Zumdahl – ch6 review
24. Calculate the internal energy change for each of the following.
a. One hundred (100.) joules of work are required to compress a gas. At the same time, the gas releases 23 J of heat
b. A prison is compressed from a volume of 8.30 L to 2.80 L against a constant pressure of 1.90 atm. In the process,
there is a heat gain by the system of 350 J.
c. A piston expands against 1.00 atm of pressure from 11.2 L to 29.1 L. In the process, 1037 J of heat is absorbed.
24.
a. ΔE = q + w = -23 J + 100. J = 77 J
b. w = -P ΔV = -1.90 atm (2.80 L - 8.30 L) = 10.5 L atm ×
101 .3 J
= 1060 J
L atm
ΔE = q + w = 350. J + 1060 = 1410 J
c. w = -P ΔV = -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
25. A sample of an ideal gas at 15.0 atm and 10.0 L is allowed to expand against a constant external pressure of
2.00 atm at a constant temperature. Calculate the work in units of kJ for the expansion. (hint: Boyle’s law applies).
25.
w = -PΔV; We need the final volume of the gas. Since T and n are constant, P1V1 = P2V2.
V2 
V1 P1
P2

10.0 L (15.0 atm)
= 75.0 L
2.00 atm
w = -PΔV = -2.00 atm (75.0 L - 10.0 L) = -130. L atm ×
101.3 J
1 kJ

L atm
1000 J
= -13.2 kJ = work
38. Consider the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
ΔH = -891 kJ
Calculate the enthalpy change for each of the following cases:
a. 1.00 g methane is burned in excess oxygen
b. 1.00 x 103 L methane gas at 740. torr and 25 ̊C is burned in excess oxygen.
38.
a. 1.00 g CH4 ×
b. n =
PV
,
RT
39.8 mol ×
1 mol CH 4
 891 kJ
= -55.5 kJ

16.04 g CH 4 mol CH 4
740 .
atm  1.00  10 3 L
760
= 39.8 mol CH4
0.08206 L atm
 298 K
mol K
 891 kJ
= 3.55 × 104 kJ
mol
45. A 30.0-g sample of water at 280. K is mixed with 50.0 g of water at 330 K. Calculate the final temperature of
the mixture assuming no heat loss to the surroundings.
45.
| Heat loss by hot water | = | Heat gain by cooler water |
The magnitude of heat loss and heat gain are equal in calorimetry problems. The only difference is the sign
(positive or negative). To avoid sign errors, keep all quantities positive and, if necessary, deduce the
correct signs at the end of the problem. Water has a specific heat capacity = s = 4.18 J/°Cg = 4.18 J/Kg
(ΔT in °C = ΔT in K).
Heat loss by hot water = s × m × ΔT =
Heat gain by cooler water =
4.18 J
× 50.0 g × (330. K - Tf)
gK
4.18 J
× 30.0 g × (Tf -280. K); Heat loss = Heat gain, so:
gK
209 J
125 J
× (330. K-Tf) =
× (Tf - 280. K), 6.90 × 104 - 209 Tf = 125 Tf - 3.50 × 104
K
K
334 Tf = 1.040 × 105, Tf = 311 K
Note that the final temperature is closer to the temperature of the more massive hot water, which is as it
should be.
46. A biology experiment requires the preparation of a water bath at 37.0 ̊C. The temperature of the cold tap water
is 22 ̊C, and the temperature of the hot tap water is 55.0 ̊C. If a student starts with 90.0 g of cold water,
what mass of hot water must be added to reach 37.0 ̊C?
46.
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) =
× 90.0 g × (37.0 °C - 22.0°C)
o
g C
g oC
mhot =
90.0 g  15.0 o C
= 75.0 g hot water needed
18.0 o C
57. The enthalpy of combustion of solid carbon to form carbon dioxide is -393.7 kJ/mol carbon, and the enthalpy of
combustion of carbon monoxide to form carbon dioxide is -283.3 kJ/mol CO. Use these data to calculate ΔH for
the reaction
2C(s) + O2(g)  2CO(g)
57.
Information given:
C(s) + O2(g) → CO2(g)
CO(g) + 1/2 O2(g) → CO2(g)
ΔH = -393.7 kJ
ΔH = -283.3 kJ
Using Hess’s Law:
2 C(s) + 2 O2(g) → 2 CO2(g)
ΔH1 = 2(-393.7 kJ)
2 CO2(g) → 2 CO(g) + O2(g)
ΔH2 = -2(-283.3 kJ)
__________________________________________________________
2 C(s) + O2(g) → 2 CO(g)
ΔH = ΔH1 + ΔH2 = -220.8 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, the value of
ΔH is multiplied by the same integer while the sign stays the same.
58. Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and
hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion
for C4H4 (-2341 Kj/MOL), C4H8 (-2755 kJ/mol), and H2 (-286 kJ/mol), calculate ΔH for the reaction.
C4H4(g) + 2H2(g)  C4H8(g)
58.
C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
C4H8(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(l)
H2(g) + 1/2 O2(g) → H2O(l)
ΔHcomb = -2341 kJ
ΔHcomb = -2755 kJ
ΔHcomb = -286 kJ
By convention, H2O(l) is produced when enthalpies of combustion are given and, since 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
75. Calculate the enthalpy of combustion per gram of liquid ethanol (C2H5OH).
75.
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
46.07 g
76. Calculate the enthalpy of combustion per gram of liquid ethanol (CH3OH). Compare your answer to that of
ethanol in previous exercise.
76.
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 vs. -29.67 kJ/g for ethanol

mol
32.04 g
Ethanol has a slightly higher fuel value than methanol.
103. The preparation of NO2(g) from N2(g) and O2(g) is an endothermic reaction:
N2(g) + O2(g) → NO2(g) (unbalanced)
The enthalpy change of the reaction for the balanced equation (with the lowest whole number coefficients) is H =
67.7 kJ. If 2.50 x 102 mL of N2(g) at 100.°C and 3.50 atm and 4.50 x 102 mL of O2(g) at 100.°C and 3.50 atm are
mixed, what amount of heat is necessary to synthesize NO2(g)?
103.
N2(g) + 2 O2(g) → 2 NO2(g)
H = 67.7 kJ
n N2 =
PV
3.50 atm  0.250 L
= 2.86 × 10 2 mol N2

0.08206  L atm
RT
 373 K
mol K
n O2 =
PV
3.50 atm  0.450 L
= 5.15 × 10 2 mol O2

0.08206  L atm
RT
 373 K
mol K
The balanced equation requires a 2:1 O2 to N2 mole ratio. The actual mole ratio is
5.15 × 10 2 /2.86 × 10 2 = 1.80; Because the actual mole ratio < required mole ratio,
O2 in the numerator is limiting.
5.15 × 10 2 mol O2 ×
2 mol NO2
= 5.15 × 10 2 mol NO2
2 mol O 2
5.15 × 10 2 mol NO2 ×
67.7 kJ
= 1.74 kJ
2 mol NO2