Chapter 7 Review, pages...

Copyright © 2012 Nelson Education Ltd. Chapter 7: Chemical Equilibrium 7-2

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.

2 N2O(g) + O2(g) ! "!# !! 4 NO(g) I 0.00 0.00 3.00 C +x +0.5x −2x E 1.00 0.5x 3.00 – 2x


Solution: x represents the change in concentration of N2O(g).

x = [N2O(g)]equilibrium ! [N2O(g)]initial= 1.00 mol/L ! 0.00 mol/L

x = 1.00 mol/L

2 equilibrium

12 equilibrium

[O (g)] 0.5

0.5(1.00 mol/L)

[O (g)] 5.00 1 0 mol/L −

=

=

= ×

x

equilibrium

equilibrium

[NO(g)] (3.00 mol/L) – 2

(3.00 mol/L) – 2(1.00 mol/L) [NO(g)] 1.00 mol/L

=

==

x

[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.

Br2(g) ! "!# !! 2 Br(g) I 1.00 0.00 C −x +2x E 1.00 − x 2x

Solution: x represents the change in concentration of Br2(g).

2x = [Br(g)]equilibrium

2x = 0.032 mol/Lx = 0.016 mol/L

2 equilibrium

12 equilibrium

[Br (g)] (1.00 mol/L) –

(1.00 mol/L) – (0.016 mol/L)

[Br (g)] 9.8 1 0 mol/L−

=

=

= ×

x

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.

23. (a) 3

22

3

[O (g)][O (g)]

K =

(b) 4

24

2

[H (g)][H O(g)]

K =


(c) 22 2 [N (g)][H O(g)]K =

(d) 2

22

[NO(g)] [Cl (g)][NOCl(g)]

K =

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) 25. Given: Cl2(g) + CO(g) ! "!# !! COCl2(g); [CO(g)]equilibrium = 1.11 × 10

−1 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.

K =[COCl2(g)]

[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. 26. Given: 2 ClF3(g) ! "!# !! Cl2(g) + 3 F2(g); [Cl2(g)]equilibrium = 1.62 × 10

−1 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.

K =

[Cl2(g)][F2(g)]3

[ClF3(g)]2

Substitute the given values for the equilibrium concentrations of the products and the equilibrium constant and solve for [ClF3(g)].

K =[Cl2(g)][F2(g)]

3

[ClF3(g)]2

3.72 ! 10"2 = (1.62 ! 10"1)(1.85 ! 10"1)3

[ClF3(g)]2

[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 = 2.76 ! 10"2

[ClF3(g)]equilibrium = ±1.66 ! 10"1 mol/L


Since a negative concentration value is not possible in the real world, use 1

3 equilibrium[ClF (g)] 1.66 1 0 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.

2 SO2(g) + O2(g) ! "!# !! 2 SO3(g) I 0.40 1.6 29.7 C +x +0.5x −x E 1.2 1.6 + 0.5x 29.7 – x

Solution: x represents the change in concentration of SO2(g).

x = [SO2(g)]equilibrium ! [SO2(g)]initial

= 1.2 mol/L ! 0.40 mol/Lx = 0.80 mol/L

2 equilibrium

2 equilibrium

[O (g)] (1.6 mol/L) 0.5

(1.6 mol/L) 0.5(0.80 mol/L)[O (g)] 2.0 mol/L

= +

= +=

x

3 equilibrium

3 equilibrium

[SO (g)] (29.7 mol/L) –

(29.7 mol/L) – (0.80 mol/L)[SO (g)] 28.9 mol/L

=

==

x

232

22

2

2

[SO (g)][SO(g)] [O (g)]

(28.9)(1.2) (2.0)2.9 1 0

=

=

= ×

K

K

Statement: The equilibrium constant, K, for this chemical reaction system is 2.9 × 102. 28. Given: PCl5(g) ! "!# !! PCl3(g) + Cl2(g); [PCl5(g)]equilibrium = 4.06 × 10

−1 mol/L; [PCl3(g)]equilibrium = 1.17 × 10−1 mol/L; K = 1.24 × 10−1 Required: [Cl2(g)]equilibrium

Analysis: K =

[PCl3(g)][Cl2(g)][PCl5(g)]


Solution:

K =[PCl3(g)][Cl2(g)]

[PCl5(g)]

1.24 ! 10"1 =(1.17 ! 10"1)[Cl2(g)]

4.06 ! 10"1

[Cl2(g)]equilibrium =(1.24 ! 10"1)(4.06 ! 10"1)

1.17 ! 10"1

[Cl2(g)]equilibrium = 4.30 ! 10"1 mol/L

Statement: The equilibrium concentration of chlorine gas, Cl2(g), is 4.30 × 10–1 mol/L. 29. Given: 2 P(s) + 3 Cl2(g) ! "!# !! PCl3(g); [PCl3(g)]equilibrium = 1.09 × 10

−1 mol/L; K = 2.74 × 10−2 Required: [Cl2(g)]equilibrium

Analysis: K =

[PCl3(g)]2

[Cl2(g)]3

Solution:

K =[PCl3(g)]

2

[Cl2(g)]3

2.74 ! 10"2 = (1.09 ! 10"1)2

[Cl2(g)]3

[Cl2(g)]3 = (1.09 ! 10

"1)2

2.74 ! 10"2

[Cl2(g)]33 = 4.33 ! 10"13

[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

Analysis: K =

[NH3(g)]2

[N2(g)][H2(g)]3

Solution:

K =[NH3(g)]

2

[N2(g)][H2(g)]3

626 = (0.46)2

(0.50)3[N2(g)]

[N2(g)]=(0.46)2

626(0.50)3

[N2(g)]equilibrium = 2.7 !10"3 mol/L

Statement: The equilibrium concentration of nitrogen gas, N2(g), is 2.7 × 10–3 mol/L.


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 ! "!# !! 2 X + 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

Solution:

Q =[CH3OH(g)]

[H2(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.


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.

2 NO(g) ! "!# !! N2(g) + O2(g) I 1.6 0.60 0.60 C −x +0.5x +0.5x E 1.4 0.60 + 0.5x 0.60 + 0.5x

Solution: x represents the change in concentration of NO(g). equilibrium initial[NO(g)] [NO(g)]

(1.6 mol/L) (1.4 mol/L)0.2 mol/L

= −

= −=

x

x

2 equilibrium

2 equilibrium

[N (g)] (0.60 mol/L) 0.5

(0.60 mol/L) 0.5(0.2 mol/L)[N (g)] 0.70 mol/L (one extra digit carried)

= +

= +=

x

[O2(g)]equilibrium = [N2(g)]equilibrium[O2(g)]equilibrium = 0.70 mol/L (one extra digit carried)

K =[N2(g)][O2(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.

CO(g) + H2O(g) ! "!# !! CO2(g) + H2(g) I 0.80 2.40 0.62 0.50 C −x −x +x +x E 0.80 − x 2.40 − x 0.92 0.50 + x


Solution: x represents the change in concentration of CO2(g).

x = [CO2(g)]equilibrium ! [CO2(g)]initial= (0.92 mol/L) ! (0.62 mol/L)

x = 0.30 mol/L

equilibrium

equilibrium

[CO(g)] (0.80 mol/L)

(0.80 mol/L) (0.30 mol/L)[CO(g)] 0.50 mol/L

= −

= −=

x

2 equilibrium

2 equilibrium

[H O(g)] (2.40 mol/L)

(2.40 mol/L) (0.30 mol/L)[H O(g)] 2.10 mol/L

= −

= −=

x

2 equilibrium

2 equilibrium

[H (g)] (0.50 mol/L)

(0.50 mol/L) (0.30 mol/L)[H (g)] 0.80 mol/L

= +

= +=

x

2 2

2

1

[CO (g)][H (g)][CO(g)][H O(g)](0.92)(0.80)(0.50)(2.10)7.0 1 0−

=

=

= ×

K

K

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!H2 (g) = 1.00!mol; K = 1.60 Required: [CO2(g)]equilibrium; [H2(g)]equilibrium; [CO(g)]equilibrium; [H2O(g)]equilibrium

Analysis: c = nV

Solution: Step 1. Calculate concentrations, c, in mol/L from the given amounts of all entities.

Calculate [CO2(g)]initial:

c = ninitialV

= 1.00!mol10.0!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.


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. CO2(g) + H2(g) ! "!# !! CO(g) + H2O(g) I 0.100 0.100 0 0 C −x −x +x +x E 0.100 − x 0.100 − x x x

Step 3. Substitute the equilibrium concentrations into the equilibrium constant equation,

and solve for the unknown.

( )( )

( )( )

2

2 22

2

[CO(g)][H O(g)][CO (g)][H (g)]

1.600.100 0.100

1.600.100 0.100

1.26 0.100

( 1.26)(0.100 – )0.126 1.26

0.126 2.260.056 mol/L

=

=− −

=− −

± =−

± =± − =

± == ±

K

xx x

xx x

xx

x xx x

xx

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.


Step 4. Calculate the equilibrium concentrations.

[CO2(g)]equilibrium = (0.100 mol/L) – x

= (0.100 mol/L)! (0.056 mol/L)

[CO2(g)]equilibrium = 4.4 " 10!2 mol/L

2 equilibrium 2 equilibrium

22 equilibrium

[H (g)] [CO (g)]

[H (g)] 4.4 1 0 mol/L −=

= ×

[CO(g)]equilibrium = x

[CO(g)]equilibrium = 5.6 !10"2 mol/L

[H2O(g)]equilibrium = x

[H2O(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.

c = ninitialV

= 0.50!mol2.00!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. I2(g) + Cl2(g) ! "!# !! 2 ICl(g) I 0.25 0.25 0.000 C −x −x +2x E 0.25 – x 0.25 – x 2x


Step 3. Substitute the equilibrium concentrations into the equilibrium constant equation, and solve for x.

2

2 22

2

[ICl(g)][I (g)][Cl (g)]

(2 )81.9(0.25 – )(0.25 – )

(2 )81.9(0.25 – )(0.25 – )29.05

0.25 –( 9.05)(0.25 – ) 2

2.26 – 9.05 22.26 11.05

0.205 mol/L

=

=

=

± =

± =± =

± == ±

K

xx x

xx x

xx

x xx x

xx

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. 2 equilibrium

22 equilibrium

[H (g)] (0.25 mol/L) –

(0.25 mol/L) – (0.205 mol/L)

[H (g)] 4.5 1 0 mol/L −

=

=

= ×

x

2 equilibrium 2 equilibrium

22 equilibrium

[I (g)] [H (g)]

[I (g)] 4.5 1 0 mol/L −=

= ×

equilibrium

1equilibrium

[HI(g)] 2

2(0.205 mol/L)

[HI(g)] 4.1 10 mol/L−

=

=

= ×

x

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.


44. Given: 2 H2S(g) ! "!# !! 2 H2(g) + S2(g); V = 1.00 L; ninitial!H2S(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.

c = ninitialV

= 0.200!mol1.00!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. 2 H2S(g) ! "!# !! 2 H2(g) + S2(g) I 0.200 0.00 0.00 C −x +x +0.5x E 0.200 – x x 0.5x


22 2

22

26

2

36

2

[H (g)] [S (g)][H S(g)]

(0.5 )4.20 1 0(0.200 – )

0.54.20 1 0(0.200 – )

−

−

=

× =

× =

K

x xx

xx

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,

2 equilibrium 2 initial

2 equilibrium

[H S(g)] [H S(g)]

0.200 mol/L (very small number)[H S(g)] 0.200 mol/L

= −

= −≈

x


Apply the hundred rule to check whether x can be assumed to be negligible.

[H2S(g)]equilibriumK

= 0.2004.20 ! 10"6

[H2S(g)]equilibriumK

= 4.76 ! 104

The ratio of 4.76 × 104 is much greater than 100, so the assumption 2 equilibrium[H S(g)] 0.200 mol/L≈ is warranted.

At equilibrium, then, the cubic equation for K can be simplified and solved.

0.5x3

(0.200 – x)2= 4.20 ! 10"6

0.5x3

(0.200)2# 4.20 ! 10"6

0.5x3 # (4.20 ! 10"6 )(0.200)2

x3 # (4.20 ! 10"6 )(0.200)2

0.5x # 3.36 ! 10"73

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. 45. 2 HCl(g) ! "!# !! H2(g) + Cl2(g); V = 1.00 L; ninitial!HCl(g) = 2.00!mol; K = 3.2 × 10

−34 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.

c = ninitialV

= 2.00!mol1.00!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


Step 2. Use an ICE table to determine the relationship between the equilibrium concentrations of the reactant and the products. 2 HCl(g) ! "!# !! H2(g) + Cl2(g) I 2.00 0.00 0.00 C −x +0.5x +0.5x E 2.00 – x 0.5x 0.5x


K =[H2(g)][Cl2(g)]

[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

±1.79 ! 10"17 = 0.5x2.00 – x

(±1.79 ! 10"17 )(2.00 – x) = 0.5x(±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

0.5+ (1.79 ! 10"17 )= x

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.


Step 4. Calculate the equilibrium concentrations. [H2 (g)]equilibrium = 0.5x

= 0.5(7.2 !10"17 !mol/L)

[H2 (g)]equilibrium = 3.6 !10"17 mol/L

[Cl2 (g)]equilibrium = [H2 (g)]equilibrium[Cl2 (g)]equilibrium = 3.6 !10

"17 mol/L

[HCl(g)]equilibrium = 2.00!mol/L – x

= (2.00!mol/L) ! (7.2 "10–17 !mol/L)[HCl(g)]equilibrium = 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 (b) Given: Zn(OH)2(s) ! "!# !! Zn

2+(aq) + 2 OH−(aq); Ksp = 7.7 × 10−17 at 25 ºC Required: molar solubility of Zn(OH)2(s) at 25 ºC Solution:

Zn(OH)2(s) ! "!# !! Zn2+(aq) + 2 OH−(aq)

I −− 0.00 0.00 C −− +x +2x E −− x 2x

Ksp = [Zn2+ (aq)][OH! (aq)]2

Ksp = (x)(2x)2

7.7!"!10!17 = 4x3

x3 = 7.7!"!10!17

4x = 1.925!"!10!173

x = 2.7!"!10!6

[Zn2+(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.


47. Given: 3 2Pb(NO )

10.0 mLV = ; [Pb(NO3)2(aq)] = 0.0040 mol/L; KCl 15.0 mLV = ; [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.

[Pb2+ (aq)]final = (0.0040 mol/L)10.0 mL25.0 mL

!

"#

$

%&

[Pb2+ (aq)]final = 1.6 ' 10(3 mol/L

[Cl! (aq)]final = (0.25 mol)15.0 mL25.0 mL

"

#$

%

&'

[Cl! (aq)]final = 1.5 ( 10!1 mol/L

Solution: PbCl2(s) ! "!# !! Pb2+(aq) + 2 Cl−(aq)

Q = [Pb2+ (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.

51. The amount of ammonia formed would be 23

x because 2 mol of ammonia form when

3 mol of hydrogen are consumed.


52. (a) According to equilibrium law, the value of K should be identical for all

concentrations of products and reactants: K =

[PCl3(g)][Cl2(g)][PCl5(g)]

.

trial 1: K = (0.23)(0.55)

0.023= 5.5

trial 2: K = (0.15)(0.37)

0.010= 5.5

trial 3: K = (0.99)(0.47)

0.085= 5.5

trial 4: K = (3.66)(1.5)

1.00= 5.5

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.


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: Ksp = [Ca

2+ (aq)][CO32 (aq)]

4.8 !10"9 = (4.0 !10"3 !mol/L)(maximum [CO32 (aq)])

maximum [CO32 (aq)]= 1.2 !10"6 !mol/L

max mass of Na2CO3(s) =1.2 ! 10-6 mol

L! 105.99 g

mol ! 5.0 L

max mass of Na2CO3(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.


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.


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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