chapter7 equilibrium pp answers

10. ProblemMethane, ethyne, and hydrogen form the following equilibrium mixture.2CH4(g) ⇀↽ C2H2(g) + 3H2(g)

While studying this reaction mixture, a chemist analysed a 4.0 L sealed flask at1700˚C. The chemist found 0.46 mol of CH4(g) , 0.64 mol of C2H2(g), and 0.92 molof H2(g). What is the value of Kc for the reaction at 1700˚C?

What Is Required?You need to calculate the value of Kc.

What Is Given?You have the chemical equation and the amount of each substance at equilibrium in a 4.0 L flask.

Plan Your StrategyStep 1 Calculate the molar concentration of each compound at equilibrium.Step 2 Write the equilibrium expression. Then substitute the equilibrium molar

concentrations into the expression.

Act on Your StrategyStep 1 The reaction takes place in a 4.0 L flask. Calculate the molar concentrations

at equilibrium.

[CH4] =0.46 mol

4.0 L= 0.115 mol/L

[C2H2] =0.64 mol

4.0 L= 0.16 mol/L

[H2] =0.92 mol

4.0 L= 0.23 mol/L

Step 2 Write the equilibrium expression. Substitute the equilibrium molar concentrations into the expression.

Kc =[C2H2][H2]

3

[CH4]2

=(0.16)(0.23)3

(0.115)2

= 0.15

Check Your SolutionThe equilibrium expression has the product terms in the numerator and the reactantterms in the denominator. The exponents in the equilibrium expression match thecorresponding coefficients in the chemical equation. The molar concentrations atequilibrium were substituted into the expression. The answer has the correct numberof significant digits.

Solutions for Practice Problems

Student Textbook pages 347–348

11. ProblemAt 25˚C the value of Kc for the following reaction is 82.I2(g) + Cl2(g) ⇀↽ 2ICl(g)

0.83 mol of I2(g) and 0.83 mol of Cl2(g) are placed in a 10 L container at 25˚C. What are the concentrations of the three gases at equilibrium?

What Is Required?You need to find [I2], [Cl2] and [ICl] at equilibrium.

What Is Given?You are given the chemical equation. You know the initial amount of each gas, the volume of the container, and the equilibrium constant.

106Chapter 7 Reversible Reactions and Chemical Equilibrium • MHR

CHEMISTRY 12

Plan Your StrategyStep 1 Calculate the initial concentrations.Step 2 Set up an ICE table. Insert the initial concentrations you calculated in

Step 1 in your ICE table. Let the change in molar concentrations of the reactants be x. Use the stoichiometry of the chemical equation to write and record expressions for the equilibrium concentrations.

Step 3 Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the expression. Solve the equilibrium expression for x.

Step 4 Calculate the equilibrium concentration of each gas.

Act on Your StrategyStep 1 The initial amount of I2 is equal to the initial amount of Cl2.

[I2] = [Cl2] =0.83 mol

10 L= 0.083 mol/L

Step 2 Set up an ICE table.

Step 3 Kc =[ICI]2

[I2][Cl2]

82 =(2x)2

(0.083 − x)(0.083 − x)

=(2x)2

(0.083 − x)2

The right side of the equation is a perfect square.

±9.06 =(2x)

(0.083 − x)Solving for both values,x = 0.068 and x = 0.11

Step 4 The value x = 0.11 is physically impossible because it is a larger concentra-tion than was initially placed in the container. To achieve equilibrium, theinitial [I2] and [Cl2] must decrease.[I2] = [Cl2] = 0.083 − 0.068 = 0.015 mol/L[ICl] = 2 × 0.068 = 0.14 mol/L

Check Your SolutionThe equilibrium expression has product concentrations in the numerator and reactantconcentrations in the denominator. Check Kc:

Kc =(0.14)2

(0.015)2 = 87

This is close to the given value, Kc = 82. The difference is due to mathematicalrounding.

12. ProblemAt a certain temperature, Kc = 4.0 for the following reaction.2HF(g) ⇀↽ H2(g) + F2(g)

A 1.0 L reaction vessel contained 0.045 mol of F2(g) at equilibrium. What was theinitial amount of HF in the reaction vessel?

Concentration (mol/L)

Initial

Change

Equilibrium

I2(g) + Cl2(g)

0.083

−x

0.083 − x

0.083

−x

0.083 − x

0

+2x

2x

2ICl(g)


CHEMISTRY 12

What Is Required?You need to find the initial amount of HF(g) placed in the reaction vessel.

What Is Given?You are given the chemical equation. You know the amount of F2(g) at equilibrium,the volume of the container, and the equilibrium constant.

Plan Your StrategyStep 1 Calculate the concentration of F2(g) at equilibrium and use the chemical

equation to find the equilibrium concentration of H2(g).Step 2 Set up an ICE table. Insert the equilibrium concentrations you calculated

in Step 1 in your ICE table. Let the change in molar concentrations of thereactants be x. Use the stoichiometry of the chemical equation to write andrecord expressions for the equilibrium concentrations.

Step 3 Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the expression. Solve the equilibrium expression for the initial [HF].

Step 4 Calculate the initial amount of HF(g).

Act on Your StrategyStep 1 The volume of the reaction vessel is 1.0 L.

Therefore,

[F2] =0.045 mol

1.0 L= 0.045 mol/L

[H2] = [F2] = 0.045 mol/LStep 2 Set up an ICE table.

Therefore, x = 0.045

Step 3 Kc =[H2][F2][HF]2

4.0 =(0.045)(0.045)

([HF] − 0.090)2

=(0.045)2

([HF] − 0.090)2


±2.0 =(0.045)

([HF] − 0.090)

Solving for both values gives[HF] = 0.112 and [HF] = 0.0675

Step 4 The value [HF] = 0.0675 is physically impossible because subtracting thechange in concentration, 0.090, results in a negative value for the initial concentration of HF(g) placed in the container. Therefore, [HF] = 0.11 mol/L.Because the volume of the reaction vessel is 1.0 L, the initial amount ofHF(g) placed in the reaction vessel was 0.11 mol.


Initial

Change

Equilibrium

2HF(g) +H2(g)

[HF]

−2x

([HF] − 2x)

0

+x

0.045

0

+x

0.045

F2(g)


CHEMISTRY 12


Kc =(0.045)2

(0.11 − 0.090)2 = 5.1

This is close to the given value, Kc = 4.0. The difference is due to mathematicalrounding.

13. ProblemA chemist was studying the following reaction.SO2(g) + NO2(g) ⇀↽ NO(g) + SO3(g)

In a 1.0 L container, the chemist added 1.7 × 10−1 mol of SO2(g) to 1.1 × 10−1 molof NO2(g). The value of Kc for the reaction at a certain temperature is 4.8. What isthe equilibrium concentration of SO3(g) at this temperature?

What Is Required?You need to find the equilibrium [SO3].

What Is Given?You are given the chemical equation. You know the equilibrium constant for the reaction, Kc = 4.8. You also know the initial amounts of SO2(g) and NO2(g), and thevolume of the container.

Plan Your StrategyStep 1 Calculate the initial concentrations of SO2(g) and NO2(g).Step 2 Set up an ICE table. Insert the equilibrium concentrations you calculated


Step 3 Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the expression. Rearrange the equilibriumexpression into the form of a quadratic equation for x. Solve the quadraticequation for x.

Step 4 Substitute x into the equilibrium line of the ICE table to find the equili-brium concentration of [SO3].


Therefore,

[SO2] =1.7 × 10−1 mol

1.0 L= 1.7 × 10−1 mol/L

[NO2] =1.1 × 10−1 mol

1.0 L= 1.1 × 10−1 mol/L


Step 3 Kc =[NO][SO3][SO2][NO2]

4.8 =(x)(x)

(0.17 − x)(0.11 − x)


Initial

Change

Equilibrium

SO2(g) + NO2(g)

1.7 × 10−1

−x

(1.7 × 10−1 − x)

1.1 × 10−1

−x

(1.1 × 10−1 − x)

0

+x

x

SO3(g)NO(g) +

0

+x

x


CHEMISTRY 12

This equation must be rearranged into a quadratic equation.3.8x2 − 1.344x + 0.08976 = 0Recall that a quadratic equation of the form ax2 + bx + c = 0 has the following solution.

x =−b ±

√b2 − 4ac

2a

Therefore,

x =−(−1.344) ±

√1.806 − 1.364

7.6

=1.344 ± 0.6648

7.6x = 0.089 and x = 0.264The value x = 0.264 is not physically possible. It would result in negativeconcentration of SO2 and NO2 at equilibrium. The concentration of eachsubstance at equilibrium is found by substituting x = 0.089 into the last lineof the ICE table.

Applying the rules for subtraction involving measured values:[SO2] = 0.08 mol/L[NO2] = 0.02 mol/L[NO] = 0.089 mol/L[SO3] = 0.089 mol/LAt equilibrium the concentration of SO3(g) is 0.089 mol/L.


Kc =(0.089)2

(0.08)(0.02)= 5

Solving the quadratic equation gave a value of x known only to two decimal places.Subtracting the value of x from [SO2] and [NO2] gave values only good to one sig-nificant figure. The calculated value of Kc is equal to the given value, 4.8, within theerror introduced by rounding.

14. ProblemPhosgene, COCl2(g) , is an extremely toxic gas. It was used during World War I.Today, it is used to manufacture pesticides, pharmaceuticals, dyes, and polymers. It isprepared by mixing carbon monoxide and chlorine gas.CO(g) + Cl2(g) ⇀↽ COCl2(g)

0.055 mol of CO(g) and 0.072 mol of Cl2(g) are placed in a 5.0 L container. At 870 K, the equilibrium constant is 0.20. What are the equilibrium concentrationsof the mixture at 870 K?

What Is Required?You need to find the equilibrium concentrations of CO, Cl2 and COCl.

What Is Given?You are given the chemical equation. You know the equilibrium constant for the reaction, Kc = 0.20. You also know the initial amounts of CO(g) and Cl2(g) , and thevolume of the container.


Equilibrium

SO2(g) + NO2(g)

(0.17 − 0.089) (0.11 − 0.089) 0.089

SO3(g)NO(g) +

0.089


CHEMISTRY 12

Plan Your StrategyStep 1 Calculate the initial concentrations of CO(g) and Cl2(g) .Step 2 Set up an ICE table. Insert the equilibrium concentrations you calculated


Step 3 Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the expression. Rearrange the equilibriumexpression into the form of a quadratic equation for x. Solve the quadraticequation for x.

Step 4 Substitute x into the equilibrium line of the ICE table to find the equilibrium concentrations of CO, Cl2, and COCl.


Therefore,

[CO] =0.055 mol

5.0 L= 0.011 mol/L

[Cl2] =0.072 mol

5.0 L= 0.0144 mol/L


Step 3 Kc =[COCl2]

[CO][Cl2]

0.20 =x

(0.011 − x)(0.0144 − x)This equation must be rearranged into a quadratic equationx2

− 5.025x + 1.58 × 10−4= 0

Recall that a quadratic equation of the form ax2+ bx + c = 0 has the

following solution.

x =−b ±

√b2 − 4ac

2a

Therefore,

x =−(−5.025) ±

√25.25 − 6.32 × 10−4

2.0

=5.025 ± 5.025

2.0

x = 5.02 or x = 0.0The value x = 5.02 is not physically possible. It would result in negativeconcentration of CO and Cl2 at equilibrium. The concentration of reactantsremains essentially unchanged with very little formation of COCl2(g) .

Check Your SolutionCalculate the equilibrium [COCl2] to check that it is a very small value.

Kc = 0.20 =[COCl2]

0.011 × 0.014

[COCl2] = 3.1 × 10−5 mol/L

The equilibrium concentration of [COCl2] is a very small value.


Initial

Change

Equilibrium

CO(g) + Cl2(g)

0.011

−x

(0.011 − x)

0.0144

−x

(0.0144 − x)

COCl2(g)

0

+x

x


CHEMISTRY 12

15. ProblemHydrogen bromide decomposes at 700 K.2HBr(g) ⇀↽ H2(g) + Br2(g) Kc = 4.2 × 10−9

0.090 mol of HBr is placed in a 2.0 L reaction vessel and heated to 700 K. What isthe equilibrium concentration of each gas?

What Is Required?You need to find [HBr], [H2], and [Br2] at equilibrium.

What Is Given?You are given the chemical equation. You know the initial amount of HBr, the volume of the container, and the equilibrium constant.

Plan Your StrategyStep 1 Calculate the initial concentration of HBr.Step 2 Set up an ICE table. Insert the initial concentrations you calculated in Step 1

in your ICE table. Let the change in molar concentrations of HBr be x. Usethe stoichiometry of the chemical equation to write and record expressionsfor the equilibrium concentrations.

Step 3 Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the expression. Solve the equilibrium expression for x.

Step 4 Calculate the equilibrium concentration of each gas.

Act on Your StrategyStep 1

[HBr] =0.090 mol

2.0 L= 0.045 mol/L


Step 3 Kc =[H2][Br2][HBr]2

4.2 × 10−9=

x2

(0.045 − 2x)2


±6.48 × 10−5=

x

0.045 − 2x

Solving for both values,x = +2.92 × 10−6 and x = −2.92 × 10−6

Step 4 The value x = −2.92 × 10−6 is physically impossible because it would resultin negative concentration of both H2 and Br2.Therefore, [H2] = [Br2] = 2.9 × 10−6 mol/L .[HBr] = 0.045 − (2 × 2.9 × 10−6) = 0.045 mol/L

Check Your SolutionThe equilibrium expression has product concentrations in the numerator and reactantconcentrations in the denominator. The equilibrium concentrations are in mol/L.Check Kc:

Kc =(2.9)2

(0.045)2 = 4.2 × 10−9

This is equal to the given value of Kc.


Initial

Change

Equilibrium

2HBr(g) +H2(g)

0.045

−2x

0.045 − 2x

0

+x

x

Br2(g)

0

+x

x


CHEMISTRY 12


Student Textbook pages 349–350

16. ProblemFor the reaction H2(g) + I2(g) ⇀↽ 2HI(g), the value of Kc is 25.0 at 1100 K and 8.0 × 102 at room temperature, 300 K. Which temperature favours the dissociationof HI(g) into its component gases?

What Is Required?Choose the temperature that favours the greater dissociation of HI.

What Is Given?Kc = 25.0 at 1100 K and 8.0 × 102 at 300 K

Plan Your StrategyHI is on the right in the chemical equation. A greater dissociation of hydrogen iodidecorresponds to a smaller value of Kc.

Act on Your StrategyThe value of Kc is smaller at 1100 K. The position of equilibrium lies more to the leftand favours the dissociation of hydrogen iodide at the higher temperature.

Check Your SolutionThe question asked to choose the conditions for a smaller concentration of a product.Kc is a fraction with product terms in the numerator, so the smallest value of Kc mustcorrespond to the smallest concentration of product.

17. ProblemThree reactions, and their equilibrium constants, are given below.

I. N2(g) + O2(g) ⇀↽ 2NO(g) Kc = 4.7 × 10−31

II. 2NO(g) + O2(g) ⇀↽ 2NO2(g) Kc = 1.8 × 10−6

III. N2O4(g) ⇀↽ 2NO2(g) Kc = 0.025Arrange these equations in the order of their tendency to form products.

What Is Required?Place the reactions in the order of their tendency to form products.

What Is Given?Kc is given for each reaction.

Plan Your StrategyThe equilibrium constant, Kc, is a ratio of product concentrations divided by reactantconcentrations. The larger the value of Kc, the greater the tendency to form products.

Act on Your StrategyThe largest value of Kc is 0.025, then 1.8 × 10−6 , and finally, 4.7 × 10−31. Fromgreatest to least tendency to form products, the order is III, II, I.

Check Your SolutionThe question asked about the tendency to form products. Kc is a ratio of product concentrations divided by reactant concentrations. The largest value of Kc

corresponds to the greatest tendency to form products.


CHEMISTRY 12

18. ProblemIdentify each reaction as essentially going to completion or not taking place.(a) N2(g) + 3Cl2(g) ⇀↽ 2NCl3(g) Kc = 3.0 × 1011

(b) 2CH4(g) ⇀↽ C2H6(g) + H2(g) Kc = 9.5 × 10−13

(c) 2NO(g) + 2CO(g) ⇀↽ N2(g) + 2CO2(g) Kc = 2.2 × 1059

What Is Required?For each reaction, determine whether the equilibrium mixture consists mostly ofproducts or reactants.

What Is Given?Kc is given for each reaction.

Plan Your StrategyThe equilibrium constant, Kc, is a ratio of product concentrations divided by reactantconcentrations. A large value of Kc corresponds to mostly products, and a small valueof Kc indicates mostly reactants.

Act on Your StrategyReaction (a): Kc = 3.0 × 1011. At equilibrium, mostly products are present and thereaction essentially goes to completion.Reaction (b): Kc = 9.5 × 10−13. At equilibrium, mostly reactants are present andthere is essentially no reaction.Reaction (c): Kc = 2.2 × 1059. At equilibrium, mostly products are present and thereaction goes to completion.

Check Your SolutionKc is a ratio of product concentrations divided by reactant concentrations. Reactionsthat go to completion have very large values for Kc. Reactions that do not take placehave very small values for Kc.

19. ProblemMost metal ions combine with other ions in solution. For example, in aqueousammonia, silver(I) ions are in equilibrium with different complex ions.[Ag(H2O)2]

+(aq) + 2NH3(aq) ⇀↽ [Ag(NH3)2]

+(aq) + 2H2O(ℓ)

At room temperature, Kc for this reaction is 1 × 107. Which of the two silver complex ions is the more stable? Explain your reasoning.

What Is Required?Decide which silver ion is more stable.

What Is Given?Kc is given for the reaction.

Plan Your StrategyThe value of Kc indicates the relative concentrations of products and reactants atequilibrium.

Act on Your StrategyKc for this reaction is 1 × 107. Therefore, the concentration of products is muchgreater than the concentration of reactants. Consequently, [Ag(NH3)2]

+(aq) must be

the more stable silver ion.

Check Your SolutionKc is a ratio of product concentrations divided by reactant concentrations. Kc for thereaction is a large value, therefore the concentration of silver ion on the product sideof the equation must be greater than the concentration of silver ion on the reactantside of the equation.


CHEMISTRY 12

20. ProblemConsider the following reaction.H2(g) + Cl2(g) ⇀↽ 2HCl(g) Kc = 2.4 × 1033 at 25˚CHCl(g) is placed into a reaction vessel. To what extent do you expect the equilibriummixture to dissociate into H2(g) and Cl2(g)?

What Is Required?Describe the extent of dissociation of HCl at 25˚C.

What Is Given?Kc = 2.4 × 1033 at 25˚C

Plan Your StrategyHCl is on the right in the chemical equation. A small dissociation of HCl corresponds to a large [HCl] and a large value of Kc.

Act on Your StrategyThe value of Kc is large. The position of equilibrium lies to the right and very littleHCl dissociates at 25˚C.

Check Your SolutionKc is a fraction with product terms in the numerator. A large value of Kc correspondsto large concentration of product terms. Because HCl is on the product side of theequation, essentially no dissociation takes place.


Student Textbook page 352

21. ProblemThe following equation represents the equilibrium reaction for the dissociation of phosgene gas.COCl2(g) ⇀↽ CO(g) + Cl2(g)

At 100˚C, the value of Kc for this reaction is 2.2 × 10−8 . The initial concentration of COCl2(g) in a closed container at 100˚C is 1.5 mol/L. What are the equilibriumconcentrations of CO(g) and Cl2(g)?

What Is Required?You need to find [CO] and [Cl2] at equilibrium.

What Is Given?You are given the chemical equation. You know the value of Kc, and the initial con-centration of COCl2(g) .

Plan Your StrategyStep 1 Divide the initial concentration of COCl2(g) by Kc to determine whether you

can ignore the change in concentration.Step 2 Set up an ICE table. Let x represent the change in [COCl2]. Step 3 Write the equilibrium expression. Substitute the equilibrium concentrations

into the equilibrium expression. Solve the equilibrium expression for x. Step 4 Calculate [CO] and [Cl2] at equilibrium.

Act on Your Strategy

Step 1initial concentration

Kc=

1.52.2 × 10−8

= 6.8 × 107

Because this is well above 500, the change in [COCl2] can be ignored.


CHEMISTRY 12


Step 3 Kc =[CO][Cl2][COCl2]

2.2 × 10−8=

(x)(x)1.5

x =

√

3.3 × 10−8

= 1.8 × 10−4

Step 4 [CO] = [Cl2] = 1.8 × 10−4 mol/L

Check Your SolutionThe assumption that x is negligible compared with the initial concentration is valid because, using the rules for subtracting measured quantities,1.5 − (1.8 × 10−4) = 1.5 . Next, check the equilibrium values:(1.8 × 10−4)2

1.5= 2.2 × 10−8

This is equal to the equilibrium constant.

22. ProblemHydrogen sulfide is a poisonous gas with a characteristic, offensive odour. It dissociates at 1400˚C, with Kc equal to 2.4 × 10−4 .H2S(g) ⇀↽ 2H2(g) + S2(g)

4.0 mol of H2S is placed in a 3.0 L container. What is the equilibrium concentrationof H2(g) at 1400˚C?

What Is Required?You need to find the concentration of H2(g) at equilibrium.

What Is Given?You are given the chemical equation. You know the value of Kc, and the initialamount of H2(g). The volume of the container is given.

Plan Your StrategyStep 1 Calculate the initial molar concentration of H2S(g).Step 2 Divide the initial concentration of H2S(g) by Kc to determine whether you

can ignore the change in concentration.Step 3 Set up an ICE table. Let x represent the change in [H2S(g)]. Step 4 Write the equilibrium expression. Substitute the equilibrium concentrations

into the equilibrium expression. Solve the equilibrium expression for x. Step 5 Calculate [H2] at equilibrium.


Step 1 [H2S] =4.0 mol3.0 L

= 1.33 mol/L

Step 2initial concentration

Kc=

1.332.4 × 10−4

= 5.5 × 103

Because this is well above 500, the change in [H2S] can be ignored.


Initial

Change

Equilibrium

COCl2(g) +CO(g)

1.5

−x

1.5 − x ≈ 1.5

0

+x

x

Cl2(g)

0

+x

x


CHEMISTRY 12


Step 4 Kc =[H2]

2[S2][H2S]

2.4 × 10−4=

(2x)2(x)1.33

x =3√

7.98 × 10−5

= 4.3 × 10−2

Step 4 [H2] = 2 × (4.3 × 10−2) = 8.6 × 10−2 mol/L

Check Your SolutionThe assumption that x is negligible compared with the initial concentration is validbecause, using the rules for subtracting measured quantities,1.3 − (4.3 × 10−2) = 1.3 . Next, check the equilibrium values:

(8.6 × 10−2)2(4.3 × 10−2)1.3

= 2.4 × 10−4

This is equal to the equilibrium constant.

23. ProblemAt a certain temperature, the value of Kc for the following reaction is 3.3 × 10−12.2NCl3(g) ⇀↽ N2(g) + 3Cl2(g)

A certain amount of nitrogen trichloride, NCl3(g) is put in a 1.0 L reaction vessel atthis temperature. At equilibrium, 4.6 × 10−4 mol of N2(g) is present. What amountof NCl3(g) was placed in the reaction vessel?

What Is Required?You need to find the initial amount of NCl3(g) placed in the reaction vessel.

What Is Given?You are given the chemical equation and the value of Kc. You know the volume of thereaction vessel. You are told the amount of N2(g) at equilibrium.

Plan Your StrategyStep 1 Calculate the equilibrium molar concentration of N2(g).Step 2 Kc is very small. Assume you can ignore the change in concentration of

NCl3(g) and set up an ICE table. Let x represent the change in [N2] . Step 3 Write the equilibrium expression. Substitute the equilibrium concentrations

into the equilibrium expression. Solve the equilibrium expression for [NCl3].Step 4 Determine the amount of NCl3(g) put in the reaction vessel.


Initial

Change

Equilibrium

H2S(g) +2H2(g)

1.33

−x

1.33 − x ≈ 1.33

0

+2x

2x

S2(g)

0

+x

x

⇀

⇀


CHEMISTRY 12

Act on Your StrategyStep 1 The volume of the reaction vessel is 1.0 L. Therefore, the equilibrium

[N2] = 4.6 × 10−4 mol/L.Step 2 Set up an ICE table.

Step 3 Kc =[N2][Cl2]

3

[NCl3]2

3.3 × 10−12=

(4.6 × 10−4)(1.38 × 10−3)3

[NCl3]2

[NCl3] =

√

3.66 × 10−1

= 6.1 × 10−1 mol/LStep 4 The reaction vessel has a volume of 1.0 L. The amount of NCl3(g) put in the

vessel must have been 6.1 × 10−1 mol.

Check Your SolutionFirst, check your assumption that x is negligible compared with the initial concentration of NCl3(g). Using the rules for subtracting measured quantities, 6.1 × 10−1

− 2(4.6 × 10−4) = 6.1 × 10−1 . Next, check the equilibrium values:

(4.6 × 10−4)(1.4 × 10−3)3

(6.1 × 10−1)2 = 3.4 × 10−12

This is equal to the equilibrium constant within mathematical rounding error.

24. ProblemAt a certain temperature, the value of Kc for the following reaction is 4.2 × 10−8 .N2(g) + O2(g) ⇀↽ 2NO(g)

0.45 mol of N2(g) and 0.26 mol of O2(g) are put in a 6.0 L reaction vessel. What isthe equilibrium concentration of NO(g) at this temperature?

What Is Required?You need to find the concentration of NO(g) at equilibrium.

What Is Given?You are given the chemical equation. You know the value of Kc, and the initialamounts of N2(g) and O2(g). The volume of the container is given.

Plan Your StrategyStep 1 Calculate the initial molar concentrations of N2(g) and O2(g).Step 2 Divide the smallest initial concentration by Kc to determine whether you can

ignore the change in concentration.Step 3 Set up an ICE table. Let x represent the change in [N2(g)]. Step 4 Write the equilibrium expression. Substitute the equilibrium concentrations

into the equilibrium expression. Solve the equilibrium expression for x. Step 5 Calculate [NO] at equilibrium.


Initial

Change

Equilibrium

2NCl3(g) +N2(g)

[NCl3]

−2x

[NCl3] − 2x ≈ [NCl3]

0

+x

4.6 × 10−4

⇀

⇀ 3Cl2(g)

0

+3x

3x = 3(4.6 × 10−4)

= 1.38 × 10−3


CHEMISTRY 12


Step 1 [N2] =0.45 mol

6.0 L= 0.075 mol/L

[O2] =0.26 mol

6.0 L= 0.043 mol/L

Step 2 The smallest initial concentration is [O2].initial concentration

Kc=

0.0434.2 × 10−8

= 1.0 × 106

Because this is well above 500, the changes in [O2] and [N2] can be ignored.Step 3 Set up an ICE table.

Step 4 Kc =[NO]2

[N2][O2]

4.2 × 10−8=

(2x)2

(0.075)(0.043)

=4x2

3.22 × 10−3

x =

√

3.39 × 10−11

= 5.8 × 10−6

Step 5 [NO] = 2 × (5.8 × 10−6) = 1.2 × 10−5 mol/L

Check Your SolutionThe assumption that x is negligible compared with the initial concentration is validbecause, using the rules for subtracting measured quantities,0.043 − (5.8 × 10−6) = 0.043. Next, check the equilibrium values:

(1.2 × 10−5)2

(0.075)(0.043)= 4.5 × 10−8

This is equal to the equilibrium constant, within the error introduced by mathematical rounding.

25. ProblemAt a particular temperature, Kc for the decomposition of carbon dioxide gas is 2.0 × 10−6 .2CO2(g) ⇀↽ 2CO(g) + O2(g)

3.0 mol of CO2 is put in a 5.0 L container. Calculate the equilibrium concentrationof each gas.

What Is Required?You need to find the concentration of CO2(g), CO(g) , and O2(g) at equilibrium.

What Is Given?You are given the chemical equation. You know the value of Kc, and the initialamount of CO2(g). The volume of the container is given.


Initial

Change

Equilibrium

N2(g) + O2(g)

0.075

−x

0.075 −x 0.075

0.043

−x

0.043 −x 0.043

2NO(g)

0

+2x

2x


CHEMISTRY 12

Plan Your StrategyStep 1 Calculate the initial molar concentration of CO2(g).Step 2 Divide the initial concentration of CO2(g) by Kc to determine whether you

can ignore the change in concentration.Step 3 Set up an ICE table. Let x represent the change in [O2(g)]. Step 4 Write the equilibrium expression. Substitute the equilibrium concentrations

into the equilibrium expression. Solve the equilibrium expression for x. Step 5 Calculate the concentration of each gas at equilibrium.


Step 1 [CO2] =3.0 mol5.0 L

= 0.60 mol/L

Step 2

initial concentrationKc

=0.60

2.0 × 10−6

= 3.0 × 105

Because this is well above 500, the change in [CO2] can be ignored.Step 3 Set up an ICE table.

Step 4 Kc =[CO]2[O2]

[CO2]2

2.0 × 10−6=

(2x)2(x)0.60

x =3√

3.0 × 10−7

= 6.7 × 10−3

Step 4 At equilibrium: [CO2] = 0.60 mol/L[CO] = 2(6.7 × 10−3) = 1.3 × 10−2 mol/L[O2] = 6.7 × 10−3 mol/L

Check Your SolutionThe assumption that x is negligible compared with the initial concentration is validbecause, using the rules for subtracting measured quantities,0.60 − 2(6.7 × 10−3) = 0.59 ≈ 0.60 . Next, check the equilibrium values:

(1.3 × 10−2)2(6.7 × 10−3)0.60

= 1.9 × 10−6

This is equal to the equilibrium constant, within the error introduced by mathematical rounding.


Student Textbook page 356

26. ProblemThe following reaction takes place inside a cylinder with a moveable piston.2NO2(g) ⇀↽ N2O4(g)


Initial

Change

Equilibrium

2CO2(g) +2CO(g)

0.60

−2x

0.60 − 2x 0.60

0

+2x

2x

O2(g)

0

+x

x


CHEMISTRY 12

chapter7 equilibrium pp answers

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