Chem 1B
Answers to Chapter 13 HW Exercises for Zumdahl’s 8th edition. (Problem numbers in parentheses are for
7th edition.)
15 (11)
Numbers of molecules at equilibrium: 4 H2O, 2 CO, 4 H2, and 4 CO2.
24 (20)
(a) Kp = 5.3 x 10-3;
27 (23)
K = 1.7 x 10-5;
30 (26)
Kp = 3.8 x 104;
32 (28)
K = 23
40.
Kc = Kp = 0.0900;
(b) Kp = 2.9 x 10-5;
(c) 190
Qp = 1.21 x 103  mixture is not at equilibrium.
(a) Qp = 1.0 x 102 > Kp  not equilibrium; reaction shifts left
(b) Qp = 0.090 = Kp  equilibrium
(c) Qp = 110 > Kp  not equilibrium; reaction shifts left
45 (39)
[SO3] = [NO] = 1.30 M;
[SO2] = [NO2] = 0.70 M;
Kc = 3.4
49 (43)
[H2]0 = 5.0 M + 6.0 M = 11.0 M;
51 (45)
[SO2] = [NO2] = 0.54 M; [SO3] = [NO] = 1.06 M.
53 (47)
PNO = 7.8 x 10-2 atm; PN2 = 0.76 atm; PO2 = 0.16 atm
59 (53)
[CO2]e = 0.39 M;
[CO] = 8.6 x 10-3 M;
61 (55)
PNH3 = 0.18 atm;
PCO2 = 0.090 atm;
Ptotal = 0.27 atm
65 (59)
(a) shift right;
(b) shift right;
(c) No effect;
(d) shift left;
(b) Shift right;
(c) No effect;
(d) Shift right
[N2]0 = 8.0 M + 2.0 M = 10.0 M;
[O2] = 4.3 x 10-3 M.
(e) No effect;
68 (62)
(a) Shifts left;
(e) Shift right
81 (65)
K = (1.5 x 1048)(1.7 x 1033) = 2.6 x 1081
88 (69)
(a) Initial PPCl5 = 1.16 atm;
(b) PPCl5 = 0.10 atm (at equilibrium)
(c) At equilibrium, PPCl3 = PCl2 = 1.06 atm, PPCl5 = 0.10 atm; Ptotal = 2.22 atm
92 (72)
(a) Doubling the volume decreases concentrations - reaction shifts to the left to increase
concentrations;
(b) Adding Ag+, reduces SCN- due to formation of AgSCN – reaction shifts left;
(c) Adding OH-, decreases Fe3+ due to formation of Fe(OH)3 precipitate – reaction shifts left;
(d) Adding Fe3+, reaction shifts right.
93 (73)
OH- reacts with H+ to form H2O, decreasing H+ and reaction shifts right to produce more H+
and CrO42- and decreasing Cr2O72-  solution changes from orange to yellow.
Chem 1B
94 (74)
The reaction: N2 + 3H2  2NH3 is exothermic
(a) Increasing the temperature causes the equilibrium to shifts left and decreasing K value, but
the reaction rate will increase – producing NH3 at shorter time. At low temperature the reaction
shift right and increasing the yields of NH3 at equilibrium. However, the reaction will be too
slow to be of any use.
(b) Removing NH3 causes reaction to shift right – making more NH3;
(c) A catalyst increases both forward and reversed reaction – no net change to concentration at
equilibrium. However, the reaction rate becomes faster and equilibrium amount of NH3 is
formed at much shorter time.
(d) Increasing pressure caused reaction to shifts right, reducing the number of gas molecules.
More NH3 will be produced.
100 (80) (a) PNO = 0.0526 atm;
(b) PNO = 0.052 atm;
PBr2 = 0.0159 atm,
PNOBr = 0.0768 atm;
PBr2 = 0.18 atm;
PNOBr = 0.25 atm
and Kp = 134
103 (83) Kp = 4.2
Doubling the volume decreases pressure by a factor of 2; PNO2 = 0.600 atm and PN2O4 = 0.17
Qp = 2.1 < Kp  reaction shifts right;
Pressures at new equilibrium: PNO2 = 0.704 atm; PN2O4 = 0.12 atm
104 (84) (a) Partial pressure of CO2 and H2O are: PCO2 = PH2O = 0.50 atm
(b) Mole of Na2CO3 = mole of CO2 =
(0.50 atm)(1.00 L)
= 0.015 mol
(0.08206 L.atm/mol .K)(398 K)
Mass of Na2CO3 = (0.015 mol)(105.99 g/mol) = 1.6 g
Mass of NaHCO3 reacted = 2 x 0.015 mol x 84.01 g/mol = 2.5 g
Mass of NaHCO3 at equilibrium = 10.0 – 2.5 = 7.5 g
(c) Mole of CO2 produced when 10.0 g of NaHCO3 is completely decomposed
= 10.0 g x
1 mol CO 2
1 mol
x
= 5.95 x 10-2 mol CO2
84.01 g 2 mol NHCO 3
Minimum volume of container needed to enable all of NaHCO3 to decompose is
V = nRT/P =
(5.95 x 10-2 mol)(0.082 06 L.atm/mol .K)(398 K)
= 3.9 L
0.50 atm
111 (91) Mole of NH4HS produced at equilibrium = 5.00 L x 0.350 mol/L = 1.75 mol
Total mole of NH4HS = 2.00 mol + 1.75 mol = 3.75 mol
Total mass of NH4HS at equilibrium = 3.75 mol x 51.12 g/mol = 192 g
Partial pressure of H2S = (0.050 mol/L)(0.08206 L.atm/K.mol)(308.2 K) = 1.3 atm