Download Chapter 12 384 12.1 A system is isolated if it exchanges neither

Chapter 12
12.1
A system is isolated if it exchanges neither energy nor matter with its surroundings,
closed if it exchanges energy but not matter, and open if it exchanges both:
(a) Open, energy and matter are constantly entering and leaving the system;
(b) (almost) isolated, because the thermos minimizes energy transfer to the coffee;
(c) open, energy and matter can flow in and out; and
(d) closed, the balloon isolates the helium from the outside.
12.2
A system is isolated if it exchanges neither energy nor matter with its surroundings,
closed if it exchanges energy but not matter, and open if it exchanges both:
(a) Closed, because it can exchange energy but not matter;
(b) (almost) isolated, because the chest minimizes energy transfer;
(c) open, because energy and matter flow in and out; and
(d) closed, because the satellite is self-contained but exchanges energy with its
surroundings.
12.3
State variables are properties that are independent of the past history of the system.
(a) not a state variable; (b) state variable; (c) state variable (mass is conserved); and
(d) not a state variable.
12.4
State variables are properties that are independent of the past history of the system.
(a) state variable; (b) not a state variable; (c) not a state variable; and (d) state variable.
12.5
A system is any part of the universe that can be defined with specified boundaries.
Among possible systems are the automobile and driver, the driver, the automobile, the
cooling system, the engine, a cylinder, the battery, etc.
12.6
A system is any part of the universe that can be defined with specified boundaries.
Among possible systems are the entire body, heart, lungs, digestive system, liver,
kidneys, etc.
12.7
Calculate the temperature change resulting from an energy input using Equation 12-1 and
C values from Table 12-1 of your textbook: q = nC∆T, which can be rearranged to
q
∆T =
nC
 1 mol 
m
(a) n =
= (10.0 g) 
 26.98 g 
 = 0.3706 mol;
MM


25.0 J
∆T =
= 2.77 K = 2.77 °C.
(0.3706 mol)(24.35 J mol -1 K -1 )
Tf = Ti + ∆T = 15.0 + 2.77 = 17.8 °C.
 1 mol 
m
= (25.0 g) 
 26.98 g 
 = 0.9266 mol;
MM


25.0 J
∆T =
= 1.108 K.
(0.9266 mol)(24.35 J mol -1 K -1 )
(b) n =
384
Chapter 12
Tf = Ti + ∆T = 295 + 1.108 = 296 K.
 1 mol 
m
= (25.0 g) 
 = 0.2317 mol;
107.9 g 
MM


25.0 J
= 4.256 K.
∆T =
(0.2317 mol)(25.351 J mol -1 K -1 )
Tf = Ti + ∆T = 295 + 4.256 = 299 K.
(c) n =
 1 mol 
m
= (25.0 g) 
 = 1.387 mol;
18.02 g 
MM


25.0 J
= 0.2394 K = 0.2394 °C.
∆T =
(1.387 mol)(75.291 J mol -1 K -1 )
Tf = Ti + ∆T = 22.0 + 0.2394 = 22.2 °C.
(d) n =
12.8
Calculate the temperature change resulting from an energy input using Equation 12-1 and
C values from Table 12-1 of your textbook: q = nC∆T, which can be rearranged to
q
∆T =
nC
 1 mol 
m
(a) n =
= (35.0 g) 
 26.98 g 
 = 1.297 mol;
MM


– 65.0 J
∆T =
== –2.058 K = –2.058 °C.
(1.297 mol)(24.35 J mol -1 K -1 )
Tf = Ti + ∆T = 65.0 + (–2.058) = 63.0 °C.
 1 mol 
m
= (50.0 g) 
 26.98 g 
 = 1.853 mol;
MM


– 65.0 J
∆T =
= –1.441 K = –1.441 °C.
(1.853 mol)(24.35 J mol -1 K -1 )
Tf = Ti + ∆T = 65.0 + (–1.441) = 63.6 °C.
(b) n =
 1 mol 
m
= (50.0 g) 
107.9 g 
 = 0.4634 mol;
MM


– 65.0 J
∆T =
= –5.533 K = –5.533 °C
(0.4634 mol)(25.351 J mol -1 K -1 )
Tf = Ti + ∆T = 65.0 + (–5.533) = 59.5 °C
(c) n =
(d) n =
 1 mol 
m
= (50.0 g) 
18.02 g 
 = 2.775 mol;
MM


385
Chapter 12
– 65.0 J
= –0.311 K.
(2.775 mol)(75.291 J mol -1 K -1 )
Tf = Ti + ∆T = 325.0 + (–0.311) = 324.7 K.
∆T =
12.9
Calculate an energy change accompanying a temperature change using Equation 12-1 and
C values from Table 12-1 of your textbook: q = nC∆T.
∆T = 95.0 – 23.0 = 72.0 °C = 72.0 K;
1000 g  1 mol 

nkettle = 1.35 kg
 = 24.2 mol Fe


 1 kg 
55.85 g 

qkettle = (24.2 mol)(25.10 J mol-1 K-1)(72.0 K) = 4.37 x 104 J;
1000 g  1 mol 

nwater = 2.75 kg
 = 153 mol water


 1 kg 
18.02 g 

qwater = (153 mol)(75.291 J mol-1 K-1)(72.0 K) = 8.29 x 105 J.
12.10 Calculate an energy change accompanying a temperature change using Equation 12-1 and
C values from Table 12-1 of your textbook: q = nC∆T.
∆T = 37.6 – 24.0 = 13.6 °C = 13.6 K;
 1 mol 
m
nsilver =
= 15.0 g
107.9 g 
 = 0.1390 mol Ag
MM


qsilver= (0.1390 mol)(25.351 J mol-1 K-1)(13.6 K) = 47.9 J;
 1 mol 
m
= 25.0 g
18.02 g 
 = 1.387 mol water
MM


qwater= (1.387 mol)(75.291 J mol-1 K-1)(13.6 K) = 1.42 x 103 J.
nwater =
12.11 When two objects at different temperatures are placed in contact, energy flows from the
warmer to the cooler object until the two objects are at the same temperature. Let that
temperature be T. The energy lost by the warmer object equals the energy gained by the
cooler object: qcool = –qwarm, and q = nC∆T.
 1 mol 
Water: n = 37.5 g 
18.02 g 
= 2.081 mol water;


q = (2.081 mol)(75.351 J mol-1 K-1)(T – 20.5 °C) = (156.8 J K-1)(T – 20.5 °C).
 1 mol 
Coin: n = (27.4 g) 
107.9 g 
= 0.2539 mol Ag;


q = (0.2539 mol)(25.351 J mol-1 K-1)(T – 100.0 °C) = (6.437 J K-1)(T – 100.0 °C).
Substitute and solve for T:
(156.8 J K-1)(T – 20.5 °C) = –(6.437 J K-1)(T – 100.0 °C).
156.8 T – 3214.5 = –6.437 T + 643.7;
163.2 T = 3858.2, and
386
Chapter 12
T = 23.6 °C.
12.12 When two objects at different temperatures are placed in contact, energy flows from the
warmer to the cooler object until the two objects are at the same temperature. Let that
temperature be T. The energy lost by the warmer object equals the energy gained by the
cooler object: qcool = –qwarm, and q = nC∆T.
1.00 g  1 mol 


Coffee: n = 85.0 mL 
 = 4.717 mol;

 1 mL 18.02 g 
75.291 J 
(T – 84.0 °C).
qcoffee = (4.717 mol) 
 1 mol K 
 1 mol 
Spoon: n = 24.7 g 
= 0.4423 mol;
55.85 g 


 25.10 J 
(T – 18.5 °C).
qspoon = (0.4423 mol) 
1 mol K 
Substitute and solve for T:
 25.10 J 
(0.4423 mol) 
(T – 18.5 °C) = –(4.717 mol)
1 mol K 
11.10 T – 205.4 = –355.15 T + 29,832;
366.25 T = 30,037, and
T = 82.0 °C.
75.291 J 

 (T – 84.0 °C).
 1 mol K 
12.13 Calculate the energy change for a temperature change using Equation 12-1 and C values
from Table 12-1 of your textbook: q = nC∆T, and
∆T = 87.6 – 21.5 = 66.1 °C = 66.1 K;
m
1.00 g  1 mol 

nwater =
= 475 mL 


 = 26.36 mol ;
MM
 1 mL 18.02 g 
75.291 J 
qwater = (26.36 mol) 
(66.1 K) = 1.31 x 105 J = 131 kJ.
1
mol
K


12.14 Calculate an energy change accompanying a temperature change using Equation 12-1 and
C values from Table 12-1 of your textbook: q = nC∆T.
∆T = 5.50 – 54.0 = –48.5 °C = –48.5 K;
 1 mol 
m
nwater =
= 145 g
18.02 g 
 = 8.047 mol
MM


75.291 J 
qwater = (8.047 mol) 
(–48.5 K) = –2.94 x 104 J = –29.4 kJ.
 1 mol K 
12.15 Expansion work can be calculated using Equation 12-3, wsys = –Pext∆Vsys . The external
pressure opposing inflation of a balloon is atmospheric pressure, 1.00 atm:
Vf = 2.5 L, Vi = 0.0 L
387
Chapter 12
∆V= 2.5 L – 0.0 L = 2.5 L
wsys = –Pext∆Vsys = –(1.00 atm)(2.5 L) = –2.5 L atm.
Convert to joules:
101.325 J 
wsys = –2.5 L atm 
= – 2.5 x 102 J.
 1 L atm 
12.16 Expansion work can be calculated using Equation 12-3, wsys = –Pext∆Vsys . The
external pressure opposing inflation of a balloon is 755 torr.
 1 atm 
= 0.993 atm:
Converted to atmospheres: 755 torr 
760 torr 
The final volume of the balloon is 19.5m3, converted to liters:
103 L 
3

19.5 m 
3  = 19,500 L
 1m 
Assuming the initial volume of the balloon is 0 L then,
wsys = –Pext∆Vsys = –(0.993 atm)(19,500L) = –19400 L atm.
Convert to joules:
101.325 J 
 = –1.97x 106 J.
wsys = –19400 L atm 
 1 L atm 
12.17 To work this problem, use data in Chemistry and Life Box. First determine the energy
difference(∆E) between chicken and beef:
6.0 kJ 
Chicken: 250 g 
 1g 
= 1500 kJ;


16 kJ 
beef: 250 g 
 1g 
 = 4000 kJ;


∆E = 4000 kJ – 1500 kJ = 2500 kJ;
According to data in Chemistry and Life Box, a 55-kg person walking 6.0 km/hr
consumes 1090 kJ/hr:
 1 hr 
t = 2500 kJ 
 = 2.3 hr.
1090 kJ 
Therefore, to consume the additional energy a person must walk:
6.0 km 
distance = 2.3 hr 
 = 1.4 km.
 1 hr 
12.18 To work this problem, use data in Chemistry and Life Box. The energy content of sugar
is 16 kJ/g, so one pound of sugar has an energy content of:
103 g 16 kJ 

0.455 kg 

 = 7.28 x 103 kJ.
 1 kg 


 1 g 
A person weighing 85 kg consumes 4245 kJ/hr when running at 16 km/hr:
388
Chapter 12
 1 hr 
t = 7.28 x 103 kJ 
= 1.7 hr.
 4245 kJ 
Therefore, to consume the energy of the sugar a person must run:
16 km 
 = 27 km.
distance = 2.3 hr 
 1 hr 
12.19 The heat capacity of a calorimeter can be found from energy and temperature data using
Equation 12-5, qcalorimeter = Ccal ∆T.
Here, the heat released by the combustion process is absorbed by the calorimeter so:
- 15.57 kJ 

qcalorimeter = –qglucose = –1.7500 g 
 = 27.25 kJ
 1g


∆T = 23.34 – 21.45 = 1.89 °C
q
27.25 kJ
= 14.4 kJ/°C.
Ccal = calorimete r =
1.89o C
∆T
12.20 The heat capacity of a calorimeter can be found from energy and temperature data using
Equation 12-5, qcalorimeter = Ccal ∆T.
Here, qcalorimeter = 1150 J
∆T = 25.25 – 23.45 = 1.80 °C
q
1150 J
Ccal = calorimete r =
= 639 J/°C = 639 J/K
∆T
1.80o C
The heat capacity of 125 mL of water is:
1.00 g  1 mol 75.291 J 

Cwater = 125 mL

 = 522 J/K


 1 mL 18.02 g  1 mol K 
Thus, the percentage due to water is:
C
522 J/K
×100% = 81.7%.
% = water =
C cal
639 J/K
12.21 (a) In a combustion reaction, the products are CO2 and H2O:
C7H6O2 + O2 → CO2 + H2O (unbalanced)
Follow standard procedures to balance the equation. Give CO2 a coefficient of 7 to
balance C, H2O a coefficient of 3 to balance H:
C7H6O2 + O2 → 7 CO2 + 3 H2O
7C + 6H + 4O → 7C + 6H + 17O
Balance O by giving O2 a coefficient of 15/2, then multiply by 2 to clear fractions:
2 C7H6O2 + 15 O2 → 14 CO2 + 6 H2O
(b) Find energy per mole from energy per gram using the molar mass (122.1 g/mol):
389
Chapter 12
− 35.61 kJ 122.1 g 
∆E (kJ/mol) = [∆E (kJ/g)][MM (g/mol)] = 
 1 mol = –3.221 x 103 kJ/mol
 1.350 g 



(c) 15 moles O2 is consumed for each 2 mol of benzoic acid, so the energy released per
mol of O2 is:
 − 3.221 x 103 kJ  2 mol acid 

E =
 = –4.295 x 102 kJ/mol.


 1 mol acid 
15 mol O 2 

12.22 (a) In a combustion reaction, the products are CO2 and H2O:
C2H2 + O2 → CO2 + H2O (unbalanced)
Follow standard procedures to balance the equation. Give CO2 a coefficient of 2 to
balance C:
C2H2 + O2 → 2CO2 + H2O
2C + 2H + 2O → 2C + 2H + 5O
Give O2 a coefficient of 5/2 to balance O, then multiply by 2 to clear fractions:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
(b) Find energy per mole from energy per gram using the molar mass (26.04 g/mol):
− 48.2 kJ  26.04 g 
∆E (kJ/mol) = [∆E (kJ/g)][MM (g/mol)] = 
 1.00 g 
 1 mol  = – 1.26 x 103 kJ/mol



(c) 5 moles O2 is consumed for every 2 moles of acetylene, so the energy released per
mol of O2 is:
 − 1.26 x 103 kJ  2 mol acetylene 

E= 
1 mol acetylene 

 5 mol O
 = –5.04 x 102 kJ/mol.



2
12.23 Calculate the energy released during combustion from the temperature increase and the
total heat capacity of the calorimeter. Then convert to molar energy using molar mass.
q = C∆T = (7.85 kJ/K)(302.04 K – 297.65 K) = 34.46 kJ
∆E = –q = –34.46 kJ
− 34.46 kJ 194.2 g 
∆E (kJ/mol) = [∆E (kJ/g)][MM (g/mol)]= 
 1.35 g 
 1 mol  = –4.96 x 103 kJ/mol



12.24 Calculate the heat capacity of the calorimeter from the electrical energy added and the
temperature change during electrical heating. Then find the energy released during
combustion from the temperature increase and the total heat capacity of the calorimeter.
Finally convert to molar energy using molar mass.
q
19.75 kJ
C = calorimete r =
= 4.68 kJ/°C
∆T
4.22o C
390
Chapter 12
 4.68 kJ 
qcombustion = C∆Tcombustion =  o
(8.47 °C) = 39.64 kJ
 1C 
- q − 39.64 kJ
=
∆E =
= – 22.65 kJ/g
m
1.75 g
The molar heat of combustion for methanol is obtained by multiplying the energy per
gram by the molar mass:
− 22.65 kJ 32.04 g 

∆E (kJ/mol) = [∆E (kJ/g)][MM (g/mol)] = 
 1 mol  = –7.26 x 102 kJ/mol
 1g



12.25 Standard enthalpy changes are calculated from Equation 12-10 using standard enthalpies
of formation, which can be found in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof(reactants)
(a) ∆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;
o
(b) ∆H reaction= 1 mol (0 kJ/mol) + 3 mol(0 kJ/mol) – 2 mol(–45.9 kJ/mol) = 91.8 kJ;
(c) ∆Horeaction= [1 mol(–2984.0 kJ/mol) + 5 mol(0kJ/mol)]
– [5 mol(–277.4 kJ/mol) + 4 mol(0 kJ/mol)] = –1597.0 kJ;
o
(d) ∆H reaction= [1 mol(–910.7 kJ/mol) + 4 mol(–92.3 kJ/mol)]
– [1mol(–687.0 kJ/mol) + 2 mol(–285.83 kJ/mol)] = –21.2 kJ.
12.26 Standard enthalpy changes are calculated from Equation 12-10 using standard enthalpies
of formation, which can be found in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
(a) ∆Horeaction= 2 mol(–704.2 kJ/mol)
– [2 mol(0 kJ/mol) + 3 mol(0 kJ/mol)] = –1408.4 kJ;
o
(b) ∆H reaction= [2 mol(–133.9 kJ/mol) + 1 mol(91.3 kJ/mol)]
– [3 mol(33.2 kJ/mol) + 1 mol(–285.83 kJ/mol)] = 9.7 kJ;
o
(c) ∆H reaction= [4 mol(–393.5 kJ/mol) + 2 mol(–285.83 kJ/mol)]
–[2 mol(227.4 kJ/mol) + 5 mol(0 kJ/mol)] = –2600.5 kJ.
12.27 Reaction energies are related to reaction enthalpies (see Prob. 12.25) through Equation
12-9:
∆Hreaction ≅ ∆Ereaction + ∆(nRT)gases
Since temperature does not change, we can rearrange the equation to:
∆Ereaction ≅∆Hreaction – RT∆(n)(gases)
Begin by calculating the change in moles of gases, then use the rearranged equation to
determine the ∆Ereaction
(a) ∆ngases = 2 – (3 + 1) = –2 mol;
10 − 3 kJ 
 8.314 J 

(
)
298K

∆Ereaction ≅ − 1411.1 kJ – (–2 mol) 
 1J 

1 mol K 


391
Chapter 12
= –1411.1 + 4.96 = –1406.1 kJ;
(b) ∆ngases = 3 + 1 – 2 = 2 mol;
10 − 3 kJ 
 8.314 J 

(
)
298K


∆Ereaction ≅ 91.8 kJ – (2 mol)
 = 91.8 – 4.96 = 86.8 kJ;
 1J 
1 mol K 


(c) all compounds are solid, ∆ngases = 0 mol;
∆Ereaction ≅ –1597.0 kJ;
(d) ∆ngases = 4 mol;
10 − 3 kJ 
 8.314 J 
(298K )
∆Ereaction ≅ –21.2 kJ – (4 mol) 
 = –21.2 + 9.91 = –11.3 kJ
 1J 
1 mol K 


12.28 Reaction energies are related to reaction enthalpies (see Prob. 12.26) through Equation
12-9:
∆Hreaction ≅ ∆Ereaction + ∆(nRT)gases
Since temperature does not change, we can rearrange the equation to:
∆Ereaction ≅∆Hreaction – RT∆(n)(gases)
Begin by calculating the change in moles of gases, then use the rearranged equation to
determine the ∆Ereaction
(a) ∆ngases = 0 – 3 = –3 mol;
10 − 3 kJ 
 8.314 J 

(
)
298K


∆Ereaction ≅ –1408.4 kJ – (–3 mol)
 1J 
= –1408.4 + 7.43
1 mol K 


= –1401.0 kJ;
(b) ∆ngases = 1 + 2 – 3 = 0 mol;
∆Ereaction ≅ –9.7 kJ;
(c) ∆ngases = 4 – (5 + 2) = –3 mol;
10 − 3 kJ 
 8.314 J 

(
)
298K


∆Ereaction ≅ –2600.5 kJ – (–3 mol)
 1J 
= –2600.5 + 7.43
1 mol K 


= –2593.1 kJ.
12.29 A formation reaction has elements in their standard states as reactants and 1 mol of a
single product:
(a) 3 K(s) + P(s) + 2 O2(g) → K3PO4(s);
(b) 2 C(graphite) + 2 H2(g) + O2(g) → CH3CO2H(l);
(c) 3 C(graphite) + 9/2 H2(g) +1/2 N2(g) → (CH3)3N(g);
(d) 2 Al(s) + 3/2 O2(g) → Al2O3(s).
392
Chapter 12
12.30 A formation reaction has elements in their standard states as reactants and 1 mol of a
single product:
(a) 4 C(graphite) + 5 H2(g) +1/2 O2(g) → C4H9OH(l);
(b) 2 Na(s) + C(graphite) + 3/2 O2(g) → Na2CO3(s);
(c) 3/2 O2(g) → O3(g);
(d) 3 Fe(s) + 2 O2(g) → Fe3O4(s).
12.31 Standard enthalpy changes are calculated from Equation 12-10 using standard enthalpies
of formation, which can be found in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
(a) ∆Horeaction = [4 mol(91.3 kJ/mol) + 6 mol(–285.83 kJ/mol)]
– [4 mol(–45.9 kJ/mol) + 5 mol(0 kJ/mol)] = –1166.2 kJ;
o
(b) ∆H reaction = [2 mol(0 kJ/mol) + 6 mol(–285.83 kJ/mol)]
– [4 mol(–45.9 kJ/mol) + 3 mol(0 kJ/mol)] = –1531.4 kJ.
12.32 Standard enthalpy changes are calculated from Equation 12-10 using standard enthalpies
of formation, which can be found in Appendix D of your textbook:
∆Horeaction= Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
(a) ∆Horeaction= 2 mol(–823.0 kJ/mol)
– [1 mol(–824.2 kJ/mol) + 3 mol(–285.83 kJ/mol)] = 35.7 kJ;
o
(b) ∆H reaction= 2 mol(–1094.3)
– [1mol(–1273.5 kJ/mol) + 3 mol(–285.83 kJ/mol)] = –57.6 kJ.
12.33 Calculate the energy released during the dissolving process from the temperature increase
and the total heat capacity of the calorimeter. Then convert to molar energy using molar
mass.
Assume that the heat capacity of the calorimeter is the heat capacity of its water contents:
 1 mol  75.291 J 
Ccal = 110.0 g
18.02 g 
 1 mol K  = 459.6 J/K = 459.6 J/°C;



q = Ccal∆T = ( 459.6 J/°C)(29.7°C –22.0°C) = 3.54 x 103 J;
remember that the heat absorbed by the calorimeter is the heat lost by the solution
process:
- q − 3.54 x 103 J
∆H =
=
= –745 J/g
m
4.75 g
− 745 J 111.0 g 
∆H (J/mol) = [∆H (J/g)][MM (g/mol)] = 
 1g 
 1 mol = –8.3 x 104 J/mol.



12.34 Calculate the energy absorbed during the dissolving process from the temperature
decrease and the total heat capacity of the calorimeter. Then convert to molar energy
using MM.
Assume that the heat capacity of the calorimeter is the heat capacity of its water contents:
393
Chapter 12
 1 mol 75.291 J 
Ccal = 50.0 g
 1 mol K = 208.9 J/K;
18.02 g 



q = Ccal∆T = (208.9 J/K)(296.36 K – 298.00 K) = –342.6 J;
- q − (− 342.6 J)
=
= 342.6 J/g
∆H =
m
1.00 g
342.6 J 122.55 g 
∆H (J/mol) = [∆H (J/g)][MM (g/mol)] = 
 1 mol  = 4.20 x 104 J/mol.
 1g 



12.35 (a) The positive value for ∆H signals that reaction to form solid MgSO4 is endothermic.
Dissolving, the reverse reaction, is exothermic, releasing energy, which is absorbed by the
water.
(b) Use the molar enthalpy and number of moles dissolving to calculate the energy released:
 1 mol 
m
-2
= 2.55 g 
n=
120.367 g 
= 2.119 x 10 mol MgSO4;
MM


- 91.3 kJ 
qwater = –qsalt = –n ∆H = –(2.119 x 10-2 mol) 
 = 1.93 kJ.
 1 mol 
(c) Use Equation 12-1, q = nC∆T, to determine the temperature change:
∆T = q ;
nC
1.00 g  1 mol 
m

n=
= (5.00 x 102 mL) 


 = 27.747 mol;
MM
 1 mL 18.02 g 
∆T =
1.93 x 103 J
= 0.924 K
(27.747 mol)(75.291 J mol -1 K -1 )
12.36 (a) The negative value for ∆H signals that reaction to form solid NH4NO3 is exothermic.
Dissolving, the reverse reaction, is endothermic, absorbing energy from the water.
(b) Use the molar enthalpy and number of moles dissolving to calculate the energy released:
 1 mol 
m
n=
= 25.0 g 
80.04 g 
 = 0.3123 mol;
MM


 21.1 kJ 
qwater = –qsalt = –n ∆H = – 0.3123 mol 
 = –6.59 kJ.
 1 mol 
(c) Use Equation 12-1, q = nC∆T, to determine the temperature change:
∆T = q ;
nC
1.00 g  1 mol 
m

n=
= 2.50 x 102 mL 


 = 13.87 mol;
MM
 1 mL 18.02 g 
∆T =
− 6.59 x 103 J
= − 6.31 K
(13.87 mol)(75.291 J mol -1 K -1 )
394
Chapter 12
12.37 The formation of a solid from ions in solution involves two opposing energy effects.
Energy is released because of the coulombic attraction between cations and anions, but
energy is absorbed to overcome the ion-dipole attractions between ions and water
molecules. Solid formation is endothermic when the sum of all ion-dipole attractions in
solution is greater than the ion-ion interactions in the solid.
12.38 The formation of a solid from ions in solution involves two opposing energy effects.
Energy is released because of the coulombic attraction between cations and anions, but
energy is absorbed to overcome the ion-dipole attractions between ions and water
molecules. Solid formation is exothermic when the ion-ion interactions in the solid are
greater than the sum of all ion-dipole attractions in solution.
12.39 The heats of phase changes indicate the strengths of intermolecular forces; the larger the
intermolecular forces, the larger the heat of the phase change.
(a) Methane has a lower heat of vaporization than ethane because, being a smaller
molecule, it has smaller dispersion forces.
(b) Ethanol has a significantly higher heat of vaporization than diethyl ether because of
strong hydrogen bonding.
(c) Argon (18 electrons) has a higher heat of fusion than methane (10 electrons) because
it has a higher polarizability due to its larger number of electrons.
12.40 The heats of phase changes indicate the strengths of intermolecular forces; the larger the
intermolecular forces, the larger the heat of the phase change.
(a) Water has a significantly higher heat of vaporization than methane because of strong
hydrogen bonding.
(b) Benzene has a higher heat of fusion than ethane because, being a larger molecule, it
has larger dispersion forces.
(c) Oxygen (12 electrons, two atoms rather than one) has a higher heat of vaporization
than argon (18 electrons, a single atom) because it has a higher polarizability due to its
larger size (two atoms rather than one).
12.41 Because E is a state function, the energy released (∆E) when one gram of gasoline burns
is the same regardless of the conditions. Work (w) is done when an automobile
accelerates but no work is done when an automobile idles. Thus,
qidle = ∆E and qaccel = ∆E – waccel, so more heat must be removed under idling
conditions (this is part of the reason why automobiles tend to overheat in traffic jams).
12.42 When a liquid droplet vaporizes in a vacuum, the opposing pressure is zero, so no work
must be done against an external pressure and ∆E = q. Thus, the heat absorbed under
these conditions is ∆E for the phase change.
12.43 To work a problem involving heat transfers, it is useful to set up a block diagram
illustrating the process. In this problem, a copper block transfers energy to ice:
395
Chapter 12
Thus, qice = –qCu,
qCu, = nCu C∆T, and qice = nice∆Hfus.
Substituting gives:
nice∆Hf = – nCu C∆T
Here are the data needed for the calculation:
 1 mol 
m
nCu =
= 12.7 g 
 = 0.1999 mol;
63.546 g 
MM


C = 24.435 J/mol K,
∆T = (0.0 °C – 200.0 °C) = –200 °C = –200 K;
∆Hf = 6.01 kJ/mol = 6.01 x 103 J/mol;
Substitute and solve for the amount of ice that melts:
− nCu C∆T
(0.1999 mol)(24.435 J mol -1 K -1 )(-200K)
nice =
=−
= 0.1625 mol
∆H f
6.01 x 103 J mol -1
Finally, convert to mass:
18.02 g 
mice = n MM = 0.1625 mol 
= 2.93 g of ice melts.
 1 mol 
12.44 The heat that is lost by the coin will be absorbed by the ice.
Thus, qice = –qAu,
qAu, = nAu C∆T, and qice = nice∆Hfus.
Substituting gives:
nice∆Hf = – nAu C∆T
Here are the data needed for the calculation:
 1 mol 
nAu = 7.65 g 
196.97 g 
= 0.0388 mol,


∆T = (0.0 + 273)K - (100.0 + 273)K = –100 K,
CAu = 25.4 J/mol K = 0.0254 kJ/mol K
Finally, convert to mass:
 25.4 J 
qAu = 0.0388 mol 
(–100 K) = –98.6 J
1 mol K 
–qAu = qice,
 1 mol 
nice = 0.0986 kJ
 = 0.0164 mol;
6.01 kJ 
Now use mass-mole calculations to determine the mass of ice that melts:
396
Chapter 12
18.02 g 
mice = 0.0164 mol 
= 0.296 g ice melted.
 1 mol 
12.45 (a) In a combustion reaction, a substance reacts with molecular oxygen to form CO2 and
H2O:
C6H12O6 + O2 → CO2 + H2O
Balance the equation using standard procedures. Give CO2 a coefficient of 6 and H2O a
coefficient of 6 to balance C and H;
C6H12O6 + O2 → 6 CO2 + 6 H2O
6C + 12H + 8 O → 6C + 12H + 18O
then give O2 a coefficient of 6 to balance O:
C6H12O6 + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
(b) To determine the molar heat of combustion, multiply the heat released (negative,
indicating an exothermic reaction) in burning one gram by the molar mass:
- 15.7 kJ 180.15g 
3
∆Hmolar = 
 1g 
 1 mol  = –2.83 x 10 kJ/mol;




(c) Use the molar heat of combustion along with Equation 12-10 and data from Appendix
D to determine the heat of formation of glucose:
o
= Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
∆H reaction
–2.83 x 103 kJ/mol = [6 mol(–393.5 kJ/mol) + 6 mol(–285.83 kJ/mol)]
– [6 mol(0 kJ/mol) + ∆H o (glucose)]
f
∆Hof (glucose) = (2.83 x 103 – 1.715 x 103 – 2.361 x 103) kJ/mol = –1.25 x 103 kJ/mol
12.46 (a) In this reaction, urea reacts with molecular oxygen to form CO2, H2O, and N2:
(NH2)2CO + O2 → CO2 + H2O + N2
Give CO2 a coefficient of 1, H2O a coefficient of 2, and N2 a coefficient of 1 to balance
C, H, and O.
(NH2)2CO + O2 → CO2 + 2 H2O + N2
2N + 4H + C + 3O → 2N + 4H + C + 4O
Then give O2 a coefficient of 3/2 to balance O (retain the fractional coefficient because a
combustion reaction refers to one mole of substance burning):
(NH2)2CO(s) + 3/2 O2(g) → CO2(g) + 2 H2O(l) + N2(g)
397
Chapter 12
(b) A heat of combustion refers to one mole of substance burning. When 1 mol of urea
burns, 2 mol of H2O form.
Thus, the heat generated per mole of H2O formed is:
 − 632.2 kJ  1 mol urea 


qH2O = 
 = –316.1 kJ.

1 mol urea  2 mol H 2 O 
(c) Use the molar heat of combustion along with Equation 12-10 and data from Appendix
D to determine the heat of formation of urea:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
–632.2 kJ/mol = [1 mol(–393.5 kJ/mol) + 2 mol(–285.83 kJ/mol) + 1 mol(0 kJ/mol)]
– [1.5 mol(0 kJ/mol) + ∆Hof (urea)]
o
∆H f (urea) = (632.2 – 393.5 – 571.66) kJ/mol = –333.0 kJ/mol
12.47 Living organisms require both matter and energy to carry out their life processes. An
isolated system receives neither matter nor energy from its surroundings, so a living
organism that is isolated will quickly die.
12.48 When a moving automobile brakes and slows, its kinetic energy of motion is reduced.
This energy is transferred to the brakes and tires as heat, causing the brakes and tires to
heat up. The net transformation is conversion of kinetic energy of the automobile into
thermal energy of the brakes and tires.
12.49 A molar heat capacity can be calculated from a temperature change using q = n C∆T:
∆T = 299.6 K – 280.0 K = 19.6 K;
 1 mol 
n = 52.5 g 
 207.2 g 
 = 0.2534 mol;


q
(100.0 J)
C=
=
= 20.1 J/mol K.
n∆T (0.2534 mol)(19.6 K)
12.50 To determine the molar heat capacity, Cm, of the rhodium metal we will need to use the
equation:
q
q = n Cm ∆T, or Cm =
.
n∆T
∆T = Tf – Ti = 273 K – 373 K = –100 K
In order to solve the equation for Cm we need to find q. To find q we must notice that the
heat lost by the rhodium metal is equal to the heat gained by the ice, or –qRh = qice. The q
gained by the ice is only in the form of phase change energy so we must use the formula:
 1 mol 6.01 kJ 
qp = n∆Hfus = 0.316 g 
18.02 g 
 1 mol  = 0.105 kJ



Using this energy and the above eqn we calculate molar heat capacity of Rh:
398
Chapter 12
 1 mol 
nRh = 4.35 g 
= 0.423 mol
102.910 g 


− 0.105 kJ
q
=
= 0.00248 kJ mol-1 K-1
Cm =
n∆T (0.423 mol)( − 100 K)
12.51 In this process, neither P nor V is constant, so we cannot calculate q directly. Instead, we
can calculate w and ∆E. An ideal gas has no intermolecular forces, so its energy does not
change when it expands or contracts. Thus, for ∆T = 0, ∆E = 0.
Calculate w using w = –Pext ∆V:
w = –(4.00 atm)(20.0 L – 30.0 L) = 40.0 L atm;
Convert to joules, recalling that 1 L atm = 101.325 J:
101.325 J 
w = (40.0 L atm) 
= 4.05 x 103 J
1
L
atm


q = ∆E – w = 0 – 4.05 x 103 J = – 4.05 x 103 J.
12.52 In this process, the pressure can be assumed to be atmospheric pressure, so this is a
constant-pressure cooling, for which ∆H = q = nCp∆T. Once ∆H has been calculated,
find ∆E using ∆H = ∆E + ∆(PV).
 1 mol 
m
-2
= 0.197 g 
nHe =
 4.003 g 
= 4.92 x 10 mol;
MM


Cp = 20.8 J/mol K;
∆T = (255 K – 325 K) = –70 K
 20.8 J 
∆H = (4.92 x 10-2 mol) 
(–70 K) = –71.7 J;
1 mol K 
 8.314 J 
∆(PV) = ∆(nRT) = nR∆T = (4.92 x 10-2 mol) 
(–70 K) = –28.6 J
1 mol K 
∆E = ∆H – ∆(PV) = –71.7 J – (–28.6 J) = –43.1 J.
12.53 Standard enthalpy changes are calculated from Equation 12-10 using standard enthalpies
of formation, which can be found in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
(a) ∆Horeaction = 2 mol(–395.7 kJ/mol)
– [2 mol(–296.8 kJ/mol) + 1 mol(0 kJ/mol)] = –197.8 kJ;
o
(b) ∆H reaction = 1 mol(11.1 kJ/mol) – 2 mol(33.2 kJ/mol) = –55.3 kJ;
(c) ∆Horeaction = 1 mol(–1675.7 kJ/mol) + 2 mol(0 kJ/mol)
– [1 mol(–824.2 kJ/mol) + 2 mol(0 kJ/mol)] = –851.5 kJ.
12.54 Standard enthalpy changes are calculated from Equation 12-10 using standard enthalpies
of formation, which can be found in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
(a) ∆Horeaction =2 mol(135.1 kJ/mol) + 6 mol(–241.83 kJ/mol)
399
Chapter 12
–[2 mol(–45.9 kJ/mol) +3 mol(0.0 kJ/mol) +2 mol(–74.6 kJ/mol)]
= –939.8 kJ;
(b) ∆Horeaction = [2 mol(–241.83 kJ/mol) + 4 mol(–393.5 kJ/mol)]
–[5 mol(0.0 kJ/mol) + 2 mol(227.4 kJ/mol)] = –2512.5 kJ;
(c) ∆Horeaction = [1mol (–166.2 kJ/mol) + 1mol(0 kJ/mol)]
– [1 mol(52.4 kJ/mol) + 1 mol(142.7 kJ/mol)] = –361.3 kJ.
12.55 An enthalpy of formation can be calculated from the heat of a reaction provided all other
enthalpies of formation are known. For this reaction, formation enthalpies of three of the
four reagents appear in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
–3677 kJ = 1 mol(–2984.0 kJ/mol) + 3 mol(–296.8 kJ/mol)
– [ ∆H o (P4S3) + 8 mol(0 kJ/mol)]
f
∆H fo (P4S3)
= 3677 kJ –2984.0 kJ – 890.4 kJ = –197 kJ
12.56 Standard enthalpy changes are calculated from Equation 12-10 using standard enthalpies
of formation, which can be found in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
First balance the overall reaction. Give PH3 a coefficient of 4 to balance P and H2O a
coefficient of 6 to balance H. This gives 16 O on the right, requiring a coefficient of 8
for O2:
4 PH3(g) + 8 O2(g) → P4O10(s) + 6 H2O(g)
o
∆H reaction = 1 mol(–2984.0 kJ/mol) + 6 mol(–241.83 kJ/mol)
– [4 mol(5.4 kJ/mol) + 8 mol(0 kJ/mol)] = –4456.6 kJ
12.57 To work a problem involving heat transfers, it is useful to set up a block diagram
illustrating the process. In this problem, the unknown metal transfers energy to water:
Thus, qmetal = –qwater = -(nC∆T)water
 1 mol 
m
nwater =
= 80.0 g 
18.02 g 
= 4.44 mol water;
MM


C = 75.291 J/mol °C,
and ∆T = (28.4 – 24.8) = 3.6 °C;
 75.291 J 
qmetal = –qwater = –(4.44 mol) 
 (3.6 °C) = –1203 J.
o
1 mol C 
400
Chapter 12
Without knowing the identity of the metal, we cannot determine nmetal, but heat capacity
values are provided in units of J/g K, so we can use q = mC∆T:
∆T = 28.4 – 100.0 = –71.6 °C = –71.6 K;
q
(− 1203 J)
C=
=
= 0.382 J / g K
m∆T (44.0 g)(-71.6 K)
This matches the heat capacity of Cu (0.385 J/gK), so the metal is copper.
12.58 To work a problem involving heat transfers, it is useful to set up a block diagram
illustrating the process. In this problem, the coin transfers energy to water:
Thus, qcoin = –qwater = (nC∆T)water
 1 mol 
m
nwater =
= 21.5 g 
18.02 g 
= 1.193 mol water;
MM


C = 75.291 J/mol °C,
and ∆T = (21.5 – 15.5) = 6.0 °C;
 75.291 J 
qcoin= -qwater = –(1.193 mol) 
 (6.0 °C) = –539 J.
o
1 mol C 
Now use q = mC∆T to calculate C in J/g K:
∆T = 21.5 – 100.0 = –78.5 °C = –78.5 K;
q
(− 539 J)
C=
=
= 0.44 J/g K
m∆T (15.5 g)(-78.5 K)
Calculate C for Ni and Ag:
 25.351 J  1 mol 

Ag: C = 


= 0.235 J/g K
 1 mol K 107.9 g 
 26.07 J  1 mol 

Ni: C = 


 = 0.444 J/g K.
1 mol K 58.69 g 
C matches that of nickel, so the coin is counterfeit.
12.59 The process shown is condensation of a gas to form a solid, the reverse of sublimation.
Sublimation always is endothermic, as energy must be provided to overcome
intermolecular forces of attraction; thus the depicted process is exothermic.
(a) ∆Hsys is negative (exothermic process);
(b) ∆Esurr is positive (surroundings must absorb the energy released in the process);
(c) ∆Euniverse = 0 (total energy always is conserved).
401
Chapter 12
12.60 The process shown is a constant-volume decomposition of diatomic molecules to atoms.
Energy must be provided to the system to break the chemical bonds holding the
molecules together, so the depicted process is endothermic.
(a) wsys = 0 because there is no volume change;
(b) qsys is positive because heat must be provided to break the bonds;
(c) ∆Esurr is negative because the surroundings provide the energy required to break the
bonds.
12.61 The enthalpy of a reaction can be calculated from Equation 12-10 and standard heats of
formation (see Appendix D):
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
For this "reaction," the calculation is simple:
∆Horeaction = 1 mol(–241.83 kJ/mol) – 1 mol(–285.83 kJ/mol) = 44.00 kJ
Compare this value with ∆Hvap = 40.79 kJ/mol. The values differ because ∆Hvap refers
o
refers
to vaporization of liquid water at the normal boiling point, 373 K, while ∆H reaction
to vaporization of liquid water under standard thermodynamic conditions, 298 K.
12.62 An enthalpy of formation can be calculated from the heat of a reaction provided all other
enthalpies of formation are known. For this reaction, formation enthalpies of four of the
five reagents appear in Appendix D of your textbook:
∆Horeaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
–1196 kJ = [1 mol(0 kJ/mol) + 6 mol(–273.3 kJ/mol) + 1 mol(0 kJ/mol)]
– [2 mol(–45.9 kJ/mol) + 2 ∆H fo (ClF3)]
∆Hof (ClF3) = ½ (1196 –1639.8 + 91.8)kJ/mol = –176.0 kJ/mol.
12.63 The enthalpy of a reaction can be calculated from Equation 12-10 and standard heats of
formation (see Appendix D):
∆Hreaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
(a) The question asks for ∆H when 1 mol of hydrazine burns, so the appropriate reaction
is:
N2H4(l) + 1/2 N2O4(g) → 3/2 N2(g) + 2 H2O(g)
∆Hreaction = [1.5 mol(0 kJ/mol) + 2 mol(–241.83 kJ/mol)]
– [1 mol(50.6 kJ/mol) + 0.5 mol(11.1 kJ/mol)] = –539.8 kJ/mol
(b) When hydrazine burns in oxygen, the reaction is:
N2H4(l) + O2(g) → N2(g) + 2 H2O(g)
∆Hreaction = [1 mol(0 kJ/mol) + 2 mol(–241.83 kJ/mol)]
– [1 mol(50.6 kJ/mol) + 1 mol(0 kJ/mol)] = –534.3 kJ/mol
402
Chapter 12
The reaction with dinitrogen tetroxide is slightly more exothermic because that
compound is slightly less stable than the elements from which it forms. Dinitrogen
tetroxide is used in preference to molecular oxygen because the reaction occurs on
contact, while oxygen would require an ignition device.
12.64 Convert heats of combustion to a per-gram basis by dividing by the molar mass of the
substance:
- 1694 kJ  1 mol 

(a) ∆Hper gram = 

 = – 28.19 kJ/g;

 1 mol 60.10 g 
- 726 kJ  1 mol 

(b) ∆Hper gram = 

 = – 22.7 kJ/g;

 1 mol 32.04 g 
- 5590 kJ  1 mol 

(c) ∆Hper gram = 

= – 48.9 kJ/g;

 1 mol 114.22 g 
Octane has the highest energy content and methanol the lowest per gram of fuel (octane's
high energy content is a main reason why it is widely used as an automobile fuel).
12.65 The difference between ∆Hreaction and ∆Ereaction can be estimated using Equation 12-9:
∆Hreaction ≅ ∆Ereaction + ∆(nRT)gases so ∆Hreaction – ∆Ereaction ≅ ∆(nRT)(gases)
(a) ∆ngases = 0, so ∆Hreaction – ∆Ereaction ≅ 0;
(b) ∆ngases = 1 mol + 1 mol – 0 mol = 2 mol gases;
 8.314 J 
∆Hreaction – ∆Ereaction ≅ (2 mol) 
(298 K) = 4.96 x 103 J;
1
mol
K


(c) The reaction is C4H9OH(l) + 6 O2(g) → 4 CO2(g) + 5 H2O(l);
∆ngases = 4 mol – 6 mol = –2 mol;
 8.314 J 
∆Hreaction – ∆Ereaction ≅ –(2 mol) 
(298 K) = –4.96 x 103 J.
1
mol
K


12.66 The difference between ∆Hreaction and ∆Ereaction can be estimated using Equation 12-9:
∆Hreaction ≅ ∆Ereaction + ∆(nRT)gases so ∆Ereaction ≅ ∆Hreaction – ∆(nRT)(gases)
For a vaporization process, T = Tvap and ∆n = 1;
10-3 kJ 
 8.314 J 
Argon: ∆Ereaction ≅ (6.3 kJ/mol) – 1 
(87 K) 
 1J 
= 5.6 kJ/mol;
1 mol K 


10-3 kJ 
 8.314 J 
Ethane: ∆Ereaction ≅ (15.5 kJ/mol) – 1 
(184 K) 
 1J 
= 14.0 kJ/mol;
1 mol K 


403
Chapter 12
10-3 kJ 
 8.314 J 
Mercury: ∆Ereaction ≅ (59.0 kJ/mol) – 1 
(630 K) 
= 53.8 kJ/mol;
 1J 
1 mol K 


∆E reaction − ∆H reaction
;
∆E reaction
(5.6 kJ / mol - 6.3 kJ / mol)
Argon: % difference = (100%)
= –13%;
(5.6 kJ / mol)
(14.0 kJ/mol - 15.5 kJ/mol)
= –11%;
Ethane: % difference = (100%)
(14.0 kJ/mol)
(53.8 kJ / mol - 59.0 kJ / mol)
Mercury: % difference = (100%)
= –9.7%;
(53.8 kJ / mol)
Argon has the highest percentage difference.
% difference = (100%)
12.67 Use the definition of each type of process to identify the reaction associated with each
process:
(a) I2(s) → I2(g);
(b) 1/2 I2(s) → I(g);
(c) 2 C(graphite) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g);
(d) Na2SO4(s) → 2 Na+(aq) + SO42-(aq).
12.68 Use the definition of each type of process to identify the reaction associated with each
process:
(a) Br2(l) → Br2(g);
(b) 1/2 Br2(l) → Br(g);
(c) C(graphite) + 2H2(g) + ½O2(g) → CH3OH(g);
(d) MgCl2(s) → Mg2+(aq) + 2Cl-(aq)
12.69 (a) This is a cooling process, for which q = nC∆T:
 1 mol 
m
n=
= (2.50 g) 
18.02 g 
= 0.1387 mol;
MM


∆T = (37.5 –100.0) °C = –62.5 °C = –62.5 K;
 75.291 J 
q = (0.1387 mol) 
 ( –62.5 K) = –653 J .
o
1 mol C 
(b) This is a phase change, for which q = n ∆Hphase, followed by the same cooling
process as in part (a):
3
- 40.79 kJ 10 J 
3
q = (0.1387 mol) 

 1 kJ 
– 653 J = –5658 J – 653 J = –6.31 x 10 J.
1
mol



Steam at 100.0 °C releases nearly ten times as much heat as boiling water.
404
Chapter 12
12.70 Balance the reactions by inspection, starting with N. Then use Equation 12-10 to
calculate heats of reaction:
∆Hreaction = Σ coeffp ∆Hof (products) – Σ coeffr ∆Hof (reactants)
(a) N2 + O2 → 2 NO;
∆Hreaction = 2 mol(91.3 kJ/mol) – [1 mol(0 kJ/mol) + 1 mol(0 kJ/mol)] = 182.6 kJ;
182.6 kJ 
∆Hmol N = 
= 91.3 kJ/mol;
 2 mol N 
(b) 2 N2O + O2 → 4 NO;
∆Hreaction = 4 mol(91.3 kJ/mol) – [2 mol(81.6 kJ/mol) + 1 mol(0 kJ/mol)] = 202.0 kJ;
 202.0 kJ 
∆Hmol N = 
= 50.5 kJ/mol;
 4 mol N 
(c) 2 NO + O2 → 2 NO2;
∆Hreaction = 2 mol(33.2 kJ/mol) – [2 mol(91.3 kJ/mol) + 1(0)] = –116.2 kJ;
- 116.2 kJ 
∆Hmol N = 
 = –58.1 kJ/mol;
 2 mol N 
(d) 4 NO2 + O2 → 2 N2O5;
∆Hreaction = 2 mol(13.3 kJ/mol) – [4 mol(33.2 kJ/mol) + 1 mol(0 kJ/mol)] = –106.2 kJ;
- 106.2 kJ 
∆Hmol N = 
= –26.6 kJ/mol
 4 mol N 
12.71 Compression work can be calculated using Equation 12-3, wsys = –Pext∆Vsys . The
external pressure is the pressure exerted by the compressor, 15.0 atm. The air is initially
at 1.00 atm and occupies a volume that can be calculated using PfVf = Pi Vi :
PV
(15.0 atm)(30.0 L)
Vi = f f =
= 450 L
Pi
(1.00 atm)
∆V = Vf – Vi = 30.0 L – 450 L = –420 L;
wsys = –Pext∆Vsys = –(15.0 atm)(–420 L) = 6.30 x 103 L atm.
Convert to joules:
101.325 J 
wsys = (6.30 x 103 L atm) 
 = 6.38 x 105 kg m2/s2 = 638 kJ.
 1 L atm 
12.72 Data in Problem 12.45 indicate that glucose combustion releases 15.7 kJ/g of energy.
First determine how much glucose would be required at 100% efficiency, then correct for
inefficiency and sugar content:
1 g glucose 100%  100 g cereal 

Mass required = 220 kJ



 = 133 g cereal
 15.7 kJ  30% 35 g glucose 
405
Chapter 12
12.73 (a)The energy required in a heating process is calculated from Equation 12-1:
nC
q=
:
T
3
10 2 cm  1.00 g  1 mol 
m
3 
6



nwater =
= (155 m ) 
3 
= 8.60 x 10 mol;

1
m
1
cm
18.02
g

MM



C = 75.291 J/mol K,
∆T = (30 – 20) = 10°C = 10 K;
75.291 J 
9
q = (8.60 x 106 mol) 
(10 K) = 6.48 x 10 J
 1 mol K 
100% 
9
(b) If heating is 80% efficient, the heat required is (6.48 x 109 J) 
= 8.1 x 10 J;
 80% 
Given that the heat of combustion of methane is –803 kJ/mol, the amount of methane
10-3 kJ  1 mol 
4
required is (8.1 x 109 J) 
 1J 
803 kJ = 1.0 x 10 mol;



16.04 g 
5
Multiply by MM to convert to grams: (1.0 x 104 mol) 
 = 1.6 x 10 g methane
 1 mol 
12.74 First calculate the energy required to heat the water; then calculate the energy per mole of
photons; finally determine how many photons are needed:
q = n C∆T;
103 mL 1.00 g  1 mol 
m


= 40.0 L 
nwater =
= 2.22 x 103 mol;






MM
 1 L  1 mL 18.02 g 
∆T = 50 °C – 25 °C = 25 °C = 25 K;
75.291 J 
q = (2.22 x 103 mol) 
(25 K) = 4.18 x 106 J;
1
mol
K


Now determine the energy per mole of photons and compare.
N A hc (6.022 x 10 23 /mol)(6.626 x 10-34 J s)(2.998 x 108 m/s)
Ephotons =
=
λ
5.15 x 10 -7 m
Ephotons = 2.32 x 105 J/mol;
1 mol 100% 
Photons required = 4.18 x 106 J 

 = 23 mol
5 
 2.32 x 10 J  80% 
12.75 To work a problem involving heat transfers, it is useful to set up a block diagram
illustrating the process. In this problem, a silver spoon absorbs energy from coffee
(water):
406
Chapter 12
Thus, qwater = –qAg,
qAg = (nCAg∆T)Ag, and qwater = (nCH2O∆T)water
For Ag,
 1 mol 
m
n=
= 99 g 
= 0.918 mol Ag;
107.9 g 
MM


CAg = 25.351 J/mol K,
and ∆T = (x – 280) K;
For water,
m
1.00 g  1 mol 

= 205 mL 
n=


 = 11.38 mol H2O;
MM
 1 mL 18.02 g 
CH2O = 75.291 J/mol K,
and ∆T = (x –350) K;
Substitute and solve for x:
 25.351 J 
75.291 J 
(0.918 mol) 
(x –280) K = – (11.38 mol) 
(x –350) K;
 1 mol K 
 1 mol K 
23.27 x – 6516 = –856.8 x + 299880,
880.1 x = 306390
x = 348 K.
The final temperature of the coffee is 348 K.
To determine whether an Al spoon would be more or less effective, compare nC for the
two spoons.
 1 mol  24.35 J 
Al: = 99 g 
 26.982 g 
1 mol K  = 89 J/K;



 1 mol  25.351 J 
Ag: = 99 g 
107.9 g 
 1 mol K = 23 J/K;



An Al spoon requires more energy per degree temperature increase, so an Al spoon
would be more effective in cooling the coffee.
12.76 To determine the mass of methane required, first determine the heat required to convert
the water into steam. The conversion of water at 25.0 °C into steam at 100.0 °C takes
place in two steps, heating to 100.0 °C and vaporization at that temperature. The total
heat required is the sum of the heats for the two steps: q = nC ∆T + n ∆Hvap:
103 g  1 mol 
m

nwater =
= 2.50 kg 

= 138.7 mol;
 1 kg 


18.02 g 
MM
407
Chapter 12
10-3 kJ 
75.291 J 
qwater = (138.7 mol)[ 
(75.0 K) 
 + 40.79 kJ/mol] = 6.44 x 103 kJ;
 1J 


 1 mol K 
qmethane = - qwater = -6.44 x 103 kJ
The heat of combustion of methane could be calculated from Equation 12-10 using data
from Appendix D. However, the value is given in Problem 12.73: ∆Hcomb = –803 kJ/mol.
 1 mol 16.043 g 
3

 = 129 g.
Mass required = - 6.44 x 10 kJ
- 803 kJ  1 mol 
12.77 The energy required in a heating process is calculated from Equation 12-1: q = nC∆T.
To determine q, we must first compute the amounts of N2 and O2 in the room, using the
PV
ideal gas equation: n =
(assume no air escapes the room as it is heated):
RT
3
10 2 cm   1 L 


4
V = (3.0 m)(5.0 m)(4.0 m) 
3
3  = 6.0 x 10 L;

1
m
10
cm




pN2 = (1 atm)(0.78) = 0.78 atm;
pO2 = (1 atm)(0.22) = 0.22 atm;
T = 15 + 273 = 288 K;
(0.78 atm)(6.0 x 10 4 L)
nN 2 =
= 1.98 x 103 mol
L atm
(0.08206 mol K )(288 K)
(0.22 atm)(6.0 x 10 4 L)
nO 2 =
= 0.559 x 103 mol
L atm
(0.08206 mol K )(288 K)
∆T = 25 – 15 = 10 °C = 10 K.
 29.125 J 
(10 K) = 5.77 x 105 J
qN2 = (1.98 x 103 mol) 
1
mol
K


 29.355 J 
qO2 = (0.559 x 103 mol) 
(10 K) = 1.64 x 105 J
 1 mol K 
qtot = qN2 + qO2 = 5.77 x 105 J+1.64 x 105 J = 7.4 x 102 kJ. (Value only has two sig. figs.
because the gas pressures are known to only two sig. figs.)
To determine the amount of methane required, we need the heat of combustion of
methane, which can be calculated from standard heats of formation using Equation 12-10
using data from Appendix D. However, the value is given in Problem 12.73:
∆Hcomb = –803 kJ/mol.
Divide the energy required by the heat of combustion to determine amount of methane
 1 mol 
required: (7.4 x 102 kJ) 
 = 0.92 mol.
803 kJ 
408
Chapter 12
16.04 g 
Multiply by molar mass to convert to grams: 0.92 mol 
= 15 g CH4 required.
 1 mol 
12.78 The expression that always is correct for q is (e) q = ∆E – w, which is derived from the
first law of thermodynamics. For the others:
(a) ∆E ≠ q when volume changes;
(b) ∆H ≠ q when pressure changes;
(c) qv ≠ q when volume changes; and
(d) qp ≠ q when pressure changes.
12.79 The process shows an expansion at constant temperature. As a gas expands, it does work
on its surroundings, but if temperature remains constant, the energy of the gas does not
change:
(a) wsys is negative (system does work on surroundings);
(b) ∆Esurr = 0 because ∆Esys = 0 and total energy is conserved;
(c) qsys is positive because wsys is negative and ∆Esys = 0.
12.80 (a) Work done on the system compresses it, so the new figure should show a smaller
volume.
(b) An exothermic reaction increases the temperature of the system, leading to expansion
and a larger volume:
12.81 First calculate the heat required to warm the bear. Then calculate the heat of combustion
of arachadonic acid. Finally, determine the amount of arachidonic acid required to do the
job:
103 g  4.184 J 

q = mC∆T = 500 kg 
o

(25 °C – 5 °C) = 4.18 x 107 J;
 1 kg 

 1 g C 

∆Hcomb = [20 mol(–393.5 kJ/mol) + 16 mol(–285.8 kJ/mol)]
– [1 mol(–636 kJ/mol) + 27 mol(0 kJ/mol)]
4
∆Hcomb = –1.18 x 10 kJ
1 kJ  1 mol
304.45 g 
3
Mass required = 4.18 x 107 J 
 3 

 = 1.1 x 10 g .
4
10
J
1.18
x
10
kJ
1
mol




12.82 First compute the quantity of gasoline consumed in traveling 1 km, then determine the
energy content of that quantity of gasoline:
409
Chapter 12
3
 1 L 10 mL  0.68 g 


1.0
km
 = 113 g


Mass of gasoline =


6.0 km  1 L  1 mL 
 48 kJ 
Energy consumed = 113 g 
= 5.4 x 103 kJ.
 1g 


12.83 To work a problem involving heat transfers, it is useful to set up a block diagram
illustrating the process. In this problem, a copper block transfers energy to water:
Thus, qwater = –qCu,
qCu = (nC∆T)Cu, and qwater = (nC∆T)water
 1 mol 
m
= 9.50 g 
63.546 g 
= 0.1495 mol Cu;


MM
C = 24.435 J/mol °C,
and ∆T = (x – 200.0) °C;
1.00 g  1 mol 
m


For water, n =
= 200 mL 

 = 11.10 mol water;
 1 mL 18.02 g 
MM
C = 75.291 J/mol °C,
and ∆T = (x – 5.00) °C;
For Cu, n =
Substitute and solve for x:
 24.435 J 
 75.291 J 
(0.1495 mol) 
 (x – 200.0) °C = – (11.10 mol) 
 (x – 5.00) °C;
o
o
1 mol C 
1 mol C 
3.653 x – 730.6 = –835.7 x + 4179,
839.4 x = 4910,
x = 5.85 °C
12.84 This problem asks us to construct a graph that summarizes the phase changes of water.
The first step is to determine the number of moles of water in 75.0 mL:
1.00 g  1 mol 


75.0 mL 

 = 4.16 mol
 1 mL 18.02 g 
Now determine the energy losses for the processes described:
0.0753 kJ 
Energy lost during liquid cooling = 4.16 mol 
 (35K) = 11 kJ
 1 mol K 
410
Chapter 12
6.01 kJ 
Energy lost during freezing = 4.16 mol 
= 25.0 kJ
 1 mol 
0.0377 kJ 
(25 K) = 3.9 kJ
Energy lost during ice cooling = 4.16 mol 
 1 mol K 
Using these values construct the graph:
12.85 To determine the thermodynamic values first calculate the number of moles of the
ammonia and the heat of vaporization:
0.81 g  1 mol 


moles NH3 = 275 mL 

= 13.1 mol
 1 mL 17.04 g 
 23.2 kJ 
q = n ∆Hvap = 13.1 mol 
 = 304 kJ
 1 mol 
Since this system is at constant pressure, ∆H = q = 304 kJ
The only work on the system is expansion work. Thus, w = -P∆V = -nRTvap
8.314 kJ 
w = -13.1 mol 
 (240 K) = -2.61 x 104 J or -26.1 kJ
 1 mol K 
Finally ∆E = ∆H - P∆V =∆H - nRTvap = 304kJ – 26.1 kJ = 2.8 x 102 kJ. (density of NH3
is known to only two significant figures)
12.86 A system should be defined in such a way that energy transfers can be conveniently
specified.
(a) Water is the system, and you are part of the surroundings. The water absorbs heat as
it evaporates, and this heat must be provided by the surroundings (your body).
(b) If our focus is on the beaker of water, it is the convenient choice for the system. The
Bunsen burner and methane then are part of the surroundings. The beaker absorbs heat
from the burning methane. (If our focus were on how a Bunsen burner operates, we
would choose the methane as the system; then the system would release heat to the
surroundings.)
411
Chapter 12
(c) The acid and base solutions inside the thermos constitute a convenient system. This is
a (nearly) isolated system, with minimum heat flow. The energy released by the reacting
species remains in the system and causes the temperature rise of the mixed solutions.
12.87 The cooling process takes place at constant pressure, so ∆H can be calculated using
∆H = nC∆T. To determine ∆E, use H = E + PV, from which ∆E = ∆H – ∆(PV):
 1 mol 
m
n=
= 1.25 g 
 = 9.40 x 10-3 mol;
132.93 g 


MM
∆T = 20.0 K – 50.0 K = –30.0 K;
 80.7 J 
(–30.0 K) = –22.8 J;
∆Hcooling = (9.40 x 10-3 mol) 
1 mol K 
10-3 L 101.325 J 
∆(PV) = P∆V = (1.00 atm)(248 – 274 mL) 
 1 L atm  = – 2.63 J;
 1 mL 



∆Ecooling = –22.8 J –(–2.63 J) = –20.2 J.
12.88 The energy that is added to a gaseous monatomic substance goes entirely into
translational energy of the atoms. The energy that is added to a gaseous diatomic
substance goes partly into translational energy and partly into rotational energy of the
molecules. Thus, a given amount of added energy increases the total translational energy
(and the temperature) of a diatomic substance by a smaller amount, and the heat
capacities of diatomic gases are larger than those of monatomic substances.
12.89 The total heat released in the reaction can be calculated from the temperature rise:
q = nwater C∆T;
1.00 g  1 mol 


n = (200.0 mL) 

 = 11.099 mol;
 1 mL 18.02 g 
∆T = 31.8 – 25.0 = 6.8 °C = 6.8 K;
10-3 kJ 
75.291 J 
q = (11.099 mol) 
(6.8 K) 
 1J 
 = 5.7 kJ;


 1 mol K 
To determine the molar heat, divide the total heat by the number of moles reacting:
q
∆Hmolar =
;
nacid
10-3 L 1.00 mol 
nacid = MV = 100.0 mL 
 1 mL 
 1 L = 0.100 mol;



 5.7 kJ 
∆Hmolar = 
= 57 kJ/mol.
0.100 mol 
12.90 This problem asks for the energy released from burning 1 gallon of ethanol. Begin by
determining the balanced chemical equation using standard procedures:
The unbalanced equation is:
C2H5OH + O2 → CO2 + H2O
412
Chapter 12
Give CO2 a coefficient of 2 to balance C, and H2O a coefficient of 3 to balance H.
C2H5OH + O2 → 2 CO2 + 3 H2O
This leaves 7 O atoms on the product side and 1 O atom in C2H5OH. Give O2 a
coefficient of 3 to balance the O atoms and obtain the balanced equation:
C2H5OH + 3 O2 → 2 CO2 + 3 H2O
To determine the energy released by burning 1 gallon of ethanol we first need the
reaction enthalpy. The reaction enthalpy can be obtained using the average bond energies
in Table 9-2 or by using the heats of formation in Appendix D. We’ll show both.
Using bond energies:
Bonds Broken
Energy (kJ/mol)
C-C
345
5 C-H
5(415)
C-O
360
O-H
460
3 O=O
3(495)
Total
4725
Bonds Formed
4 C=O
6 H-O
Energy kJ/mol
4(-800)
6(-460)
-5960
∆Hreaction = 4725 kJ/mol – 5960 kJ/mol = –1235 kJ/mol
Using heats of formation:
∆Hreaction = [3 mol(-241.83 kJ/mol) + 2 mol(-393.5 kJ/mol)]
– [3 mol(0 kJ/mol) + 1 mol(-277.6 kJ/mol)] = –1235 kJ/mol
Note that both methods yield the same heat of reaction.
Now use stoichiometric calculations to determine the energy for 1 gallon of ethanol.
3.785 L 103 mL 0.787 g  1 mol 1.235 x 103 kJ 
4

1 gallon 
1 gallon 

 46.08 g 

 1L 
 1 mL 

 = 7.98 x 10 kJ
1
mol






413