Chapter 14 Chemical Equilibrium 545

if a reaction has an equal number of moles of gas on both sides of the chemical equation, then a change involume produces no effect on the equilibrium.adding an inert gas to the mixture at a fixed volume has no effect on the equilibrium.

14.20 If the temperature of a system at equilibrium is changed, the system shifts in a direction to counter thatchange. So, if the temperature is increased, the reaction shifts in the direction that tends to decrease the tem-perature and vice versa.In an exothermic chemical reaction,increasing the temperature causes the reaction to shift left and the value of the equilibrium constant decreases,decreasing the temperature causes the reaction to shift right and the value of the equilibrium constant increases.In an endothermic chemical reaction,increasing the temperature causes the reaction to shift right and the equilibrium constant increases,decreasing the temperature causes the reaction to shift left and the equilibrium constant decreases.

Problems by Topic

uilibrium and the Equilibrium Constant

The equilibrium constant is defined as the concentrations of the products raised to their stoichiometric coef-ficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

(a)

(b)

(c)

TS

[SbCl3][Cl2]

[SbOsl

[NO]2[Br2]

[BrNO]2

[CSzMH,]4

[CH4][H2S]2

[C02]2

[C0]2[02]

(a)y The equilibrium constant is defined as the concentrations of the products raised to their stoichiometriccoefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

K =[H2S]2

(b)

14.23

14.24

The equilibrium constant is defined as the concentrations of the products raised to their stoichiometriccoefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

[COC12]~ [CO][C12]

With an equilibrium constant of 1.4 x 10 ~5, the value of the equilibrium constant is small, therefore, the con-centration of reactants will be greater than the concentration of products. This is independent of the initialconcentration of the reactants and products.

Figure a at equilibrium has 8 C2H4C12/ 2 C12, and 2 C2H4.Figure b at equilibrium has 6 C2H4Br2/ 4 Br2, and 4 C2H4.Figure c at equilibrium has 3 C2H4I2, 7 1̂ , and 7 C2H4.Since the equilibrium constant is concentration of products/concentration of reactants, the equilibrium sit-uation which has the largest concentration of products will have the largest equilibrium constant. Therefore,

546 Chapter 14 Chemical Equilibrium

14.25

14.26

14.28

14.29

(i) has 10 H2 and 10 I2

(ii) has 7 H2 and 712 and 6 HI(iii) has 5 H2 and 512 and 10 HI(iv) has 4 H2 and 4 I2 and 12 HI(v) has 3 H2 and 3 I2 and 14 HI(vi) has 3 H2 and 3 I2 and 14 HI

(a) Concentration of (v) and (vi) are the same, so the system reached equilibrium at (v).

(b) If a catalyst was added to the system, the system would reach the conditions at (v) sooner since a catalystspeeds up the reaction but does not change the equilibrium conditions.

(c) The final figure (vi) would have the same amount of reactants and products since a catalyst speedsup the reaction, but does not change the equilibrium concentrations.

The equilibrium constant gives us the ratio of products to reactants at equilibrium, it does not say how longit takes to reach equilibrium. So after 15 minutes, if the smaller equilibrium constant has more products,then the kinetics of that reaction are faster.

(a)

(b)

(c)

If you reverse the reaction, invert the equilibrium constant. So K

The reactants will be favored.

J_

K,,1

2.26 x 104= 4.42 x 10"

If you multiply the coefficients in the equation by a factor, raise the equilibrium constant to thesame factor.

So K = (Kp)1/2 = (2.26 x 104)1/2 = 1.50 x 102. The products will be favored.

Begin with the reverse of the reaction and invert the equilibrium constant.

, = — = = 447y1(r5; K.

K

(a)

(b)

•p 2.26 x 104

Then multiply the reaction by 2 and raise the value of Kreverse to the 2nd power.

K = (Kreverse)2 = (4.42 xlO~5)2 = 1.96 x 10~9.The reactants will be favored.

The reaction is multiplied by 1/2, so raise the value of the equilibrium constant to 1/2.K = (Kp)

1/2 = (2.2 x 106)1/2 = 1.5 x 103. The products will be favored.

The reaction is multiplied by 3, so raise the value of the equilibrium constant to 3.

K = (Kpf = (2.2 x 106)3 = 1.1 x 1019. The products will be favored.

(c) Begin with the reverse of the reaction and invert the equilibrium constant.1

= 4.5 x 10-7

2.2 x 106

Then multiply the reaction by 2 and raise the value of Kreverse to the 2nd power.

K = (^reverse)2 = (4-5 x 10~7)2 = 2.1 x 10~13. The reactants will be favored.

To find the equilibrium constant for reaction 3, you need to combine reactions 1 and 2 to get reaction 3. Beginby reversing reaction 2, then multiply reaction 1 by 2 and add the two new reactions. When you add reac-tions you multiply the values of K.

i -i

= 4.76xlO~31

2~NQ(g) + Br2 (g) «=? 2 NOBr(g)

KP 2.1X1030

= (Kp)2 = (5.3)2 = 28.09

N2(g) + 02(g) + Br2 (g) ;± 2 NOBr(g) K = = (4.76xlO~J1)(28.09) = 1.3xlO~

Chapter 14 Chemical Equilibrium 547

14.30/ To find the equilibrium constant for reaction 3, you need to combine reactions 1 and 2 to get reaction 3. Beginby multiplying reaction 1 by 2 and then reverse reaction 2 and add the two new reactions. When you addreactions you multiply the values of K.

KI = (Kp)2 = (0.0334)2 = 1.1 x 10~3

1 1

2 A(s) *± B(g) + 2-€(g)

B(g) + 2-€(g) ±5 3 D(g)

2 A(s) ^ 3 D(g) K' = K-i K2 = (1.1 x 1(T3)(0.425) = 4.68 x 1( 4

and Heterogeneous Equilibria(a) Given: Kp = 6.26 xW'22 T = 298K Find: Kc

Conceptual Plan: Kp — » Kc

Kp = Kc(RT)a"

Solution: An = mol product gas - mol reactant gas = 2-1 = 1

- = 2.56x10-*

(0.08206 x 298 K)'mol -K

Check: Substitute into the equation and confirm that you get the original value of Kp

= 6.26 xlO'22= (2.56 x . 08206mol-K

(b) Given: Kp = 7.7 x 1024 T = 298K Find: Kc

Conceptual Plan: Kp -> Kc

Kp = KC(RT)^

Solution: An = mol product gas - mol reactant gas = 4-2 = 2

= 1.3x10-(008206 x298K)2

mol-KCheck: Substitute into the equation and confirm that you get the original value of Kp

= KC(RT)A" = (1 .3 x . 08206mol -K x 298)2 = 7.7 x 1024

(c) Given: Kp = 81.9 T = 298K Find: Kc

Conceptual Plan: Kp -* Kc

Kp = Kr(RT)A"

Solution: An = mol product gas - mol reactant gas = 2-2 = 0KP 81.9

Kc = - - = 81.9(RTr (0.08206^*5* 298 K)°

mol -KCheck: Substitute into the equation and confirm that you get the original value of Kp.

Kv = Kc(RT)*n = (81 .9X0.08206^^ x 298)° = 81.9mol-K

Given: Kc= 5.9 x KT3 T = 298K Find: Kp

Conceptual Plan: Kp —> Kc

Solution: An = mol product gas - mol reactant gas = 2-1 = 1

Kn = Kc(RT)&n = 5.9 x 10~3(0.08206 L ' ̂ to x 298 K)1 = 1.4xlO-1

mol-KCheck: Substitute into the equation and confirm that you get the original value of Kp.

(0.08206mol-K

^ 298 K)1

548 Chapter 14 Chemical Equilibrium

(b) Given: Kc = 3.7 x 108 T = 298K Find: Kp

Conceptual Plan: Kp -> Kc

Solution: An = mol product gas - mol reactant gas = 2 - 4 = - 2

KB = KC(RT)A" = 3.7 x 108(0.08206 ' atm x 298 K)~ 2 = 6.2 x 105

mol-KCheck: Substitute into the equation and confirm that you get the original value of Kp.

^P 6.2 x 105

Kc =(RT)A" (0.08206 atm

= 3.7 x 108

mol-Kx298K) -2

(c) Given: Kc = 4.10 x 10 ~31 T = 298K Find: Kp

Conceptual Plan: Kp -» Kc

Kp = KC(RT)4"

Solution: An = mol product gas - mol reactant gas = 2-2 = 0

Kv = KC(RT)A" = 4.10 x 10 (0.08206 L ' atm x 298 K)° = 4.10xlO~3 1

v C . .mol-K

Check: Substitute into the equation and confirm that you get the original value of Kp.K,, 4.10x10 -31

(RT)An

(0.08206= 4.10xlO~31

mol-Kx 298 K)°

[HC03][OH-]14.33 /(a) Since H2O is a liquid, it is omitted from the equilibrium expression. Kecf =

(b) Since KC1O3 and KC1 are both solids, they are omitted from the equilibrium expression. Keci = [O2]3

[H30+][F-]

(c) Since H2O is a liquid, it is omitted from the equilibrium expression. Ke!j =

(d) Since H2O is a liquid, it is omitted from the equilibrium expression. Ke(f =

[HF]

[NH4+][OH']

[NH3]

ince PC13 is a liquid, it is omitted from the equilibrium expression. Kei) =[C12]

[elating the Equilibrium Constant to Equilibrium Concentrations andJiquilikrium Partial Pressures

[ 14.35 /Given: At equilibrium: [CO] = 0.105M, [H2] = 0.114M, [CH3OH] = 0.185M Find: JQ\ / Conceptual Plan: Balanced reaction —> equilibrium expression —> Kc

VX StSolution: Kc =[CH3OH] (0.185)

= 136

14.36

[CO][H2]2 (0.105)(0.114)2

Check: The answer is reasonable since the concentration of products is greater than the concentration ofreactants and the equilibrium constant should be greater than 1.

Given: At equilibrium: [NH3] = 0.278M, [H2S] = 0.355M Find: KCConceptual Plan: Balanced reaction —> equilibrium expression —» Kc

Solution: Since NH4HS is a solid, it is omitted from the equilibrium expression.Kc = [NH3][H2S] = (0.278)(0.355) = 0.0987

Check: The answer is reasonable since the concentration of products is less than 1M, the equilibrium con-stant is less than 1.

14.37 At 500K: Given: At equilibrium: [N2] = 0.115M, [H2] = 0.105M, and [NH3] = 0.439 Find: Kc

Chapter 14 Chemical Equilibrium 549

Conceptual Plan: Balanced reaction — > equilibrium expression — > Kc

[NH3]2 (0.439)2

SoluHon: Kc = — - — = - - = 1.45 x 103

[N2][H2]3 (0.115)(0.105)3

Check: The value is reasonable since the concentration of products is greater than the concentration of reactants.At 575K: Given: At equilibrium: [N2] = 0.110M, [NH3] = 0.128, Kc = 9.6 Find: [H2]Conceptual Plan: Balanced reaction — * equilibrium expression — > [H2]

[NH3]2 (0.128)2

Solution: Kc = -1— 9.6 = -^— x = 0.249[N2][H2]

3 (O.llO)W3

Check: Plug the value for x back into the equilibrium expression, and check the value.

g 6 = (0.128)2

(0.110)(0.249)3

At 775K: Given: At equilibrium: [N2] = 0.120M, [H2] = 0.140, Xc = 0.0584 Find: [NH3]Conceptual Plan: Balanced reaction * equilibrium expression — * [NH3]

risiH i2 (x}^Solution: Kc = — - — 0.0584 = - x = 0.00439

[N2][H2]3 (0.120)(0.140)3

Check: Plug the value for x back into the equilibrium expression, and check the value.(0.00439)2

0.0584 = — i—(0.120)(0.140)3

14.38 At 25°C: Given: At equilibrium: [H2] = 0.0355M, [I2] = 0.0388M, and [HI] = 0.922 Find: Kc

Conceptual Plan: Balanced reaction — » equilibrium expression — » Kc

[HI]2 (0.922)2

(0.0355X0.0388)Check: The value is reasonable since the concentration of products is greater than the concentration of reactants.At 340°C: Given: At equilibrium: [I2] = 0.0455, and [HI] = 0.387, Xc = 9.6 Find: [H2]Conceptual Plan: Balanced reaction — > equilibrium expression — > [H2]

[HI]2 (0.387)2

Check: Plug the value for x back into the equilibrium expression, and check the value.(0.387)2

(0.343X0.0455)At 445°C: Given: At equilibrium: [H2] = 0.0485M, [I2] = 0.0468, Kc = 50.2 Find: [HI]Conceptual Plan: Balanced reaction * equilibrium expression * [HI]

(0.0485X0.0468)Check: Plug the value for x back into the equilibrium expression, and check the value.

(0.338)2

50.2 =(0.0485)(0.0468)

14.39 Given: PNO = 108 torr; PBr2 = 126 torr, Kp = 28.4 Find: PNOBr

Conceptual Plan: torr — * arm and then balanced reation — > equilibrium expression — * P1 atm

760 torr

Solution: PNO = 108 torr x —p^- = 0.1421 atm PBr2 = 126 torr x—^- — = 0.1658p2 ^2

NOBr 28.4 = - x = 0.308 atm = 234 torr.(0.1421)2(0.1658)

Check: Plug the value for x back into the equilibrium expression, and check the value.

28.3 =(0.308)2

(0.142ir(0.1658)

550 Chapter 14 Chemical Equilibrium

14.40

14.43

Given: Pso2 = 137 torr; PCi2 = 285 torr, Kp = 2.91 x 103 Find:Conceptual Plan: torr —> atm and then balanced reation —> equilibrium expression

1 atmPso2ci2

760 ton-

Solution: Pso2 = 137 torr x

PSO/CI,

1 atm760 torr

= 0.1803 atm Pd2 = 285 torr x1 atm

= 0.3750

2.91 x 103 =(0.1803)(0.3750)

760 torr

x = 2.32 x 10~5 atm = 0.0177 torrrso2ci2 (x)Check: Plug the value for x back into the equilibrium expression, and check the value.

, (0.1803)(0.3750)2.91xl03 = - -

(2.32 x 10~5)

Given: [Fe3+]initial = 1.0 x 10 - 3 M; [SCN - l^i = 8.0 x 10 - 4 M; [FeSCN2+]eq = 1.7 x 10 ~ *M Find:Conceptual Plan: 1. Prepare ICE table

2. Calculate concentration change for known value3. Calculate concentration changes for other reactants/products4. Determine equilibrium concentration5. Write the equilibrium expression and determine Kc

Solution: ?e3+(aq) + SCN "(at?) *=> FeSCN3+(fl<?)[SCN-] [FeSCN3+]8.0x10^ 0.00

+1.7x10-*

[Fe3+]I l.OxlO-3

C -1.7x10-* -1.7x10-*E 8.3 xl(T* 6.3x10-*

[FeSCN2+]1.7x10"

(1.7xKT4)

(8.3 x 10~4)(6.3 x 10~4)= 3.3 x 102

Given: [SO2Cl2]initia] = 0.020 M; [Cl2]eq = 1.2 x 10~2M Find:Conceptual Plan:

Solution:

1. Prepare ICE table2. Calculate concentration change for known value3. Calculate concentration changes for other reactants/products4. Determine equilibrium concentration5. Write the equilibrium expression and determine Kc

S02C12(£) ^ S02(g) + Cl2(g)

ICE

Kc =

[S02C12]0.020

-1.2xlO~ 2

0.0080[sojicy[S02C12]

[soj [cy0.00 0.00

+1.2xlO"2+1.2xlO"2

1.2xlO~21.2xlO-2

(1.2xlO"2)(1.2xlO~2)

(0.0080)= 0.018

Given: 3.67 L flask, 0.763 g H2 initial, 96.9 g I2 initial, 90.4 g HI equilibrium Find: Kc

Conceptual Plan: g —» mol —> M and theng n

n = — - M = —molar mass V

1. Prepare ICE table2. Calculate concentration change for known value3. Calculate concentration changes for other reactants/products4. Determine equilibrium concentration5. Write the equilibrium expression and determine Kc

ImolH-) 0.3785 mol H2Solution: 0.763 g-H? x „ Mf ^T r = 0.3785 mol H2 ^^-r = 0.103 M. This is an initial concentration

96.

2.

1 mol I2= 0.3818 mol I2

3.67 L

0.3818 mol I2

= 0.7068 mol HI

3.67 L

0.7068 mol HI3.67 L

= 0.104 M. This is an initial concentration

= 0.193 M. This is an equilibrium

concentration

Chapter 14 Chemical Equilibrium 551

ICE

r :

H2(g) -[H2]0.103

-0.09630.0067[HI]2

H I2 (£) *=> 2 HI(g)

[I2] [HI]0.104 0.00

-0.0963 +0.1930.0077 0.193

(0.193)2

Since HI gained 0.193M, then H2 and I2 had to lose 0.193/2= 0.0963M from the stoichiometry

9

of the balanced reaction.

14.44

[H2][I2] (0.0065)(0.0075)

Given: 5.19 L flask, 26.9 g CO initial, 2.34 g H2 initial, 8.65 g CH3OH equilibrium Find:Conceptual Plan: g — > mol — * M and then

molar mass V1. Prepare ICE table2. Calculate concentration change for known value3. Calculate concentration changes for other reactants/products4. Determine equilibrium concentration5. Write the equilibrium expression and determine Kc

1 mol CO 0.9604 mol CO1U1

ICE

K

28.01 g-€0"lmolH2

1 mol CH3OH

CO(g) + 2H2(g) fc»CH3OH(g)[CO] [H2] [CH3OH]0.185 0.224 0.00

-0.0520 -0.104 +0.05200.133 0.120 0.0520

[CH3OH] (0.0520)

^^ c 1Q T — U.10O IViO. -L z/ L

1.161 mol H2

0.2700 mol CH3OHC 1 Q T

Since CH3OH gained 0.0520 M, CO had to lose 0.0520since the stoichiometry is 1:1 and H2 had to lose 0.104

5 since the stoichiometry is 2:1.

MM

[CO][H2]2 (0.133)(0.120)2

The Reaction Quotient and Reaction Direction

14.45 Given: Kc = 8.5 x 10 ~3; [NH3] = 0.166M; [H2S] = 0.166M Find: Will solid form or decompose?Conceptual Plan: Calculate Q — •» compare Q and Kc

Solution: Q = [NH3][H2S] = (0.166)(0.166) = 0.0276Q = 0.0276 and Kc = 8.5 x 10 ~3 so Q > K,. and the reaction will shift to the left, so more solid will form.

Given: Kp = 2.4 x 10"4; PH2 = 0.112atm; P^ = 0.055atm; PHzs = 0.445atm Find: Is the reaction at equilibrium?Conceptual Plan: Calculate Q — » compare Q and Kp

Solution: Q = ̂ ^ = (°-112)2(a055) = 3.48 x 10-3 Q = 3.48 x 10 - 3 and K 2A x 1Q -AP2H2s (0.445)2

so Q > Kp, the system is not at equilibrium, and the reaction will shift to the left.

Given: 6.55 g Ag2SO4, 1.5 L solution, i<Cc = 1.1 x 10 " 5 Find: Will more solid dissolve?Conceptual Plan: g Ag2SO4 -» mol Ag2SO4 -» [Ag2SO4] -» [Ag+],[SO4"

2] -» calculate Q and compare to Kc.1 mol Ag2S04 , , _ mol Ag2SO4 + 2 _

14.47

Solution:

311.81g vol solution

mol Ag2SO4 \1 = 0.0210 mol Ag2SO4311.810.0210 mol Ag2SO4

- = 0.0140 M Ag2SO41.5 L solution[Ag+] = 2[Ag2SO4] = 2(0.0140 M) = 0.0280 M [SO2'] = [Ag2SO4] = 0.0140 M

Q = [Ag+]2[SO|-] = (0.0280)2(0.0140) = 1.1 x 10~5

552 Chapter 14 Chemical Equilibrium

Q = Kc so the system is at equilibrium and is a saturated solution. Therefore, if more solid is added/s-^/' — N it will not dissolve.

j ' 14.48 J Given: Kp = 6.7 at 298 K; 2.25 L flask; NO2 = 0.055 mol; N2O4 = 0.082 mol Find: Is the reaction at equilibrium?V y Conceptual Plan: Kp — » Kc and mol — * M and then calculate Q and compare to Kc.

K

Kp (6.7)Solution: Kc = - = =164

{DT\&n L • arm[(.0821- -X298K)]-1

mol-K/ 0.082 mol \

= [N204] V 2.25 L )

[NO]2 / 0.055 mol V2.25 L /

Q < Kc, the reaction is not at equilibrium and will shift to the right.

Finding Equilibrium Concentrations from Initial Concentrations and theEquilibrium Constant

14.49 (a) Given: [A] = 1.0 M, [B] = 0.0 Kc = 4.0; a = l,b = l Find: [A], [B] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kv predict the direction of thereaction, represent the change with x, sum the table, determine the equilibrium values, put theequilibrium values in the equilibrium expression, and solve for*. Determine [A] and [B].Solution: A(g) ̂ B(g)

[A] [B]I 1.0 0.00C - x xE l-x x

Q = = — = 0 Q < K therefore, the reaction will proceed to the right by x.[A]

Kc = — = ——— = 4.0 x = 0.80[A] (1 - x)

[A] = 1 - 0.80 = 0.20M [B] = 0.80M.0.80

Check: Plug the values into the equilibrium expression: Kc = —— = 4.0.

(b) Given: [A] = 1.0 M, [B] = 0.0 JQ = 4.0; a = 2, b = 2 Find: [A], [B] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kv predict the direction of thereaction, represent the change with x, sum the table, determine the equilibrium values, put theequilibrium values in the equilibrium expression, and solve for x. Determine [A] and [B].Solution: 2 A(g) ^ 2 B(g)

[A] [B]I 1.0 0.00C -2x 2xE l-2x 2x

Q = = - = 0 Q < K therefore, the reaction will proceed to the right by x.

Kc =.

[A]2 (1 - 2*)2

[A] = 1 - 2(0.33) = 0.33 = 0.33M [B] = 2(0.33) = 0.66 = 0.66M(0.66)2

Check: Plug the values into the equilibrium expression: Kc = = 4.0.

554 Chapter 14 Chemical Equilibrium

(c) Given: [A] = 1.0 M, [B] = 1.0, [C] = 0.0 Kc = 5.0; a = 2,b = l,c = l Find:[A],[B],[C] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and Ka predict the direction of thereaction, represent the change with x, sum the table, determine the equilibrium values, put theequilibrium values in the equilibrium expression, and solve for x Determine [A], [B], and [C].Solution: 2 A(g) +B(g) ±5 C(g)

[A] [B] [C]I 1.0 1.0 0.0C -2x -x xE l-2x l-x x

Q = — - — = T77T77 = 0 Q < K therefore, the reaction will proceed to the right by x.

= 5.0[A]2[B] (1 - 2*)2(1 - x)

- 20x3 + 40*2 - 26x + 5.0 = 0. Solve by using successive approximations or a cubic equationcalculator found on the internet.x = 0.3396 = 0.34[A] = 1 - 2x = 0.32M; [B] = 1 - x = 0.66M; [C] = x = 0.34M.

(0.34)Check: Plug the values into the equilibrium expression: K, = - - - = 5.0.

(0.32)2(0.66)

14.51 Given: [N2O4] = 0.0500 M, [NO2] = 0.0 Kc = 0.513 Find: [N2O4], [NO2] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kv predict the direction of the reaction,represent the change with x, sum the table, determine the equilibrium values, put the equilibrium valuesin the equilibrium expression, and solve for x. Determine [N2O4] and [NO2].Solution:

N204(£) ^2N02(£)[N204] [NQJ

I 0.0500 0.00C -x 2xE 0.0500 -x 2x

Q = - = — — — - = 0 Q < K therefore, the reaction will proceed to the right by x.[N2O41 0 .0500[NO2]

2 (2x)2(0.0500 - x)Kc = - -^- = — 4X2 + 0.513* - 0.02565 = 0

[N2O4] = 0.513-b ± Vb2 - 4ac -0.513 ± V(0.513)2 - 4(4)(-0.02565) -0.513 ±

2a 2(4) 2(4)x = - 0.1667 or x = 0.0385, therefore, x = 0.0385.[N2O4] = 0.0500 - 0.03850 = 0.0115M [NO2] = 2x = 2(0.03850) = 0.0770M.

(0.0770)2

Check: Plug the values into the equilibrium expression: Kc = = 0.515.U.U J. i<J

Given: [CO] = 0.1500 M, [C12] = 0.175, [COC12] = 0.0 Kc = 255 Find: [CO], [C12], [COC12] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and KC, predict the direction of the reaction,represent the change with x, sum the table, determine the equilibrium values, put the equilibrium valuesin the equilibrium expression, and solve for x. Determine [CO], [C12]. and [COC12].Solution:

CO (g) + C12 (g) fc> COC12 (g)[CO] tci2] [cocy

I 0.1500 0.175 0.0C -x -x xE 0. 1500 -x 0.1 75- x x

O — — --- Q- = - - - 0 Q « K Ihei'eforc, die reaction will proceed to the right by x.[CO][Cy (0 .1500)(0.175)

= 2 = - ± - = 255 255X2 - 83.875X + 6.69375 = 0c [COJtCl2] (0.1500 - x)(0.175 - x)

Chapter 14 Chemical Equilibrium 555

-b ± Vfr2 - 4ac _ -(-83.875) ± VH33.875)2 - 4(255)(6.69375)2a 2(255)

x = 0.1927 or x = 0.1362, therefore, x = 0.1362.[CO] = 0.1500 - 0.1362 = 0.01377 M[C12] = 0.175 - 0.1362 = 0.03878 M[COC12] = x = 0.1362M

(0.1362)Check: Plug the values into the equilibrium expression: Kc = v^nm n?Qia\ ~(0. 01 377)(0. 03878)

14.53 Given: [CO] = 0.20 M, [CO2] = 0.0 Kc = 4.0 x 103 Find: [CO2] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and Ka predict the direction of the reaction,represent the change with x, sum the table, determine the equilibrium values, put the equilibrium valuesin the equilibrium expression, and solve for x. Determine [CO], [C12], and [COC12].Solution:

NiO(s) +CO (g) *=> Ni(s) + C02 (g)[CO] [C02]

I 0.20 0.0C -x xE 0.10 -x x

Q = - = - = 0 Q < K therefore, the reaction will proceed to the right by x.[CO] (0.20)

= 4-° x lo3 4-°x = 0.199[CO2] = 0.199 MCheck: Since the equilibrium constant is so large, the reaction goes essentially to completion, therefore, it is

\reasonable that the concentration of product is 0.199 M.

iven: [CO] = 0.110M M, [H2O] = 0. 11 OM, [CO2] = 0.0, [H2] = 0.0 Kc = 102Find: [CO], [H2O], [CO2], [H2] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and Kv predict the direction of the reaction,represent the change with x, sum the table, determine the equilibrium values, put the equilibrium valuesin the equilibrium expression, and solve for x. Determine [CO], [C12], and [COC12]-Solution:

(g) +H20(g) ^ C02(g)+H2(g)[H2]

Q < K therefore, the reaction will proceed to the right by x.

IcE

AC ~

/

C0(g) +H20(g) fc*[CO] [H20]0.110 0.110- x -x0.110-x 0.110-x[C02][H2] 0

[CO][H20] (0.110)([C02][H2] (

[CO][H2O] (0.110 -

MM

C02 (g) +[C02]0.0 OJX X

X X

n0.110) •x)(x)x)(0.110 - x)

An?(0.110 - xXO.110 - x)

= ±10.09950.110 - x± 10.0995(0.110 - x) = xx = 0.10009 or x = 0.12209, therefore, x = 0.100.[CO] = [H2O] = 0.110 - 0.100 = 0.010 = 0.010 M

[H2] = [CO2] = x = 0.100 = 0.100 M(0.100X0.100) 2

Check: Plug the values into the equilibrium expression: Kc = 010)(Q 01Q) = 1.0 x 10 ; within

1% of true value, answers are valid.

Chapter 14 Chemical Equilibrium

14.55

14.56

Given: [HC2H3O2] = 0.210M, [H3O+] = 0.0, [C2H3O2-] = 0.0, Kc = 1.8 x 10 ~5

Find: [HC2H3O2], [H2O+], [C2H3O2-] at equilibrium

Conceptual Plan: Prepare an ICE table, calculate Q, compare Q and K0 predict the direction of the reaction,represent the change with x, sum the table, determine the equilibrium values, put the equilibrium valuesin the equilibrium expression, and solve for x. Determine [CO], [Cy, and [COC12J.Solution:

HC2H3O2(a<7) +H2O(/)[HC2H302] [H20]

I 0.210 0.0 0.0C - x x xE 0.210-x x x

H3O+ (aq)

[H30+] [C2H302-]

Q =[H30

+][C2H302-] = 0

[HC2H302] (0.210)[H30

+][C2H302] (x)(*C [HC2H302] (0.210 - x)

assume x is small compared to 0.210.

x2 = 0.210(1.8 xlO~5)0.00194

= 0 Q < K therefore, the reaction will proceed to the right by x.

\

= 1.8xlO~5

x = 0.00194 check assumption:0.210

x 100 = 0.92%; assumption valid.

[H3O+] = [C2H3O2] = 0.00194 M

[HC2H3O2] = 0.210 - 0.00194 = 0.2081 = 0.208M

Check: Plug the values into the equilibrium expression: Kc =(0.00194)(0.00194)

= 1.81 x 10~5;(0.208)

the answer is the same to 2 significant figures with the true value, answers are valid.

Given: [SO2C12] = 0.175M M, [SO2] = 0.0, [C12] = 0.0 Kc = 2.99 x 10 ~7

Find: [SO2C12], [SO2], [C12] at equilibriumConceptual Plan: Prepare an ICE table, calculate Q, compare Q and K0 predict the direction of the reaction,represent the change with x, sum the table, determine the equilibrium values, put the equilibrium valuesin the equilibrium expression, and solve for x. Determine [SC^Cy, [802], and [02].Solution:

C12 (g) ^ S02 (g) + Cl2fc)[C12]0.0X

x

= 0 Q < K therefore, the reaction will proceed to the right by x.

= 2.99xlO~7

[SO2C12] (0.175 - x)assume x is small compared to 0.175.

x2 = 2.99xlO~7(0.175)

x = 2.287 x 10~4 check assumption:

[SO2] = [C12] = x = 2.287x HT4 = 2.29x 10~4M

[SO2C12] = 0.175 - 2.287 x 10~4 = 0.1747 = 0.175MCheck: Plug the values into the equilibrium expression:

ICE

Q

V

S02C12 fe)[S02C12]0.175-x0.175-x[S02][C12]

[S02C12][S02][C12]

*5 S02(g)

[S02]0.0X

x0

(0.175)

WW

2.287x10'

0.175i-41

x 100 = 0.13%; assumption valid.

Kc =(0.2.287 x 10~4 )(2.287 x 10~4 )

= 2.996 xlO~7 = 3.00x10(0.175)

this is within 0.01 x 10~7 of the true value, therefore, answers are valid.

Chapter 14 Chemical Equilibrium 557

Given: PBl2 = 755 torr, Pch = 735 torr, PBlC1 = 0.0, Kp = 1.11 x 10 ~ 4 Find: PBrCi at equilibriumConceptual Plan: Torr —> atm and then: Prepare an ICE table, calculate Q, compare Q and Kv predict thedirection of the reaction, represent the change with x, sum the table, determine the equilibrium values,put the equilibrium values in the equilibrium expression, and solve for x. DetermineSolution:

PBr2 == 755 torr x— == 0.9934 atm Pch ---- 735 torr x-— = 0.9671 atm

Br2 (g) + Cl2(g) ^ 2 BrCl (g)P p p*DT2 ^-'2 BrCl

I 0.9934 0.9671 0.0C -x -* 2xE 0.9934-* 0.9671- x 2x

PBTCI 0Q = ——-— = = 0 Q < K therefore, the reaction will proceed to the right by x.

1 Qf2C12 (U.:7:7.34)(U.yo/l)

= l.llxlO~4

JWci2 (0.9934 - jr)(0.9671 - x)assume x is small compared to 0.9934 and 0.9671.

(2x)2

= l.l lxlO~4 4X2 = 1.066xlO~(0.9934)(0.9671)x = 0.00516 atm = 3.92 torrPBTCI = 2* = 2(3.92 torr) = 7.84 torr

Check: Plug the values into the equilibrium expression:

KC ~ (0.9934 - 0.00516)(0.99671 - 0.00516) "this is within 0.01 x 10~4 of the true value, therefore, answers are valid.

Given: Pco = 1344 torr, PH2o = 1766 torr, PCo2 = 0.0, PH2 = 0.0, Kp = 0.0611 Find: PCO2/ Pn2 at equilibriumConceptual Plan: Torr —» atm and then prepare an ICE table, calculate Q, compare Q and Ka predict thedirection of the reaction, represent the change with x, sum the table, determine the equilibrium values,put the equilibrium values, in the equilibrium expression, and solve for x. Determine PCOZ

anc^ ^H2-1 atm 1 atm

Solution: PCo = 1344 torr x — - = 1.7684 atm Ph,w _760 torr 760 torr

CO(g) +H2O(?) *=» CO2(g)+ H2(£)PCO PH2O

PC02 PH2

I 1.7684 2.323.7 0.00.0C - * -x x xE 1.7684-x 2.3237-* * *

O = —- = = 0 Q < K therefore, the reaction will proceed to the rieht by *.n n (1.7684)(2.3237)

(1.7684 - *)(2.3237 - x)

x2 = (0.0611)(4.1092 - 4.0921* + x2)x2 = (0.25107 - 0.2500* + 0.0611*2)0.9389*2 + 0.2500* - 0.25107 = 0

= 0.

-b ± Vb2 - 4ac -0.2500 ± V(0.2500)2 - 4(0.9389)( -0.25107)2« 2(0.9389)

x = 0.4008 or - 0.6671 so: x = 0.4008PCo2 = PHZ = x = 0.4008 arm = 305 torr

Check: Plug the values into the equilibrium expression:

Kp = - - - - 0.06108 = 0.061;p (1.7684 - 0.4008)(2.3237 - 0.4008)the answers are the same to two significant figures with the true value, therefore, answers are valid .

Chapter 14 Chemical Equilibrium 561

(c) COC12 is removed from the reaction mixture: Removing the COC12 decreases the concentration ofCOC12 and causes the reaction to shift to the right.

14.62/ Given: 2BrNO(g) ±5 2NO(g) + Br2(g) at equilibrium Find: What is the effect of each of the following?

) NO is added to the reaction mixture: Adding NO increases the concentration of NO and causes thereaction to shift to the left.

(b) BrNO is added to the reaction mixture: Adding BrNO increases the concentration of BrNO and causesthe reaction to shift to the right.

(c) Br2 is removed from the reaction mixture: Removing Br2 decreases the concentration of Br2 and causesthe reaction to shift to the right.

14.63 Given: 2KClO3(s) *=> 2KCl(s) + 3O2(g) at equilibrium Find: What is the effect of each of the following?

(a) O2 is removed from the reaction mixture: Removing the O2 decreases the concentration of O2 andcauses the reaction to shift to the right.

(b) KC1 is added to the reaction mixture: Adding KC1 does not cause any change in the reaction. KC1 is asolid and the concentration remains constant so the addition of more solid does not change the equi-librium concentration.

(c) KC1O3 is added to the reaction mixture: Adding KC1O3 does not cause any change in the reaction.KC1O3 is a solid and the concentration remains constant so the addition of more solid does not changethe equilibrium concentration.

(d) O2 is added to the reaction mixture: Adding O2 increases the concentration of O2 and causes the reac-tion to shift to the left.

Given: C(s) + H2O(g) *=> CO(g) + H2(g) Find: What is the effect of each of the following?

(a) C is added to the reaction mixture: Adding C does not cause any change in the reaction. C is a solidand the concentration remains constant so the addition of more solid does not change the equilib-rium concentration.

(b) H2O is condensed and removed from the reaction mixture: Removing the H2O decreases the concen-tration of H2O and causes the reaction to shift to the left.

(c) CO is added to the reaction mixture: Adding CO increases the concentration of CO and causes thereaction to shift to the left.

(d) H2 is removed from the reaction mixture: Removing the H2 decreases the concentration of H2 andcauses the reaction to shift to the right.

(a) Given: I2(g) *=» 2I(g) at equilibrium Find: the effect of increasing the volume.The chemical equation has 2 moles of gas on the right and 1 mole of gas on the left. Increasing the vol-ume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (towardthe side with more moles of gas particles).

(b) Given: 2H2S(g) *=> 2H2(g) + S2(g) Find: the effect of decreasing the volume.The chemical equation has 3 moles of gas on the right and 2 moles of gas on the left. Decreasing thevolume of the reaction mixture increases the pressure and causes the reaction to shift to the left(toward the side with fewer moles of gas particles).

(c) Given: I2(g) + Q2(g) ^5 2lCl(g) Find: the effect of decreasing the volume.The chemical equation has 2 moles of gas on the right and 2 moles of gas on the left. Decreasing thevolume of the reaction mixture increases the pressure but causes no shift in the reaction because themoles are equal on both sides.

(a) Given: CO(g) + H2O(#) *=> CO2(g) + H2(g) Find: the effect of decreasing the volume.The chemical equation has 2 moles of gas on the right and 2 moles of gas on the left. Decreasing the

14.66

562 Chapter 14 Chemical Equilibrium

volume of the reaction mixture increases the pressure but causes no shift in the reaction because themoles are equal on both sides.

(b) Given: PCl3(g) + Cl2(g) ^5 PC15(#) Find: the effect of increasing the volume.The chemical equation has 2 moles of gas on the left and 1 mole of gas on the right. Increasing the vol-ume of the reaction mixture decreases the pressure and causes the reaction to shift to the left (towardthe side with more moles of gas particles).

(c) Given: CaCO3(s) ^» CaO(s) + CO2(g) Find: the effect of increasing the volume.The chemical equation has 1 mole of gas on the right and 0 moles of gas on the left. Increasing the vol-ume of the reaction mixture decreases the pressure and causes the reaction to shift to the right(toward the side with more moles of gas particles).

14.67 Given: C(s) + CO2(g) *=> 2CO(g) is endothermic. Find: the effect of increasing the temperature.Since the reaction is endothermic we can think of the heat as a reactant: Increasing the temperature is equiv-alent to adding a reactant causing the reaction to shift to the right. This will cause an increase in the concen-tration of products and a decrease in the concentration of reactant; therefore, the value of K will increase.Find: the effect of decreasing the temperatureSince the reaction is endothermic we can think of the heat as a reactant: Decreasing the temperature is equiv-alent to removing a reactant causing the reaction to shift to the left. This will cause a decrease in the concen-tration of products and an increase in the concentration of reactants; therefore, the value of K will decrease.

14.68 Given: C6H12O6(s) + 6 O2(g) ** 6CO2(g) + 6 H2O(g) is exothermic.Find: the effect of increasing the temperatureSince the reaction is exothermic we can think of the heat as a product: Increasing the temperature is equiv-alent to adding a product causing the reaction to shift to the left. This will cause a decrease in the concen-tration of products and an increase in the concentration of reactant; therefore, the value of K will decrease.Find: the effect of decreasing the temperatureSince the reaction is exothermic we can think of the heat as a product: Decreasing the temperature is equiva-lent to removing a product causing the reaction to shift to the right. This will cause an increase in the concen-tration of products and a decrease in the concentration of reactants; therefore, the value of K will increase.

Given: C(s) + 2H2(g) is exothermic. Find: Determine which will favor CH,̂

(a) Adding more C to the reaction mixture does NOT favor CHj. Adding C does not cause any changein the reaction. C is a solid and the concentration remains constant so the addition of more solid doesnot change the equilibrium concentration.

(b) Adding more H2 to the reaction mixture favors CH4. Adding H2 increases the concentration of H2

causing the reaction to shift to the right.

(c) Raising the temperature of the reaction mixture does NOT favor CH^ Since the reaction is exother-mic we can think of heat as a product, raising the temperature is equivalent to adding a product caus-ing the reaction to shift to the left.

(d) Lowering the volume of the reaction mixture favors CH^ The chemical equation has 1 mole of gas onthe right and 2 moles of gas on the left. Decreasing the volume of the reaction mixture increases the pres-sure and causes the reaction to shift to the right (toward the side with fewer moles of gas particles).

(e) Adding a catalyst to the reaction mixture does NOT favor CH^ A catalyst added to the reaction mix-ture only speeds up the reaction, it does not change the equilibrium concentration.

(f) Adding neon gas to the reaction mixture does NOT favor CH4. Adding an inert gas to a reaction mix-ture at a fixed volume has no effect on the equilibrium.

14.70 Given: C(s) + H2O(g) ^=> CO(g) + H2(g) is endothermic. Find: Determine which will favor H2.

(a) Adding more C to the reaction mixture does NOT favor H2. Adding C does not cause any change inthe reaction. C is a solid and the concentration remains constant so the addition of more solid doesnot change the equilibrium concentration.