Chapter 7 Review, pages 480–485
Knowledge
1. (b)
2. (b)
3. (a)
4. (c)
5. (c)
6. (b)
7. (c)
8. True
9. False. A reversible chemical reaction produces the same set of equilibrium
concentrations of reactants and products in the forward direction as in the reverse
direction.
10. False. The equilibrium constant for any chemical reaction system varies with
temperature.
11. True
12. False. Fritz Haber was able to increase the rate of synthesis of ammonia gas, NH3(g),
from gaseous hydrogen, H2(g), and nitrogen, N2(g), by adding a catalyst to the reaction.
13. True
14. True
15. False. If the reaction quotient, Q, is less than the equilibrium constant, K, the system
will shift toward products to achieve equilibrium.
16. True
17. (a) (iii)
(b) (iv)
(c) (ii)
(d) (i)
Understanding
18. Reaction A: Products are favoured; Reaction B: Reactants are favoured.
19. The equilibrium position of a chemical reaction is the point at which the
concentrations of reactants and products in an equilibrium system stop changing.
20. Given: [NO(g)]initial = 3.00 mol/L; [N2O(g)]initial = 0.00 mol/L;
[O2(g)]initial = 0.00 mol/L; [N2O(g)]equilibrium = 1.00 mol/L
Required: [NO(g)]equilibrium; [O2(g)]equilibrium
Analysis: Use an ICE table to determine the relationship between the equilibrium
concentrations of the reactants and the product.
I
C
E
2 N2O(g)
0.00
+x
1.00
+
!!
O2(g) #
!"
!
0.00
+0.5x
0.5x
Copyright © 2012 Nelson Education Ltd.
4 NO(g)
3.00
−2x
3.00 – 2x
Chapter 7: Chemical Equilibrium
7-2
Solution: x represents the change in concentration of N2O(g).
x = [N 2O(g)]equilibrium ! [N 2O(g)]initial
= 1.00 mol/L ! 0.00 mol/L
x = 1.00 mol/L
[O 2 (g)]equilibrium = 0.5 x
= 0.5(1.00 mol/L)
[O 2 (g)]equilibrium = 5.00 ×10 −1 mol/L
[NO(g)]equilibrium = (3.00 mol/L) – 2 x
= (3.00 mol/L) – 2(1.00 mol/L)
[NO(g)]equilibrium = 1.00 mol/L
[NO(g)]equilibrium = 3.00 mol/L – 2x
Statement: The equilibrium concentration of dinitrogen monoxide gas is 1.00 mol/L; the
equilibrium concentration of oxygen gas is 5.00 × 10–1 mol/L; and the equilibrium
concentration of nitric oxide gas is 1.00 mol/L.
21. Given: [Br2(g)]initial = 1.00 mol/L; [Br(g)]initial = 0.00 mol/L;
[Br(g)]equilibrium = 0.032mol/L
Required: [Br2(g)]equilibrium
Analysis: Use an ICE table to determine the relationship between the equilibrium
concentrations of the reactant and the product.
I
C
E
!!
Br2(g) #
!"
! 2 Br(g)
1.00
0.00
−x
+2x
1.00 − x
2x
Solution: x represents the change in concentration of Br2(g).
2x = [Br(g)]equilibrium
2x = 0.032 mol/L
x = 0.016 mol/L
[Br2 (g)]equilibrium = (1.00 mol/L) – x
= (1.00 mol/L) – (0.016 mol/L)
[Br2 (g)]equilibrium = 9.8×10 −1 mol/L
Statement: The equilibrium concentration of diatomic bromine gas is 9.8 × 10–1 mol/L.
22. In a homogeneous equilibrium, all reactants and products are in the same state, but in
a heterogeneous equilibrium, reactants and products are present in at least two different
states.
3
[O (g)]
23. (a) K = 2
[O3 (g)]2
(b) K =
4
[H 2 (g)]
[H 2O(g)]4
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-3
(c) K = [N 2 (g)][H 2O(g)]2
2
[NO(g)] [Cl 2 (g)]
(d) K =
2
[NOCl(g)]
!!
24. (a) N2(g) + O2(g) #
!"
! 2 NO(g)
!!
(b) 2 NOBr(g) #
!"
! 2 NO(g) + Br2(g)
!!
(c) 3 H2(g) + CO(g) #
!"
! CH4(g) + H2O(g)
!!
(d) CS2(g) + 4 H2(g) #
!"
! CH4(g) + 2H2S(g)
−1
!!
25. Given: Cl2(g) + CO(g) #
!"
! COCl2(g); [CO(g)]equilibrium = 1.11 × 10 mol/L;
[Cl2(g)] equilibrium = 1.03 × 10−1 mol/L; [COCl2(g)]= 1.17 × 10−1 mol/L
Required: K
Solution: Write the equilibrium law equation using the balanced chemical equation.
Then, substitute the equilibrium concentrations into the equilibrium law equation and
solve for K.
[COCl2 (g)]
K=
[Cl2 (g)][CO(g)]
(1.17 ! 10"1 )
=
(1.03 ! 10"1 )(1.11 ! 10"1 )
K = 1.2 ! 101
Statement: The equilibrium constant, K, for the reaction is 1.2 × 101.
−1
!!
26. Given: 2 ClF3(g) #
!"
! Cl2(g) + 3 F2(g); [Cl2(g)]equilibrium = 1.62 × 10 mol/L;
[F2(g)]equilibrium = 1.85 × 10−1 mol/L; K = 3.72 × 10−2
Required: [ClF3(g)]equilibrium
Solution:
Write the equilibrium law equation using the balanced chemical equation.
[Cl2 (g)][F2 (g)]3
K=
[ClF3 (g)]2
Substitute the given values for the equilibrium concentrations of the products and the
equilibrium constant and solve for [ClF3(g)].
[Cl2 (g)][F2 (g)]3
K=
[ClF3 (g)]2
3.72 ! 10"2 =
(1.62 ! 10"1 )(1.85 ! 10"1 )3
[ClF3 (g)]2
(1.62 ! 10"1 )(1.85 ! 10"1 )3
3.72 ! 10"2
[ClF3 (g)]2 = 2.76 ! 10"2
[ClF3 (g)]2 =
[ClF3 (g)]2 = 2.76 ! 10"2
[ClF3 (g)]equilibrium = ±1.66 ! 10"1 mol/L
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-4
Since a negative concentration value is not possible in the real world, use
[ClF3 (g)]equilibrium = 1.66×10−1 mol/L
Statement: The equilibrium concentration of chlorine trifluoride gas, ClF3(g), is
1.66 × 10–1 mol/L.
!!
27. Given: 2 SO2(g) + O2(g) #
!"
! 2 SO3(g); [SO2(g)]initial = 0.40 mol/L;
[O2(g)]initial = 1.6 mol/L; [SO3(g)]initial = 29.7 mol/L; [SO2(g)]equilibrium = 1.2 mol/L
Required: K
Analysis: Use an ICE table to determine the relationship between the equilibrium
concentrations of the reactants and the product.
I
C
E
2 SO2(g)
0.40
+x
1.2
+
!!
O2(g) #
!"
!
1.6
+0.5x
1.6 + 0.5x
2 SO3(g)
29.7
−x
29.7 – x
Solution: x represents the change in concentration of SO2(g).
x = [SO 2 (g)]equilibrium ! [SO 2 (g)]initial
= 1.2 mol/L ! 0.40 mol/L
x = 0.80 mol/L
[O 2 (g)]equilibrium = (1.6 mol/L) + 0.5 x
= (1.6 mol/L) + 0.5(0.80 mol/L)
[O 2 (g)]equilibrium = 2.0 mol/L
[SO3 (g)]equilibrium = (29.7 mol/L) – x
= (29.7 mol/L) – (0.80 mol/L)
[SO3 (g)]equilibrium = 28.9 mol/L
[SO3 (g)]2
K=
[SO(g)]2 [O 2 (g)]
=
(28.9) 2
(1.2)2 (2.0)
K = 2.9 ×102
Statement: The equilibrium constant, K, for this chemical reaction system is 2.9 × 102.
−1
!!
28. Given: PCl5(g) #
!"
! PCl3(g) + Cl2(g); [PCl5(g)]equilibrium = 4.06 × 10 mol/L;
[PCl3(g)]equilibrium = 1.17 × 10−1 mol/L; K = 1.24 × 10−1 Required: [Cl2(g)]equilibrium
[PCl3 (g)][Cl2 (g)]
Analysis: K =
[PCl5 (g)]
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-5
K=
Solution:
[PCl3 (g)][Cl2 (g)]
[PCl5 (g)]
(1.17 ! 10"1 )[Cl2 (g)]
4.06 ! 10"1
(1.24 ! 10"1 )(4.06 ! 10"1 )
=
1.17 ! 10"1
= 4.30 ! 10"1 mol/L
1.24 ! 10"1 =
[Cl2 (g)]equilibrium
[Cl2 (g)]equilibrium
Statement: The equilibrium concentration of chlorine gas, Cl2(g), is 4.30 × 10–1 mol/L.
−1
!!
29. Given: 2 P(s) + 3 Cl2(g) #
!"
! PCl3(g); [PCl3(g)]equilibrium = 1.09 × 10 mol/L;
K = 2.74 × 10−2
Required: [Cl2(g)]equilibrium
[PCl3 (g)]2
Analysis: K =
[Cl2 (g)]3
K=
Solution:
2.74 ! 10"2 =
[PCl3 (g)]2
[Cl2 (g)]3
(1.09 ! 10"1 )2
[Cl2 (g)]3
(1.09 ! 10"1 )2
[Cl2 (g)] =
2.74 ! 10"2
3
3
[Cl2 (g)]3 = 3 4.33 ! 10"1
[Cl2 (g)]equilibrium = 7.57 ! 10"1 mol/L
Statement: The equilibrium concentration of chlorine gas is 7.57 × 10–1 mol/L.
!!
30. Given: N2(g) + 3 H2(g) #
!"
! 2 NH3(g); [H2(g)]equilibrium = 0.50 mol/L;
[NH3(g)]equilibrium = 0.46 mol/L; K = 626
Required: [N2(g)]equilibrium
[NH 3 (g)]2
Analysis: K =
[N 2 (g)][H 2 (g)]3
Solution:
[NH 3 (g)]2
K=
[N 2 (g)][H 2 (g)]3
626 =
[N 2 (g)] =
(0.46)2
(0.50)3[N 2 (g)]
(0.46)2
626(0.50)3
[N 2 (g)]equilibrium = 2.7 ! 10"3 mol/L
Statement: The equilibrium concentration of nitrogen gas, N2(g), is 2.7 × 10–3 mol/L.
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-6
!!
31. (a) H2(g) + I2(g) #
!"
! 2 HI(g) + energy
(b) If more iodine gas is added after the reaction reaches equilibrium, the reaction will
shift to the right because the reactant concentration is higher than at equilibrium.
(c) If thermal energy is added after the reaction reaches equilibrium, the reaction will
shift to the left to absorb excess thermal energy.
32. If the volume of the container is decreased, the reaction will shift to the left because
the decrease in volume increases pressure, and shifting to the side with fewer molecules
relieves the stress.
33. If more sulfur trioxide gas is added after the reaction reaches equilibrium, the reaction
will shift to the left because the product concentration is higher than at equilibrium.
!!
34. (a) 3 Z #
!"
! 2X+Y
(b) If more of compound Z is added at 10 min, the concentrations will shift to a new
equilibrium in which all three concentrations are higher because the addition of more
reactant shifts the equilibrium toward the products.
35. Catalysts are added to some industrial processes that involve an equilibrium because
the catalyst causes the reaction to proceed more quickly in both directions. If product is
removed, the catalyzed reaction produces new product faster.
36. The addition of argon will shift the reaction to the left because addition of argon
increases the pressure and shifts the reaction toward the side with fewer molecules.
37. (a) mostly reactants
(b) mostly products
(c) mostly products
(d) mostly reactants
38. The reaction quotient, Q, is the product of the concentrations of the products, raised
to the power of their coefficients, divided by the product of the concentrations of the
reactants, raised to the power of their coefficients, for a chemical reaction that is not
necessarily at equilibrium. The equilibrium constant, K, is the same ratio at equilibrium
conditions.
!!
39. Given: 2 H2(g) + CO(g) #
!"
! CH3OH(g); [H2(g)] = 0.25 mol/L;
[CO(g)] = 0.25 mol/L; [CH3OH(g)] = 0.040 mol/L; K = 10.5
Required: Is reaction at equilibrium? If not, direction of reaction to reach equilibrium
[CH 3OH(g)]
Solution: Q =
[H 2 (g)]2 [CO(g)]
=
(0.040)
(0.25)2 (0.25)
Q = 2.6
Statement: The value of Q is 2.6. Q is less than K, so the product concentrations are less
than at equilibrium; the reaction will shift toward the right, more methanol.
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-7
!!
40. Given: 2 NO(g) #
!"
! N2(g) + O2(g); [NO(g)]initial = 1.6 mol/L;
[N2(g)]initial = 0.60 mol/L; [O2(g)]initial = 0.60 mol/L; [NO(g)]equilibrium = 1.4 mol/L
Required: K
Analysis: Use an ICE table to determine the equilibrium concentrations of the reactants
and products and then use those values to calculate K.
I
C
E
!!
2 NO(g) #
!"
!
1.6
−x
1.4
N2(g)
+
0.60
+0.5x
0.60 + 0.5x
O2(g)
0.60
+0.5x
0.60 + 0.5x
Solution: x represents the change in concentration of NO(g).
x = [NO(g)]equilibrium − [NO(g)]initial
= (1.6 mol/L) − (1.4 mol/L)
x = 0.2 mol/L
[N 2 (g)]equilibrium = (0.60 mol/L) + 0.5 x
= (0.60 mol/L) + 0.5(0.2 mol/L)
[N 2 (g)]equilibrium = 0.70 mol/L (one extra digit carried)
[O 2 (g)]equilibrium = [N 2 (g)]equilibrium
[O 2 (g)]equilibrium = 0.70 mol/L (one extra digit carried)
K=
=
[N 2 (g)][O 2 (g)]
[NO(g)]2
(0.70)(0.70)
(1.4)2
K = 2.5 ! 10"1
Statement: The equilibrium constant, K, for the chemical reaction system is 2.5 × 10–1.
41. Given: [CO(g)]initial = 0.80 mol/L; [H2O(g)]initial = 2.40 mol/L;
[CO2(g)]initial = 0.62 mol/L; [H2(g)]initial = 0.50 mol/L; [CO2(g)]equilibrium = 0.92 mol/L
Required: K
!!
Analysis: CO(g) + H2O(g) #
!"
! H2(g) + CO2(g)
Use an ICE table to determine the equilibrium concentrations of the reactants and
products and then use those values to calculate K.
I
C
E
CO(g) +
0.80
−x
0.80 − x
!!
H2O(g) #
!"
!
2.40
−x
2.40 − x
Copyright © 2012 Nelson Education Ltd.
CO2(g)
0.62
+x
0.92
+
H2(g)
0.50
+x
0.50 + x
Chapter 7: Chemical Equilibrium
7-8
Solution: x represents the change in concentration of CO2(g).
x = [CO 2 (g)]equilibrium ! [CO 2 (g)]initial
= (0.92 mol/L) ! (0.62 mol/L)
x = 0.30 mol/L
[CO(g)]equilibrium = (0.80 mol/L) − x
= (0.80 mol/L) − (0.30 mol/L)
[CO(g)]equilibrium = 0.50 mol/L
[H 2O(g)]equilibrium = (2.40 mol/L) − x
= (2.40 mol/L) − (0.30 mol/L)
[H 2O(g)]equilibrium = 2.10 mol/L
[H 2 (g)]equilibrium = (0.50 mol/L) + x
= (0.50 mol/L) + (0.30 mol/L)
[H 2 (g)]equilibrium = 0.80 mol/L
[CO 2 (g)][H 2 (g)]
K=
[CO(g)][H 2O(g)]
=
(0.92)(0.80)
(0.50)(2.10)
K = 7.0 ×10−1
Statement: The equilibrium constant, K, for the chemical reaction system is 7.0 × 10–1.
!!
42. Given: CO2(g) + H2(g) #
!"
! CO(g) + H2O(g); V = 10.0 L;
ninitial!CO2 (g) = 1.00!mol; ninitial!H 2 (g) = 1.00!mol; K = 1.60
Required: [CO2(g)]equilibrium; [H2(g)]equilibrium; [CO(g)]equilibrium; [H2O(g)]equilibrium
n
Analysis: c =
V
Solution:
Step 1. Calculate concentrations, c, in mol/L from the given amounts of all entities. Calculate [CO2(g)]initial:
n
c = initial
V
1.00!mol
=
10.0!L
c = 0.100!mol/L
[CO2(g)]initial = 0.100 mol/L
Using the same formula,
[H2(g)]initial = 0.100 mol/L
Since there is no carbon monoxide gas or hydrogen gas initially, [CO(g)]initial
and [H2(g)]initial are both 0.0 mol/L.
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-9
Step 2. Use an ICE table to determine the relationship between the equilibrium
concentrations of the reactants and the products. Because the value of K is
greater than 1, the product concentrations will increase and the reactant
concentrations will decrease.
I
C
E
CO2(g) +
0.100
−x
0.100 − x
!!
H2(g) #
!"
!
0.100
−x
0.100 − x
CO(g) +
0
+x
x
H2O(g)
0
+x
x
Step 3. Substitute the equilibrium concentrations into the equilibrium constant equation,
and solve for the unknown.
K=
1.60 =
1.60 =
[CO(g)][H 2O(g)]
[CO 2 (g)][H 2 (g)]
x2
( 0.100 − x )( 0.100 − x )
x2
( 0.100 − x )( 0.100 − x )
x
0.100 − x
(±1.26)(0.100 – x) = x
±0.126 − 1.26 x = x
±1.26 =
±0.126 = 2.26 x
x = ±0.056 mol/L
Substituting the negative value of x into the expressions for equilibrium concentrations in
the ICE table results in a negative concentration value for CO(g) and H2O(g), which is
not possible in the real world. Therefore, use x = 0.056 mol/L in the equilibrium
concentration calculations.
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-10
Step 4. Calculate the equilibrium concentrations.
[CO 2 (g)]equilibrium = (0.100 mol/L) – x
= (0.100 mol/L) ! (0.056 mol/L)
[CO 2 (g)]equilibrium = 4.4 " 10 !2 mol/L
[H 2 (g)]equilibrium = [CO 2 (g)]equilibrium
−2
[H 2 (g)]equilibrium = 4.4 ×10 mol/L
[CO(g)]equilibrium = x
[CO(g)]equilibrium = 5.6 ! 10 "2 mol/L
[H 2O(g)]equilibrium = x
[H 2O(g)]equilibrium = 5.6 ! 10 "2 mol/L
Statement: The equilibrium concentration for both carbon dioxide gas and hydrogen gas
is 4.4 × 10–2 mol/L. The equilibrium concentration for both carbon monoxide gas and
water vapour is 5.6 × 10–2 mol/L.
!!
43. Given: I2(g) + Cl2(g) #
!"
! 2 ICl(g); V = 2.00 L; ninitial!I2 (g) = 0.50!mol;
ninitial!Cl2 (g) = 0.50!mol; K = 81.9
Required: [I2(g)]equilibrium; [Cl2(g)]equilibrium; [ICl(g)]equilibrium
Solution:
Step 1. Convert the given initial amounts of iodine gas and chlorine gas to concentration
in mol/L by dividing by the volume.
n
c = initial
V
0.50!mol
=
2.00!L
c = 0.25!mol/L
[I2(g)]initial = 0.25 mol/L
[Cl2(g)]initial = 0.25 mol/L
Since there is no iodine monochloride gas initially, [ICl(g)]initial = 0.00 mol/L.
Step 2. Use an ICE table to determine the relationship between the equilibrium
concentrations of the reactants and the product.
I
C
E
I2(g) +
0.25
−x
0.25 – x
!!
Cl2(g) #
!"
!
0.25
−x
0.25 – x
Copyright © 2012 Nelson Education Ltd.
2 ICl(g)
0.000
+2x
2x
Chapter 7: Chemical Equilibrium
7-11
Step 3. Substitute the equilibrium concentrations into the equilibrium constant equation,
and solve for x.
[ICl(g)]2
K=
[I 2 (g)][Cl2 (g)]
81.9 =
81.9 =
(2 x) 2
(0.25 – x)(0.25 – x)
(2 x) 2
(0.25 – x)(0.25 – x)
2x
0.25 – x
(±9.05)(0.25 – x) = 2 x
±9.05 =
±2.26 – 9.05 x = 2 x
±2.26 = 11.05 x
x = ±0.205 mol/L
Substituting the negative value of x into the expressions for equilibrium
concentrations in the ICE table results in a negative concentration value for
ICl(g), which is not possible in the real world. Therefore, use x = 0.205 mol/L in
the equilibrium concentration calculations.
Step 4. Calculate the equilibrium concentrations.
[H 2 (g)]equilibrium = (0.25 mol/L) – x
= (0.25 mol/L) – (0.205 mol/L)
[H 2 (g)]equilibrium = 4.5×10−2 mol/L
[I 2 (g)]equilibrium = [H 2 (g)]equilibrium
−2
[I 2 (g)]equilibrium = 4.5×10 mol/L
[HI(g)]equilibrium = 2 x
= 2(0.205 mol/L)
[HI(g)]equilibrium = 4.1×10−1 mol/L
Statement: The equilibrium concentration for both hydrogen gas and iodine gas is
4.5 × 10–2 mol/L, and the equilibrium concentration for iodine monochloride gas is
4.1 × 10–1 mol/L.
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-12
!!
44. Given: 2 H2S(g) #
!"
! 2 H2(g) + S2(g); V = 1.00 L; ninitial!H 2 S(g) = 0.200!mol;
K = 4.20 × 10−6 Required: [S2(g)]equilibrium
Solution:
Step 1. Convert the given initial amount of hydrogen sulfide gas to concentration in
mol/L by dividing by the volume.
n
c = initial
V
0.200!mol
=
1.00!L
c = 0.200!mol/L
[H2S(g)]initial = 0.200 mol/L
Since there is no hydrogen gas or sulfur gas initially,
[H2(g)]initial = 0.00 mol/L and [S2(g)]initial = 0.00 mol/L
Step 2. Use an ICE table to determine the relationship between the equilibrium
concentrations of the reactant and the products.
I
C
E
!!
2 H2S(g) #
!"
! 0.200
−x
0.200 – x
2 H2(g)
0.00
+x
x
+
S2(g)
0.00
+0.5x
0.5x
Step 3. Substitute the equilibrium concentrations into the equilibrium constant equation,
and solve for x.
[H 2 (g)]2 [S2 (g)]
K=
[H 2S(g)]2
4.20 ×10−6 =
x 2 (0.5 x)
(0.200 – x) 2
0.5 x 3
(0.200 – x) 2
The equilibrium constant value is very small compared to the initial
concentration of hydrogen sulfide gas. Therefore, very little hydrogen sulfide
gas will decompose at this temperature. From this, assume that the value of x
will also be so small as to be negligible.
If x is negligible, then 2x will be a very small number. Therefore,
[H 2S(g)]equilibrium = [H 2S(g)]initial − x
4.20 ×10−6 =
= 0.200 mol/L − (very small number)
[H 2S(g)]equilibrium ≈ 0.200 mol/L
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-13
Apply the hundred rule to check whether x can be assumed to be negligible.
[H 2S(g)]equilibrium
0.200
=
K
4.20 ! 10"6
[H 2S(g)]equilibrium
= 4.76 ! 104
K
The ratio of 4.76 × 104 is much greater than 100, so the assumption
[H 2S(g)]equilibrium ≈ 0.200 mol/L is warranted.
At equilibrium, then, the cubic equation for K can be simplified and solved.
0.5x 3
= 4.20 ! 10"6
2
(0.200 – x)
0.5x 3
# 4.20 ! 10"6
2
(0.200)
0.5x 3 # (4.20 ! 10"6 )(0.200)2
(4.20 ! 10"6 )(0.200)2
x #
0.5
3
x # 3 3.36 ! 10"7
x # 6.95 ! 10"3
Step 4. Calculate the equilibrium concentration of S2(g).
[S2 (g)]equilibrium = 0.5x
= 0.5(6.95 ! 10 "3 mol/L)
[S2 (g)]equilibrium = 3.48 ! 10 "3 mol/L
Statement: The equilibrium concentration of sulfur gas, S2(g), is 3.48 × 10–3 mol/L.
−34
!!
45. 2 HCl(g) #
!"
! H2(g) + Cl2(g); V = 1.00 L; ninitial!HCl(g) = 2.00!mol; K = 3.2 × 10
Required: [H2(g)]equilibrium; [Cl2(g)]equilibrium; [HCl(g)]equilibrium
Solution:
Step 1. Convert the given initial amount of hydrogen chloride gas to concentration in
mol/L by dividing by the volume.
n
c = initial
V
2.00!mol
=
1.00!L
c = 2.00!mol/L
[HCl(g)]initial = 2.00 mol/L
Since there is no hydrogen gas or chlorine gas initially,
[H2(g)]initial = 0.00 mol/L and [Cl2(g)]initial = 0.00 mol/L
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Chapter 7: Chemical Equilibrium
7-14
Step 2. Use an ICE table to determine the relationship between the equilibrium
concentrations of the reactant and the products.
I
C
E
!!
2 HCl(g) #
!"
! 2.00
−x
2.00 – x
H2(g)
0.00
+0.5x
0.5x
+
Cl2(g)
0.00
+0.5x
0.5x
Step 3. Substitute the equilibrium concentrations into the equilibrium constant equation,
and solve for x.
[H (g)][Cl2 (g)]
K= 2
[HCl(g)]2
3.2 ! 10"34 =
(0.5x)(0.5x)
(2.00 – x)2
3.2 ! 10"34 =
(0.5x)(0.5x)
(2.00 – x)2
0.5x
2.00 – x
"17
(±1.79 ! 10 )(2.00 – x) = 0.5x
±1.79 ! 10"17 =
(±3.58 ! 10"17 ) – (1.79 ! 10"17 )x = 0.5x
±3.58 ! 10"17 = 0.5x + (1.79 ! 10"17 )x
±3.58 ! 10"17
=x
0.5 + (1.79 ! 10"17 )
x = ±7.2 ! 10"17 mol/L
Substituting the negative value of x into the expressions for equilibrium
concentrations in the ICE table results in negative concentration values for H2(g)
and Cl2(g), which is not possible in the real world. Therefore, use
x = 7.2 × 10–17 mol/L in the equilibrium concentration calculations.
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-15
Step 4. Calculate the equilibrium concentrations.
[H 2 (g)]equilibrium = 0.5x
= 0.5(7.2 ! 10 "17 !mol/L)
[H 2 (g)]equilibrium = 3.6 ! 10 "17 mol/L
[Cl2 (g)]equilibrium = [H 2 (g)]equilibrium
[Cl2 (g)]equilibrium = 3.6 ! 10 "17 mol/L
[HCl(g)]equilibrium = 2.00!mol/L – x
[HCl(g)]equilibrium
= (2.00!mol/L) ! (7.2 " 10 –17 !mol/L)
= 2.0 mol/L
Statement: The equilibrium concentration of both hydrogen gas and chlorine gas is
3.6 × 10–17 mol/L, and the equilibrium concentration hydrogen chloride gas is
2.0 × 10–17 mol/L.
46. (a) Ksp = [Zn2+(aq)][OH−(aq)]2
2+
−
−17
!!
(b) Given: Zn(OH)2(s) #
!"
! Zn (aq) + 2 OH (aq); Ksp = 7.7 × 10 at 25 ºC
Required: molar solubility of Zn(OH)2(s) at 25 ºC
Solution: !!
Zn(OH)2(s) #
!"
! I −−
C −−
E −−
Zn2+(aq)
0.00
+x
x
2+
!
+
2 OH−(aq)
0.00
+2x
2x
K sp = [Zn (aq)][OH (aq)]
2
K sp = (x)(2x)2
7.7!"!10 !17 = 4x 3
7.7!"!10
x =
4
!17
3
x = 3 1.925!"!10 !17
x = 2.7!"!10 !6
[Zn (aq)] = molar solubility of Zn(OH)2(s)
molar solubility of Zn(OH)2(s) = 2.7 × 10−6 mol/L
Statement: The molar solubility of zinc hydroxide at 25 ºC is 2.7 × 10−6 mol/L.
2+
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Chapter 7: Chemical Equilibrium
7-16
47. Given: VPb(NO3 )2 = 10.0 mL ; [Pb(NO3)2(aq)] = 0.0040 mol/L; VKCl = 15.0 mL ;
[KCl(aq)] = 0.25 mol/L; Ksp for PbCl2(s) = 1.2 × 10–5
Required: To predict whether lead(II) chloride will precipitate
Analysis: Ions present in the mixture: Pb2+(aq), NO3−(aq), K+(aq), and Cl−(aq).
The ratio of [Pb2+(aq)] to [Pb(NO3)2(aq)] is 1:1.
The ratio of [Cl−(aq)] to [KCl(aq)] is 1:1.
! 10.0 mL $
[Pb2+ (aq)]final = (0.0040 mol/L) #
&
" 25.0 mL %
[Pb2+ (aq)]final = 1.6 ' 10(3 mol/L
" 15.0 mL %
[Cl! (aq)]final = (0.25 mol) $
'
# 25.0 mL &
[Cl! (aq)]final = 1.5 ( 10!1 mol/L
2+
−
!!
Solution: PbCl2(s) #
!"
! Pb (aq) + 2 Cl (aq)
Q = [Pb 2+ (aq)][Cl! (aq)]2
= (1.6!"!10 !3 )(1.5!"!10 !1 )2
Q = 3.6!"!10 !5
Ksp of PbI2(s) = 1.2 × 10−5
Q is greater than Ksp, so the equilibrium will shift to the left and a precipitate will form.
Statement: A precipitate of solid lead(II) chloride, PbCl2(s), will form when the two
solutions are mixed.
Analysis and Application
48. A bottle of pop stays carbonated longer if the bottle cap is replaced after opening
because, when the cap is placed on the bottle, a dynamic equilibrium forms and the
carbon dioxide concentration remains constant. If the cap is left off the bottle, carbon
dioxide escapes to the atmosphere so the equilibrium shifts away from carbon dioxide in
solution.
49. When bubbles are produced and a colour change occurs when two solutions are
mixed, and then after a few minutes, the bubbles stop and the colour has reached a
constant shade, the constancy of the colour may suggest that an equilibrium has been
reached. However, it may also suggest that the reaction has gone to completion and one
of the reaction products is coloured. Based on the evidence, it is not possible to say
whether or not equilibrium has been reached.
50. The equilibrium constant for dissolution of fluoroapatite, Ca5(PO4)3F(s). is lower than
that of hydroxyapatite, Ca5(PO4)3(OH)(s), because the fluoride ion replaces the
hydroxide, forming a precipitate.
2
51. The amount of ammonia formed would be x because 2 mol of ammonia form when
3
3 mol of hydrogen are consumed.
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Chapter 7: Chemical Equilibrium
7-17
52. (a) According to equilibrium law, the value of K should be identical for all
[PCl3 (g)][Cl2 (g)]
concentrations of products and reactants: K =
.
[PCl5 (g)]
(0.23)(0.55)
= 5.5
0.023
(0.15)(0.37)
= 5.5
trial 2: K =
0.010
(0.99)(0.47)
= 5.5
trial 3: K =
0.085
trial 1: K =
(3.66)(1.5)
= 5.5
1.00
Since the value of K is constant for all observed equilibrium concentrations, the data is
consistent with the equilibrium law.
(b) The equilibrium constant for this reaction is 5.5.
53. Temperatures lower than 500 °C are not used in industrial production of ammonia
because at lower temperatures and pressures, there would be fewer effective collisions, so
the rate of the formation of ammonia would be lower.
54. In a chemical manufacturing process that involves a system at equilibrium, the
reactants are continually added and the products are continually removed in order to
maximize the yield of the product. If additional reactants were not added, the reaction
would slow, and eventually stop. If the products were not removed, the reaction mixture
would reach an equilibrium state at which no additional product would be formed.
Removing the products shifts the equilibrium and ensures the reaction continues.
55. The equilibrium of myoglobin (M) binding oxygen can be represented by the
equation
!!
M + O2 #
!"
! MO2.
The equilibrium of hemoglobin (H) binding oxygen can be represented by the equation
!!
H + O2 #
!"
! HO2.
Since myoglobin binds more tightly to oxygen than hemoglobin, we know that the right
side of the myglobin equilibrium is more favoured than the right side of the hemoglobin
equilibrium. Therefore, the equilibrium constant for hemoglobin binding should be
smaller.
56. The engineer’s suggestion that money might be saved during the industrial production
of sulfuric acid by eliminating the catalyst and raising the temperature of the reaction
system is invalid because eliminating the catalyst will reduce the rate of product
formation and increasing the temperature will shift the equilibrium toward the reactants,
because the reaction is exothermic.
57. The magnitude of the values of the equilibrium constant, K, for chemical reaction
systems in which very little product is present at equilibrium will be much less than 1
because the product concentrations are in the numerator and are much lower than the
denominator, which is based on the higher reactant concentrations.
trial 4: K =
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-18
58. Given: V = 5.0 L; [Ca2+(aq)] = 4.0 × 10−3 mol/L; Ksp = 4.8 × 10−9
Required: maximum mass of sodium carbonate that can be added without forming a
precipitate
!!
Analysis: Ca2+(aq) + CO32(aq) #
!"
! CaCO3(s)
Solution:
K sp = [Ca 2+ (aq)][CO 23 (aq)]
4.8 ! 10 "9 = (4.0 ! 10 "3 !mol/L)(maximum [CO 23 (aq)])
maximum [CO 23 (aq)] = 1.2 ! 10 "6 !mol/L
max mass of Na 2CO3 (s) =
1.2 ! 10-6 mol
L
!
105.99 g
mol
! 5.0 L
max mass of Na 2CO3 (s) = 6.4 ! 10"4 g
Statement: The maximum mass of sodium carbonate that can be added to the given
volume of water without causing any precipitate to form is 6.4 × 10–4 g.
59. Answers may vary. Sample answer: Barium nitrate and sodium sulfate will decrease
the solubility of barium sulfate when added to a solution because each compound has an
ion in common with barium sulfate.
60. The solubility of most compounds increases as the temperature of a solution
increases. Heated ground water dissolves more minerals. As the water cools, the
maximum concentration of dissolved substances it can hold decreases, causing excess
solutes to precipitate.
Evaluation
61. Answers may vary. Sample answer: I disagree with the statement that “unsaturated
solutions are in dynamic equilibrium, but saturated solutions are not” because a dynamic
equilibrium only occurs when the rate of dissolution and the rate of precipitation are
equal. This is only true when the solution is saturated and there is solid present. The
diagram below shows a saturated solution in dynamic equilibrium. Some reactants and
some products are present, and the reaction is proceeding in both directions.
Copyright © 2012 Nelson Education Ltd.
Chapter 7: Chemical Equilibrium
7-19
62. Answers may vary. Sample answer: No, I do not think that chemical engineers should
focus on identifying catalysts for all industrial processes. The appropriateness and use of
a catalyst depends on the chemical reaction or process. Some reactions occur very rapidly
on their own, without catalysts. Spending time and money finding catalysts for these
reactions would not be helpful and would be a waste of valuable resources. Conversely,
the use of a catalyst is beneficial if the catalyzed process involves fewer steps or
chemicals. This helps reduce cost and waste.
63. No, I do not believe the person who wrote this has a complete understanding of
equilibria. While increasing the temperature and the pressure will shift the equilibrium to
one side or the other, the names “product” and “reactant” depend on your goal for the
reaction, not the reaction itself. For example, in the reaction
!!
Ca2+(aq) + CO32-−(aq) #
!"
! CaCO3(s)
the CaCO3(s) may be the product, or it may be the reactant.
Reflect on Your Learning
64. Answers may vary.
65. Answers may vary. Sample answer: The information in this chapter can help me
understand how toxins such as carbon monoxide can interfere with normal body
functions, how refrigeration can help slow food spoilage, and the role carbon dioxide
plays in forming acid rain. My understanding of chemical equilibrium will help me
remember to put the milk back in the refrigerator after I have used it, and will also make
me more cautious around substances that may be toxic.
66. Answers may vary. Sample answer: I found it difficult to understand the effects that
changes in temperature or pressure could have on equilibrium. Writing all the elements
involved in an equilibrium in a graphic organizer such as a drawing of a balance could
help me to visualize the role each one could play. I also found it difficult to use the
equilibrium constant and initial concentration to determine the equilibrium concentration
of reactants and products. More practice doing these calculations and comparing my
solutions to those of a classmate could help me remember the steps I need to use in each
calculation.
Research
67. Answers may vary. Students’ posters should include a description of the role
ammonia plays in the nitrogen cycle, including providing nitrogen to plants which then
provide nitrogen to animals.
68. Answers may vary. Students’ answers should include the following information:
• There is an equilibrium between dissolved nitrogen and nitrogen gas.
• Increased pressure favours the dissolved nitrogen; then, as the diver ascends, the
equilibrium shifts toward nitrogen gas, causing bubbles to form in the blood.
• These bubbles of nitrogen gas in blood are the cause of the physiological effects known
as the bends.
• Symptoms include joint pain, itching rashes, and staggering.
• Treatment includes placing the victim in a hyperbaric, or high-pressure, chamber to
allow the reaction between nitrogen gas and dissolved nitrogen to reach equilibrium very
slowly.
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Chapter 7: Chemical Equilibrium
7-20
69. Answers may vary. Sample answer: The chemical equation involved in dissolving
calcium carbonate in rocks such as limestone is: CaCO3(s) + H2O(l) → Ca(HCO3)2(aq).
When the solution comes into contact with air, the equilibrium shifts and it is reversed,
causing the calcium carbonate to precipitate out of solution and form deposits dripping
off cave roofs or accumulating on cave floors.
70. Answers may vary. Students should identify the equilibrium between the anhydrous
and hexahydrate forms of cobalt(II) chloride as an indicator of level of water vapour in
the air surrounding the cobalt(II) chloride. The colour changes from red-purple
(hexahydrate) to blue (anhydrous) as water is lost. The direction of the equilibrium is
affected by humidity. Sample products may include novelty humidity indicators.
71. Answers may vary. Sample answer: Approximately 83 % of ammonia worldwide is
used for fertilizer. Other uses of ammonia include metal production, livestock feed
additive, manufacture of nitric acid, neutralization of acids in waste water and stack
gases, and household cleaners.
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Chapter 7: Chemical Equilibrium
7-21