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Chemistry 122 [Tyvoll]
Spring 2008
Chapter 14 Homework Solutions for Moore
Problems # – 14, 18, 20, 24, 25, 28, 31, 33, 38
14. Strategy and Explanation: For the expression of the equilibrium constant, use this form:
KC = [products]/[reactants]; if a stoichiometric coefficient precedes a species, that number is
used as the mathematical power of the concentration in the expression. Remember that the
equilibrium state is not affected by the relative quantity of any solids or liquids, so those
materials do not show up in the equilibrium expression. The answer for (b) does not have SiO2
(s) in the expression, the answer for (c) does not have MgO (s) or MgSO4 (s) in the expression,
and the answer for (d) does not have PbS (s) or PbO (s) in the expression.
(a) KC = [O3]2/[O2]3
(c) KC = 1/([SO2][O2]1/2)
(b) KC = [H2O]2/([SiH4][O2]2)
(d) KC = [SO2]2/ [O2]3
18. Strategy and Explanation: Follow the same instructions given in the solution to Question 14.
The answer for (a) does not have CaSO4•5 H2O (s) and CaSO4•3 H2O (s) in the expression, the
answer for (b) does not have SiO2 (s) in the expression, and the answer for (c) does not have
H2O (l) in the expression.
(a) KC = [O2]/([N2O4][O3])
(c) KC = [H2O(g)]
(b) KC = [CO2][N2]2/[N2O]2
20. Strategy and Explanation: The two equations are related, so we will use the information in
Section 14.2 to identify how their equilibrium constants are related. Multiplying the first
equation by a constant factor of 3 gives the second equation.
(1) N2H4 (g) + 4/3 ClF3 (g) ' 4 HF (g) + N2 (g) + 2/3 Cl2 (g)
(2) 3 N2H4 (g) + 4 ClF3 (g) ' 12 HF (g) + 3 N2 (g) + 2 Cl2 (g)
This change means we need to raise the K from the first reaction to the power of 3. Therefore
the second equation’s equilibrium constant is related to the first equation’s equilibrium
constant as represented in choice (c), Kp3 = Kp'.
24. Strategy and Explanation: Follow the same instructions given in the solution to Question 14.
The answer for (b) does not have SiO2(s) in the expression, the answer for (c) does not have
MgO(s) or MgSO4(s) in the expression, and the answer for (d) does not have PbS(s) or PbO(s)
in the expression.
(a) KP = PO32/PO23
(c) KP = 1/(PSO2PO21/2)
(b) KP = PH2O2/(PSiH4PO22)
(d) KP = PSO22/ PO23
25. Strategy and Explanation: Given a phase change reaction and the vapor pressure of the gaseous
product, find the value of KC. Write the KP expression for the reaction and plug in the known
value. Then use the relationship between KP and KC (Equation 14.5), with R = 0.08206
Latm/molK, and Δn = change in the number of moles of gas in the reaction, to get KC.
H2O (l) ' H2O (g)
KP = PH2O(g)= 0.467 atm
Now, assume temperature is 80.°C + 273.15 = 353 K and Δn = 1 mol H2O gas product – 0 mol
gas reactants = 1.
KP = KC(RT)Δn
So, KC = KP(RT)−Δn = (0.467 atm){(0.08206 Latm/molK)(353 K)}−1
KC = 0.0161
Answer: KC = 0.0161
28. Strategy and Explanation: Given the equation for a dimerization reaction and the concentrations
of the gaseous reactants and product at equilibrium, find the value of KC. Write the KC
expression for the reaction and substitute the known values.
2 A (g) ' B (g)
KC = [B]/[A]2
(a) KC = (0.74 M)/(0.74 M)2 = 1.4
(b) KC = (2.0 M)/(2.0 M)2 = 0.50
(c) KC = (0.01 M)/(0.01 M)2 = 100
Because the reactant concentration is squared, it makes sense that the value of
KC gets larger as the concentrations get smaller.
31. Strategy and Explanation: Given an equation for a reaction and the moles of the gaseous
reactants and products at equilibrium in a known volume, find the value of KP. Write the KC
expression for the reaction, calculate the concentrations of the gases, and plug them into the
expression to get the value of KC. Use the relationship between KP and KC (also described
above in the solution to Question 25) to get KP:
KP = KC(RT)Δn
H2 (g) + CO2 (g) ' H2O (g) + CO (g)
KC = ([H2O][CO])/([H2][CO2]) = {0.11 M)(0.11 M)}/{0.087 M)(0.087 M)} = 1.6
where molarities = mol since volume was 1.0 L; t = 986 °C + 273.15 = 1259 K
Δn = 1 mol H2O (g) + 1 mol CO (g) – 1 mol H2 (g) – 1 mol CO2 (g) = 0
KP = KC(RT)Δn = KC(RT)0 = KC = 1.6
Answer: Kp = 1.6
33. Strategy and Explanation: Given an equation for a reaction, the initial moles of the gaseous
reactant in a known volume, and the equilibrium concentration of the gaseous reactant, find the
value of KC. Following the procedure given in Section 14.3, write the equation and construct a
reaction table. Put all known concentrations in the table. There is no NO2 (g) present, initially.
Initially, CN2O4 = 0.010 mol N2O4/2.0 L = 0.0050 M and, at equilibrium, [N2O4] = 0.00090 M.
KC = [NO2]2/[N2O4] = ??
Construct the ICE table: (reaction proceeds left to right since initial CNO2 = 0)
I
C
E
[N2O4]
0.0050
−x
0.00090
[NO2]
0
+ 2x
2x
Now, for [N2O4], 0.0050 − x = 0.00090; x = 0.0050 − 0.00090 = 0.0041 M
2x = 2(0.0041) = 0.0082 M = [NO2] (at equilibrium)
KC = [NO2]2/[N2O4] = (0.0082)2 / (0.00090) = 0.075
The smaller product concentration suggests this reaction is reactant-favored, so it makes sense that
the KC is smaller than 1.
Answer: Kc = 0.075
38. Strategy and Explanation: Given a table of equations with values of KC and KP, order a set
equations in order of most reactant-favored to most product-favored. Look up the given
equation or a related equation, determine the size of Kc, using techniques described at the end
of Section 14.2, as needed. KC values larger than 1 are product-favored. The smaller KC is, the
more reactant-favored the reaction is. The larger KC is, the more product-favored the reaction
is. Order the equations from smallest KC to largest KC.
(a) 2 NO2 (g) ' N2O4 (g) is the sixth reaction in Table 14.1, so KC = 1.7 × 102. KC is larger
than 1, so the reaction is product-favored.
(b) H2CO3 (aq) ' H+ (aq) + HCO3– (aq) is the same as the eleventh reaction in Table 14.1
because the aqueous H+ ion reacts with H2O to form H3O+ (as described in Section 5.2,
page 171), so KC = 4.2 × 10–7. KC is smaller than 1, so the reaction is not product-favored.
(c) AgI (s) ' Ag+ (aq) + I– (aq) is the last reaction in Table 14.1, so KC = 1.5 × 10–16. KC is
smaller than 1, so the reaction is not product-favored.
Therefore, the order is (c) 1.5 × 10–16, then (b) 4.2 × 10–7, then (a) 1.7 × 102.
Answer: Reaction (a) is product-favored. Most reactant-favored (c), then (b), then (a).