AP Chemistry Chapter 6 Answers – Zumdahl
6.31
This is an endothermic reaction so heat must be absorbed in order to convert reactants into
products. The high temperature environment of internal combustion engines provides the heat.
6.33
a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an endothermic
process.
b. Heat is released as CH4 is burned, so this is an exothermic process.
c. Heat is released to the water (it gets hot) as H2SO4 is added, so this is an exothermic process.
d. Heat must be added (absorbed) to boil water, so this is an endothermic process.
6.35
4Fe(s) + 3O2(g) → 2Fe2O3(s) ∆H = -1652 kJ; Note that 1652 kJ of heat are released when 4 mol
Fe react with 3 mol O2 to produce 2 mol Fe2O3.
a. 4.00 mol Fe x (-1652 kJ/4 mol Fe) = -1650 kJ; 1650 kJ of heat released
b. 1.00 mL Fe2O3 x (-1652 kJ/2 mol Fe2O3) = -826 kJ; 826 kJ of heat released
c. 1.00 g Fe x (1 mol Fe/55.85 g) x (-1652 kJ/4 mol Fe) = -7.39 kJ; 7.39 kJ of heat released
d. 10.0 g Fe x (1 mol Fe/55.85 g) = 0.179 mol Fe; 2.00 g O2 x (1 mol O2/32.00 g O2) = 0.0625
mol O2
0179 mol Fe/0.0625 mol O2 = 2.86; The balanced equation requires a 4 mol Fe/3mol O2 =
1.33 mol ratio. O2 is limiting since the actual mol Fe/mol O2 ratio is greater than the required
mol ratio.
0.0625 mol O2 x (-1652 kJ/3 mol O2) = -34.4 kJ; 34.4 kJ of heat release
6.37
From Sample Exercise 6.3, q = 1.3 x 108 J. Since the heat transfer process is only 60.% efficient,
the total energy required is: 1.3 x 108 J (100. J/60. J)= 2.2 x 108 J
mass C3H8 = 2.2 x 108 J x (1 mol C3H8/2221 x 103 J) x (44.09 g C3H8/1 mol C3H8) =
4.4 x 103 g C3H8
6.41
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:
energy = s x m x ∆T = (4.18 J/g oC) x 25.0 g x (37.0oC – 15.0oC) = 2.30 x 103 J
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:
∆T = energy/s x m = (10.7 kJ x 1000 J/1 kJ)/(0.14 J/g oC x 550. g) = 140oC
6.43
s = specific heat capacity = q/(m x ∆T) = 133 J/(5.00 g x (55.1 – 25.2)oC) = 0.890 J/oC g
From Table 6.1, the substance is aluminum.
6.45
| 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/oC g = 4.18 J/K g (∆T in oC = ∆T in K).
Heat loss by hot water = s x m x ∆T = 4.18 J/g K x 50.0 g x (330. K – Tf)
Heat gain by cooler water = 4.18 J/g K x 30.0 g x (Tf – 280. K); Heat loss = Heat gain, so:
209 J/K x (330. K – Ff) = 125 J/K x (Tf – 280. K), 6.90 x 104 – 209Tf = 125Tf – 3.50 x 104
334Tf = 1.040 x 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
6.47
Heat loss by Al + heat loss by Fe = heat gain by water; Keeping all quantities positive to avoid
sign error:
0.89 J/g oC x 5.00 g Al x (100.0oC – Tf) + 0.45 J/goC x 10.0 g Fe x (100.0 – Tf)
= 4.18 J/g oC x 97.3 g H2O x (Tf – 22.0oC)
4.5(100.0 – Tf) + 4.5(100.0 – Tf) = 407(Tf – 22.0), 450 – 4.5Tf + 450 – 4.5Tf =
407Tf – 8950
416Tf = 9850, Tf = 23.7oC
6.49
Heat gain by water = heat loss by metal = s x m x ∆T where s = specific heat capacity.
Heat gain = 4.18 J/g oC x 150.0 g x (18.3oC – 15.0oC) = 2100 J
A common error in calorimetry problems is sign errors. Keeping all quantities positive helps
eliminate sign errors.
heat loss = 2100 J = s x 150.0 g x (75.0oC – 18.3oC), s = 2100 J/(150.0 g x 56.7oC) = 0.25 J/g oC
6.51
50.0 x 10-3 L x 0.100 mol/L = 5.00 x 10-3 mol of both AgNO3 and HCl are reacted. Thus,
5.00 x 10-3 mol of AgCl will be produced since there is a 1:1 mole ratio between reactants.
Heat lost by chemicals = Heat gained by solution
Heat gain = 4.18 J/goC x 100.0 g x (23.40 – 22.60)oC = 330 J
Heat loss = 330 J; This is the heat evolved (exothermic reaction) when 5.00 x 10-3 mol of AgCl
is produced. So q = -330 J and ∆H (heat evolved per mol AgCl formed) is negative with a value
of:
∆H = -330 J/5.00 x 10-3 mol x 1 kJ/1000 J = -66 kJ/mol
Note: Sign errors are common with calorimetry problems. However, the correct sign for ∆H can
easily be determined from the ∆T data, i.e., if ∆T of the solution increases, then the reaction is
exothermic since heat was released, and if ∆T of the solution decreases, then the reaction is
endothermic since the reaction absorbed heat from the water. For calorimetry problems, keep all
quantities positive until the end of the calculation, then decide the sign for ∆H. This will help
eliminate sign errors.
6.55
a. heat gain by calorimeter = heat loss by CH4 = 6.79 x 1 mol CH4/16.04 g x 802 kJ/1 mol =
340. kJ
heat capacity of calorimeter = 340. kJ/10.8oC = 31.5 kJ/oC
b. heat loss by C2H2 = heat gain by calorimeter = 16.9oC x 31.5 kJ/oC = 532 kJ
∆Ecomb = -532 kJ/12.6g C2H2 x 26.04 g/1 mol C2H2 = -1.10 x 103 kJ/mol