Chapter 13
Chemical Equilibrium

13.1 (a) K_c = (b) K_c =

13.2 (a) K_c = = 7.9 x 10⁴

(b) K_c = = 1.3 x 10^-5

13.3 The ratio in the equilibrium mixture is 0.333. Only mixture (3) has this same ratio and is also at equilibrium.

13.4 K_p = = 9.48

13.5 2 NO(g) + O₂ ¾ 2 NO₂(g); n = 2 - 3 = -1
K_p = K_c(RT)ⁿ, K_c = K_p(1/RT)
at 500 K: K_p = (6.9 x 10⁵)[(0.0821)(500)]^-1 = 1.7 x 10⁴
at 1000 K: K_c = (1.3 x 10^-2) = 1.1

13.6 (a) K_c = , K_p = (b) K_c = [H₂]²[O₂], K_p =

13.7 K_c = 1.2 x 10^-42. Since K_c is very small, the equilibrium mixture contains mostly H₂ molecules. H is in periodic group 1A. A very small value of K_c is consistent with strong bonding between 2 H atoms, each with one valence electron.

13.8 The container volume of 5.0 L must be included to calculate molar concentrations.
(a) Q_c = = 890
Because Q_c < K_c, the reaction is not at equilibrium. The reaction will proceed to the right to reach equilibrium.

(b) Q_c = = 1.6 x 10⁷
Because Q_c > K_c, the reaction is not at equilibrium. The reaction will proceed to the left to reach equilibrium.

13.9 (a) (2) (b) (1), reverse; (3), forward

13.10 K_c = = 1.2 x 10^-42

(a) [H] = = 3.5 x 10^-22 M

(b) H atoms = (3.5 x 10^-22 mol/L)(1.0 L)(6.022 x 10²³ atoms/mol) = 210 H atoms
H₂ molecules = (0.10 mol/L)(1.0 L)(6.022 x 10²³ molecules/mol) = 6.0 x 10²² H₂ molecules

13.11 CO(g) + H₂O(g) ¾ CO₂(g) + H₂(g)

K_c = 4.24 =
Take the square root of both sides and solve for x.
; 2.06 = ; x = 0.101
At equilibrium, [CO₂] = [H₂] = x = 0.101 M
[CO] = [H₂O] = 0.150 - x = 0.150 - 0.101 = 0.049 M

13.12 N₂O₄(g) ¾  2 NO₂(g)

K_c = 4.64 x 10^-3 =
4x² + (4.64 x 10^-3)x - (2.32 x 10^-4) = 0
Use the quadratic formula to solve for x.
x =
x = -0.008 22 and 0.007 06
Discard the solution that uses the negative square root (-0.008 22) because it will lead to negative concentrations and that is impossible.
[N₂O₄] = 0.0500 - x = 0.0500 - 0.007 06 = 0.0429 M
[NO₂] = 2x = 2(0.007 06) = 0.0141 M

13.13 N₂O₄(g) ¾ 2 NO₂(g)
Q_c = = 0.0450; Q_c > K_c
The reaction will approach equilibrium by going from right to left.
N₂O₄(g) ¬ 2 NO₂(g)

K_c = 4.64 x 10^-3 =
4x² - 0.1246x + (8.072 x 10^-4) = 0
Use the quadratic formula to solve for x.
x =
x = 0.0220 and 0.009 19
Discard the solution that uses the positive square root (0.0220) because it will lead to negative concentration of NO₂, and that is impossible.
[N₂O₄] = 0.0200 + x = 0.0200 + 0.009 19 = 0.0292 M
[NO₂] = 0.0300 - 2x = 0.0300 - 2(0.009 19) = 0.0116 M

13.14 (a) CO(reactant) added, H₂ concentration increases.
(b) CO₂ (product) added, H₂ concentration decreases.
(c) H₂O (reactant) removed, H₂ concentration decreases.
(d) CO₂ (product) removed, H₂ concentration increases.
At equilibrium, Q_c = K_c = . If some CO₂ is removed from the equilibrium mixture, the numerator in Q_c is decreased, which means that Q_c < K_c and the reaction will shift to the right, increasing the H₂ concentration.

13.15 (a) Because there are 2 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by a change in pressure. The number of moles of reaction products remains the same.

(b) Because there are 2 mol of gas on the left side and 1 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to products). The number of moles of reaction products increases.

(c) Because there is 1 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to reactants). The number of moles of reaction product decreases.

13.16

13.17 Le Châtelier's principle predicts that a stress of added heat will be relieved by net reaction in the direction that absorbs the heat. Since the reaction is endothermic, the equilibrium will shift from left to right (K_c will increase) with an increase in temperature. Therefore, the equilibrium mixture will contain more of the offending NO, the higher the temperature.

13.18 The reaction is exothermic. As the temperature is increased the reaction shifts from right to left. The amount of ethyl acetate decreases.
K_c =
As the temperature is decreased, the reaction shifts from left to right. The product concentrations increase, and the reactant concentrations decrease. This corresponds to an increase in K_c.

13.19 There are more AB(g) molecules at the higher temperature. The equilibrium shifted to the right at the higher temperature, which means the reaction is endothermic.

13.20 (a) A catalyst does not affect the equilibrium composition. The amount of CO remains the same.

(b) The reaction is exothermic. An increase in temperature shifts the reaction toward reactants. The amount of CO increases.

(c) Because there are 3 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to products). The amount of CO decreases.

(d) An increase in pressure as a result of the addition of an inert gas (with no volume change) does not affect the equilibrium composition. The amount of CO remains the same.

(e) Adding O₂ increases the O₂ concentration and shifts the reaction toward products. The amount of CO decreases.

13.21 (a) Because K_c is so large, k_f is larger than k_r.
(b) K_c = ; k_r = = 2.5 x 10^-28 M^-1 s^-1
(c) Because the reaction is exothermic, E_a (forward) is less than E_a (reverse). Consequently, as the temperature decreases, k_r decreases more than k_f decreases, and therefore K_c = increases.

13.22 Hb + O₂ ¾ Hb(O₂)
If CO binds to Hb, Hb is removed from the reaction and the reaction will shift to the left resulting in O₂ being released from Hb(O₂). This will decrease the effectiveness of Hb for carrying O₂.

13.23 The equilibrium shifts to the left because at the higher altitude the concentration of O₂ is decreased.

13.24 There are 26 electrons.

13.25 The partial pressure of O₂ in the atmosphere is 0.2095 atm.
PV = nRT
n = = 4.28 x 10^-3 mol O₂
4.28 x 10^-3 mol O₂ x = 2.58 x 10²¹ O₂ molecules

Understanding Key Concepts

13.26 (a) (1) and (3) because the number of A and B's are the same in the third and fourth box.
(b) K_c = = 1.5
(c) Because the same number of molecules appear on both sides of the equation, the volume terms in K_c all cancel. Therefore, we can calculate K_c without including the volume.

13.27 (a) A₂ + C₂ ¾ 2 AC (most product molecules)

(b) A₂ + B₂ ¾ 2 AB (fewest product molecules)

13.28 (a) Only reaction (3), = 2, is at equilibrium.
(b) = 15 for reaction (1). Because Q_c > K_c, the reaction will go in the reverse direction to reach equilibrium.
= 1/3 for reaction (2). Because Q_c < K_c, the reaction will go in the forward direction to reach equilibrium.

13.29 (a) A₂ + 2 B ¾ 2 AB
(b) The number of AB molecules will increase, because as the volume is decreased at constant temperature, the pressure will increase and the reaction will shift to the side of fewer molecules to reduce the pressure.

13.30 When the stopcock is opened, the reaction will go in the reverse direction because there will be initially an excess of AB molecules.

13.31 As the temperature is raised, the reaction proceeds in the reverse direction. This is consistent with an exothermic reaction where "heat" can be considered as a product.

13.32 (a) AB ¾ A + B
(b) The reaction is endothermic because a stress of added heat (higher temperature) shifts the AB ¾ A + B equilibrium to the right.
(c) If the volume is increased, the pressure is decreased. The stress of decreased pressure will be relieved by a shift in the equilibrium from left to right, thus increasing the number of A atoms.

13.33 Heat + BaCO₃(s) ¾ BaO(s) + CO₂(g)

(a) (b)

13.34 (a) (b) (c)

13.35 This equilibrium mixture has a K_c and is less than 1. This means that k_f < k_r.

Additional Problems

Equilibrium Expressions and Equilibrium Constants

13.36 (a) (b) (c) K_c =

13.37 (a) (b) (c)

13.38 (a) K_p = , n = 2 and K_p = K_c(RT)²
(b) K_p = , n = -2 and K_p = K_c(RT)^-2
(c) K_p = , n = 0 and K_p = K_c

13.39 (a) K_p = , n = -1 and K_p = K_c(RT)^-1
(b) K_p = , n = 0 and K_p = K_c
(c) K_p = , n = 1 and K_p = K_c(RT)

13.40

13.41 K_c =

13.42

13.43

13.44 The two reactions are the reverse of each other.
K_c(reverse) = = 1.3 x 10⁸

13.45 The two reactions are the reverse of each other.
K_p(reverse) = = 1.99 x 10^-2

13.46 K_c = = 0.058

13.47 K_p = = 52.0

13.48 The container volume of 2.00 L must be included to calculate molar concentrations.
Initial [HI] = 9.30 x 10^-3 mol/2.00 L = 4.65 x 10^-3 M = 0.004 65 M
H₂(g) + I₂(g) ¾ 2 HI(g)

x = [H₂] = [I₂] = 6.29 x 10^-4 M = 0.000 629 M
[HI] = 0.004 65 - 2x = 0.004 65 - 2(0.000 629) = 0.003 39 M
K_c = = 29.0

13.49 (a) K_c =
(b) CH₃CO₂H(aq) ¾ H⁺(aq) + CH₃CO₂^-(aq)

K_c = = 1.8 x 10^-5

13.50 (a)
(b) CH₃CO₂H(soln) + C₂H₅OH(soln) ¾ CH₃CO₂C₂H₅(soln) + H₂O(soln)

x = 0.65 mol; 1.00 - x = 0.35 mol; K_c = = 3.4
Because there are the same number of molecules on both sides of the equation, the volume terms in K_c cancel. Therefore, we can calculate K_c without including the volume.

13.51 CH₃CO₂C₂H₅(soln) + H₂O(soln) ¾ CH₃CO₂H(soln) + C₂H₅OH(soln)
K_c(hydrolysis) = = 0.29

13.52 n = 1 and K_p = K_c(RT) = (0.575)(0.082 06)(500) = 23.6

13.53 2 SO₂(g) + O₂(g) ¾ 2 SO₃(g); n = 2 - (2 + 1) = -1 and K_p = 3.30
K_c = K_p = 271

13.54 K_p = = 0.0313 atm; n = 1
K_c = K_p = 1.28 x 10^-3

13.55 = 1.3 x 10^-4 atm
K_p = = 1.3 x 10^-4; n = 1 - 0 = 1, T = 27^oC = 300 K
K_c = K_p = 5.3 x 10^-6

13.56 (a) ,     (b) ,
(c) K_c = [SO₃], (d) K_c = [Ba²⁺][SO₄^2-]

13.57 (a) K_c = , K_p = (b) K_c =

(c) K_c = , K_p = (d) K_c = [CO₂], K_p =

Using the Equilibrium Constant

13.58 (a) Because K_c is very large, the equilibrium mixture contains mostly product.
(b) Because K_c is very small, the equilibrium mixture contains mostly reactants.

13.59 (a) proceeds hardly at all toward completion
(b) goes almost all the way to completion

13.60 (a) Because K_c is very small, the equilibrium mixture contains mostly reactant.
(b) Because K_c is very large, the equilibrium mixture contains mostly product.
(c) Because K_c = 1.8, the equilibrium mixture contains an appreciable concentration of both reactants and products.

13.61 (a) Because K_c is very large, the equilibrium mixture contains mostly product.
(b) Because K_c = 7.5 x 10^-3, the equilibrium mixture contains an appreciable concentration of both reactants and products.
(c) Because K_c is very small, the equilibrium mixture contains mostly reactant.

13.62 K_c = 1.2 x 10⁸² is very large. When equilibrium is reached, very little if any ethanol will remain because the reaction goes to completion.

13.63 Because K_c is very small, pure air will contain very little O₃ (ozone) at equilibrium.
3 O₂(g) ¾ 2 O₃(g); K_c = = 1.7 x 10^-56; [O₂] = 8 x 10^-3 M
[O₃] = = 9 x 10^-32 M

13.64 The container volume of 10.0 L must be included to calculate molar concentrations.
Q_c = = 7.6 x 10^-2; K_c = 2.5 x 10^-3
The reaction is not at equilibrium because Q_c > K_c. The reaction will proceed from right to left to reach equilibrium.

13.65 Q_c = = 0.69; K_c = 4.7
The reaction is not at equilibrium because Q_c < K_c. The reaction will proceed from left to right to reach equilibrium.

13.66 K_c = = 0.29; At equilibrium, [N₂] = 0.036 M and [H₂] = 0.15 M
[NH₃] = = 5.9 x 10^-3 M

13.67 K_c = 2.7 x 10² = ; Because [SO₃] = [SO₂], then 2.7 x 10² =
[O₂] = 3.7 x 10^-3 M

13.68 N₂(g) + O₂(g) ¾ 2 NO(g)

K_c = 1.7 x 10^-3 =
Take the square root of both sides and solve for x.
; 4.1 x 10^-2 = ; x = 2.8 x 10^-2
At equilibrium, [NO] = 2x = 2(2.8 x 10^-2) = 0.056 M
[N₂] = [O₂] = 1.40 - x = 1.40 - (2.8 x 10^-2) = 1.37 M

13.69 N₂(g) + O₂(g) ¾ 2 NO(g)

K_c = = 1.7 x 10^-3 =
4x² + (4.8 x 10^-3)x - (2.1 x 10^-3) = 0
Use the quadratic formula to solve for x.
x =
x = -0.0235 and 0.0223
Discard the solution that uses the negative square root (x = -0.0235) because it gives a negative NO concentration and that is impossible.
[N₂] = 2.24 - x = 2.24 - 0.0223 = 2.22 M
[O₂] = 0.56 - x = 0.56 - 0.0223 = 0.54 M; [NO] = 2x = 2(0.0223) = 0.045 M

13.70 PCl₅(g) ¾ PCl₃(g) + Cl₂(g)

K_c =
x² + (5.8 x 10^-2)x - 0.00928 = 0
Use the quadratic formula to solve for x.

x = 0.071 and -0.129
Discard the solution that uses the negative square root (x = -0.129) because it gives negative concentrations of PCl₃ and Cl₂ and that is impossible.
[PCl₃] = [Cl₂] = x = 0.071 M; [PCl₅] = 0.160 - x = 0.160 - 0.071 = 0.089 M

13.71 Q_c = = 0.020, Q_c < K_c therefore the reaction proceeds from reactants to products to reach equilibrium.

K_c =
x² + 0.198x - (7.60 x 10^-3) = 0
Use the quadratic formula to solve for x.

x = 0.033 and -0.231
Discard the solution that uses the negative square root (x = -0.231) because it gives negative concentrations of PCl₃ and Cl₂ and that is impossible.
[PCl₃] = 0.100 + x = 0.100 + 0.033 = 0.133 M
[Cl₂] = 0.040 + x = 0.040 + 0.033 = 0.073 M
[PCl₅] = 0.200 - x = 0.200 - 0.033 = 0.167 M

13.72 (a) K_c = ; x = 6.8 moles CH₃CO₂C₂H₅
Note that the volume cancels because the same number of molecules appear on both sides of the chemical equation.

(b) CH₃CO₂H(soln) + C₂H₅OH(soln) ¾ CH₃CO₂C₂H₅(soln) + H₂O(soln)

K_c = 3.4 =
2.4x² - 37.4x + 34 = 0
Use the quadratic formula to solve for x.
x =
x = 0.969 and 14.6
Discard the solution that uses the positive square root (x = 14.6) because it leads to negative concentrations and that is impossible.
mol CH₃CO₂H = 1.00 - x = 1.00 - 0.969 = 0.03 mol
mol C₂H₅OH = 10.00 - x = 10.00 - 0.969 = 9.03 mol
mol CH₃CO₂C₂H₅ = mol H₂O = x = 0.97 mol

13.73 When equal volumes of two solutions are mixed together, their concentrations are cut in half.
CH₃Cl(aq) + OH^-(aq) ¾ CH₃OH(aq) + Cl^-(aq)

K_c = ; Because K_c is very large, x << 0.05.
; x = 5 x 10^-18
[CH₃Cl] = x = 5 x 10^-18 M; [OH^-] = [CH₃OH] = [Cl^-] 0.05 M

13.74 ClF₃(g) ¾ ClF(g) + F₂(g)

K_p = = 0.140 = ; solve for x.
x² + 0.140x - 0.2058 = 0
Use the quadratic formula to solve for x.
x =
x =
x = 0.389 and - 0.529
Discard the solution that uses the negative square root (- 0.529) because it gives negative partial pressures and that is impossible.
= x = 0.389 atm
= 1.47 - x = 1.47 - 0.389 = 1.08 atm

13.75 Fe₂O₃(s) + 3 CO(g) ¾ 2 Fe(s) + 3 CO₂(g)

K_p = = 19.9 = ; take the cube root of both sidea and solve for x.
= 2.71 =
2.65 - 8.13x = 3x
2.65 = 11.13x
x = 0.238 atm
= 0.978 - 3x = 0.978 - 3(0.238) = 0.264 atm
= 3x = 3(0.238) = 0.714 atm

Le Châtelier's Principle

13.76 (a) Cl^- (reactant) added, AgCl(s) increases
(b) Ag⁺ (reactant) added, AgCl(s) increases
(c) Ag⁺ (reactant) removed, AgCl(s) decreases
(d) Cl^- (reactant) removed, AgCl(s) decreases
Disturbing the equilibrium by decreasing [Cl^-] increases Q_c (Q_c = ) to a value greater than K_c. To reach a new state of equilibrium, Q_c must decrease, which means that the denominator must increase; that is, the reaction must go from right to left, thus decreasing the amount of solid AgCl.

13.77 (a) ClNO (product) added, NO₂ concentration decreases
(b) NO (reactant) added, NO₂ concentration increases
(c) NO (reactant) removed, NO₂ concentration decreases
(d) ClNO₂ (reactant) added, NO₂ concentration increases
Adding ClNO₂ decreases the value of Q_c (Q_c = ). To reach a new state of equilibrium, the reaction must go from left to right, thus increasing the concentration of NO₂.

13.78 (a) Because there are 2 mol of gas on the left side and 3 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to reactants). The number of moles of reaction products decreases.

(b) Because there are 2 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by a change in pressure. The number of moles of reaction product remains the same.

(c) Because there are 2 mol of gas on the left side and 1 mol of gas on the right side of the balanced equation, the stress of an increase in pressure is relieved by a shift in the reaction to the side with fewer moles of gas (in this case, to products). The number of moles of reaction products increases.

13.79 As the volume increases, the pressure decreases at constant temperature.

(a) Because there is 1 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in volume (decrease in pressure) is relieved by a shift in the reaction to the side with the larger number of moles of gas (in this case, to products).

(b) Because there are 3 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in volume (decrease in pressure) is relieved by a shift in the reaction to the side with the larger number of moles of gas (in this case, to reactants).

(c) Because there are 3 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by an increase in volume (decrease in pressure). There is no net reaction in either direction.

13.80 CO(g) + H₂O(g) ¾ CO₂(g) + H₂(g) H^o = -41.2 kJ
The reaction is exothermic. [H₂] decreases when the temperature is increased.
As the temperature is decreased, the reaction shifts to the right. [CO₂] and [H₂] increase, [CO] and [H₂O] decrease, and K_c increases.

13.81 Because H^o is positive, the reaction is endothermic.
heat + 3 O₂(g) ¾ 2 O₃(g)
K_c =
As the temperature increases, heat is added to the reaction, causing a shift to the right. The [O₃] increases, and the [O₂] decreases. This results in an increase in K_c.

13.82 (a) HCl is a source of Cl^- (product), the reaction shifts left, the equilibrium CoCl₄^2- increases.
(b) Co(NO₃)₂ is a source of Co(H₂O)₆²⁺ (product), the reaction shifts left, the equilibrium CoCl₄^2- increases.
(c) All concentrations will initially decrease and the reaction will shift to the right, the equilibrium CoCl₄^2- decreases.
(d) For an exothermic reaction, the reaction shifts to the left when the temperature is increased, the equilibrium CoCl₄^2- increases.

13.83 (a) Fe(NO₃)₃ is a source of Fe³⁺. Fe³⁺ (reactant) added; the FeCl²⁺ concentration increases.
(b) Cl^- (reactant) removed; the FeCl²⁺ concentration decreases.
(c) An endothermic reaction shifts to the right as the temperature increases; the FeCl²⁺ concentration increases.
(d) A catalyst does not affect the composition of the equilibrium mixture; no change in FeCl²⁺ concentration.

13.84 (a) The reaction is exothermic. The amount of CH₃OH (product) decreases as the temperature increases.
(b) When the volume decreases, the reaction shifts to the side with fewer gas molecules. The amount of CH₃OH increases.
(c) Addition of an inert gas (He) does not affect the equilibrium composition. There is no change.
(d) Addition of CO (reactant) shifts the reaction toward product. The amount of CH₃OH increases.
(e) Addition or removal of a catalyst does not affect the equilibrium composition. There is no change.

13.85 (a) An endothermic reaction shifts to the right as the temperature increases. The amount of acetone increases.
(b) Because there is 1 mol of gas on the left side and 2 mol of gas on the right side of the balanced equation, the stress of an increase in volume (decrease in pressure) is relieved by a shift in the reaction to the side with the larger number of moles of gas (in this case, to products). The amount of acetone increases.
(c) The addition of Ar (an inert gas) with no volume change does not affect the composition of the equilibrium mixture. The amount of acetone does not change.
(d) H₂ (product) added; the amount of acetone decreases.
(e) A catalyst does not affect the composition of the equilibrium mixture. The amount of acetone does not change.

Chemical Equilibrium and Chemical Kinetics

13.86 A + B ¾ C
rate_f = k_f[A][B] and rate_r = k_r[C]; at equilibrium, rate_f = rate_r
k_f[A][B] = k_r[C];

13.87 An equilibrium mixture that contains large amounts of reactants and small amounts of products has a small K_c. A small K_c has k_f < k_r (c).

13.88 K_c = = 210

13.89 K_c = ; k_r = = 6 x 10^-22 M^-1s^-1

13.90 k_r increases more than k_f, this means that E_a (reverse) is greater than E_a (forward). The reaction is exothermic when E_a (reverse) > E_a (forward).

13.91 k_f increases more than k_r, this means that E_a (forward) is greater than E_a (reverse). The reaction is endothermic when E_a (forward) > E_a (reverse).

General Problems

13.92 (a) [N₂O₄] = = 0.125 M

N₂O₄(g) ¾ 2 NO₂(g)

At equilibrium, [N₂O₄] = 0.125 - (0.793)(0.125) = 0.0259 M
[NO₂] = (2)(0.793)(0.125) = 0.198 M
K_c = = 1.51
n = 2 - 1 = 1 and K_p = K_c(RT)ⁿ; K_p = K_c(RT) = (1.51)(0.0821)(400) = 49.6

(b)

13.93 K_c is very large. The reaction goes essentially to completion.
K_c = ; [CO₂] = = 6.2 x 10^-25 M

13.94 K_c = = 0.291
At equilibrium, [N₂] = 1.0 x 10^-3 M and [H₂] = 2.0 x 10^-3 M
[NH₃] = = 1.5 x 10^-6 M

13.95 (a) K_p = = 7.83 at 1500 K
P_F = = 1.25 atm

(b) F₂(g) ¾   2 F(g)

2y = 1.25 so y = 0.625
x - y = 0.200; x = 0.200 + y = 0.200 + 0.625 = 0.825
f = = 0.758

(c) The shorter bond in F₂ is expected to be stronger. However, because of the small size of F, repulsion between the lone pairs of the two halogen atoms are much greater in F₂ than in Cl₂.

13.96 2 HI(g) ¾ H₂(g) + I₂(g)
Calculate K_c. K_c = = 0.0206
[HI] = = 0.40 M

K_c = 0.0206 =
Take the square root of both sides, and solve for x.
; 0.144 = ; x = 0.045
At equilibrium, [H₂] = [I₂] = x = 0.045 M; [HI] = 0.40 - 2x = 0.40 - 2(0.045) = 0.31 M

13.97 Note the container volume is 5.00 L
[H₂] = [I₂] = 1.00 mol/5.00 L = 0.200 M
[HI] = 2.50 mol/5.00 L = 0.500 M

K_c =
Take the square root of both sides, and solve for x.
; 11.4 = ; x = 0.133
[H₂] = [I₂] = 0.200 - x = 0.200 - 0.133 = 0.067 M
[HI] = 0.500 + 2x = 0.500 + 2(0.133) = 0.766 M

13.98 [H₂O] = = 1.20 M
C(s) + H₂O(g) ¾  CO(g) + H₂(g)

K_c =
x² + (3.0 x 10^-2)x - 0.036 = 0
Use the quadratic formula to solve for x.
x =
x = 0.176 and -0.206
Discard the solution that uses the negative square root (x = -0.206) because it leads to negative concentrations and that is impossible.
[CO] = [H₂] = x = 0.18 M; [H₂O] = 1.20 - x = 1.20 - 0.18 = 1.02 M

13.99 (a) Because K_p is larger at the higher temperature, the reaction has shifted toward products at the higher temperature, which means the reaction is endothermic.

(b) (i) Increasing the volume causes the reaction to shift toward the side with more mol of gas (product side). The equilibrium amounts of PCl₃ and Cl₂ increase while that of PCl₅ decresases.
(ii) If there is no volume change, there is no change in equilibrium concentrations.
(iii) Addition of a catalyst does not affect the equilibrium concentrations.

13.100 A decrease in volume (a) and the addition of reactants (c) will affect the composition of the equilibrium mixture, but leave the value of K_c unchanged.
A change in temperature (b) affects the value of K_c.
Addition of a catalyst (d) or an inert gas (e) affects neither the composition of the equilibrium mixture nor the value of K_c.

13.101 (a) Addition of a solid does not affect the equilibrium composition. There is no change in the number of moles of CO₂.
(b) Adding a product causes the reaction to shift toward reactants. The number of moles of CO₂ decreases.
(c) Decreasing the volume causes the reaction to shift toward the side with fewer mol of gas (reactant side). The number of moles of CO₂ decreases.
(d) The reaction is endothermic. An increase in temperature shifts the reaction toward products. The number of moles of CO₂ increases.

13.102 2 monomer ¾ dimer
(a) In benzene, K_c = 1.51 x 10²

K_c =
604x² - 61.4x +1.51 = 0
Use the quadratic formula to solve for x.
x =
x = 0.0600 and 0.0417
Discard the solution that uses the positive square root (x = 0.0600) because it gives a negative concentration of the monomer and that is impossible.
[monomer] = 0.100 - 2x = 0.100 - 2(0.0417) = 0.017 M; [dimer] = x = 0.0417 M
= 2.5

(b) In H₂O, K_c = 3.7 x 10^-2
2 monomer ¾ dimer

K_c =
0.148x² - 1.0148x + 0.000 37 = 0
Use the quadratic formula to solve for x.
x =
x = 6.86 and 3.7 x 10^-4
Discard the solution that uses the positive square root (x = 6.86) because it gives a negative concentration of the monomer and that is impossible.
[monomer] = 0.100 - 2x = 0.100 - 2(3.7 x 10^-4) = 0.099 M; [dimer] = x = 3.7 x 10^-4 M
= 0.0038

(c) K_c for the water solution is so much smaller than K_c for the benzene solution because H₂O can hydrogen bond with acetic acid, thus preventing acetic acid dimer formation. Benzene cannot hydrogen bond with acetic acid.

13.103 C(s) + CO₂(g) ¾  2 CO(g)

[CO] = 2x = 7.00 x 10^-2 M; x = 0.0350 M
(a) [CO₂] = 0.0750 - x = 0.0750 - 0.0350 = 0.0400 M
(b) K_c = = 0.122

13.104 (a) [PCl₅] = 1.000 mol/5.000 L = 0.2000 M
PCl₅(g) ¾   PCl₃(g) + Cl₂(g)

K_c = = = 0.573
n = 1 and K_p = K_c(RT) = (0.573)(0.082 06)(500) = 23.5
(b) Q_c = = = 0.18
Because Q_c < K_c, the reaction proceeds to the right to reach equilibrium.

PCl₅(g) ¾ PCl₃(g) + Cl₂(g)

K_c = = 0.573 = ; solve for x.
x² + 1.323x - 0.1965 = 0
x =
x = - 1.458 and 0.135
Discard the solution that uses the negative square root (- 1.458) because it will lead to negative concentrations and that is impossible.
[PCl₅] = 0.500 - x = 0.500 - 0.135 = 0.365 M
[PCl₃] = 0.150 + x = 0.150 + 0.135 = 0.285 M
[Cl₂] = 0.600 + x = 0.600 + 0.135 = 0.735 M

13.105 Q_p = = 1.00, Q_p < K_p therefore the reaction proceeds from reactants to products to reach equilibrium.

PCl₅(g) ¾  PCl₃(g) + Cl₂(g)

K_p =
x² + 4.92x - 1.26 = 0
Use the quadratic formula to solve for x.

x = 0.245 and -5.165
Discard the solution that uses the negative square root (x = -5.165) because it gives negative partial pressures and that is impossible.
= 3.00 - x = 3.00 - 0.245 = 2.76 atm
= 2.00 + x = 2.00 + 0.245 = 2.24 atm
= 1.50 + x = 1.50 + 0.245 = 1.74 atm
P_total = + + = 2.76 + 2.24 + 1.74 = 6.74 atm

13.106 (a)
(b) K_p = 12; n = 1; K_c = K_p = 0.19
(c) C₄H₁₀(g) ¾ C₂H₆(g) + C₂H₄(g)

K_p = 12 = ; x² + 12x - 600 = 0
Use the quadratic formula to solve for x.
x =
x = -31.22 and 19.22
Discard the solution that uses the negative square root (x = -31.22) because it leads to negative concentrations and that is impossible.
% C₄H₁₀ converted = = 38%
P_total = = (50 - x) + x + x = (50 - 19) + 19 + 19 = 69 atm
(d) A decrease in volume would decrease the % conversion of C₄H₁₀.

13.107 (a) Because n = 0, K_p = K_c = 1.0 x 10⁵
(b) K_p = 1.0 x 10⁵ = ; P_cyclopropane = = 5.0 x 10^-5 atm

(c) The ratio of the two concentrations is equal to K_c. The ratio (K_c) cannot be changed by adding cyclopropane. The individual concentrations can change but the ratio of concentrations can't. Because there is one mole of gas on each side of the balanced equation, the composition of the equilibrium mixture is unaffected by a decrease in volume. The ratio of the two concentrations will not change.
(d) Because K_c is large, k_f > k_r.
(e) Because the C-C-C bond angles are 60^o and the angles between sp³ hybrid orbitals are 109.5^o, the hybrid orbitals are not oriented along the bond directions. Their overlap is therefore poor, and the C-C bonds are correspondingly weak.

13.108 (a) K_p = 3.45; n = 1; K_c = K_p = 0.0840
(b) [(CH₃)₃CCl] = 1.00 mol/5.00 L = 0.200 M
(CH₃)₃CCl(g) ¾  (CH₃)₂C=CH₂(g) + HCl(g)

K_c = 0.0840 = ; x² + 0.0840x - 0.0168 = 0
Use the quadratic formula to solve for x.
x =
x = -0.178 and 0.094
Discard the solution that uses the negative square root (x = -0.178) because it leads to negative concentrations and that is impossible
[(CH₃)₂C=CCH₂] = [HCl] = x = 0.094 M
[(CH₃)₃CCl] = 0.200 - x = 0.200 - 0.094 = 0.106 M

(c) K_p = 3.45
(CH₃)₃CCl(g) ¾ (CH₃)₂C=CH₂(g) + HCl(g)

K_p = 3.45 =
x² - 4.45x + 0.240 = 0
Use the quadratic formula to solve for x.
x =
x = 0.055 and 4.40
Discard the solution that uses the positive square root (x = 4.40) because it leads to a negative partial pressures and that is impossible.
P_{t-butyl chloride} = x = 0.055 atm; P_isobutylene = 0.400 - x = 0.400 - 0.055 = 0.345 atm
P_HCl = 0.600 - x = 0.600 - 0.055 = 0.545 atm

13.109 (a) The Arrhenius equation gives for the forward and reverse reactions
and
Addition of a catalyst decreases the activation energies by E_a, so
=

and =
Therefore, the rate constants for both the forward and reverse reactions increase by the same factor, .

(b) The equilibrium constant is given by K_c =
where E = E_a,f - E_a,r. Addition of a catalyst decreases the activation energies by E_a.
So, K_c =
K_c is unchanged because of cancellation of , the factor by which the two rate constants increase.

13.110 The activation energy (E_a) is positive, and for an exothermic reaction, E_a,r > E_a,f.
k_f = A_f , k_r = A_r
K_c =

(E_a,r - E_a,f) is positive, so the exponent is always positive. As the temperature increases, the exponent, , decreases and the value for K_c decreases as well.

Multi-Concept Problems

13.111 (a) H₂O, 18.015 amu; 125.4 g H₂O x = 6.96 mol H₂O
Given that mol CO = mol H₂O = 6.96 mol
= 40.0 atm
CO(g) + H₂O(g) ¾  CO₂(g) + H₂(g)

K_p = = 9.50

(b) 31.4 g H₂O x = 1.743 mol H₂O
= 10.0 atm
has been increased by 10.0 atm; a new equilibrium will be established.

CO(g) + H₂O(g) ¾  CO₂(g) + H₂(g)

K_p =
8.50x² - 341.6x + 931.34 = 0
Use the quadratic formula to solve for x.
x =
x = 37.25 and 2.94
Discard the solution that uses the positive square root (x = 37.25) because it leads to a negative partial pressures and that is impossible.
= 9.80 - x = 9.80 - 2.94 = 6.86 atm
= 19.8 - x = 19.8 - 2.94 = 16.9 atm
= 30.2 + x = 30.2 + 2.94 = 33.1 atm
= 5.76 mol H₂
= 3.47 x 10²⁰ H₂ molecules/cm³

13.112 (a) CO₂, 44.01 amu; CO, 28.01 amu
79.2 g CO₂ x = 1.80 mol CO₂
CO₂(g) + C(s) ¾ 2 CO(g)

total mass of gas in flask = (16.3 g/L)(5.00 L) = 81.5 g
81.5 = (1.80 - x)(44.01) + (2x)(28.01)
81.5 = 79.22 - 44.01x + 56.02x; 2.28 = 12.01x; x = 2.28/12.01 = 0.19

= 1.80 - x = 1.80 - 0.19 = 1.61 mol CO₂; = 2x = 2(0.19) = 0.38 mol CO
= 26.4 atm
= 6.24 atm
K_p = = 1.47

(b) At 1100K, the total mass of gas in flask = (16.9 g/L)(5.00 L) = 84.5 g
84.5 = (1.80 - x)(44.01) + (2x)(28.01)
84.5 = 79.22 - 44.01x + 56.02x; 5.28 = 12.01x; x = 5.28/12.01 = 0.44
= 1.80 - x = 1.80 - 0.44 = 1.36 mol CO₂; = 2x = 2(0.44) = 0.88 mol CO

= 24.6 atm
= 15.9 atm
K_p = = 10.3

(c) In agreement with Le Châtelier's principle, the reaction is endothermic because K_p increases with increasing temperature.

13.113 CO₂, 44.01 amu; CO, 28.01 amu; BaCO₃, 197.34 amu
1.77 g CO₂ x = 0.0402 mol CO₂
CO₂(g) + C(s) ¾ 2 CO(g)

3.41 g BaCO₃ x x = 0.0173 mol CO₂
mol CO₂ = 0.0173 = 0.0402 - x; x = 0.0229
mol CO = 2x = 2(0.0229) = 0.0458 mol CO
= 1.562 atm
= 4.134 atm
K_p = = 11.0

13.114 (a) N₂O₄, 92.01 amu
14.58 g N₂O₄ x = 0.1585 mol N₂O₄
PV = nRT
= 5.20 atm

N₂O₄(g) ¾ 2 NO₂(g)

= (5.20 - x) + (2x) = 9.15 atm
5.20 + x = 9.15 atm
x = 3.95 atm
= 5.20 - x = 5.20 - 3.95 = 1.25 atm
= 2x = 2(3.95) = 7.90 atm
K_p = = 49.9
n = 1 and K_c = K_p = 1.52

(b) H^o_rxn = [2 H^o_f(NO₂)] - H^o_f(N₂O₄)
H^o_rxn = [(2 mol)(33.2 kJ/mol)] - [(1mol)(9.16 kJ/mol)] = 57.2 kJ
PV = nRT
moles N₂O₄ reacted = n = = 0.1203 mol N₂O₄
q = (57.24 kJ/mol N₂O₄)(0.1203 mol N₂O₄) = 6.89 kJ

13.115 C₁₀H₈(s) ¾ C₁₀H₈(g)
(a) K_c = [C₁₀H₈] = 5.40 x 10^-6
room volume = 8.0 ft x 10.0 ft x 8.0 ft = 640 ft²
room volume = 640 ft² x x x = 18122.8 L
C₁₀H₈ molecules = (5.40 x 10^-6 mol/L)(18122.8 L)(6.022 x 10²³ molecules/mol)
= 5.89 x 10²² C₁₀H₈ molecules

(b) C₁₀H₈, 128.17 amu
mol C₁₀H₈ = (5.40 x 10^-6 mol/L)(18122.8 L) = 0.0979 mol C₁₀H₈
mass C₁₀H₈ = 0.0979 mol C₁₀H₈ x = 12.55 g C₁₀H₈
moth ball: r = 12.0 mm/2 = 6.0 mm = 0.60 cm
volume of moth ball = (4/3)r³ = (4/3)(0.60cm)³ = 0.905 cm³
mass of moth ball = (0.905 cm³/moth ball)(1.16 g/cm³) = 1.05 g/moth ball
number of moth balls = = 12 moth balls

13.116 The atmosphere is 21% (0.21) O₂; = (0.21)= 0.199 atm
2 O₃(g) ¾ 3 O₂(g)
K_p = ; = 2.46 x 10^-30 atm
vol = 10 x 10⁶ m³ x = 1.0 x 10¹⁰ L
= 1.0 x 10^-21 mol O₃
O₃ molecules = 1.0 x 10^-21 mol O₃ x = 6.0 x 10² O₃ molecules