Charpter 12
12.15 ANS
(a) This inference can be made because the equilibrium symbol implies this fact.
(b) This is not implied by equilibrium. Although concentrations do not change at
equilibrium, this does not imply that they are all the same.
(c) This is not implied either. Chemical equilibrium can be approached from either
“direction” – starting with all reactants or with all products are just two possible
starting points.
(a) K3 = K1xK2 = (2.3 x 106)x(1.8 x 1010) = 4.1 x 1016.
(b) For the reverse reaction, K3-1 = 2.4 x 10-17.
The 3rd reaction equation can be obtained by (eq. 1) – (eq. 2). Therefore,
K = K1/K2 = (4.0 x 10-12)/(5.9 x 106) = 6.8 x 10-19.
Assume at equilibrium, the concentration of H2 is equal to that of I2,
[H2] = [I2] = x, and [HI] = (0.500-2x) mol L-1
K = [HI(g)]2/[H2(g)][I2(g)] = (0.500- 2x)2/x2 = 161
x = 0.0340 mol L-1
[H2] = [I2] = 0.0340 mol L-1
[HI] = 0.500-2x = 0.432 mol L-1.
Cl2(g)
+
PCl3(g) Æ PCl5(g) ……..K = 51
Initial : 0.125 (M)
0.012 (M)
0 (M)
Final: 0.125-x (M) 0.012-x (M)
x (M)
K = 51 = x/[(0.125-x) (0.012-x)]
51 (0.00150 – 0.137x + x2) = x
51x2 – 8.0x + 0.076 = 0
x = 0.010 M = [PCl5]
[Cl2] = 0.125-x = 0.115 M.
[PCl3] = 0.012-x = 0.002 M.
12.55 ANS
(a) no effect on equilibrium concentrations.
(b) shifts toward products.
(c) shifts toward products.
(d) does not affect the equilibrium.
AgCl(s) Æ Ag+(aq) + Cl-(aq)
Initial: solid
0M
0M
Final:
solid
x
x
Ksp = [Ag-][Cl-] = 1.8 x 10-10, [Ag-] = [Cl-] = x,
x = 1.3 x 10-5 M = molar solubility of AgCl.
(a)
BaSO4(g) Æ Ba2+(aq) + SO4-(aq)
Initial: solid
0M
0M
Final:
solid
x
x
2+
4-10
Ksp = [Ba ][SO ] = 1.1 x 10 ,
[Ba2+] = [SO4-] = x, x2 = 1.1 x 10-10, x = 1.0 x 10-5 M = molar solubility of BaSO4.
(b)
MW of BaSO4 = 233.40 g mol-1.
Solubility in grams per 100 g of water = (1.0 x 10-5 mol)x(233.450 g mol-1)/[(1 L)x(1
1000g/L)- (1.0 x 10-5 mol)x(233.450 g mol-1)] = 2.3 x 10-4 g per 100 g of water.
(a) in pure water,
Cu(OH)2 Æ Cu2+(aq) + 2OH-(aq)
Initial: Solid
0M
0M
Final:
Solid
x
2x
2+
- 2
3
Ksp = [Cu ][2OH ] = 4x = 1.6 x 10-19.
Solubility of Cu(OH)2 is 3.4 x 10-7 M.
(b) in 0.01 M NaOH.
NaOH is a strong base and it would totally dissociate to Na+ and OH- in water.
The concentration of [OH-] = 0.01 M.
Cu(OH)2 Æ Cu2+(aq) + 2OH-(aq)
Initial: Solid
0M
0.01M
Final:
Solid
x
(0.01 + 2x) M
[x][0.01 + 2x]2 = 1.6 x 10-19.
Assume x << 0.01, [0.01 + 2x] ~ 0.01
0.0001x = 1.6 x 10-19, x = 1.6 x 10-15 M.
HBrO Æ H+ + BrO-.........Ka = ?
initial: 0.10
0
0
final:
0.10-x
x
x
+
+
pH = 6.4 = -log [H ], [H ] = 10-6.4 = 4.0 x 10-7 M.
Ka = [4.0 x 10-7]x[4.0 x 10-7]/[0.10 – 4.0 x 10-7] = 1.6 x 10-12.
12.85 ANS
PCl5(g) Æ PCl3(g) + Cl2(g)
K = 7.7 x10-3
(a) Because the value of K is less than 1, it suggests that equilibrium lies toward
reactants. Therefore, theΔGo should be a positive number.
(b) ΔGo = -RT ln(K)
At T = 298 K, ΔGo = -(8.314 J mol-1 K-1)x(298 K)ln(7.7x10-3) = 12.1 kJ mol-1.