.
13.25 Reactions have negative entropy changes when products are more ordered than reactants.
(a) There is a relatively large negative value because of the reduction in number of moles
of gaseous substances: 4 moles of gaseous reactants are converted to 2 moles of gaseous
products;
(b) There is a relatively large negative value because of the reduction in number of moles
of gaseous substances: 3 moles of gaseous reactants are converted to 2 moles of gaseous
products;
(c) This reaction has a near-zero entropy change because all reagents are relatively
ordered solid substances;
(d) There is a relatively large negative value because of the reduction in number of moles
of gaseous substances: 4 moles of gaseous reactants are converted to 2 moles of gaseous
products.
13.26 Reactions have negative entropy changes when products are more ordered than reactants.
(a) There is a relatively large negative value because of the reduction in number of moles
of gaseous substances: 3 moles of gaseous reactants are converted to 0 moles of gaseous
products;
(b) There is a negative value because of the reduction in number of moles of gaseous
substances: 2 moles of gaseous reactants are converted to 1 mole of gaseous products;
(c) There is a negative value because of the reduction in number of moles of gaseous
substances: 3 moles of gaseous reactants are converted to 2 moles of gaseous products;
(d) This reaction has a small entropy change because all reagents are solid substances.
13.27 Entropies of substances at pressures different from 1.00 atm or concentrations different
from 1 M are calculated using Equation 13-4: S(p ≠ 1) = So – R ln p or
S(c ≠ 1) = So – R ln c. Take into account amounts different from 1 mol by multiplying by
the number of moles.
419
Chapter 13
(a) S = (2.50 mol)[154.843 J/mol K – (8.314 J/mol K)ln(0.25 atm)] = 416 J/K;
(b) S = (0.75 mol)[238.9 J/mol K – (8.314 J/mol K)ln(2.75 atm)] = 1.7 x 102 J/K;
(c) Calculate the entropy of each component separately, using ni = Xi ntot and pi = Xi ptot:
SN2 = (0.78)(0.45 mol){191.61 J/mol K – (8.314 J/mol K)ln[(0.78)(1.00 atm)]} = 68 J/K;
SO2 =(0.22)(0.45 mol){205.152 J/mol K – (8.314 J/mol K)ln[(0.22)(1.00 atm)]} = 22 J/K;
S = SN2 + SO2 = 9.0 x 101 J/K.
13.28 Entropies of substances at pressures different from 1.00 atm or concentrations different
from 1 M are calculated using Equation 13-4: S(p ≠ 1) = So – R ln p or
S(c ≠ 1) = So – R ln c. Take into account amounts different from 1 mol by multiplying by
the number of moles.
(a) S = (1.00 mol)[130.680 J/mol K – (8.314 J/mol K) ln(5.0 atm)] = 117 J/K;
(b) S = (0.25 mol)[229.2 J/mol K – (8.314 J/mol K) ln(0.10 atm)] = 62 J/K;
(c) Calculate the entropy of each component separately, then add:
(1.00 mol)(375 atm)
nH2 =
= 0.750 mol;
125 atm + 375 atm
nN2 = 1.00 – 0.750 = 0.250 mol;
SH2 = (0.750 mol){130.680 J/mol K – (8.314 J/mol K)ln(375 atm)} = 61.1 J/K;
SN2 = (0.250 mol){191.61 J/mol K – (8.314 J/mol K)ln(125 atm)} = 37.9 J/K;
S = SN2 + SO2 = 99.0 J/K.
13.29 False statements can be made true in various ways; we choose the simplest change that
corrects each statement:
(a) ∆G system < 0 for any spontaneous process at constant T and P.
(b) The free energy of a system decreases in any spontaneous process at constant T and P.
(c) ∆G = ∆H – ∆(TS).
13.30 False statements can be made true in various ways; we choose the simplest change that
corrects each statement:
(a) In any spontaneous process at constant T and P, the free energy of the system
decreases.
(b) ∆G system < 0 for any spontaneous process at constant T and P.
(c) ∆G = ∆H – ∆(TS)
13.31 Standard free energy changes are calculated from Equation 13-8 using standard free
energies of formation, which can be found in Appendix D of your textbook:
o
∆Grxn
= Σcoeffp ∆G fo (products) – Σ coeffr ∆G fo (reactants)
(a) ∆Goreaction= 2 mol(–16.4 kJ/mol) – [1 mol(0 kJ/mol) + 3 mol(0 kJ/mol)] = –32.8 kJ;
(b) ∆Goreaction = 2 mol(163.2 kJ/mol) – 3 mol(0 kJ/mol) = 326.4 kJ.
(c) ∆Goreaction = [1 mol(0) + 2 mol(–211.7 kJ/mol)]
420
Chapter 13
(d)
∆Goreaction=
– [1 mol(–217.3 kJ/mol) + 2 mol(0 kJ/mol)] = –206.1 kJ;
[2 mol(–394.4 kJ/mol) + 2 mol(–237.1 kJ/mol)]
– [1 mol(68.4 kJ/mol) + 3 mol(0 kJ/mol)] = –1331.4 kJ.
13.32 Standard free energy changes are calculated from Equation 13-8 using standard free
energies of formation, which can be found in Appendix D of your textbook:
∆Goreaction = Σcoeffp ∆G o (products) – Σ coeffr ∆G o (reactants)
f
f
o
(a) ∆G reaction = 2 mol(–237.1 kJ/mol) – [2 mol(0 kJ/mol) + 1 mol(0 kJ/mol)] = –474.2 kJ;
(b) ∆Goreaction = 1 mol(–50.5 kJ/mol) – [1 mol(0 kJ/mol) + 2 mol(0 kJ/mol)] = –50.5 kJ;
(c) ∆Goreaction = [2 mol(–394.4 kJ/mol) + 3 mol(–237.1 kJ/mol)]
– [1 mol(–174.8 kJ/mol) + 3 mol(0 kJ/mol)] = –1325.3 kJ;
(d) ∆Goreaction = [1 mol(–1582.3 kJ/mol) + 2 mol(0 kJ/mol)]
– [1 mol(–742.2 kJ/mol) + 2 mol(0 kJ/mol)] = –840.1 kJ.
13.33 Use Equation 13-10 to estimate the standard free energy change at a temperature different
from 298 K: ∆Goreaction, T ≅ ∆Horeaction, 298 - T∆Soreaction, 298. Standard entropy changes are
calculated in Problem 13.23, but standard enthalpy changes need to be calculated from
standard enthalpies of formation:
(a) ∆Horeaction = 2 mol(–45.9 kJ/mol) – [1 mol(0 kJ/mol) +3 mol(0 kJ/mol)] = – 91.8 kJ;
10 − 3 kJ 
o

∆Greaction,
≅
–91.8
kJ
–
(
)
425
K
−
198.1
J/K
T
 1J 
 = –7.6 kJ;


(b) ∆Horeaction = 2 mol(142.7 kJ/mol) – 3 mol(0 kJ/mol) = 285.4 kJ;
10 − 3 kJ 
o

∆Greaction,
≅
285.4
kJ
–
(
)
425
K
−
1
37
.
7
J/K
T
 1J 
= 343.9 kJ;


(c) ∆Horeaction = [1 mol(0 kJ/mol) + 2 mol(-239.7 kJ/mol)]
– [1 mol(-277.4 kJ/mol) + 2 mol(0 kJ/mol)] = –202.0 kJ;
10 − 3 kJ 
o

(
)
∆Greaction,
≅
–202.0
kJ
–
425
K
12
.
3
J/K
T
 1J 
 = –207.2 kJ;


(d) ∆Horeaction = [2 mol(-393.5 kJ/mol) + 2 mol(-285.83 kJ/mol)]
– [1 mol(52.4 kJ/mol) + 3 mol(0 kJ/mol)] = –1411.1 kJ;
10 − 3 kJ 
o
(
)
∆Greaction,
≅
–1411.1
kJ
–
425 K − 267.3 J/K 
T
 1J 
 = –1297 kJ.


13.34 Use Equation 13-12 to estimate the standard free energy change at a temperature different
o
o
from 298 K: ∆Greaction = ∆Greaction
+ RT ln Q. ∆Greaction
values are calculated in Problem
13.32.



1
(a) RT ln  1  = (8.314 x 10-3 kJ/mol K)(298 K) ln 
= 10.3 kJ;

(0.25 atm) 2 (0.25 atm) 

2
 pH p 


 2 O2 
∆Greaction = –474.2 kJ + 10.3 kJ = –463.9 kJ;
421
Chapter 13
 0.25 atm  = 3.43 kJ;
p 
-3
(b) RT ln 
 CH4  = (8.314 x 10 kJ/mol K)(298 K)ln 

(0.25 atm) 2 
 pH2 


 2 
∆Greaction = –50.5 kJ + 3.43 kJ = –47.1 kJ;
(0.25 atm) 2 
p2 
-3
(c) RT ln 
 CO 2  = (8.314 x 10 kJ/mol K)(298 K)ln 
 = 3.43 kJ;
 (0.25 atm)3 
3
 pO 


 2 
∆Greaction = –1325.3 kJ + 3.43 kJ = –1321.9 kJ;
(d) RT ln (1) = (8.314 x 10-3 kJ/mol K)(298 K)(0) = 0 kJ;
∆Greaction = –840.1 kJ.
13.35 Use Equations 12-10, 13-5, and 13-8 to calculate standard thermodynamic values:
o
o
o
= Σ coeffp ∆H f (products) – Σ coeffr ∆H f (reactants)
∆H reaction
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
o
= Σcoeffp ∆G fo (products) – Σ coeffr ∆G fo (reactants)
∆Greaction
o
= [1 mol(–365.6 kJ/mol) + 1 mol(–285.83 kJ/mol)]
∆H reaction
– [2 mol(–45.9 kJ/mol) + 2 mol(0 kJ/mol)] = –559.6 kJ;
∆Soreaction = [1 mol(151.1 J/mol K) + 1 mol(69.95 J/mol K)]
– [2 mol(192.8 J/mol K) + 2 mol(205.152 J/mol K)] = –574.9 J/K;
o
∆Greaction = [1 mol(–183.9 kJ/mol) + 1 mol(–237.1 kJ/mol)]
– [2 mol(–16.4 kJ/mol) + 2 mol(0 kJ/mol)] = –388.2 kJ.
13.36 Use Equations 12-10, 13-5, and 13-8 to calculate standard thermodynamic values:
o
= Σ coeffp ∆H fo (products) – Σ coeffr ∆H fo (reactants)
∆H reaction
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
o
= Σcoeffp ∆G fo (products) – Σ coeffr ∆G fo (reactants)
∆Greaction
o
= [1 mol(–333.1 kJ/mol) + 1 mol(–285.83 kJ/mol)]
∆H reaction
– [1 mol(–393.5 kJ/mol) + 2 mol(–45.9 kJ/mol)] = –133.6 kJ;
∆Soreaction = [1 mol(105 J/mol K) + 1 mol(69.95 J/mol K)]
– [1 mol(213.8 J/mol K) + 2 mol(192.8 J/mol K)] = –424 J/K;
o
∆Greaction = [1 mol(–198 kJ/mol) + 1 mol(–237.1 kJ/mol)]
– [1 mol(–394.4 kJ/mol) + 2 mol(–16.4 kJ/mol)] = –7.9 kJ.
422
Chapter 13
13.37 A phase diagram is a plot of P vs. T, and phase boundary lines meet at the triple point and
cross the horizontal line at P = 1 atm at the normal freezing and boiling points:
13.38 A phase diagram is a plot of P vs. T, and phase boundary lines meet at the triple point and
cross the horizontal line at P = 1 atm at the normal freezing and boiling points:
13.39 Phase diagrams provide "road maps" allowing us to determine what phase changes occur
as temperature and pressure vary in particular ways.
(a) At T = 400 K, P = 1.00 atm, Br2 is a gas. As it cools under a pressure of 1.00 atm, it
liquefies at 331.9 K and solidifies at 265.9 K.
(b) At T = 265.8 K, P = 1.00 x 10-3 atm, Br2 is a gas. Compressing it at this temperature
causes it to liquefy at about P = 6 x 10-2 atm and solidify at about 0.5 atm.
423
Chapter 13
(c) P = 2.00 x 10-2 atm is below the triple point of Br2. Heating a solid at this pressure
from 250 to 400 K causes sublimation to the vapor at about 265 K.
13.40 Phase diagrams provide "road maps" allowing us to determine what phase changes occur
as temperature and pressure vary in particular ways.
(a) At T = 125 K, P = 1.00 atm, O2 is a gas. As it cools under a pressure of 1.00 atm, it
liquefies at 90.2 K and solidifies at 55 K.
(b) At T = 54.5 K, P = 1.00 x 10-2 atm, O2 is a gas. Compressing it at this temperature
causes it to liquefy at about P = 25 x 10-2 atm and solidify at about 0.8 atm.
(c) P = 1.00 x 10-3 atm is below the triple point of O2. Heating a solid at this pressure
from 25 to 100 K causes sublimation to the vapor at about 20 K.
13.41 Coupled reactions share a common intermediate that transfers spontaneity (or energy)
from one reaction to the other. In the case of the seesaw the first "reaction" would be
child 1 starting on the ground and rising into the air. The second "reaction" would be
child 2 starting in the air and falling to the ground. In this case child 2 would be the
spontaneous reaction, child 1 would be non-spontaneous (things tend to fall down, not
float up), so as child 2 is falling he will cause the first child to rise. When they are at
equal heights (the common intermediate) they will switch spontaneity (or energy).
424
Chapter 13
13.42 A pair of weights and a pulley are a particularly simple example of a coupled system.
The lifting of the 1-kg weight is "non-spontaneous" in that it cannot occur without
outside intervention. The falling of the 2-kg weight is spontaneous. By linking the two
together by a rope over a pulley, the spontaneous process is coupled to the nonspontaneous, and the 1-kg weight is driven uphill by the falling 2-kg weight.
13.43 The reactions that are coupled in this example are the following:
ATP + H2O → ADP + H3PO4
∆G° = –30.6 kJ
C6H12O6 (fructose) + C6H12O6 (glucose) → C12H22O11 + H2O ∆G° = 23.0 kJ;
Coupled reaction:
ATP + C6H12O6 (fructose) + C6H12O6 (glucose) → ADP + C12H22O11
∆G° = 23.0 kJ – 30.6 kJ = –7.6 kJ.
13.44 The reactions that are coupled in this example are the following:
glutamic acid + NH3 → glutamine + H2O
∆G° = 14 kJ
CH3COOPO3H2 + H2O → CH3COOH + H3PO4 ∆G° = – 46.9 kJ;
Coupled net reaction:
glutamic acid + CH3COOPO3H + NH3 → glutamine + CH3COOH + H3PO4
∆G° = 14 kJ – 46.9 kJ = –33 kJ.
13.45 (a) Energy is always conserved, so ∆Euniverse = 0;
(b) The teaspoon warms, so ∆Eteaspoon > 0;
(c) This is a spontaneous process, so ∆Suniverse > 0;
(d) The water cools, so ∆Swater < 0; and
(e) The teaspoon warms, so qteaspoon > 0.
13.46 (a) Energy is always conserved, so ∆Euniverse = 0;
(b) The pie cools, so ∆Epie < 0;
(c) This is a spontaneous process, so ∆Suniverse > 0;
(d) The pie cools, so ∆Spie < 0; and
(e) The pie cools, so qpie < 0.
13.47 Entropy depends on amount, phase, temperature, and concentration. The order is:
(0.5 mol, l, 298 K) < (1 mol, l, 298 K) < (1 mol, l, 373 K)
< (1 mol, g, 1 atm, 373 K) < (1 mol, g, 0.1 atm, 373 K).
13.48 Entropy depends on amount, phase, temperature, and concentration. Total amount is the
same for each of these samples. The order is:
liquid < (g, 331.9 K, 1.00 atm) < (g, 331.9 K, 0.10 atm) < (atomic gas, 331.9 K, 0.10
atm).
13.49 Spontaneity is determined by free energy, heat flow by enthalpy, and change in order by
entropy: Use Equations 12-10, 13-5, and 13-8 to calculate standard thermodynamic values:
425
Chapter 13
o
= Σ coeffp ∆H fo (products) – Σ coeffr ∆H fo (reactants)
∆H reaction
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
o
= Σcoeffp ∆G o (products) – Σ coeffr ∆G o (reactants)
∆Greaction
f
f
o
(a) ∆Greaction
= [2 mol(0 kJ/mol) + 3 mol(–237.1 kJ/mol)]
– [1 mol(–1582.3 kJ/mol) + 3 mol(0 kJ/mol)] = 871.0 kJ.
The reaction is not spontaneous.
o
(b) ∆H reaction
= [2 mol(0 kJ/mol) + 3 mol(–285.83 kJ/mol)]
– [1 mol(–1675.7 kJ/mol) + 3 mol(0 kJ/mol)] = 818.2 kJ.
The reaction absorbs heat.
(c) ∆Soreaction = [2 mol(28.3 J/mol K) + 3 mol(69.95 J/mol K)]
– [1 mol(50.9 J/mol K) + 3 mol(130.680 J/mol K)] = –176.5 J/K.
Entropy decreases, so the products are more ordered than the reactants.
13.50 Spontaneity is determined by free energy, heat flow by enthalpy, and change in order by
entropy: Use Equations 12-10, 13-5, and 13-8 to calculate standard thermodynamic values:
o
= Σ coeffp ∆H fo (products) – Σ coeffr ∆H fo (reactants)
∆H reaction
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
o
= Σcoeffp ∆G fo (products) – Σ coeffr ∆G fo (reactants)
∆Greaction
o
(a) ∆Greaction
= [1 mol(–1015.4 kJ/mol) + 4 mol(0 kJ/mol)]
– [3 mol(0 kJ/mol) + 4 mol(–237.1 kJ/mol)] = –67 kJ.
The reaction is spontaneous.
o
(b) ∆H reaction
= [1 mol(–1118.4 kJ/mol) + 4 mol(0 kJ/mol)]
– [3 mol(0 kJ/mol) + 4 mol(–285.83 kJ/mol)] = 24.9 kJ.
The reaction absorbs heat.
(c) ∆Soreaction = [1 mol(146.4 J/mol K) + 4 mol(130.680 J/mol K)]
– [3 mol(27.3 J/mol K) + 4 mol(69.95 J/mol K)] = 307.4 J/K.
Entropy increases, so the products are less ordered than the reactants.
13.51 Use Equation 13-5 to calculate the entropy change of a reaction:
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
(a) ∆Soreaction = 2 mol(162 J/mol K)
– [2 mol(101 J/mol K) + 1 mol(205.152 J/mol K)] = – 83 J/K;
o
(b) ∆S reaction = [2 mol(87.4 J/mol K) + 6 mol(223.1 J/mol K)]
– [4 mol(142.3 J/mol K) + 3 mol(205.152 J/mol K)] = 328.7 J/K;
o
(c) ∆S reaction = [6 mol(210.8 J/mol K) + 6 mol(69.95 J/molK)]
– [3 mol(121.2 J/mol K) + 4 mol(238.9 J/mol K)] = 365.3 J/K.
426
Chapter 13
13.52 Use Equation 13-5 to calculate the entropy change of a reaction:
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
(a) ∆Soreaction= [1 mol(197.7 J/mol K) + 1 mol(130.680 J/mol K)]
– [1 mol(5.7 J/mol K) + 1 mol(188.835 J/mol K)] = 133.8 J/K;
(b) ∆Soreaction = 1 mol(213.8 J/mol K)
– [1 mol(5.7 J/mol K) + 1 mol(205.152 J/mol K)] = 2.9 J/K;
o
(c) ∆S reaction = 1 mol(197.7 J/mol K)
– [1 mol(5.7 J/mol K) + 0.5 mol(205.152 J/molK)] = 89.4 J/K;
o
(d) ∆S reaction= [1 mol(213.8 J/mol K) + 1 mol(130.680 J/mol K)]
– [1 mol(197.7 J/mol K) +1 mol(188.835 J/mol K)] = – 42.1 J/K.
13.53 Use Equation 13-10 to calculate the standard free energy change of a reaction:
o
= Σcoeffp ∆G o (products) – Σ coeffr ∆G o (reactants)
∆Greaction
f
f
(a) ∆Goreaction = 2 mol(– 3 kJ/mol) – [2 mol(17 kJ/mol) + 1 mol(0 kJ/mol)] = – 40 kJ;
(b) ∆Goreaction = [2 mol(–742.2 kJ/mol) + 6 mol(0)]
– [4 mol(–334.0 kJ/mol) + 3 mol(0 kJ/mol)] = – 148.4 kJ;
o
(c) ∆G reaction = [6 mol(87.6 kJ/mol) + 6 mol(–237.1 kJ/mol)]
– [3 mol(149.3 kJ/mol) + 4 mol(163.2 kJ/mol)] = – 1997.7 kJ.
13.54 Use Equation 13-10 to calculate the standard free energy change of a reaction:
o
o
∆Goreaction = Σcoeffp ∆G f (products) – Σ coeffr ∆G f (reactants)
(a) ∆Goreaction = [1 mol(–137.2 kJ/mol) + 1 mol(0 kJ/mol)]
– [1 mol(0 kJ/mol) + 1 mol(–228.72 kJ/mol)] = 91.5 kJ;
o
(b) ∆G reaction = 1 mol(–394.4 kJ/mol) – [1 mol(0 kJ/mol) + 1 mol(0 kJ/mol)] = –394.4
kJ;
(c) ∆Goreaction = 1 mol(–137.2 kJ/mol)
– [1 mol(0 kJ/mol) + 0.5 mol(0 kJ/mol)] = –137.2 kJ;
o
(d) ∆G reaction= [1 mol(–394.4 kJ/mol) + 1 mol(0 kJ/mol)]
–[1 mol(–137.2 kJ/mol) + 1 mol(–228.72 kJ/mol)] = –28.5 kJ.
13.55 To determine ∆G° at non-standard temperature, calculate ∆S° and ∆H° using Equations
13-5 and 12-10, then use Equation 13-10. The standard entropy changes are calculated in
Problem 13.51.
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
o
o
o
≅ ∆H reaction,
- T ∆S reaction,
∆Greaction,
T
298
298
(a)
427
Chapter 13
∆Horeaction = 2 mol(– 104 kJ/mol) – [2 mol(–67 kJ/mol) + 1 mol(0 kJ/mol)] = – 74 kJ;
−3
 − 83 J 10 kJ 

∆Soreaction = (623.2 K) 

 = -52 kJ;

 1 K  1 J 
o
= (–74 kJ) – (–52 kJ) = – 22 kJ.
∆Greaction,
350
(b)
∆Horeaction = [2 mol(–824.2 kJ/mol) + 6 mol(0 kJ/mol)]
– [4 mol(–399.5 kJ/mol) + 3 mol(0 kJ/mol)] = –50.4 kJ;
−3
328.7 J 10 kJ 


∆Soreaction = (623.2 K) 

 = 204.8 kJ ;
 1 K  1 J 
o
= (–50.4 kJ) – (204.8 kJ) = –255.2 kJ.
∆Greaction,
350
(c)
∆Horeaction = [6 mol(91.3 kJ/mol) + 6 mol(–285.83 kJ/mol)]
– [3 mol(50.6 kJ/mol) + 4 mol(142.7 kJ/mol)] = – 1889.8 kJ.
−3
 − 365.3 J 10 kJ 
o

∆S reaction = (623.2 K) 


 = -227.7 kJ
 1 K  1 J 
o
= (–1889.8 kJ) – (-227.7 kJ) = –1662.1 kJ.
∆Greaction,
350
13.56 To determine ∆G° at non-standard temperature, calculate ∆S° and ∆H° using Equations
13-5 and 12-10, then use Equation 13-10. The standard entropy changes are calculated in
Problem 13.51, and 200.0 °C = 473.2 K.
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
o
o
∆Horeaction = Σ coeffp ∆H f (products) – Σ coeffr ∆H f (reactants)
o
o
o
≅ ∆H reaction,
- T ∆S reaction,
∆Greaction,
T
298
298
(a)
∆Horeaction = [1 mol(–110.5 kJ/mol) + 1 mol(0 kJ/mol)]
– [1 mol(0 kJ/mol) + 1 mol(–241.83 kJ/mol)] = 131.3 kJ;
−3
133.8 J 10 kJ 
o


∆S reaction = (473.2 K) 

 = 63.3 kJ.
 1 K  1 J 
o
= (131.3 kJ) – (63.3 kJ) = 68.0 kJ;
∆Greaction,
473
(b)
∆Horeaction = 1 mol(–393.5 kJ/mol) – [1 mol(0 kJ/mol) + 1 mol(0 kJ/mol)] = –393.5 kJ;
−3
 2.9 J 10 kJ 


∆Soreaction = (473.2 K) 

 = 1.4 kJ;
 1 K  1 J 
428
Chapter 13
o
= (–393.5 kJ) – (1.4 kJ) = –394.9 kJ.
∆Greaction,
473
(c)
∆Horeaction =1 mol(–110.5 kJ/mol) – [1 mol(0 kJ/mol) + 0.5 mol(0 kJ/mol)] = –110.5 kJ;
−3
89.4 J 10 kJ 


∆Soreaction = (473.2 K) 

 = 42.3 kJ;
 1 K  1 J 
o
= (–110.5 kJ) – (42.3 kJ) = –152.8 kJ.
∆Greaction,
473
(d)
∆Horeaction = [1 mol(–393.5 kJ/mol) + 1 mol(0 kJ/mol)]
– [1 mol(–110.5 kJ/mol) + 1 mol(–241.83 kJ/mol)] = –41.2 kJ.
−3
 − 42.1 J 10 kJ 
o


∆S reaction = (473.2 K) 

 = -19.9 kJ;
 1 K  1 J 
o
= (–41.2 kJ) – (-19.9) = –21.3 kJ.
∆Greaction,
473
13.57 The figure shows a chemical reaction in which a set of diatomic molecules fragment into
atoms with no change in volume or temperature.
(a) There is no volume change, so wsys = 0;
(b) Energy must be supplied to break the chemical bonds, so qsys > 0; and
(c) Because qsys > 0, qsurr < 0, so ∆Ssurr < 0.
13.58
The figure shows a condensation in which a gaseous substance becomes solid.
(a) Energy is released in this phase change, and this energy must be transferred to the
surroundings, so qsys < 0;
(b) Because qsys < 0, ∆Ssys < 0;
(c) ∆Euniverse = 0 because energy is conserved.
13.59 A reaction is thermodynamically feasible if ∆G < 0.
(a) ∆G° = [2 mol(–73.5 kJ/mol) + 1 mol(87.6 kJ/mol)]
– [3 mol(51.3 kJ/mol) + 1 mol(–237.1 kJ/mol)] = 23.8 kJ;
This is not thermodynamically feasible under standard conditions.
2
(p

HNO 3 ) (pNO )
(b) ∆G = ∆G° + RT ln Q,
Q= 
;
3
(p
)
NO


2
−
6
2
-6
(10 ) (10 ) 
∆G = 23.8 kJ + (0.008314 kJ/mol K)(298 K) ln 
 = 23.8 kJ – 102.69 kJ
(1)3


∆G = – 78.9 kJ; thermodynamically feasible under these conditions.
429
Chapter 13
13.60 The standard free energy change for the reaction between H2S and O2 can be calculated
using Equation 13-10 and data from Appendix D:
∆Goreaction = Σcoeffp ∆Gfo (products) – Σ coeffr ∆Gfo (reactants)
∆Goreaction = [8 mol(–237.1 kJ/mol) + 1 mol(0 kJ/mol)]
– [8 mol(–33.4 kJ/mol) + 1 mol(0 kJ/mol)] = –1629.6 kJ.
The glucose synthesis reaction forms glucose from CO2 and H2O:
6 CO2 + 6 H2O → C6H12O6 + 6 O2
∆Goreaction = 2870 kJ
To drive this reaction "uphill" requires coupling with a process with –∆Goreaction > 2870
2870
kJ. The H2S reaction releases this much free energy if it occurs
= 1.76 times as
1630
often as the glucose synthesis reaction. On a molecular level, this requires 16 mol of H2S
to be consumed for every 1 mol of glucose produced, so the overall coupled reaction is:
16 H2S + 8 O2 + 6 CO2 + 6 H2O → 16 H2O + 2 S8 + C6H12O6 + 6 O2, or
16 H2S + 2 O2 + 6 CO2 → 10 H2O + 2 S8 + C6H12O6,
∆Goreaction = 2 mol(–1630 kJ/mol) + 2870 kJ = –390 kJ.
13.61 Entropy changes for constant-temperature processes can be calculated using
q
Equation 13-1, ∆S = T , and q for water freezing can be found using q = – n∆Hfus:
T
 1 mol 
m
nwater =
= 155 g 
18.02 g 
 = 8.60 mol;
MM


Twater = (0.0 + 273.15 K) = 273.2 K;
3
6.01 kJ 10 J 
3
qwater = –8.60 mol 

 1 kJ 
 = –51.7 x 10 J;
1
mol



Tsurr = (–20.0 + 273.15 K) = 253.2 K;
- 51.7 x 103 J
= –189 J/K.
(a) ∆Swater =
273.2 K
(b) The energy released by the water will be absorbed by the surrounding so:
qsurr = –qwater = 51.7 x 103 J/K
51.7 x 103 J
∆Ssurr =
= 204 J/K;
253.2 K
∆Suniverse = ∆Swater + ∆Ssurr = –189 + 204 = 15 J/K.
(c) Cooling the ice to –20 °C is an irreversible change (because the temperatures of ice
and freezer are not the same except at the end of the process), for which the
430
Chapter 13
∆Suniverse > 0. Thus, the cooling must generate an additional entropy increase for the
universe.
13.62 Entropy changes for constant-temperature processes can be calculated using
q
Equation 13-1, ∆S = T , and q for water boiling can be found using q = n∆Hvap:
T
 1 mol 
m
nwater =
= 25.0 g 
 = 1.387 mol;
18.02 g 
MM


Twater = (100. + 273.15 K) = 373 K;
3
 40.79 kJ 10 J 
3
qwater = 1.387 mol 

 = 56.6 x 10 J;
 1 kJ 
1
mol



Tsurr = (315 + 273.15 K) = 588 K;
56.6 x 103 J
= 152 J/K;
∆Swater =
373 K
- 56.6 x 103 J
∆Ssurr =
= –96 J/K;
588 K
∆Suniverse = 152 J/K – 96 J/K = 56 J/K.
13.63 Spontaneity is determined by free energy, heat flow by enthalpy, and change in order by
entropy. Use Equations 12-10, 13-5, and 13-10 to calculate standard thermodynamic values:
∆Horeaction = Σ coeffp ∆H fo (products) – Σ coeffr ∆H fo (reactants)
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
o
o
∆Goreaction = Σcoeffp ∆G f (products) – Σ coeffr ∆G f (reactants)
(a) ∆Goreaction = [1 mol(–261.905 kJ/mol) + 1 mol(–131.0 kJ/mol)]
– 1 mol(–384.1 kJ/mol) = –8.8 kJ. The reaction is spontaneous.
o
(b) ∆H reaction = [1 mol(–240.3 kJ/mol) + 1 mol(–167.1 kJ/mol)]
– 1 mol(–411.2 kJ/mol) = 3.8 kJ. The reaction absorbs heat.
(c) ∆Soreaction = [1 mol(58.5 J/mol K) + 1 mol(56.5 J/mol K)]
– 1 mol(72.1 J/mol K) = 42.9 J/K.
Entropy increases, so the amount of order decreases during this reaction.
13.64 Spontaneity is determined by free energy, heat flow by enthalpy, and change in order by
entropy. Use Equations 12-10, 13-5, and 13-10 to calculate standard thermodynamic values:
∆Horeaction = Σ coeffp ∆H fo (products) – Σ coeffr ∆H fo (reactants)
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
∆Goreaction = Σcoeffp ∆G fo (products) – Σ coeffr ∆G fo (reactants)
(a) ∆Goreaction = [1 mol(77.107 kJ/mol) + 1 mol(–131.0 kJ/mol)]
431
Chapter 13
– 1 mol(–109.8 kJ/mol) = 55.9 kJ. The reaction is not spontaneous.
(b) ∆Horeaction = [1 mol(105.8 kJ/mol) + 1 mol(–167.1 kJ/mol)]
– 1 mol(–127.0 kJ/mol) = 65.7 kJ. The reaction absorbs heat.
(c) ∆Soreaction = [1 mol(73.4 J/molK) + 1 mol(56.5 J/molK)]
– 1 mol(96.3 J/molK) = 33.6 J/K.
Entropy increases, so the amount of order decreases during this reaction.
The significantly greater amount of heat absorbed when AgCl dissolves compared to
NaCl dissolving makes solution of AgCl non-spontaneous under standard conditions and
accounts for the fact that AgCl is insoluble while NaCl is soluble.
13.65 Energy "transactions" in the body are carried out using the ATP-ADP energy-storing
reaction. When a cell needs energy, it "spends" ATP; when fat or carbohydrates
("capital") are consumed, the energy is stored by converting ADP to ATP ("buying"
ATP). Like money, ATP is readily transported from place to place and is readily
converted into "buying power."
13.66 The total amount of disorder in the universe must increase in any actual process (second
law of thermodynamics). The generation of electricity represents an increase in order
which must be accompanied by generation of greater disorder in the surroundings. A
nuclear power plant generates this disorder by transferring heat to the surroundings.
13.67 A phase diagram provides a "road map" that allows us to determine the temperatures and
pressures at which phase changes occur.
(a) A constant temperature process is a vertical line on the phase diagram. T = 70 K is in
between the triple point (0.124 atm, 63 K) and normal boiling point (1 atm, 77 K) of
nitrogen. At 70 K, 1 atm, nitrogen is liquid. When the pressure falls below about 0.5
atm, the liquid boils, and at 70 K, 0.1 atm, the substance is gaseous.
(b) A constant temperature process is a vertical line on the phase diagram. T = 298 K is
within the gaseous region at all pressures, so compression from 1.00 atm to 50.0 atm
leaves the substance gaseous.
(c) A constant pressure process is a horizontal line on the phase diagram, and 1 atm
represents "normal" pressure, shown on the figure as a dotted line. When the temperature
of the gas reaches 77 K, the gas liquefies, and when the temperature reaches 63 K, the
liquid solidifies. At 50 K, N2 is solid.
13.68 Phase diagrams provide "road maps" allowing us to determine what phase changes occur
as temperature and pressure vary in particular ways.
(a) At T = 298 K, P = 1.00 atm, CO2 is a gas. Compression at this temperature causes the
gas to liquefy at about P = 6 atm, and the substance remains a liquid at 50.0 atm.
(b) At T = 195 K, P = 6.00 atm, CO2 is a solid. The pressure is higher than the triple
point, heating at this pressure causes the solid to melt at about 217 K and vaporize at
about 300 K. At T = 350 K, P = 6.00 atm, the substance is a gas.
(c) At T = 298 K, P = 1.00 atm, CO2 is a gas. Cooling at the pressure causes the gas to
condense to solid at the sublimation point, 195 K. The substance is a solid at 50 K.
432
Chapter 13
13.69 Ammonia is a toxic gas, while urea and ammonium nitrate both are relatively non-toxic
solids. Thus the transport and application of ammonia entail significant risks to humans.
Even in aqueous solution, ammonia is highly irritating, as anyone knows who has used
strong ammonia as a cleanser.
13.70 The four most important nitrogen-containing chemicals are ammonia (NH3), nitric acid
(HNO3), urea [(NH2)2CO], and ammonium nitrate (NH4NO3). All these chemicals are
used in the fertilizer industry, as feedstocks (nitric acid and ammonia) or fertilizers (urea,
ammonium nitrate, and ammonia).
13.71 According to your textbook, ∆G°glucose = –2870 kJ/mol and
∆G°palmitic acid = –9790 kJ/mol. Divide molar quantities by molar masses to obtain free
energy per gram:
- 2870 kJ 1 mol 

Glucose: ∆Gper gram = 


 = –15.9 kJ/g;
 1 mol 180 g 
- 9790 kJ 1 mol 

Palmitic acid: ∆Gper gram = 


 = –38.2 kJ/g;
 1 mol  256 g 
Palmitic acid has the higher energy content per gram.
13.72 Use Equations 12-10, 13-5, and 13-10 to calculate standard thermodynamic values:
∆Horeaction = Σ coeffp ∆H fo (products) – Σ coeffr ∆H fo (reactants)
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
∆Goreaction, T ≅ ∆Horeaction, 298 - T∆Soreaction, 298
∆Horeaction = [1 mol(–110.5 kJ/mol) + 3 mol(0 kJ/mol)]
– [1 mol(–74.6 kJ/mol) + 1 mol(–241.83 kJ/mol)] = 205.9 kJ;
∆Soreaction = [1 mol(197.7 J/mol K) + 3 mol(130.680 J/mol K)]
– [1 mol(186.3 J/mol K) + 1 mol(188.835 J/mol K)] = 214.6 J/K
-3
 214.6 J 10 kJ 


(a) ∆Goreaction, 298 = 205.9 kJ – 298K 

 = 141.9 kJ
 1 K  1 J 
-3
 214.6 J 10 kJ 

(b) ∆Goreaction, 1300 = 205.9 kJ –1300 K 

= – 73.1 kJ

 1 K  1 J 
13.73 The process that takes place is vaporization of the liquid bromine to form bromine vapor.
(a) At the end of the process, all the bromine is in the gas phase, with the molecules well
separated, and the piston has moved back:
433
Chapter 13
(b) The process is reversible, since the temperatures of system and surroundings are the
same and the external and internal pressures both are 1.0 atm (bromine is at its boiling
point). Thus, ∆Suniverse = 0.
13.74 (a) Water falling from high to low altitude is a spontaneous process that drives the turbine
and "organizes" electrons into an electric current, a non-spontaneous process.
(b) Gasoline burns in the engine in a spontaneous process that drives the pump and
pushes water uphill, a non-spontaneous process.
13.75 The attractive forces in CaO are substantially larger than the attractive forces in KCl, so
the individual Ca2+ and O2- ions cannot as easily vibrate in the crystal. Thus the amount
of vibrational disorder is substantially less in CaO than in KCl.
13.76 Use Equations 12-10 and 13-10 to calculate standard thermodynamic values:
∆Horeaction = Σ coeffp ∆H o (products) – Σ coeffr ∆H o (reactants)
f
f
o
o
o
∆G reaction = Σcoeffp ∆G f (products) – Σ coeffr ∆G f (reactants)
CCl4:
∆Horeaction = [1 mol(–393.5 kJ/mol) + 4 mol(102.5 kJ/mol)]
– [1 mol(–128.2 kJ/mol) + 5 mol(0 kJ/mol)] = 144.7 kJ;
o
∆G reaction = [1 mol(–394.4 kJ/mol) + 4 mol(120.5 kJ/mol)]
– [1 mol(–65.21 kJ/mol) + 5 mol(0 kJ/mol)] = 152.8 kJ;
CS2:
∆Horeaction = [1 mol(–393.5 kJ/mol) + 2 mol(–296.8 kJ/mol)]
– [1 mol(89.0 kJ/mol) + 3 mol(0 kJ/mol)] = –1076.1 kJ;
o
∆G reaction = [1 mol(–394.4 kJ/mol) + 2 mol(–300.2 kJ/mol)]
– [1 mol(64.6 kJ/mol) + 3 mol(0 kJ/mol)] = –1059.2 kJ.
The calculations indicate that combustion of CS2 is highly spontaneous, while that of
CCl4 is not. Thus CCl4 is not a fire hazard, but plants using CS2 must be extremely
cautious about fire hazards, as this solvent has a similar flammability as gasoline.
434
Chapter 13
13.77 Make use of thermodynamic definitions to identify the state functions:
(a) qv = ∆E; (b) qp = ∆H; (c) qT = T∆S.
13.78 (a) When dry ice at its sublimation temperature is placed in water at its freezing
temperature, energy flows as heat from molecules of H2O to molecules of CO2, allowing
CO2 molecules to sublime into the gas phase and causing H2O molecules to form solid
ice. The CO2 system undergoes its phase transition at 195 K, while the H2O system
undergoes its phase transition at 273.15 K.
(b) Entropy changes for constant-temperature processes are calculated using
q
Equation 13-1, ∆S = T , and q for dry ice subliming can be found using q = n∆Hsubl:
T
 1 mol 
m
ndry ice =
= 12.5 g 
 = 0.284 mol;
 44.01 g 
MM


3
 25.2 kJ 10 J 
3
qdry ice= 0.284 mol 

 1 kJ 
 = 7.16 x 10 J;
1
mol



7.16 x 103 kJ
= 36.7 J/K;
195 K
- 7.16 x 10 3 kJ
= –26.2 J/K;
∆Swater =
273.15 K
∆Suniverse = 36.7 – 26.2 = 10.5 J/K.
∆Sdry ice =
13.79 (a) Whenever a substance cools, ∆S is negative.
(b) Whenever a gas is compressed (at constant T), ∆S is negative.
(c) Whenever two substances mix, disorder increases and ∆S is positive.
13.80 (a) Maximum entropy results when the distribution of molecules is most disordered:
(b) Minimum entropy results when the distribution of molecules is most ordered, that is
when both substances exist as crystalline solids:
435
Chapter 13
13.81 Use bond energy, intermolecular forces, and order-disorder to predict signs for ∆H and
∆S:
(a) Bonds form, so energy is released and ∆H is negative. The number of independent
particles decreases, so order increases and ∆S is negative.
(b) When a solid melts, energy must be added; q is positive, ∆H is positive, and ∆S is
positive.
(c) Combustion reactions release energy, so we expect ∆H to be negative. The number of
molecules of gas remains the same during this reaction, so ∆S is expected to be small, but
it is not possible to predict whether it is positive or negative without doing actual
calculations.
13.82 The transfer of heat from low temperature to high temperature is a non-spontaneous
process, because the entropy change is negative. This process must be coupled with one
that generates a positive entropy change in the surroundings, namely the transfer of heat
to the surroundings.
13.83 Standard conditions refers to 1 atm, 298 K, all solutes at 1 M. In a cell, temperature is
37 °C (310 K) and no solutes are present at 1 M. In particular, hydronium ion
concentration is around 10-7 M, far from standard conditions.
13.84 Opening the refrigerator door will result in overall heating rather than cooling, because in
order to remove heat from its contents, the refrigerator must transfer additional heat to the
surroundings. Coolers such as air conditioners reduce the internal temperature at the cost
of increasing the external temperature by transferring more heat to the surroundings than
is removed from the system.
13.85 According to your textbook, the complete oxidation of 1 mol of palmitic acid produces
130 ATP molecules. The reactions are:
ADP + H3PO4 → ATP + H2O ∆Go = +30.6 kJ
Palmitic acid: C15H31CO2H + 23 O2 → 16 CO2 + 16 H2O ∆Go = –9790 kJ
Thus the amount of energy stored is (130)(30.6 kJ) = 3978 kJ and the efficiency is:
energy stored
3978 kJ
×100% =
Efficiency =
×100% = 40.6 %.
energy released
9790 kJ
436
Chapter 13
Metabolism of 1 mol of palmitic acid generates (9790 kJ) – (3978 kJ) = 5612 kJ of heat.
5612 kJ  1 mol 

On a per gram basis, this is qper gram = 

= 21.89 kJ/g.

 1 mol  256.42 g 
Evaporation of 75 g of water requires
 1 mol  40.79 kJ 
2
q = n∆Hvap = 75 g 
 1 mol = 1.7 x 10 kJ.
18.02 g 




 1g 
Mass required = 1.7 x 102 kJ 
 = 7.8 g.
 21.89 kJ 
13.86 The standard free energies of formation of oxides and sulfides are generally highly negative,
so reactions of elements with elemental oxygen or sulfur are highly spontaneous. Thus, most
elements have long since reacted to form oxides and/or sulfides. For nitrogen, however, the
standard free energies of formation of oxides are generally positive, so molecular nitrogen
does not spontaneously react to form oxides.
13.87 Both of these processes are spontaneous. Thus, assuming that each of these processes
occurs at constant T and P (perhaps more reasonable for the firefly than for lightning),
both must have ∆G < 0.
13.88 The heat flow described in this problem is from body + water to the atmosphere. Here is
a flow diagram of the process:
The key feature is that heat is transferred from the system at 37.5 °C to the surroundings
at 23.5 °C.
Vaporization involves the release of energy:
 1 mol  − 40.79 kJ 103 J 
3
q = n ∆Hvap = 1.0 g 
18.02 g 
 1 mol 
 1 kJ 
 = -2.3 x 10 J.




− 2.3 x 103 J
= – 7.4 J/K;
37.5 + 273.15 K
2.3 x 103 J
∆Ssurr =
= 7.7 J/K;
23.5 + 273.15 K
∆Stotal = 7.7 J/K – 7.4 J/K = 0.3 J/K.
∆Ssys =
437
Chapter 13
13.89 Your molecular picture should show all three phases simultaneously present, with
transfers of atoms occurring among all phases:
13.90 The HCl molecule is heteronuclear, while H2 and Cl2 both are homonuclear. The
homonuclear molecules are more symmetric than the heteronuclear, so they have fewer
possible rotational orientations in space. This leads to less rotational disorder and
accounts for the difference.
13.91 (a) Use Equations 12-10, 13-5, and 13-10 to calculate standard thermodynamic values:
o
o
∆Horeaction = Σ coeffp ∆H f (products) – Σ coeffr ∆H f (reactants)
∆Soreaction = ΣcoeffpSo(products) – Σcoeffr So(reactants)
∆Goreaction = Σcoeffp ∆G o (products) – Σ coeffr ∆G o (reactants)
f
f
o
∆H reaction = 2 mol(142.7 kJ/mol) – 3 mol(0 kJ/mol) = 285.4 kJ;
∆Soreaction = 2 mol(238.9 J/molK) – 3 mol(205.152 J/molK) = –137.7 J/K
∆Goreaction = 2 mol(163.2 kJ/mol) – 3 mol(0 kJ/mol) = 326.4 kJ.
(b) ∆G° = ∆H° – T ∆S°; because ∆Horeaction is positive and ∆Soreaction is negative, ∆Goreaction
is positive at all temperatures.
(c) Calculate the pressure of O3 at which ∆G = 0; the reaction will be spontaneous at any
lower pressure:
2
2
∆G = ∆Goreaction + RT ln  pO3  = ∆Goreaction + RT ln  pO3  = 0;
 3 

3


 pO 2 

(0.20) 

2
326.4 kJ + (8.314 x 10-3 kJ/mol K)(298 K) ln  pO3  = 0;

3
(0.20) 


2
326.4 kJ
ln  pO3  = –
= – 131.7;
-3 kJ

3
(8.314
x
10
)(298
K)
mol
K

(0.20) 

 p2 
 O 3  = 6.1 x 10-58;
(0.20)3 



438
Chapter 13
pO3 = 2.2 x 10-30 atm.
(d) The calculations show that the production of ozone is not spontaneous, regardless of
temperature and pressure. In the upper atmosphere, the reaction is coupled with the
absorption of light by O2 molecules, which deposits energy in the system and breaks
O–O bonds. Once O atoms are present, the formation of ozone becomes spontaneous.
13.92 The bonding network in graphite is two-dimensional, leaving planar arrays of carbon
atoms free to take up random positions relative to one another. The bonding network in
diamond is three-dimensional, leaving very little possibility for random positions. Thus,
graphite is substantially more disordered than diamond. Because the chemical formula of
buckminster-fullerene is C60, one mole of buckminster-fullerene contains 60 mol of C.
Thus the molar entropy of fullerene is roughly 60 times as great as the molar entropy of
diamond or graphite (on a per-gram basis, the entropy of buckminster-fullerene is
between that of diamond and graphite, because the C60 units are highly ordered but there
is some randomness in the way the C60 units stack in a crystal).
−3
 28.9 J 10 kJ 


13.93 (a) ∆Go = ∆Ho – T ∆So = 5.65 kJ/mol – 298 K 

 = – 2.96 kJ/mol;
1 mol K  1 J 
(b) The freezing point is the temperature at which
−3
 28.9 J 10 kJ 


∆G° = 0 = (5.65 kJ/mol) – T 


1 mol K  1 J 
3
5.65 kJ 1 mol K 10 J 

T= 



 = 196 K.
 1 mol  28.9 J  1 kJ 
13.94 In general, we expect the standard entropies of gaseous substances to increase with the
number of atoms per molecule; thus the lower entropy of methane (5 atoms) than
ammonia (4 atoms) is surprising. Methane, however, is highly symmetric, so it has fewer
different possible rotational orientations in space. This leads to less rotational disorder
and accounts for the difference.
13.95 (a) At sufficiently low temperature, every substance adopts the form that is most highly
ordered. The stable form of phosphorus at low temperature is P4β, so this is the more
ordered crystalline structure.
(b) A phase change that occurs when temperature falls involves the release of heat
(consider gaseous water condensing or liquid water freezing). Thus ∆H for this process
is negative. The more ordered form of phosphorus is P4β, so for this process ∆S is
negative.
439