CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
The Nature of Energy
15.
Ball A: PE = mgz = 2.00 kg ×
196 kg m 2
9.81 m
×
10.0
m
=
= 196 J
s2
s2
At point I: All this energy is transferred to ball B. All of B's energy is kinetic energy at this
point. Etotal = KE = 196 J. At point II, the sum of the total energy will equal 196
J.
At point II: PE = mgz = 4.00 kg ×
9.81 m
× 3.00 m = 118 J
s2
KE = Etotal  PE = 196 J − 118 J = 78 J
16.
Plot a represents an exothermic reaction. In an exothermic process, the bonds in the product
molecules are stronger (on average) than those in the reactant molecules. The net result is
that the quantity of energy Δ(PE) is transferred to the surroundings as heat when reactants are converted to products.
For an endothermic process (plot b), energy flows into the system as heat to increase the
potential energy of the system. In an endothermic process, the products have higher potential
energy (weaker bonds on average) than the reactants.
17.
Path-dependent functions for a trip from Chicago to Denver are those quantities that depend
on the route taken. One can fly directly from Chicago to Denver or one could fly from
Chicago to Atlanta to Los Angeles and then to Denver. Some path-dependent quantities are
miles traveled, fuel consumption of the airplane, time traveling, airplane snacks eaten, etc.
State functions are path independent; they only depend on the initial and final states. Some
state functions for an airplane trip from Chicago to Denver would be longitude change,
latitude change, elevation change, and overall time zone change.
18.
a. ΔE = q + w = 23 J + 100. J = 77 J
b. w = PΔV = 1.90 atm(2.80 L  8.30 L) = 10.5 L atm ×
ΔE = q + w = 350. J + 1060 = 1410 J
363
101.3 J
= 1060 J
L atm
364
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
c. w = PΔV = 1.00 atm(29.1 L  11.2 L) = 17.9 L atm ×
101.3 J
= 1810 J
L atm
ΔE = q + w = 1037 J  1810 J = 770 J
20.8 J
× 39.1 mol × (38.0  0.0)°C = 30,900 J
C mol
= 30.9 kJ
101.3 J
w = PΔV = 1.00 atm × (998 L  876 L) = 122 L atm ×
= 12,400 J = 12.4 kJ
L atm
ΔE = q + w = 30.9 kJ + (12.4 kJ) = 18.5 kJ
19.
q = molar heat capacity × mol × ΔT = 20.
ΔE = q + w, 102.5 J = 52.5 J + w, w = 155.0 J ×
o
1 L atm
= 1.530 L atm
101.3 J
w = PΔV, 1.530 L atm = 0.500 atm × ΔV, ΔV = 3.06 L
ΔV = Vf – Vi, 3.06 L = 58.0 L  Vi, Vi = 54.9 L = initial volume
21.
H2O(g)  H2O(l); ΔE = q + w; q = 40.66 kJ; w = PΔV
Volume of 1 mol H2O(l) = 1.000 mol H2O(l) ×
18.02 g
1 cm3
= 18.1 cm3 = 18.1 mL

mol
0.996 g
w = PΔV = 1.00 atm × (0.0181 L  30.6 L) = 30.6 L atm ×
101.3 J
= 3.10 × 103 J
L atm
= 3.10 kJ
ΔE = q + w = 40.66 kJ + 3.10 kJ = 37.56 kJ
22.
Only when there is a volume change can PV work be done. In pathway I (steps 1 + 2), only
the first step does PV work (step 2 has a constant volume of 30.0 L). In pathway II (steps 3 +
4), only step 4 does PV work (step 3 has a constant volume of 10.0 L).
101.3 J
L atm
= 4.05 × 103 J
101.3 J
Pathway II: w = PΔV = 1.00 atm(30.0 L  10.0 L) = 20.0 L atm ×
L atm
= 2.03 × 103 J
Pathway I: w = PΔV = 2.00 atm(30.0 L  10.0 L) = 40.0 L atm ×
Note: The sign is minus () because the system is doing work on the surroundings (an
expansion).
We get different values of work for the two pathways; both pathways have the same initial
and final states. Because w depends on the pathway, work cannot be a state function.
CHAPTER 9
23.
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
365
Step 1: ΔE1 = q + w = 72 J + 35 J = 107 J; step 2: ΔE2 = 35 J  72 J = 37 J
ΔEoverall = ΔE1 + ΔE2 = 107 J  37 J = 70. J
24.
In this problem, q = w = 950. J.
950. J ×
1 L atm
= 9.38 L atm of work done by the gases
101.3 J
w = PΔV, 9.38 L atm =
 650.
atm × (Vf  0.040 L), Vf  0.040 = 11.0 L, Vf = 11.0 L
760
Properties of Enthalpy
25.
One should try to cool the reaction mixture or provide some means of removing heat since
the reaction is very exothermic (heat is released). The H2SO4(aq) will get very hot and
possibly boil unless cooling is provided.
26.
a. 1.00 mol H2O ×
 572 kJ
= 286 kJ; 286 kJ of heat released
2 mol H 2 O
b. 4.03 g H2 ×
1 mol H 2
 572 kJ

= 572 kJ; 572 kJ of heat released
2.016 g H 2
2 mol H 2
c. 186 g O2 ×
1 mol O 2
 572 kJ

= 3320 kJ; 3320 kJ of heat released
32.00 g O 2
mol O 2
d.
n H2 =
PV
1.0 atm  2.0  108 L
=
= 8.2 × 106 mol H2
0
.
08206
L
atm
RT
 298 K
K mol
8.2 × 106 mol H2 ×
27.
 572 kJ
= 2.3 × 109 kJ; 2.3 × 109 kJ of heat released
2 mol H 2
4 Fe(s) + 3 O2(g)  2 Fe2O3(s) ΔH = -1652 kJ; note that 1652 kJ of heat is released when
4 mol Fe reacts with 3 mol O2 to produce 2 mol Fe2O3.
a. 4.00 mol Fe ×
 1652 kJ
= 1650 kJ; 1650 kJ of heat released
4 mol Fe
b. 1.00 mol Fe2O3 ×
c. 1.00 g Fe ×
 1652 kJ
= 826 kJ; 826 kJ of heat released
2 mol Fe 2 O 3
1 mol Fe
 1652 kJ

= 7.39 kJ; 7.39 kJ of heat released
55.85 g
4 mol Fe
366
CHAPTER 9
d. 10.0 g Fe ×
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
1 mol O 2
1 mol Fe
= 0.179 mol Fe; 2.00 g O2 ×
= 0.0625 mol O2
55.85 g
32.00 g
0.179 mol Fe/0.0625 mol O2 = 2.86; the balanced equation requires a 4 mol Fe/3 mol O2
= 1.33 mole ratio. O2 is limiting since the actual mole Fe/mole O2 ratio is greater than the
required mole ratio.
0.0625 mol O2 ×
28.
 1652 kJ
= 34.4 kJ; 34.4 kJ of heat released
3 mol O 2
a. The combustion of gasoline releases heat, so this is an exothermic process.
b. H2O(g) → H2O(l); heat is released when water vaper condenses, so this is an exothermic
process.
c. To convert a solid to a gas, heat must be absorbed, so this is an endothermic process.
d. Heat must be added (absorbed) in order to break a bond, so this is an endothermic
process.
29.
H = E + PV at constant P; from the definition of enthalpy, the difference between H
and E at constant P is the quantity PV. Thus when a system at constant P can do pressurevolume work, then H ≠ E. When the system cannot do PV work, then H = E at constant
pressure. An important way to differentiate H from E is to concentrate on q, the heat flow;
the heat flow by a system at constant pressure equals H, and the heat flow by a system at
constant volume equals E.
30.
H = E + PV; from this equation, H > E when V > 0, H < E when V < 0, and H
= E when V = 0. Concentrate on the moles of gaseous products versus the moles of
gaseous reactants to predict V for a reaction.
a. There are 2 moles of gaseous reactants converting to 2 moles of gaseous products, so
V = 0. For this reaction, H = E.
b. There are 4 moles of gaseous reactants converting to 2 moles of gaseous products, so
V < 0 and H < E.
c. There are 9 moles of gaseous reactants converting to 10 moles of gaseous products, so
V > 0 and H > E.
31.
When a liquid is converted into gas, there is an increase in volume. The 2.5 kJ/mol quantity is
the work done by the vaporization process in pushing back the atmosphere.
32.
From Example 9.1, q = 1.3 × 108 J. Because the heat transfer process is only 60.%
efficient, the total energy required is: 1.3 × 108 J × (100. J/60. J) = 2.2 × 108 J
Mass C3H8 = 2.2 × 108 J ×
1 mol C3H8
44.09 g C3 H8

= 4.4 × 103 g C3H8
3
mol C3H8
2221  10 J
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
The Thermodynamics of Ideal Gases
33.
Pathway I:
Step 1: (5.00 mol, 3.00 atm, 15.0 L)  (5.00 mol, 3.00 atm, 55.0 L)
w = PΔV = (3.00 atm)(55.0  15.0 L) = 120. L atm
w = 120. L atm 
101.3 J
1 kJ

= 12.2 kJ
L atm 1000 J
ΔH = qp = nCpΔT = nCp 
C Δ(PV)
Δ(PV)
 p
;; Δ(PV) = (P2V2  P1V1)
nR
R
For an ideal monatomic gas: Cp =
5
5
 5  Δ(PV)
R; ΔH =  R 
= Δ(PV)
2
2
2  R
5
5
Δ(PV) = (165  45.0) L atm = 300. L atm
2
2
101.3 J
1 kJ

ΔH = qp = 300. L atm ×
= 30.4 kJ
L atm
1000 J
ΔH = qp =
ΔE = q + w = 30.4 kJ  12.2 kJ = 18.2 kJ
Step 2: (5.00 mol, 3.00 atm, 55.0 L) → (5.00 mol, 6.00 atm, 20.0 L)
 3   Δ(PV)  3
ΔE = nCvΔT = n  R  
  ΔPV
 2   nR  2
3
ΔE = (120.  165) L atm = 67.5 L atm (Carry an extra significant figure.)
2
101.3 J
1 kJ

ΔE = 67.5 L atm ×
= 6.8 kJ
L atm 1000 J
 5   Δ(PV)  5
ΔH = nCpΔT = n  R  
  ΔPV
 2   nR  2
5
ΔH = (45 L atm) = 113 L atm (Carry an extra significant figure.)
2
101.3 J
1 kJ

ΔH = 113 L atm ×
= 11.4 = 11 kJ
L atm 1000 J
w = PextΔV = (6.00 atm)(20.0  55.0) L = 210. L atm
101.3 J
1 kJ

= 21.3 kJ
L atm 1000 J
ΔE = q + w, 6.8 kJ = q + 21.3 kJ, q = 28.1 kJ
w = 210. L atm ×
367
368
CHAPTER 9
Summary:
Path I
q
w
ΔE
ΔH
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
Step 1
Step 2
30.4 kJ
12.2 kJ
18.2 kJ
30.4 kJ
28.1 kJ
21.3 kJ
6.8 kJ
11 kJ
Total
2.3 kJ
9.1 kJ
11.4 kJ
19 kJ
Pathway II:
Step 3: (5.00 mol, 3.00 atm, 15.0 L)  (5.00 mol, 6.00 atm, 15.0 L)
ΔE = qv =
3
5
Δ(PV) = (90.0  45.0) L atm = 67.5 L atm
2
2
ΔE = qv = 67.5 L atm ×
101.3 J
1 kJ

= 6.84 kJ
L atm 1000 J
w = PΔV = 0 because ΔV = 0
ΔH = ΔE + Δ(PV) = 67.5 L atm + 45.0 L atm = 112.5 L atm = 11.40 kJ
Step 4: (5.00 mol, 6.00 atm, 15.0 L)  (5.00 mol, 6.00 atm, 20.0 L)
 5   Δ(PV)  5
ΔH = qp = nCpΔT =  R  
  ΔPV
 2   nR  2
ΔH = 5
(120. - 90.0) L atm = 75 L atm
2
ΔH = qp = 75 L atm ×
101.3 J
1 kJ

= 7.6 kJ
L atm 1000 J
w = PΔV =  (6.00 atm)(20.0  15.0) L = 30. L atm
w = 30. L atm ×
101.3 J
1 kJ

= 3.0 kJ
L atm 1000 J
ΔE = q + w = 7.6 kJ  3.0 kJ = 4.6 kJ
Summary:
Path II
q
w
ΔE
ΔH
Step 3
Step 4
6.84 kJ
0
6.84 kJ
11.40 kJ
7.6 kJ
3.0 kJ
4.6 kJ
7.6 kJ
Total
14.4 kJ
3.0 kJ
11.4 kJ
19.0 kJ
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
369
State functions are independent of the particular pathway taken between two states; path
functions are dependent on the particular pathway. In this problem, the overall values of ΔH and ΔE for the two pathways are the same; hence ΔH and ΔE are state functions. The overall values of q and w for the two pathways are different; hence q and w are path functions.
34.
For a monoatomic gas, Cv = (3/2)R and Cp = (5/2)R.
Step 1: (2.00 mol, 10.0 atm, 10.0 L)  (2.00 mol, 10.0 atm, 5.0 L)
 5   Δ(PV)  5
ΔH = qp = nCpΔT = n  R  
  ΔPV
 2   nR  2
ΔH = qp =
5
(50.  100.) = 125 L atm  101.3 J L1 atm1 = 12.7 kJ
2
(We will carry all calculations to 0.1 kJ.)
w = PΔV = (10.0 atm)(5.0  10.0) L = 50. L atm = 5.1 kJ
ΔE = q + w = 12.7 + 5.1 = 7.6 kJ
Step 2: (2.00 mol, 10.0 atm, 5.0 L)  (2.00 mol, 20.0 atm, 5.0 L)
ΔE = qv = nCvT =
3
3
Δ(PV) = (100  50.) = 75 L atm = 7.6 kJ;; w = 0 since ΔV = 0
2
2
ΔH = ΔE + Δ(PV) = 75 L atm + 50. L atm = 125 L atm = 12.7 kJ
Step 3: (2.00 mol, 20.0 atm, 5.0 L) → (2.00 mol, 20.0 atm, 25.0 L)
ΔH = qp =
5
5
Δ(PV) = (500.  100) = 1.0 × 103 L atm = 101.3 kJ
2
2
w = PΔV = (20.0 atm)(25.0  5.0) L = 400. L atm = 40.5 kJ
ΔE = q + w = 101.3  40.5 = 60.8 kJ
Summary:
q
w
ΔE
ΔH
35.
Step 1
Step 2
Step 3
Total
12.7 kJ
5.1 kJ
7.6 kJ
12.7 kJ
7.6 kJ
0
7.6 kJ
12.7 kJ
101.3 kJ
40.5 kJ
60.8 kJ
101.3 kJ
96.2 kJ
35.4 kJ
60.8 kJ
101.3 kJ
Consider the constant volume process first.
n = 1.00 × 103 g ×
1 mol
44.60 J
44.60 J
 o
= 33.3 mol C2H6; Cv =
30.07 g
K mol
C mol
370
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
ΔE = nCvΔT = (33.3 mol)(44.60 J °C1 mol1)(75.0  25.0°C) = 74,300 J = 74.3 kJ
ΔE = q + w;; since ΔV = 0, w = 0;; ΔE = qv = 74.3 kJ
ΔH = ΔE + ΔPV = ΔE + nRΔT
ΔH = 74.3 kJ + (33.3 mol)(8.3145 J K1 mol1)(50.0 K)(1 kJ/1000 J)
ΔH = 74.3 kJ + 13.8 kJ = 88.1 kJ
Now consider the constant pressure process.
qp = ΔH = nCpΔT = (33.3 mol)(52.92 J K1 mol1)(50.0 K)
qp = 88,100 J = 88.1 kJ = ΔH
w = PΔV = nRΔT = (33.3 mol)(8.3145 J K1 mol1)(50.0 K) = 13,800 J = 13.8 kJ
ΔE = q + w = 88.1 kJ  13.8 kJ = 74.3 kJ
Summary
q
ΔE
ΔH
w
36.
88.0 g N2O ×
Constant V
74.3 kJ
74.3 kJ
88.1 kJ
0
Constant P
88.1 kJ
74.3 kJ
88.1 kJ
13.8 kJ
1 mol N 2 O
= 2.00 mol N2O
44.02 g N 2 O
At constant pressure, qp = ΔH.
ΔH = nCpΔT = (2.00 mol)(38.70 J °C1 mol1)(55°C  165°C)
ΔH = 8510 J = 8.51 kJ = qp
w = PΔV = nRΔT = (2.00 mol)(8.3145 J K1 mol1)(110. K) = 1830 J = 1.83 kJ
ΔE = q + w = 8.51 kJ + 1.83 kJ = 6.68 kJ
Calorimetry and Heat Capacity
37.
Specific heat capacity is defined as the amount of heat necessary to raise the temperature of
one gram of substance by one degree Celsius. Therefore, H2O(l) with the largest heat
capacity value requires the largest amount of heat for this process. The amount of heat for
H2O(l) is:
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
371
4.18 J
× 25.0 g × (37.0°C  15.0°C) = 2.30 × 103 J
o
Cg
The largest temperature change when a certain amount of energy is added to a certain mass of
substance will occur for the substance with the smallest specific heat capacity. This is Hg(l),
and the temperature change for this process is:
energy = s × m × ΔT = 38.
1000 J
10.7 kJ 
energy
kJ = 140°C

ΔT = 0.14 J
sm
 550. g
o
Cg
0.24 J
0.24 J

a. s = specific heat capacity = o
since ΔT(K) = ΔT(°C).
Kg
Cg
Energy = s × m × ΔT = b. Molar heat capacity =
c. 1250 J =
39.
0.24 J
× 150.0 g × (298 K  273 K) = 9.0 × 102 J
Kg
0.24 J 107.9 g Ag


o
mol Ag
Cg
o
26 J
C mol
1250
0.24 J
× m × (15.2°C  12.0°C), m =
= 1.6 × 103 g Ag
o
0.24  3.2
Cg
In calorimetry, heat flow is determined into or out of the surroundings. Because ΔE univ = 0 by
the first law of thermodynamics, ΔEsys = ΔEsurr; what happens to the surroundings is the
exact opposite of what happens to the system. To determine heat flow, we need to know the
heat capacity of the surroundings, the mass of the surroundings that accepts/donates the heat,
and the change in temperature. If we know these quantities, qsurr can be calculated and then
equated to qsys (qsurr = qsys). For an endothermic reaction, the surroundings (the calorimeter
contents) donates heat to the system. This is accompanied by a decrease in temperature of the
surroundings. For an exothermic reaction, the system donates heat to the surroundings (the
calorimeter), so temperature increases.
qP = ΔH;; qV = ΔE;; a coffee-cup calorimeter is at constant (atmospheric) pressure. The heat
released or gained at constant pressure is ΔH. A bomb calorimeter is at constant volume. The heat released or gained at constant volume is ΔE.
40.
Heat released to water = 5.0 g H2 ×
Heat gain by water = 1.10 × 103 J =
50. J
120. J
+ 10. g methane ×
= 1.10 × 103 J
g methane
g H2
4.18 J
o
Cg
× 50.0 g × T
T = 5.26°C, 5.26°C = Tf  25.0°C, Tf = 30.3°C
372
41.
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
Heat gained by water = heat lost by nickel = s × m × ΔT, where s = specific heat capacity.
Heat gain =
4.18 J
o
Cg
× 150.0 g × (25.0°C  23.5°C) = 940 J
A common error in calorimetry problems is sign errors. Keeping all quantities positive helps
to eliminate sign errors.
Heat loss = 940 J =
42.
0.444 J
o
Cg
× mass × (99.8  25.0) °C, mass =
940
= 28 g
0.444  74.8
Heat gain by water = heat loss by metal = s × m × ΔT, where s = specific heat capacity.
4.18 J
× 150.0 g × (18.3°C - 15.0°C) = 2100 J
o
Cg
A common error in calorimetry problems is sign errors. Keeping all quantities positive helps
to eliminate sign errors.
Heat gain =
Heat loss = 2100 J = s × 150.0 g × (75.0°C  18.3°C), s =
43.
2100 J
56.7 C  150.0 g
= 0.25 J °C1 g1
o
Heat loss by hot water = heat gain by cold water; keeping all quantities positive to avoid sign
errors:
4.18 J
4.18 J
× mhot × (55.0°C  37.0°C) = o
× 90.0 g × (37.0 °C  22.0°C)
o
Cg
Cg
mhot =
44.
90.0 g  15.0o C
= 75.0 g hot water needed
18.0o C
Heat loss by Al + heat loss by Fe = heat gain by water; keeping all quantities positive to
avoid sign error:
0.89 J
0.45 J
× 5.00 g Al × (100.0°C  Tf) + o
× 10.00 g Fe × (100.0  Tf)
o
Cg
Cg
4.18 J
= o
× 97.3 g H2O × (Tf  22.0°C)
Cg
4.5(100.0  Tf) + 4.5(100.0  Tf) = 407(Tf  22.0), 450  (4.5)Tf + 450  (4.5)Tf
= 407Tf  8950
416Tf = 9850, Tf = 23.7°C
45.
Heat lost by solution = heat gained by KBr; mass of solution = 125 g + 10.5 g = 136 g
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
373
Note: Sign errors are common with calorimetry problems. However, the correct sign for
ΔH can easily be obtained from the ΔT data. When working calorimetry problems, keep all quantities positive (ignore signs). When finished, deduce the correct sign for ΔH. For this problem, T decreases as KBr dissolves, so ΔH is positive;; the dissolution of KBr is endothermic (absorbs heat).
Heat lost by solution =
ΔH in units of J/g = 4.18 J
× 136 g × (24.2°C  21.1°C) = 1800 J = heat gained by KBr
o
Cg
1800 J
= 170 J/g
10.5 g KBr
ΔH in units of kJ/mol = 46.
170 J
119.0 g KBr
1 kJ


= 20. kJ/mol
g KBr
mol KBr
1000 J
NH4NO3(s)  NH4+(aq) + NO3(aq) ΔH = ?;; mass of solution = 75.0 g + 1.60 g = 76.6 g
Heat lost by solution = heat gained as NH4NO3 dissolves. To help eliminate sign errors, we
will keep all quantities positive (q and ΔT) and then deduce the correct sign for ΔH at the end of the problem. Here, because temperature decreases as NH4NO3 dissolves, heat is absorbed
as NH4NO3 dissolves, so this is an endothermic process (ΔH is positive).
Heat lost by solution =
4.18 J
× 76.6 g × (25.00  23.34)°C = 532 J = heat gained as
o
Cg
NH NO dissolves
4
ΔH = 47.
80.05 g NH 4 NO3
532 J
1 kJ


= 26.6 kJ/mol NH4NO3 dissolving
1.60 g NH 4 NO3
mol NH 4 NO3
1000 J
Because ΔH is exothermic, the temperature of the solution will increase as CaCl2(s)
dissolves. Keeping all quantities positive:
heat loss as CaCl2 dissolves = 11.0 g CaCl2 
heat gained by solution = 8.08 × 103 J =
Tf  25.0°C =
48.
3
1 mol CaCl 2
81.5 kJ

= 8.08 kJ
110.98 g CaCl 2
mol CaCl 2
4.18 J
× (125 + 11.0) g × (Tf  25.0°C)
o
Cg
8.08  103
= 14.2°C, Tf = 14.2°C + 25.0°C = 39.2°C
4.18  136
0.100 L ×
0.500 mol HCl
= 5.00 × 102 mol HCl
L
0.300 L ×
0.100 mol Ba (OH) 2
= 3.00 × 102 mol Ba(OH)2
L
374
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
To react with all the HCl present, 5.00 × 102/2 = 2.50 × 102 mol Ba(OH)2 is required.
Because 0.0300 mol Ba(OH)2 is present, HCl is the limiting reactant.
118 kJ
= 2.95 kJ of heat is evolved by reaction
2 mol HCl
4.18 J
Heat gained by solution = 2.95 × 103 J = o
× 400.0 g × ΔT
Cg
5.00 × 102 mol HCl ×
ΔT = 1.76°C = Tf  Ti = Tf  25.0°C, Tf = 26.8°C
49.
50.0 × 103 L × 0.100 mol/L = 5.00 × 103 mol of both AgNO3 and HCl are reacted. Thus
5.00 × 103 mol of AgCl will be produced because there is a 1 : 1 mole ratio between
reactants.
Heat lost by chemicals = heat gained by solution
Heat gain =
4.18 J
× 100.0 g × (23.40  22.60)°C = 330 J
o
Cg
Heat loss = 330 J; this is the heat evolved (exothermic reaction) when 5.00 × 103 mol of
AgCl is produced. So q = 330 J and ΔH (heat per mol AgCl formed) is negative with a value of:
ΔH =  330 J
1 kJ

= 66 kJ/mol
3
5.00  10 mol 1000 J
Note: Sign errors are common with calorimetry problems. However, the correct sign for ΔH can be determined easily from the ΔT data;; i.e., if ΔT of the solution increases, then the reaction is exothermic because heat was released, and if ΔT of the solution decreases, then the reaction is endothermic because the reaction absorbed heat from the water. For
calorimetry problems, keep all quantities positive until the end of the calculation and then
decide the sign for ΔH. This will help eliminate sign errors.
50.
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
We have a stoichiometric mixture. All of the NaOH and HCl will react.
0.10 L ×
1.0 mol
= 0.10 mol of HCl is neutralized by 0.10 mol NaOH.
L
Heat lost by chemicals = heat gained by solution
Volume of solution = 100.0 + 100.0 = 200.0 mL
Heat gain =
1.0 g 

3
  200.0 mL 
 × (31.3 – 24.6)C = 5.6 × 10 J = 5.6 kJ
mL
Cg 

4.18 J
o
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
375
Heat loss = 5.6 kJ; this is the heat released by the neutralization of 0.10 mol HCl. Because
the temperature increased, the sign for ΔH must be negative, i.e., the reaction is exothermic. For calorimetry problems, keep all quantities positive until the end of the calculation and then
decide the sign for ΔH. This will help eliminate sign errors.
ΔH = 51.
 5.6 kJ
= 56 kJ/mol
0.10 mol
a. C12H22O11(s) + 12 O2(g)  12 CO2(g) + 11 H2O(l)
b. A bomb calorimeter is at constant volume, so heat released = qv = ΔE:
ΔE =  24.00 kJ 342.30 g

= 5630 kJ/mol C12H22O11
1.46 g
mol
c. ΔH = ΔE + Δ(PV) = ΔE + Δ(nRT) = ΔE + ΔnRT, where Δn = moles of gaseous products – moles of gaseous reactants.
For this reaction, Δn = 12 12 = 0, so ΔH = ΔE = 5630 kJ/mol.
52.
First, we need to get the heat capacity of the calorimeter from the combustion of benzoic
acid. Heat lost by combustion = heat gained by calorimeter.
26.42 kJ
= 4.185 kJ
g
4.185 kJ
Heat gain = 4.185 kJ = Ccal × ΔT, Ccal =
= 1.65 kJ/°C
2.54o C
Heat loss = 0.1584 g ×
Now we can calculate the heat of combustion of vanillin. Heat loss = heat gain.
Heat gain by calorimeter =
1.65 kJ
o
C
× 3.25°C = 5.36 kJ
Heat loss = 5.36 kJ, which is the heat evolved by combustion of the vanillin.
Ecomb =
53.
 5.36 kJ
 25.2 kJ 152.14 g

= 25.2 kJ/g; Ecomb =
= 3830 kJ/mol
0.2130 g
g
mol
a. Heat gain by calorimeter = heat loss by CH4 = 6.79 g CH4 
Heat capacity of calorimeter =
1 mol CH 4 802 kJ

16.04 g
mol
= 340. kJ
340. kJ
= 31.5 kJ/°C
10.8 o C
b. Heat loss by C2H2 = heat gain by calorimeter = 16.9°C ×
31. 5 kJ
= 532 kJ
o
C
A bomb calorimeter is at constant volume, so heat released = qv = ΔE:
376
CHAPTER 9
ΔEcomb =
54.
A(l)  A(g)
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
 532 kJ
26.04 g

= 1.10 × 103 kJ/mol
12.6 g C 2 H 2
mol C 2 H 2
ΔHvap = 30.7 kJ
w = PΔV = ΔnRT, where Δn = nproducts  nreactants = 1  0 = 1
w = (1 mol)(8.3145 J K1 mol1)(80. + 273 K) = 2940 J; because pressure is constant:
ΔE = qp + w = ΔH + w = 30.7 kJ + (2.94 kJ) = 27.8 kJ
Hess’s Law
55.
C4H4(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
ΔHcomb = 2341 kJ
C4H8(g) + 6 O2(g)  4 CO2(g) + 4 H2O(l)
ΔHcomb = 2755 kJ
H2(g) + 1/2 O2(g)  H2O(l)
ΔHcomb = 286 kJ
By convention, H2O(l) is produced when enthalpies of combustion are given, and because per
mole quantities are given, the combustion reaction refers to 1 mole of that quantity reacting
with O2(g).
Using Hess’s Law to solve:
C4H4(g) + 5 O2(g)  4 CO2(g) + 2 H2O(l)
ΔH1 = 2341 kJ
4 CO2(g) + 4 H2O(l)  C4H8(g) + 6 O2(g)
ΔH2 =  (2755 kJ)
2 H2(g) + O2(g)  2 H2O(l)
ΔH3 = 2(286 kJ)
___________________________________________________________________
C4H4(g) + 2 H2(g)  C4H8(g)
ΔH = ΔH1 + ΔH2 + ΔH3 = 158 kJ
56.
ClF + 1/2 O2  1/2 Cl2O + 1/2 F2O
ΔH = 1/2 (167.4 kJ)
1/2 Cl2O + 3/2 F2O  ClF3 + O2
ΔH = 1/2 (341.4 kJ)
F2 + 1/2 O2  F2O
ΔH = 1/2 (43.4 kJ)
____________________________________________________________
ClF(g) + F2(g)  ClF3(g)
ΔH = 108.7 kJ
57.
C6H4(OH)2  C6H4O2 + H2
ΔH = 177.4 kJ
H2O2  H2 + O2
ΔH = (191.2 kJ)
2 H2 + O2  2 H2O(g)
ΔH = 2(241.8 kJ)
2 H2O(g)  2 H2O(l)
ΔH = 2(43.8 kJ)
________________________________________________________________
C6H4(OH)2(aq) + H2O2(aq)  C6H4O2(aq) + 2 H2O(l) ΔH = 202.6 kJ
58.
We want ΔH for N2H4(l) + O2(g)  N2(g) + 2 H2O(l). It will be easier to calculate ΔH for the combustion of four moles of N2H4 because we will avoid fractions.
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
377
9 H2 + 9/2 O2  9 H2O
ΔH = 9(286 kJ)
3 N2H4 + 3 H2O  3 N2O + 9 H2
ΔH = 3(317 kJ)
2 NH3 + 3 N2O  4 N2 + 3 H2O
ΔH = 1010. kJ
N2H4 + H2O  2 NH3 + 1/2 O2
ΔH =  (143 kJ)
_____________________________________________________
4 N2H4(l) + 4 O2(g)  4 N2(g) + 8 H2O(l)
ΔH = 2490. kJ
For N2H4(l) + O2(g)  N2(g) + 2 H2O(l) ΔH =  2490. kJ
= 623 kJ
4
Note: By the significant figure rules, we could report this answer to four significant figures.
However, because the ΔH values given in the problem are only known to ±1 kJ, our final
answer will at best be ±1 kJ.
59.
2 N2(g) + 6 H2(g)  4 NH3(g)
ΔH = 2(92 kJ)
6 H2O(g)  6 H2(g) + 3 O2(g)
ΔH = 3(-484 kJ)
___________________________________________________
2 N2(g) + 6 H2O(g)  3 O2(g) + 4 NH3(g) ΔH = 1268 kJ
No, because the reaction is very endothermic (requires a lot of heat to react), it would not be a
practical way of making ammonia because of the high energy costs required.
60.
P4O10  P4 + 5 O2
ΔH = (2967.3 kJ)
10 PCl3 + 5 O2  10 Cl3PO
ΔH = 10(285.7 kJ)
6 PCl5  6 PCl3 + 6 Cl2
ΔH = 6(84.2 kJ)
P4 + 6 Cl2  4 PCl3
ΔH = 1225.6 kJ
_______________________________________________________
P4O10(s) + 6 PCl5(g)  10 Cl3PO(g)
ΔH = 610.1 kJ
61.
2 C + 2 O2  2 CO2
ΔH = 2(394 kJ)
H2 + 1/2 O2  H2O
ΔH = 286 kJ
2 CO2 + H2O  C2H2 + 5/2 O2
ΔH =  (1300.kJ)
________________________________________________
2 C(s) + H2(g)  C2H2(g)
ΔH = 226 kJ
Note: The enthalpy change for a reaction that is reversed is the negative quantity of the
enthalpy change for the original reaction. If the coefficients in a balanced reaction are
multiplied by an integer, then the value of ΔH is multiplied by the same integer.
62.
To avoid fractions, let's first calculate ΔH for the reaction:
6 FeO(s) + 6 CO(g)  6 Fe(s) + 6 CO2(g)
6 FeO + 2 CO2  2 Fe3O4 + 2 CO
2 Fe3O4 + CO2  3 Fe2O3 + CO
3 Fe2O3 + 9 CO  6 Fe + 9 CO2
6 FeO(s) + 6 CO(g)  6 Fe(s) + 6 CO2(g)
ΔH = 2(18 kJ)
ΔH =  (39 kJ)
ΔH = 3(23 kJ)
ΔH = 66 kJ
378
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
So for: FeO(s) + CO(g)  Fe(s) + CO2(g)
63.
CaC2 
CaO + H2O 
2 CO2 + H2O 
Ca + 1/2 O2 
2 C + 2 O2 
ΔH = ΔH =  (62.8 kJ)
ΔH = 653.1 kJ
ΔH =  (1300. kJ)
ΔH = 635.5 kJ
ΔH = 2(393.5 kJ)
Ca + 2 C
Ca(OH)2
C2H2 + 5/2 O2
CaO
2 CO2
CaC2(s) + 2 H2O(l)  Ca(OH)2(aq) + C2H2(g)
64.
NO + O3  NO2 + O2
3/2 O2  O3
O  1/2 O2
NO(g) + O(g)  NO2(g)
 66 kJ
= 11 kJ
6
ΔH = 713 kJ
ΔH = 199 kJ
ΔH = 1/2 (427 kJ)
ΔH = 1/2 (495 kJ)
ΔH = 233 kJ
Standard Enthalpies of Formation
65.
The change in enthalpy that accompanies the formation of one mole of a compound from its
elements, with all substances in their standard states, is the standard enthalpy of formation for
a compound. The reactions that refer to H of are:
Na(s) + 1/2 Cl2(g)  NaCl(s); H2(g) + 1/2 O2(g)  H2O(l)
6 C(graphite, s) + 6 H2(g) + 3 O2(g)  C6H12O6(s); Pb(s) + S(s) + 2 O2(g)  PbSO4(s)
66.
a. CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
H = ?
Utilizing Hess’s law: Reactants  Standard State Elements H = Ha + Hb = 75 + 0 = 75 kJ
Standard State Elements  Products
H = Hc + Hd = –394 – 572 = –966 kJ
__________________________________________________________________
Reactants  Products
H = 75 – 966 = –891 kJ
b. The standard enthalpy of formation for an element in its standard state is given a value of
zero. To assign standard enthalpy of formation values for all other substances, there
needs to be a reference point from which all enthalpy changes are determined. This
reference point is the elements in their standard state which is defined as the zero point.
So when using standard enthalpy values, a reaction is broken up into two steps. The first
step is to calculate the enthalpy change necessary to convert the reactants to the elements
in their standard state. The second step is to determine the enthalpy change that occurs
when the elements in their standard state go to form the products. When these two steps
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
379
are added together, the reference point (the elements in their standard state) cancels out
and we are left with the enthalpy change for the reaction.
c. This overall reaction is just the reverse of all the steps in the part a answer. So H =
+966 – 75 = 891 kJ. Products are first converted to the elements in their standard state
which requires 966 kJ of heat. Next, the elements in the standard states go to form the
original reactants [CH4(g) + 2 O2(g)] which has an enthalpy change of 75 kJ. All of the
signs are reversed because the entire process is reversed.
67.
  416 kJ 
4 Na(s) + O2(g)  2 Na2O(s), ΔH° = 2 mol 
 = 832 kJ
 mol 
2 Na(s) + 2 H2O(l)  2 NaOH(aq) + H2(g)

  470. kJ   
  286 kJ 
ΔH° = 2 mol 
   2 mol 
 = 368 kJ
 mol   
 mol 

2Na(s) + CO2(g)  Na2O(s) + CO(g)

  416 kJ 
  110.5 kJ  
  393.5 kJ 
ΔH° = 1 mol 
  1 mol 
  1 mol 
 = 133 kJ
 mol 
 mol  
 mol 

In reactions 2 and 3, sodium metal reacts with the "extinguishing agent." Both reactions are
exothermic and each reaction produces a flammable gas, H2 and CO, respectively.
68.
a. 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g); ΔH° =
 n pΔHof , products   n r ΔHof, reactants

 90. kJ 
  242 kJ   
  46 kJ 
ΔH° = 4 mol 
  6 mol 
   4 mol 
 = 908 kJ
 mol 
 mol   
 mol 

2 NO(g) + O2(g)  2 NO2(g)

 34 kJ   
 90. kJ 
ΔH° = 2 mol 
   2 mol 
 = 112 kJ
 mol   
 mol 

3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g)

  207 kJ 
 90. kJ 
ΔH° = 2 mol 
  1 mol 

 mol 
 mol 


 34 kJ 
  286 kJ  
 3 mol 
  1 mol 
  = 140. kJ
 mol 
 mol  

380
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
b. 12 NH3(g) + 15 O2(g)  12 NO(g) + 18 H2O(g)
12 NO(g) + 6 O2(g)  12 NO2(g)
12 NO2(g) + 4 H2O(l)  8 HNO3(aq) + 4 NO(g)
4 H2O(g)  4 H2O(l)
_________________________________________________
12 NH3(g) + 21 O2(g)  8 HNO3(aq) + 4 NO(g) + 14 H2O(g)
The overall reaction must be exothermic because each step is exothermic.
69.
In general: ΔH° =  n p ΔH of , products 
 n r ΔHof, reactants , and all elements in their standard
state have ΔH of = 0 by definition.
a. The balanced equation is: 2 NH3(g) + 3 O2(g) + 2 CH4(g)  2 HCN(g) + 6 H2O(g)
ΔH° = (2 mol HCN × ΔH of , HCN + 6 mol H2O(g)  ΔH of , H 2O )
 (2 mol NH3 × ΔH of ,
NH 3
+ 2 mol CH4 × ΔH of , CH4 )
ΔH° = [2(135.1) + 6(242)]  [2(46) + 2(75)] = 940. kJ
b. Ca3(PO4)2(s) + 3 H2SO4(l)  3 CaSO4(s) + 2 H3PO4(l)

  1433 kJ 
  1267 kJ 
ΔH° = 3 mol CaSO 4 (s)
  2 mol H 3PO4 (l)

 mol 
 mol 


  4126 kJ 
  814 kJ 
 1 mol Ca 3 (PO4 ) 2 (s)
  3 mol H 2SO 4 (l)

 mol 
 mol 

ΔH° = 6833 kJ  (-6568 kJ) = 265 kJ
c. NH3(g) + HCl(g)  NH4Cl(s)
ΔH° = (1 mol NH4Cl × ΔH of ,
NH 4Cl )
 (1 mol NH3 × ΔH of ,
NH 3
+ 1 mol HCl × ΔH of , HCl )


  46 kJ 
  92 kJ 
  314 kJ 
ΔH° = 1 mol 
  1 mol 

  1 mol 
 mol 
 mol 
 mol 


ΔH° = 314 kJ + 138 kJ = 176 kJ
d. The balanced equation is: C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(g)


  393.5 kJ 
  242 kJ  
  278 kJ 
ΔH° = 2 mol 
  3 mol 
   1 mol 

 mol 
 mol  
 mol 


ΔH° = 1513 kJ  (278 kJ) = 1235 kJ
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
381
e. SiCl4(l) + 2 H2O(l)  SiO2(s) + 4 HCl(aq)
Because HCl(aq) is H+(aq) + Cl(aq), ΔH of = 0 - 167 = -167 kJ/mol.

  167 kJ 
  911 kJ  
ΔH° = 4 mol 
  1 mol 

 mol 
 mol  


  687 kJ 
  286 kJ  
 1 mol 
  2 mol 

 mol 
 mol  

ΔH° = 1579 kJ  (1259 kJ) = 320. kJ
f.
MgO(s) + H2O(l)  Mg(OH)2(s)

  925 kJ   
  602 kJ 
  286 kJ  
ΔH° = 1 mol 
   1 mol 
  1 mol 

 mol   
 mol 
 mol  

ΔH° = 925 kJ  (888 kJ) = 37 kJ
70.
3 Al(s) + 3 NH4ClO4(s)  Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

  242 kJ 
 90. kJ 
  704 kJ 
  1676 kJ 
ΔH° = 6 mol 
  3 mol 
  1 mol 
  1 mol 

mol
mol
mol






 mol 


  295 kJ  
 3 mol 
  = 2677 kJ
 mol  

71.
5 N2O4(l) + 4 N2H3CH3(l)  12 H2O(g) + 9 N2(g) + 4 CO2(g)

  242 kJ 
  393.5 kJ 
ΔH° = 12 mol 
  4 mol 

 mol 
 mol 


  20. kJ 
 54 kJ  
 5 mol 
  4 mol 
  = 4594 kJ
 mol 
 mol  

72.
For Exercise 70, a mixture of 3 mol Al and 3 mol NH4ClO4 yields 2677 kJ of energy. The
mass of the stoichiometric reactant mixture is:
26.98 g  
117.49 g 

 3 mol 
   3 mol 
 = 433.41 g
mol  
mol 

For 1.000 kg of fuel: 1.000 × 103 g ×
 2677 kJ
= 6177 kJ
433.41 g
382
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
In Exercise 71, we get 4594 kJ of energy from 5 mol of N2O4 and 4 mol of N2H3CH3. The
92.02 g  
46.08 g 

mass is:  5 mol 
   4 mol 
 = 644.42 kJ
mol  
mol 

For 1.000 kg of fuel: 1.000 × 103 g ×
 4594 kJ
= 7129 kJ
644.42 g
Thus we get more energy per kilogram from the N2O4/N2H3CH3 mixture.
73.
2 ClF3(g) + 2 NH3(g)  N2(g) + 6 HF(g) + Cl2(g)
ΔH° = (6 ΔH of , HF )  (2 ΔH of , ClF3  2ΔH of ,
NH3
ΔH° = 1196 kJ
)
  271 kJ 
  46 kJ 
o
1196 kJ = 6 mol 
  2 ΔH f , ClF3  2 mol 

 mol 
 mol 
1196 kJ = 1626 kJ  2 ΔH of , ClF3 + 92 kJ, ΔH of , ClF3 =
74.
C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(l)
(1626  92  1196) kJ  169 kJ
=
2 mol
mol
ΔH° = 1411 kJ
ΔH° = 1411.1 kJ = 2(393.5) kJ + 2(285.8) kJ  ΔH of , C2 H 4
1411.1 kJ = 1358.6 kJ  ΔH of , C2 H 4 , ΔH of , C2 H 4 = 52.5 kJ/mol
75.
a. ΔH° = 3 mol(227 kJ/mol)  1 mol(49 kJ/mol) = 632 kJ
b. Because 3 C2H2(g) is higher in energy than C6H6(l), acetylene will release more energy
per gram when burned in air.
76.
a. C2H4(g) + O3(g)  CH3CHO(g) + O2(g)
ΔH° = 166 kJ  (143 kJ + 52 kJ) = 361 kJ
b. O3(g) + NO(g)  NO2(g) + O2(g)
ΔH° = 34 kJ  (90. kJ + 143 kJ) = 199 kJ
c. SO3(g) + H2O(l)  H2SO4(aq)
ΔH° = 909 kJ  [396 kJ + (-286 kJ)] =
227 kJ
d. 2 NO(g) + O2(g)  2 NO2(g)
ΔH° = 2(34) kJ  2(90.) kJ = 112 kJ
Energy Consumption and Sources
77.
CO(g) + 2 H2(g)  CH3OH(l)
ΔH° = 239 kJ  (110.5 kJ) = 129 kJ
CHAPTER 9
78.
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
383
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
ΔH° = [3(393.5 kJ) + 4(286 kJ)]  (104 kJ) = 2221 kJ/mol C3H8
 2221kJ
1 mol
 50.37 kJ


versus 47.7 kJ/g for octane (Example 9.8)
mol
44.096 g
g
The two fuel values are very close. An advantage of propane is that it burns more cleanly.
The boiling point of propane is -42°C. Thus it is more difficult to store propane, and there
are extra safety hazards associated with using high-pressure compressed-gas tanks.
79.
Mass of H2O = 1.00 gal ×
3.785 L 1000 mL 1.00 g


= 3790 g H2O
gal
L
mL
Energy required (theoretical) = s × m × ΔT = 4.18 J
× 3790 g × 10.0 °C = 1.58 × 105 J
o
Cg
For an actual (80.0% efficient) process, more than this quantity of energy is needed since heat
is always lost in any transfer of energy. The energy required is:
1.58 × 105 J ×
100. J
= 1.98 × 105 J
80.0 J
Mass of C2H2 = 1.98 × 105 J ×
80.
1 mol C 2 H 2
26.04 g C 2 H 2

= 3.97 g C2H2
3
mol C 2 H 2
1300.  10 J
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
ΔH° = [2(393.5 kJ) + 3(286 kJ)]  (278 kJ) = 1367 kJ/mol ethanol
 1367 kJ
1 mol

= 29.67 kJ/g
mol
46.068 g
81.
CH3OH(l) + 3/2 O2(g)  CO2(g) + 2 H2O(l)
ΔH° = [393.5 kJ + 2(286 kJ)]  ( 239 kJ) = 727 kJ/mol CH3OH
 727 kJ
1 mol

= 22.7 kJ/g versus 29.67 kJ/g for ethanol
mol
32.04 g
Ethanol has a higher fuel value than methanol.
82.
The molar volume of a gas at STP is 22.42 L (from Chapter 5).
4.19 × 106 kJ ×
1 mol CH 4
22.42 L CH 4

= 1.05 × 105 L CH4
891 kJ
mol CH 4
384
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
Additional Exercises
83.
a. aluminum oxide = Al2O3; 2 Al(s) + 3/2 O2(g)  Al2O3(s)
b. C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
c. Ba(OH)2(aq) + 2 HCl(aq)  2 H2O(l) + BaCl2(aq)
d. 2 C(graphite, s) + 3/2 H2(g) + 1/2 Cl2(g)  C2H3Cl(g)
e. C6H6(l) + 15/2 O2(g)  6 CO2(g) + 3 H2O(l)
Note: ΔHcomb values assume 1 mole of compound combusted.
f.
NH4Br(s)  NH4+(aq) + Br(aq)
84.
The specific heat capacities are: 0.89 J °C1 g1 (Al) and 0.45 J °C1 g1 (Fe). Al would be
the better choice. It has a higher heat capacity and a lower density than Fe. Using Al, the
same amount of heat could be dissipated by a smaller mass, keeping the mass of the amplifier
down.
85.
HNO3(aq) + KOH(aq)  H2O(l) + KNO3 (aq)
ΔH = 56 kJ
0.2000 L ×
0.400 mol HNO 3
= 8.00 × 102 mol HNO3
L
0.1500 L ×
0.500 mol KOH
= 7.50 × 102 mol KOH
L
Because the balanced reaction requires a 1 : 1 mole ratio between HNO3 and KOH, and
because fewer moles of KOH are actually present as compared with HNO3, KOH is the
limiting reagent.
7.50 × 102 mol KOH ×
 56 kJ
= 4.2 kJ; 4.2 kJ of heat is released.
mol KOH
86.
|qsurr| = |qsolution + qcal|; we normally assume that qcal is zero (no heat gain/loss by the calorimeter). However, if the calorimeter has a nonzero heat capacity, then some of the heat
absorbed by the endothermic reaction came from the calorimeter. If we ignore qcal, then qsurr is
too small, giving a calculated H value that is less positive (smaller) than it should be.
87.
The specific heat of water is 4.18 J °C1 g1, which is equal to 4.18 kJ °C1 kg1
4.18 kJ
= 4.18 kJ/°C
o
C kg
This is the portion of the heat capacity that can be attributed to H2O.
We have 1.00 kg of H2O, so: 1.00 kg ×
Total heat capacity = Ccal + C H 2O ,
88.
Ccal = 10.84  4.18 = 6.66 kJ/°C
Heat released = 1.056 g × 26.42 kJ/g = 27.90 kJ = heat gain by water and calorimeter
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
385
 4.18 kJ
  6.66 kJ

 0.987 kg  ΔT    o
 ΔT 
Heat gain = 27.90 kJ =  o
C

 C kg
 
27.90 = (4.13 + 6.66)ΔT = (10.79)ΔT, ΔT = 2.586°C
2.586°C = Tf  23.32°C, Tf = 25.91°C
89.
H2(g) + 1/2 O2(g)  H2O(l) ΔH° = ΔH of , H 2O(l) = 285.8 kJ
H2O(l)  H2(g) + 1/2 O2(g) ΔH° = 285.8 kJ
ΔE° = ΔH°  PΔV = ΔH°  ΔnRT
 1 kJ 

ΔE° = 285.8 kJ  (1.50  0 mol)(8.3145 J K1 mol1)(298 K) 
 1000 J 
ΔE° = 285.8 kJ  3.72 kJ = 282.1 kJ
90.
a. 2 SO2(g) + O2(g) → 2 SO3(g); w = PΔV; because the volume of the piston apparatus
decreased as reactants were converted to products (V < 0), w is positive (w > 0).
b. COCl2(g)  CO(g) + Cl2(g); because the volume increased (V > 0), w is negative
(w < 0).
c. N2(g) + O2(g)  2 NO(g); because the volume did not change (V = 0), no PV work is
done (w = 0).
In order to predict the sign of w for a reaction, compare the coefficients of all the product
gases in the balanced equation to the coefficients of all the reactant gases. When a balanced
reaction has more moles of product gases than moles of reactant gases (as in b), the reaction
will expand in volume (ΔV positive), and the system does work on the surroundings. When a balanced reaction has a decrease in the moles of gas from reactants to products (as in a), the
reaction will contract in volume (ΔV negative), and the surroundings will do compression work on the system. When there is no change in the moles of gas from reactants to products
(as in c), ΔV = 0 and w = 0.
91.
w = PΔV;; Δn = moles of gaseous products  moles of gaseous reactants. Only gases can do
PV work (we ignore solids and liquids). When a balanced reaction has more moles of
product gases than moles of reactant gases (Δn positive), the reaction will expand in volume
(ΔV positive), and the system will do work on the surroundings. For example, in reaction c,
Δn = 2  0 = 2 moles, and this reaction would do expansion work against the surroundings.
When a balanced reaction has a decrease in the moles of gas from reactants to products (Δn negative), the reaction will contract in volume (ΔV negative), and the surroundings will do
compression work on the system, e.g., reaction a, where Δn = 0  1 = 1. When there is no
change in the moles of gas from reactants to products, ΔV = 0 and w = 0, e.g., reaction b,
where Δn = 2  2 = 0.
When ΔV > 0 (Δn > 0), then w < 0, and the system does work on the surroundings (c and e).
386
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
When ΔV < 0 (Δn < 0), then w > 0, and the surroundings do work on the system (a and d).
When ΔV = 0 (Δn = 0), then w = 0 (b).
92.
H = E + PV, ΔH = ΔE + Δ(PV), ΔE = ΔH  Δ(PV)
Assume that H2O(g) is ideal. Δ(PV) = Δ(nRT) = RTΔn at constant T
We go from n = 0 to n = 1, thus Δn = 1. [H2O(l) is a liquid.]
ΔE = ΔH  (8.3145 J K1 mol1)(298 K)(1 mol) = ΔH  2480 J
ΔH = 242 kJ/mol  (286 kJ/mol) = 44 kJ/mol
ΔE = 44 kJ  2.48 kJ, ΔE = 41.52 kJ = 42 kJ
93.
I(g) + Cl(g)  ICl(g)
ΔH = (211.3 kJ)
1/2 Cl2(g)  Cl(g)
ΔH = 1/2(242.3 kJ)
1/2 I2(g)  I(g)
ΔH = 1/2(151.0 kJ)
1/2 I2(s)  1/2 I2(g)
ΔH = 1/2(62.8 kJ)
_______________________________________________________________
1/2 I2(s) + 1/2 Cl2(g)  ICl(g)
ΔH = 16.8 kJ/mol = ΔH of , ICl
94.
400 kcal ×
4.18 kJ
= 1.67 × 103 kJ ≈ 2 × 103 kJ
kcal

1 kg  9.81 m 
2.54 cm
1m 
 
 = 160 J  200 J
  8 in 

PE = mgz = 180 lb 
2
2.205 lb 
m
100 cm 
s


200 J of energy is needed to climb one step. The total number of steps to climb are:
2 × 106 J ×
95.
1 step
= 1 × 104 steps
200 J
Na2SO4(aq) + Ba(NO3)2(aq)  BaSO4(s) + 2 NaNO3(aq)
1.00 L ×
ΔH = ?
2.00 mol
0.750 mol
= 2.00 mol Na2SO4; 2.00 L ×
= 1.50 mol Ba(NO3)2
L
L
The balanced equation requires a 1 : 1 mole ratio between Na2SO4 and Ba(NO3)2. Because we
have fewer moles of Ba(NO3)2 present, it is limiting and 1.50 mol BaSO4 will be produced
[there is a 1 : 1 mole ratio between Ba(NO3)2 and BaSO4].
Heat gain by solution = heat loss by reaction
Mass of solution = 3.00 L ×
1000 mL 2.00 g

= 6.00 × 103 g
1L
mL
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
Heat gain by solution =
387
6.37 J
× 6.00 × 103 g × (42.0  30.0) °C = 4.59 × 105 J
o
Cg
Because the solution gained heat, the reaction is exothermic; q = 4.59 × 105 J for the
reaction.
H =
96.
 4.59  105 J
= 3.06 × 105 J/mol = 306 kJ/mol
1.50 mol BaSO 4
a. N2(g) + 3 H2(g)  2 NH3(g); from the balanced equation, 1 molecule of N2 will react
with 3 molecules of H2 to produce 2 molecules of NH3. So the picture after the reaction
should only have 2 molecules of NH3 present. Another important part of your drawing
will be the relative volume of the product container. The volume of a gas is directly
proportional to the number of gas molecules present (at constant T and P). In this
problem, 4 total molecules of gas were present initially (1 N2 + 3 H2). After reaction,
only 2 molecules are present (2 NH3). Because the number of gas molecules decreases
by a factor of 2 (from 4 total to 2 total), the volume of the product gas must decrease by a
factor of 2 as compared to the initial volume of the reactant gases. Summarizing, the
picture of the product container should have 2 molecules of NH3 and should be at a
volume which is one-half the original reactant container volume.
b. w = PV; here the volume decreased, so V is negative. When V is negative, w is
positive. As the reactants were converted to products, a compression occurred which is
associated with work flowing into the system (w is positive).
97.
ΔEoverall = ΔEstep 1 + ΔEstep 2; this is a cyclic process, which means that the overall initial state
and final state are the same. Because ΔE is a state function, ΔEoverall = 0 and ΔEstep 1 = 
ΔEstep 2.
ΔEstep 1 = q + w = 45 J + (10. J) = 35 J
ΔEstep 2 = ΔEstep 1 = 35 J = q + w, 35 J = 60. J + w, w = 25 J
Challenge Problems
98.
There are five parts to this problem. We need to calculate:
1. q required to heat H2O(s) from 30.oC to 0oC; use the specific heat capacity of H2O(s)
2. q required to convert 1 mol H2O(s) at 0 oC into 1 mol H2O(l) at 0 oC; use Hfusion
3. q required to heat H2O(l) from 0 oC to 100. oC; use the specific heat capacity of H2O(l)
4. q required to convert 1 mol H2O(l) at 100. oC into 1 mol H2O(g) at 100. oC;
use Hvaporization
5. q required to heat H2O(g) from 100. oC to 140. oC; use the specific heat capacity of
H2O(g)
388
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
We will sum up the heat required for all five parts, and this will be the total amount of heat
required to convert 1.00 mol of H2O(s) at  30. oC to H2O(g) at 140. oC.
q1 = 2.03 J oC1 g1 × 18.02 g × [0 – ( 30.)] oC = 1.1 × 103 J
q2 = 1.00 mol × 6.01 × 103 J/mol = 6.01 × 103 J
q3 = 4.18 J °C1 g1 × 18.02 g × (100. – 0) oC = 7.53 × 103 J
q4 = 1.00 mol × 40.7 × 104 J/mol = 4.07 × 104 J
q5 = 2.02 J °C1 g1 × 18.02 g × (140. – 100.) oC = 1.5 × 103 J
qtotal = q1 + q2 + q3 + q4 + q5 = 5.68 × 104 J = 56.8 kJ
99.
Molar heat capacity of H2O(l) = 4.184 J K1 g1(18.015 g/mol) = 75.37 J K1 mol1
Molar heat capacity of H2O(g) = 2.02 J K1 g1(18.015 g/mol) = 36.4 J K1 mol1
Using Hess’s law and the equation ΔH = nCpΔT:
ΔH1 = 1 mol(75.37 J K1 mol1)(75.0 K)(1 kJ/1000 J)
= 5.65 kJ
H2O(l, 373.2 K)  H2O(g, 373.2 K) ΔH2 = 1 mol(40.66 kJ/mol) = 40.66 kJ
H2O(g, 373.2 K)  H2O(g, 298.2 K) ΔH3 = 1 mol(36.4 J K1 mol1)(75.0 K)(1 kJ/1000 J)
= 2.73 kJ
H2O(l, 298.2 K)  H2O(l, 373.2 K)
H2O(l, 298.2 K)  H2O(g, 298.2 K)
ΔH vap, 298.2 K = ΔH1 + ΔH2 + ΔH3 = 43.58 kJ/mol
Using ΔH of values in Appendix 4 (which are determined at 25 C):
Hvap = 242 kJ  (286 kJ) = 44 kJ
To two significant figures, the two calculated Hvap values agree (as they should).
100.
Energy needed =
20.  103 g C12H 22O11
1 mol C12H 22O11
5640 kJ
= 3.3 × 105 kJ/h


h
342.3 g C12H 22O11
mol
Energy from sun = 1.0 kW/m2 = 1000 W/m2 =
10,000 m2 
1000 J 1.0 kJ

s m2
s m2
1.0 kJ 60 s 60 min


= 3.6 × 107 kJ/h
min
h
s m2
Percent efficiency =
energy used per hour
3.3  105 kJ
× 100 =
× 100 = 0.92%
totalenergy per hour
3.6  107 kJ
CHAPTER 9
101.
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
Energy used in 8.0 hours = 40. kWh =
Energy from the sun in 8.0 hours =
389
40. kJ h 3600 s

= 1.4 × 105 kJ
s
h
10. kJ 60 s 60 min


× 8.0 h = 2.9 × 104 kJ/m2
2
min
h
sm
Only 15% of the sunlight is converted into electricity:
0.15 × (2.9 × 104 kJ/m2) × area = 1.4 × 105 kJ, area = 32 m2
102.
If the gas is monoatomic:
Cv =
3
5
R = 12.47 J K1 mol1 and Cp = R = 20.79 J K1 mol1
2
2
If the gas is behaving ideally, then Cp  Cv = R = 8.3145 J K1 mol1.
At constant volume: qv = 2079 J = nCvΔT
Cv =
2079 J
2079 J

= 20.79 J K1 mol1
nΔT
(1 mol)(400.0  300.0 K)
Because Cv ≠ 3/2 R = 12.47 J, the gas is not a monoatomic gas.
At constant pressure: qp = nCpΔT
qp = ΔE  w = 1305 J  (150. J) = 1455 J (Gas expansion, so system does work on
surroundings.)
qp
1455 J

Cp =
= 29.1 J K1 mol1
nΔT
(1 mol)(600.0  550.0 K)
Cp  Cv = 29.1  20.79 = 8.3 J K1 mol1 = R
The gas is behaving ideally since Cp  Cv = R.
103.
For an isothermal (constant T) process involving the expansion or compression of a gas, E =
nCvT = 0 (H is also zero). Because E = q + w = 0, q = w = (PexV). So if the
expansion or compression occurs against some nonzero external pressure, w  0 and q  0.
Instead, q = w = PexV (if the gas expands or contracts against some constant pressure).
390
104.
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
a. Using Hess's law and the equation ΔH = nCpΔT:
CH3Cl(248°C) + H2(248°C) → CH4(248°C) + HCl(248°C)
ΔH1 = -83.3 kJ
CH3Cl(25°C)  CH3Cl(248°C)
ΔH2 = 1 mol(48.5 J °C1 mol1)(223°C)
= 10,800 J = 10.8 kJ
H2(25°C)  H2(248°C)
ΔH3 = 1(28.9)(223)(1 kJ/1000 J) = 6.44 kJ
CH4(248°C)  CH4(25°C)
ΔH4 = 1(41.3)(223)(1/1000) = 9.21 kJ
HCl(248°C)  HCl(25°C)
ΔH5 = 1(29.1)(223)(1/1000) = 6.49 kJ
CH3Cl(25°C) + H2(25°C)  CH4(25°C) + HCl(25°C)
ΔH° = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5
ΔH° = 83.3 kJ + 10.8 kJ + 6.44 kJ  9.21 kJ  6.49 kJ = 81.8 kJ
b. ΔH° = (ΔH of , CH4  ΔH of , HCl )  ( ΔH of , CH3Cl + ΔH of , H 2 )
81.8 kJ = (75 kJ  92 kJ)  (ΔH of , CH3Cl + 0), ΔH of , CH3Cl = 85 kJ/mol
105.
H2O(s)  H2O(l) ΔH = ΔHfus; for 1 mol of supercooled water at 15.0°C (or 258.2 K),
ΔHfus, 258.2 K = 10.9 kJ/2.00 mol = 5.45 kJ/mol. Using Hess’s law and the equation ΔH = nCpΔT:
H2O(s, 273.2 K)  H2O(s, 258.2 K)
H2O(s, 258.2 K)  H2O(l, 258.2 K)
H2O(l, 258.2 K)  H2O(l, 273.2 K)
ΔH1 = 1 mol(37.5 J K1 mol1)(15.0 K)
= 563 J = 0.563 kJ
ΔH2 = 1 mol(5.45 kJ/mol) = 5.45 kJ
ΔH3 = 1 mol(75.3 J K1 mol1)(15.0 K)
= 1130 J = 1.13 kJ
______________________________________________________________________________________________________________
H2O(s, 273. 2 K)  H2O(l, 273.2 K)
ΔHfus, 273.2 = ΔH1 + ΔH2 + ΔH3
ΔHfus, 273.2 = 0.563 kJ + 5.45 kJ + 1.13 kJ, ΔHfus, 273.2 = 6.02 kJ/mol
106.
a. The gas will flow out of the 4.0 L bulb into the 20.0-L bulb. Eventually, the gas will be
evenly dispersed throughout the two bulbs. The pressure will be the same inside both bulbs.
The moles of gas (n) and the temperature (T) are constant. However, as the gas expands from
4.0 L to 24.0 L, the pressure will decrease.
Pinitial = Pi =
nRT
2.4 mol(0.08206L atm K 1 mol1 )(305 K)
=
= 15 atm
Vi
4.0 L
Pfinal = Pf =
Pi Vi
15 atm(4.0 L)
=
= 2.5 atm
24.0 L
Vf
b. Assuming the gas behaves ideally, then E = nCvT and H = nCpT. Because T = 0, E
= 0 and H = 0. From E = q + w = 0, q = w. Here we have an expansion of a gas where w
= PexV. The gas is expanding against a vacuum (Pex = 0), so w = 0. We call this a free
expansion of a gas. Work can only occur in the expansion of a gas when the gas expands
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
391
against a certain nonzero external pressure. Because w = 0, q = w = 0. For a free expansion
of an ideal gas at a constant temperature, E = 0, H = 0, q = 0, and w = 0. If the expansion
occurs against some nonzero constant external pressure, then E = 0, H = 0, q = w =
PexV.
c. The driving force is the natural tendency of processes to spontaneously proceed toward states
that have the highest probability of existing. In this problem the gas mixed evenly throughout
both bulbs is the most probable (likely) state to occur. This driving force is related to entropy,
which will be discussed in detail in Chapter 10.
Marathon Problems
107.
X  CO2(g) + H2O(l) + O2(g) + A(g) ΔH = 1893 kJ/mol
(unbalanced)
To determine X, we must determine the moles of X reacted, the identity of A, and the moles of A
produced. For the reaction at constant P (ΔH = q):
 q H 2O  q rxn = 4.184 J °C-1 g-1(1.000 × 104 g)(29.52 - 25.00 °C)(1 kJ/1000 J)
qrxn = 189.1 kJ (carrying extra significant figures)
Because ΔH = 1893 kJ/mol for the decomposition reaction, and because only -189.1 kJ of heat
was released for this reaction, 189.1 kJ × (1 mol X/1893 kJ) = 0.100 mol X were reacted.
Molar mass of X =
22.7 g X
= 227 g/mol
0.100 mol X
From the problem, 0.100 mol X produced 0.300 mol CO2, 0.250 mol H2O, and 0.025 mol O2.
Therefore, 1.00 mol X contains 3.00 mol CO2, 2.50 mol H2O, and 0.25 mol O2.
 18.0 g 
 44.0 g 
1.00 mol X = 227 g = 3.00 mol CO2 

 + 2.50 mol H2O 
 mol 
 mol 
 32.0 g 
+ 0.25 mol O2 
 + (mass of A)
 mol 
Mass of A in 1.00 mol X = 227 g  132 g  45.0 g  8.0 g = 42 g A
To determine A, we need the moles of A produced. The total moles of gas produced can be
determined from the gas law data provided in the problem. Because H2O(l) is a product, we need
to subtract PH 2O from the total pressure.
ntotal =
PV
; Ptotal = Pgases + PH 2O ; Pgases = 778 torr  31 torr = 747 torr
RT
392
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
 1L 
 = 12.0 L
1V = height × area;; area = πr2; V = (59.8 cm)(π)(8.00 cm)2 
3 
 1000 cm 
T = 273.15 + 29.52 = 302.67 K
 1 atm 
 (12.0 L)
747 torr
760 torr 
PV

ntotal =
=
= 0.475 mol = mol CO2 + mol O2 + mol A
0.08206L atm
RT
(302.67 K )
K mol
Mol A = 0.475 mol total  0.300 mol CO2  0.025 mol O2 = 0.150 mol A
Because 0.100 mol X reacted, 1.00 mol X would contain 1.50 mol A, which from a previous
calculation represents 42 g A.
Molar mass of A =
42 g A
= 28 g/mol
1.50 mol A
Because A is a gaseous element, the only element that is a gas and has this molar mass is N2(g).
Thus A = N2(g).
a. Now we can determine the formula of X.
X  3 CO2(g) + 2.5 H2O(l) + 0.25 O2(g) + 1.5 N2(g). For a balanced reaction, X =
C3H5N3O9, which, for your information, is nitroglycerine.
 1 atm 
 (12.0 L  0) = 12.3 L atm
b. w = PΔV = 778 torr 
 760 torr 
 8.3145 J K 1 mol1
12.3 L atm 
1
1
 0.08206L atm K mol

 = 1250 J = 1.25 kJ, w = 1.25 kJ


c. ΔE = q + w, where q = ΔH since at constant pressure. For 1 mol of X decomposed:
w = 1.25 kJ/0.100 mol = 12.5 kJ/mol
ΔE = ΔH + w = 1893 kJ/mol + (12.5 kJ/mol) = 1906 kJ/mol
108.
 2 x  y/2 
CxHy + 
 → x CO2 + y/2 H2O
2


[x(393.5) + y/2 (242)]  ΔH oC x H y = 2044.5,  (393.5)x  121y  ΔH C x H y = 2044.5
dgas =
P  MM
, where MM = average molar mass of CO2/H2O mixture
RT
CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
0.751 g/L =
1.00 atm  MM
, MM of CO2/H2O mixture = 29.1 g/mol
0.08206L atm
 473 K
K mol
Let a = mol CO2 and 1.00  a = mol H2O (assuming 1.00 total moles of mixture)
(44.01)a + (1.00  a) × 18.02 = 29.1; solving: a = 0.426 mol CO2 ; mol H2O = 0.574 mol
y
0.574
y
 2 , 2.69  , y = (2.69)x
Thus:
0.426
x
x
For whole numbers, multiply by three, which gives y = 8, x = 3. Note that y = 16, x = 6 is
possible, along with other combinations. Because the hydrocarbon is less dense than Kr, the
molar mass of CxHy must be less than the molar mass of Kr (83.80 g/mol). Only C3H8 works.
2044.5 = 393.5(3)  121(8)  ΔH oC3H8 , ΔH oC3H8 = 104 kJ/mol
393