Contributed by:

The key highlights are:

1. The concept of equilibrium

2. A system at equilibrium

3. The equilibrium constant

4. Relationship between Kc and Kp

5. What does the value of k mean/

6. Manipulating equilibrium constants

7. Heterogeneous equilibrium

8. Equilibrium calculations

9. The reaction quotient

10. Le-Chatelier's Principle

11. Haber's Process

12. The effect of changes in temperature

1. The concept of equilibrium

2. A system at equilibrium

3. The equilibrium constant

4. Relationship between Kc and Kp

5. What does the value of k mean/

6. Manipulating equilibrium constants

7. Heterogeneous equilibrium

8. Equilibrium calculations

9. The reaction quotient

10. Le-Chatelier's Principle

11. Haber's Process

12. The effect of changes in temperature

1.
Chemistry, The Central Science, 10th edition

Theodore L. Brown; H. Eugene LeMay, Jr.; and

Bruce E. Bursten

Chapter 15

Chemical Equilibrium

John D. Bookstaver

St. Charles Community College

Equilibrium

St. Peters, MO

2006, Prentice Hall

Theodore L. Brown; H. Eugene LeMay, Jr.; and

Bruce E. Bursten

Chapter 15

Chemical Equilibrium

John D. Bookstaver

St. Charles Community College

Equilibrium

St. Peters, MO

2006, Prentice Hall

2.
The Concept of Equilibrium

play the movie

Chemical equilibrium occurs when a

reaction and its reverse reaction proceed at Equilibrium

the same rate.

play the movie

Chemical equilibrium occurs when a

reaction and its reverse reaction proceed at Equilibrium

the same rate.

3.
The Concept of Equilibrium

• As a system

approaches equilibrium,

both the forward and

reverse reactions are

occurring.

• At equilibrium, the

forward and reverse

reactions are

proceeding at the same

rate.

Equilibrium

• As a system

approaches equilibrium,

both the forward and

reverse reactions are

occurring.

• At equilibrium, the

forward and reverse

reactions are

proceeding at the same

rate.

Equilibrium

4.
A System at Equilibrium

Once equilibrium is

achieved, the

amount of each

reactant and product

remains constant.

Equilibrium

Once equilibrium is

achieved, the

amount of each

reactant and product

remains constant.

Equilibrium

5.
Depicting Equilibrium

In a system at equilibrium, both the

forward and reverse reactions are being

carried out; as a result, we write its

equation with a double arrow

N2O4 (g) 2 NO2 (g)

Equilibrium

In a system at equilibrium, both the

forward and reverse reactions are being

carried out; as a result, we write its

equation with a double arrow

N2O4 (g) 2 NO2 (g)

Equilibrium

6.
The

Constant

Equilibrium

Constant

Equilibrium

7.
The Equilibrium Constant

• Forward reaction:

N2O4 (g) 2 NO2 (g)

• Rate law:

Rate = kf [N2O4]

Equilibrium

• Forward reaction:

N2O4 (g) 2 NO2 (g)

• Rate law:

Rate = kf [N2O4]

Equilibrium

8.
The Equilibrium Constant

• Reverse reaction:

2 NO2 (g) N2O4 (g)

• Rate law:

Rate = kr [NO2]2

Equilibrium

• Reverse reaction:

2 NO2 (g) N2O4 (g)

• Rate law:

Rate = kr [NO2]2

Equilibrium

9.
The Equilibrium Constant

• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomes

kf [NO2]2

=

kr [N2O4]

Equilibrium

• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomes

kf [NO2]2

=

kr [N2O4]

Equilibrium

10.
The Equilibrium Constant

The ratio of the rate constants is a

constant at that temperature, and the

expression becomes

kf [NO2]2

Keq = =

kr [N2O4]

Equilibrium

The ratio of the rate constants is a

constant at that temperature, and the

expression becomes

kf [NO2]2

Keq = =

kr [N2O4]

Equilibrium

11.
The Equilibrium Constant

• To generalize this expression, consider

the reaction

aA + bB cC + dD

• The equilibrium expression for this

reaction would be

[C]c[D]d

Kc =

[A]a[B]b Equilibrium

• To generalize this expression, consider

the reaction

aA + bB cC + dD

• The equilibrium expression for this

reaction would be

[C]c[D]d

Kc =

[A]a[B]b Equilibrium

12.
What Are the Equilibrium

Expressions for These Equilibria?

Equilibrium

Expressions for These Equilibria?

Equilibrium

13.
SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions

Write the equilibrium expression for Kc for the following reactions:

Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each.

Plan: Using the law of mass action, we write each expression as a quotient having the product concentration

terms in the numerator and the reactant concentration terms in the denominator. Each term is raised to the

power of its coefficient in the balanced chemical equation.

PRACTICE EXERCISE

Write the equilibrium-constant expression, Kc for

Equilibrium

Write the equilibrium expression for Kc for the following reactions:

Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each.

Plan: Using the law of mass action, we write each expression as a quotient having the product concentration

terms in the numerator and the reactant concentration terms in the denominator. Each term is raised to the

power of its coefficient in the balanced chemical equation.

PRACTICE EXERCISE

Write the equilibrium-constant expression, Kc for

Equilibrium

14.
The Equilibrium Constant

Because pressure is proportional to

concentration for gases in a closed

system, the equilibrium expression can

also be written

(PC)c (PD)d

Kp =

(PA)a (PB)b

Equilibrium

Because pressure is proportional to

concentration for gases in a closed

system, the equilibrium expression can

also be written

(PC)c (PD)d

Kp =

(PA)a (PB)b

Equilibrium

15.
Relationship between Kc and Kp

• From the ideal gas law we know that

PV = nRT

• Rearranging it, we get

n

P= RT

V

Equilibrium

• From the ideal gas law we know that

PV = nRT

• Rearranging it, we get

n

P= RT

V

Equilibrium

16.
Relationship between Kc and Kp

Plugging this into the expression for Kp

for each substance, the relationship

between Kc and Kp becomes

Kp = Kc (RT)n

Where

n = (moles of gaseous product) − (moles of gaseous reactant)

Equilibrium

Plugging this into the expression for Kp

for each substance, the relationship

between Kc and Kp becomes

Kp = Kc (RT)n

Where

n = (moles of gaseous product) − (moles of gaseous reactant)

Equilibrium

17.
SAMPLE EXERCISE 15.2 Converting Between Kc and Kp

In the synthesis of ammonia from nitrogen and hydrogen,

K c = 9.60 at 300°C. Calculate Kp for this reaction at this temperature.

Analyze: We are given Kc for a reaction and asked to calculate Kp.

Plan: The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must

determine n by comparing the number of moles of product with the number of moles of reactants (Equation

Solve: There are two moles of gaseous products (2 NH3) and four moles of gaseous reactants (1 N2 + 3 H2).

Therefore, n = 2 – 4 = –2. (Remember that functions are always based on products minus reactants.) The

temperature, T, is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.0821 L-atm/mol-K. Using

Kc = 9.60, we therefore have

PRACTICE EXERCISE

For the equilibrium is 4.08 10–3 at 1000 K. Calculate the value for Kp.

Answer: 0.335

Equilibrium

In the synthesis of ammonia from nitrogen and hydrogen,

K c = 9.60 at 300°C. Calculate Kp for this reaction at this temperature.

Analyze: We are given Kc for a reaction and asked to calculate Kp.

Plan: The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we must

determine n by comparing the number of moles of product with the number of moles of reactants (Equation

Solve: There are two moles of gaseous products (2 NH3) and four moles of gaseous reactants (1 N2 + 3 H2).

Therefore, n = 2 – 4 = –2. (Remember that functions are always based on products minus reactants.) The

temperature, T, is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.0821 L-atm/mol-K. Using

Kc = 9.60, we therefore have

PRACTICE EXERCISE

For the equilibrium is 4.08 10–3 at 1000 K. Calculate the value for Kp.

Answer: 0.335

Equilibrium

18.
Equilibrium Can Be Reached from

Either Direction

As you can see, the ratio of [NO2]2 to [N2O4]

remains constant at this temperature no matter

what the initial concentrations of NO2 and N2O4 are.

Equilibrium

Either Direction

As you can see, the ratio of [NO2]2 to [N2O4]

remains constant at this temperature no matter

what the initial concentrations of NO2 and N2O4 are.

Equilibrium

19.
Equilibrium Can Be Reached from

Either Direction

This is the data from

the last two trials from

the table on the

previous slide.

Equilibrium

Either Direction

This is the data from

the last two trials from

the table on the

previous slide.

Equilibrium

20.
Equilibrium Can Be Reached from

Either Direction

It does not matter whether we start with N2

and H2 or whether we start with NH3. We will

have the same proportions of all three

substances at equilibrium.

Equilibrium

Either Direction

It does not matter whether we start with N2

and H2 or whether we start with NH3. We will

have the same proportions of all three

substances at equilibrium.

Equilibrium

21.
What Does the Value of K Mean?

• If K >> 1, the reaction

is product-favored;

product predominates

at equilibrium.

Equilibrium

• If K >> 1, the reaction

is product-favored;

product predominates

at equilibrium.

Equilibrium

22.
What Does the Value of K Mean?

• If K >> 1, the reaction

is product-favored;

product predominates

at equilibrium.

• If K << 1, the reaction is

reactant-favored;

reactant predominates

at equilibrium.

Equilibrium

• If K >> 1, the reaction

is product-favored;

product predominates

at equilibrium.

• If K << 1, the reaction is

reactant-favored;

reactant predominates

at equilibrium.

Equilibrium

23.
SAMPLE EXERCISE 15.3 Interpreting the Magnitude of an Equilibrium Constant

The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen:

The value for the equilibrium constant for this reaction at 25°C is Kc = 1 10–30. Describe the feasibility of

fixing nitrogen by forming NO at 25°C.

Analyze: We are asked to comment on the utility of a reaction based on the magnitude of its equilibrium

Plan: We consider the magnitude of the equilibrium constant to determine whether this reaction is feasible for

the production of NO.

Solve: Because Kc is so small, very little NO will form at 25°C. The equilibrium lies to the left, favoring the

reactants. Consequently, this reaction is an extremely poor choice for nitrogen fixation, at least at 25°C.

PRACTICE EXERCISE

For the reaction at 298 K and Kp = 54 at 700 K. Is the formation of HI

favored more at the higher or lower temperature?

Answer: The formation of product, HI, is favored at the lower temperature because Kp is larger at the lower

Equilibrium

The reaction of N2 with O2 to form NO might be considered a means of “fixing” nitrogen:

The value for the equilibrium constant for this reaction at 25°C is Kc = 1 10–30. Describe the feasibility of

fixing nitrogen by forming NO at 25°C.

Analyze: We are asked to comment on the utility of a reaction based on the magnitude of its equilibrium

Plan: We consider the magnitude of the equilibrium constant to determine whether this reaction is feasible for

the production of NO.

Solve: Because Kc is so small, very little NO will form at 25°C. The equilibrium lies to the left, favoring the

reactants. Consequently, this reaction is an extremely poor choice for nitrogen fixation, at least at 25°C.

PRACTICE EXERCISE

For the reaction at 298 K and Kp = 54 at 700 K. Is the formation of HI

favored more at the higher or lower temperature?

Answer: The formation of product, HI, is favored at the lower temperature because Kp is larger at the lower

Equilibrium

24.
Manipulating Equilibrium Constants

The equilibrium constant of a reaction in

the reverse reaction is the reciprocal of

the equilibrium constant of the forward

reaction.

[NO2]2

N2O4 (g) 2 NO2 (g) Kc = = 0.212 at 100C

[N2O4]

[N2O4] 1

2 NO2 (g) N2O4 (g) K c = =

[NO2]2 0.212

= 4.72 at 100C

Equilibrium

The equilibrium constant of a reaction in

the reverse reaction is the reciprocal of

the equilibrium constant of the forward

reaction.

[NO2]2

N2O4 (g) 2 NO2 (g) Kc = = 0.212 at 100C

[N2O4]

[N2O4] 1

2 NO2 (g) N2O4 (g) K c = =

[NO2]2 0.212

= 4.72 at 100C

Equilibrium

25.
Manipulating Equilibrium Constants

The equilibrium constant of a reaction that has

been multiplied by a number is the equilibrium

constant raised to a power that is equal to that

number.

[NO2]2

N2O4 (g) 2 NO2 (g) Kc = = 0.212 at 100C

[N2O4]

[NO2]4

2 N2O4 (g) 4 NO2 (g) Kc = = (0.212)2 at 100C

[N2O4]2

Equilibrium

The equilibrium constant of a reaction that has

been multiplied by a number is the equilibrium

constant raised to a power that is equal to that

number.

[NO2]2

N2O4 (g) 2 NO2 (g) Kc = = 0.212 at 100C

[N2O4]

[NO2]4

2 N2O4 (g) 4 NO2 (g) Kc = = (0.212)2 at 100C

[N2O4]2

Equilibrium

26.
Manipulating Equilibrium Constants

The equilibrium constant for a net

reaction made up of two or more steps

is the product of the equilibrium

constants for the individual steps.

Equilibrium

The equilibrium constant for a net

reaction made up of two or more steps

is the product of the equilibrium

constants for the individual steps.

Equilibrium

27.
SAMPLE EXERCISE 15.5 Combining Equilibrium Expressions

Given the following information,

determine the value of Kc for the reaction

Analyze: We are given two equilibrium equations and the corresponding equilibrium constants and are asked

to determine the equilibrium constant for a third equation, which is related to the first two.

Plan: We cannot simply add the first two equations to get the third. Instead, we need to determine how to

manipulate the equations in order to come up with the steps that will add to give us the desired equation.

Solve: If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant

(raising to the power 2), we get

Reversing the second equation and again making the corresponding change to its equilibrium constant (taking

the reciprocal) gives

Now we have two equations that sum to give the net equation, and we can multiply the individual Kc values to

get the desired equilibrium constant.

Equilibrium

Given the following information,

determine the value of Kc for the reaction

Analyze: We are given two equilibrium equations and the corresponding equilibrium constants and are asked

to determine the equilibrium constant for a third equation, which is related to the first two.

Plan: We cannot simply add the first two equations to get the third. Instead, we need to determine how to

manipulate the equations in order to come up with the steps that will add to give us the desired equation.

Solve: If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant

(raising to the power 2), we get

Reversing the second equation and again making the corresponding change to its equilibrium constant (taking

the reciprocal) gives

Now we have two equations that sum to give the net equation, and we can multiply the individual Kc values to

get the desired equilibrium constant.

Equilibrium

28.
SAMPLE EXERCISE 15.5 continued

PRACTICE EXERCISE

Given that, at 700 K, Kp = 54.0 for the reaction for the

reaction determine the value of Kp for the reaction

Equilibrium

PRACTICE EXERCISE

Given that, at 700 K, Kp = 54.0 for the reaction for the

reaction determine the value of Kp for the reaction

Equilibrium

29.
Equilibrium

Equilibrium

Equilibrium

30.
The Concentrations of Solids and

Liquids Are Essentially Constant

Both can be obtained by dividing the

density of the substance by its molar

mass—and both of these are constants

at constant temperature.

Equilibrium

Liquids Are Essentially Constant

Both can be obtained by dividing the

density of the substance by its molar

mass—and both of these are constants

at constant temperature.

Equilibrium

31.
The Concentrations of Solids and

Liquids Are Essentially Constant

Therefore, the concentrations of solids

and liquids do not appear in the

equilibrium expression

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Kc = [Pb2+] [Cl−]2

Equilibrium

Liquids Are Essentially Constant

Therefore, the concentrations of solids

and liquids do not appear in the

equilibrium expression

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Kc = [Pb2+] [Cl−]2

Equilibrium

32.
CaCO3 (s) CO2 (g) + CaO(s)

As long as some CaCO3 or CaO remain in the

system, the amount of CO2 above the solid

will remain the same.

Equilibrium

As long as some CaCO3 or CaO remain in the

system, the amount of CO2 above the solid

will remain the same.

Equilibrium

33.
SAMPLE EXERCISE 15.6 Writing Equilibrium-Constant Expressions

for Heterogeneous Reactions

Write the equilibrium-constant expression for Kc for each of the following reactions:

Analyze: We are given two chemical equations, both for heterogeneous equilibria, and asked to write the

corresponding equilibrium-constant expressions.

Plan: We use the law of mass action, remembering to omit any pure solids, pure liquids, and solvents from the

Solve: (a) The equilibrium-constant expression is

Because H2O appears in the reaction as a pure liquid, its concentration does not appear in the equilibrium-

constant expression.

(b) The equilibrium-constant expression is

Because SnO2 and Sn are both pure solids, their concentrations do not appear in the equilibrium-constant

PRACTICE EXERCISE

Write the following equilibrium-constant expressions:

Equilibrium

for Heterogeneous Reactions

Write the equilibrium-constant expression for Kc for each of the following reactions:

Analyze: We are given two chemical equations, both for heterogeneous equilibria, and asked to write the

corresponding equilibrium-constant expressions.

Plan: We use the law of mass action, remembering to omit any pure solids, pure liquids, and solvents from the

Solve: (a) The equilibrium-constant expression is

Because H2O appears in the reaction as a pure liquid, its concentration does not appear in the equilibrium-

constant expression.

(b) The equilibrium-constant expression is

Because SnO2 and Sn are both pure solids, their concentrations do not appear in the equilibrium-constant

PRACTICE EXERCISE

Write the following equilibrium-constant expressions:

Equilibrium

34.
SAMPLE EXERCISE 15.7 Analyzing a Heterogeneous Equilibrium

Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of

attaining the equilibrium (a) pure CaCO3, (b) CaO and a CO2 pressure greater

than the value of Kp, (c) some CaCO3 and a CO2 pressure greater than the value of Kp, (d) CaCO3 and CaO?

Analyze: We are asked which of several combinations of species can establish an equilibrium between

calcium carbonate and its decomposition products, calcium oxide and carbon dioxide.

Plan: In order for equilibrium to be achieved, it must be possible for both the forward process and the reverse

process to occur. In order for the forward process to occur, there must be some calcium carbonate present. In

order for the reverse process to occur, there must be both calcium oxide and carbon dioxide. In both cases,

either the necessary compounds may be present initially, or they may be formed by reaction of the other

Solve: Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present.

(a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equilibrium pressure of CO2 is attained.

There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium. (b) CO2 continues to

combine with CaO until the partial pressure of the CO2 decreases to the equilibrium value. (c) There is no CaO

present, so equilibrium can’t be attained because there is no way the CO2 pressure can decrease to its

equilibrium value (which would require some of the CO2 to react with CaO). (d) The situation is essentially the

same as in (a): CaCO3 decomposes until equilibrium is attained. The presence of CaO initially makes no

PRACTICE EXERCISE

When added to Fe3O4(s) in a closed container, which one of the following substances—H 2(g), H2O(g), O2(g)

—will allow equilibrium to be established in the reaction Equilibrium

Answers: only H2(g)

Each of the following mixtures was placed in a closed container and allowed to stand. Which is capable of

attaining the equilibrium (a) pure CaCO3, (b) CaO and a CO2 pressure greater

than the value of Kp, (c) some CaCO3 and a CO2 pressure greater than the value of Kp, (d) CaCO3 and CaO?

Analyze: We are asked which of several combinations of species can establish an equilibrium between

calcium carbonate and its decomposition products, calcium oxide and carbon dioxide.

Plan: In order for equilibrium to be achieved, it must be possible for both the forward process and the reverse

process to occur. In order for the forward process to occur, there must be some calcium carbonate present. In

order for the reverse process to occur, there must be both calcium oxide and carbon dioxide. In both cases,

either the necessary compounds may be present initially, or they may be formed by reaction of the other

Solve: Equilibrium can be reached in all cases except (c) as long as sufficient quantities of solids are present.

(a) CaCO3 simply decomposes, forming CaO(s) and CO2(g) until the equilibrium pressure of CO2 is attained.

There must be enough CaCO3, however, to allow the CO2 pressure to reach equilibrium. (b) CO2 continues to

combine with CaO until the partial pressure of the CO2 decreases to the equilibrium value. (c) There is no CaO

present, so equilibrium can’t be attained because there is no way the CO2 pressure can decrease to its

equilibrium value (which would require some of the CO2 to react with CaO). (d) The situation is essentially the

same as in (a): CaCO3 decomposes until equilibrium is attained. The presence of CaO initially makes no

PRACTICE EXERCISE

When added to Fe3O4(s) in a closed container, which one of the following substances—H 2(g), H2O(g), O2(g)

—will allow equilibrium to be established in the reaction Equilibrium

Answers: only H2(g)

35.
Equilibrium

36.
Equilibrium Calculations

A closed system initially containing

1.000 x 10−3 M H2 and 2.000 x 10−3 M I2

At 448C is allowed to reach equilibrium. Analysis

of the equilibrium mixture shows that the

concentration of HI is 1.87 x 10−3 M. Calculate Kc at

448C for the reaction taking place, which is

H2 (g) + I2 (g) 2 HI (g)

Equilibrium

A closed system initially containing

1.000 x 10−3 M H2 and 2.000 x 10−3 M I2

At 448C is allowed to reach equilibrium. Analysis

of the equilibrium mixture shows that the

concentration of HI is 1.87 x 10−3 M. Calculate Kc at

448C for the reaction taking place, which is

H2 (g) + I2 (g) 2 HI (g)

Equilibrium

37.
What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

At 1.87 x 10-3

Equilibrium

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

At 1.87 x 10-3

Equilibrium

38.
[HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At 1.87 x 10-3

Equilibrium

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At 1.87 x 10-3

Equilibrium

39.
Stoichiometry tells us [H2] and [I2]

decrease by half as much

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 1.87 x 10-3

Equilibrium

decrease by half as much

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 1.87 x 10-3

Equilibrium

40.
We can now calculate the equilibrium

concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

Equilibrium

concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

Equilibrium

41.
…and, therefore, the equilibrium constant

[HI]2

Kc =

[H2] [I2]

(1.87 x 10-3)2

=

(6.5 x 10-5)(1.065 x 10-3)

= 51

Equilibrium

[HI]2

Kc =

[H2] [I2]

(1.87 x 10-3)2

=

(6.5 x 10-5)(1.065 x 10-3)

= 51

Equilibrium

42.
SAMPLE EXERCISE 15.8 Calculating K When All Equilibrium Concentrations Are Known

A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472°C. The

equilibrium mixture of gases was analyzed and found to contain 7.38 atm H 2 , 2.46 atm N2 , and 0.166 atm NH3.

From these data, calculate the equilibrium constant Kp for the reaction

Analyze: We are given a balanced equation and equilibrium partial pressures and are asked to calculate the

value of the equilibrium constant.

Plan: Using the balanced equation, we write the equilibrium-constant expression. We then substitute the

equilibrium partial pressures into the expression and solve for Kp.

PRACTICE EXERCISE

An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25°C: [HC 2H3O2] =

1.65 10–2 M; [H+] = 5.44 10–4 M; and [C2H3O2–] = 5.44 10–4 M. Calculate the equilibrium constant Kc for the

ionization of acetic acid at 25°C. The reaction is

Answer: 1.79 10–5 Equilibrium

A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472°C. The

equilibrium mixture of gases was analyzed and found to contain 7.38 atm H 2 , 2.46 atm N2 , and 0.166 atm NH3.

From these data, calculate the equilibrium constant Kp for the reaction

Analyze: We are given a balanced equation and equilibrium partial pressures and are asked to calculate the

value of the equilibrium constant.

Plan: Using the balanced equation, we write the equilibrium-constant expression. We then substitute the

equilibrium partial pressures into the expression and solve for Kp.

PRACTICE EXERCISE

An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25°C: [HC 2H3O2] =

1.65 10–2 M; [H+] = 5.44 10–4 M; and [C2H3O2–] = 5.44 10–4 M. Calculate the equilibrium constant Kc for the

ionization of acetic acid at 25°C. The reaction is

Answer: 1.79 10–5 Equilibrium

43.
SAMPLE EXERCISE 15.9 Calculating K from Initial and Equilibrium Concentrations

A closed system initially containing 1.000 10–3 M H2 and 2.000T10–3 M I2 at 448°C is allowed to reach

equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 10–3 M. Calculate

Kc at 448°C for the reaction taking place, which is

Analyze: We are given the initial concentrations of H2 and I2 and the equilibrium concentration of HI. We

are asked to calculate the equilibrium constant Kc for the reaction

Plan: We construct a table to find equilibrium concentrations of all species and then use the equilibrium

concentrations to calculate the equilibrium constant.

Solve: First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also

provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the

chemical equation as the heading for the table.

Equilibrium

A closed system initially containing 1.000 10–3 M H2 and 2.000T10–3 M I2 at 448°C is allowed to reach

equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 10–3 M. Calculate

Kc at 448°C for the reaction taking place, which is

Analyze: We are given the initial concentrations of H2 and I2 and the equilibrium concentration of HI. We

are asked to calculate the equilibrium constant Kc for the reaction

Plan: We construct a table to find equilibrium concentrations of all species and then use the equilibrium

concentrations to calculate the equilibrium constant.

Solve: First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also

provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the

chemical equation as the heading for the table.

Equilibrium

44.
SAMPLE EXERCISE 15.9 continued

Second, we calculate the change in concentration of HI, which is the difference between the equilibrium values

and the initial values:

Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H 2] and

Fourth, we calculate the equilibrium concentrations of H2 and I2 , using the initial concentrations and the

changes. The equilibrium concentration equals the initial concentration minus that consumed:

The completed table now looks like this (with equilibrium concentrations in blue for emphasis):

Equilibrium

Second, we calculate the change in concentration of HI, which is the difference between the equilibrium values

and the initial values:

Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H 2] and

Fourth, we calculate the equilibrium concentrations of H2 and I2 , using the initial concentrations and the

changes. The equilibrium concentration equals the initial concentration minus that consumed:

The completed table now looks like this (with equilibrium concentrations in blue for emphasis):

Equilibrium

45.
SAMPLE EXERCISE 15.9 continued

Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is

Finally, now that we know the equilibrium concentration of each reactant and product, we can use the

equilibrium-constant expression to calculate the equilibrium constant.

Comment: The same method can be applied to gaseous equilibrium problems, in which case partial

pressures are used as table entries in place of molar concentrations.

PRACTICE EXERCISE

Sulfur trioxide decomposes at high temperature in a sealed container:

Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3

partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.

Answer: 0.338

Equilibrium

Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is

Finally, now that we know the equilibrium concentration of each reactant and product, we can use the

equilibrium-constant expression to calculate the equilibrium constant.

Comment: The same method can be applied to gaseous equilibrium problems, in which case partial

pressures are used as table entries in place of molar concentrations.

PRACTICE EXERCISE

Sulfur trioxide decomposes at high temperature in a sealed container:

Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3

partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.

Answer: 0.338

Equilibrium

46.
The Reaction Quotient (Q)

• To calculate Q, one substitutes the

initial concentrations on reactants and

products into the equilibrium

expression.

• Q gives the same ratio the equilibrium

expression gives, but for a system that

is not at equilibrium.

Equilibrium

• To calculate Q, one substitutes the

initial concentrations on reactants and

products into the equilibrium

expression.

• Q gives the same ratio the equilibrium

expression gives, but for a system that

is not at equilibrium.

Equilibrium

47.
If Q = K,

the system is at equilibrium.

Equilibrium

the system is at equilibrium.

Equilibrium

48.
If Q > K,

there is too much product and the

equilibrium shifts to the left.

Equilibrium

there is too much product and the

equilibrium shifts to the left.

Equilibrium

49.
If Q < K,

there is too much reactant, and the

equilibrium shifts to the right.

Equilibrium

there is too much reactant, and the

equilibrium shifts to the right.

Equilibrium

50.
SAMPLE EXERCISE 15.10 Predicting the Direction of Approach to Equilibrium

At 448°C the equilibrium constant Kc for the reaction

is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448°C if we start with 2.0

10–2 mol of HI, 1.0 10–2 mol of H2, and 3.0 10–2 mol of I2 in a 2.00-L container.

Analyze: We are given a volume and initial molar amounts of the species in a reaction and asked to

determine in which direction the reaction must proceed to achieve equilibrium.

Plan: We can determine the starting concentration of each species in the reaction mixture. We can then

substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient,

Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us

in which direction the reaction will proceed.

Solve: The initial concentrations are

The reaction quotient is therefore

Equilibrium

At 448°C the equilibrium constant Kc for the reaction

is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448°C if we start with 2.0

10–2 mol of HI, 1.0 10–2 mol of H2, and 3.0 10–2 mol of I2 in a 2.00-L container.

Analyze: We are given a volume and initial molar amounts of the species in a reaction and asked to

determine in which direction the reaction must proceed to achieve equilibrium.

Plan: We can determine the starting concentration of each species in the reaction mixture. We can then

substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient,

Qc. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us

in which direction the reaction will proceed.

Solve: The initial concentrations are

The reaction quotient is therefore

Equilibrium

51.
SAMPLE EXERCISE 15.10 continued

Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to

reach equilibrium; the reaction will proceed from left to right as it moves toward equilibrium.

PRACTICE EXERCISE

At 1000 K the value of Kp for the reaction is 0.338. Calculate the value for

Qp , and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures

Answer: Qp = 16; Qp > Kp, and so the reaction will proceed from right to left, forming more SO3.

Equilibrium

Because Qc < Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to

reach equilibrium; the reaction will proceed from left to right as it moves toward equilibrium.

PRACTICE EXERCISE

At 1000 K the value of Kp for the reaction is 0.338. Calculate the value for

Qp , and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures

Answer: Qp = 16; Qp > Kp, and so the reaction will proceed from right to left, forming more SO3.

Equilibrium

52.
SAMPLE EXERCISE 15.11 Calculating Equilibrium Concentrations

For the Haber process, at 500°C. In an equilibrium

mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is

the partial pressure of NH3 in this equilibrium mixture?

Analyze: We are given an equilibrium constant, Kp , and the equilibrium partial pressures of two of the three

substances in the equation (N2 and H2), and we are asked to calculate the equilibrium partial pressure for the

third substance (NH3).

Plan: We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that

are known. Then we can solve for the only unknown in the equation.

Solve: We tabulate the equilibrium pressures as follows:

Because we do not know the equilibrium pressure of NH3 , we represent it with a variable, x. At equilibrium the

pressures must satisfy the equilibrium-constant expression:

We now rearrange the equation to solve for x:

Equilibrium

For the Haber process, at 500°C. In an equilibrium

mixture of the three gases at 500°C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is

the partial pressure of NH3 in this equilibrium mixture?

Analyze: We are given an equilibrium constant, Kp , and the equilibrium partial pressures of two of the three

substances in the equation (N2 and H2), and we are asked to calculate the equilibrium partial pressure for the

third substance (NH3).

Plan: We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that

are known. Then we can solve for the only unknown in the equation.

Solve: We tabulate the equilibrium pressures as follows:

Because we do not know the equilibrium pressure of NH3 , we represent it with a variable, x. At equilibrium the

pressures must satisfy the equilibrium-constant expression:

We now rearrange the equation to solve for x:

Equilibrium

53.
SAMPLE EXERCISE 15.11 continued

Comment: We can always check our answer by using it to recalculate the value of the equilibrium constant:

PRACTICE EXERCISE

At 500 K the reaction has Kp = 0.497. In an equilibrium mixture at 500 K, the

partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the

equilibrium mixture?

Answer: 1.22 atm

Equilibrium

Comment: We can always check our answer by using it to recalculate the value of the equilibrium constant:

PRACTICE EXERCISE

At 500 K the reaction has Kp = 0.497. In an equilibrium mixture at 500 K, the

partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the

equilibrium mixture?

Answer: 1.22 atm

Equilibrium

54.
SAMPLE EXERCISE 15.12 Calculating Equilibrium Concentrations

from Initial Concentrations

A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium

constant Kc for the reaction

at 448°C is 50.5. What are the equilibrium concentrations of H2 , I2 , and HI in moles per liter?

Analyze: We are given the volume of a container, an equilibrium constant, and starting amounts of reactants

in the container and are asked to calculate the equilibrium concentrations of all species.

Plan: In this case we are not given any of the equilibrium concentrations. We must develop some relationships

that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that

outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial concentrations.

Solve: First, we note the initial concentrations of H2 and I2 in the 1.000-L flask:

Second, we construct a table in which we tabulate the initial concentrations:

Equilibrium

from Initial Concentrations

A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium

constant Kc for the reaction

at 448°C is 50.5. What are the equilibrium concentrations of H2 , I2 , and HI in moles per liter?

Analyze: We are given the volume of a container, an equilibrium constant, and starting amounts of reactants

in the container and are asked to calculate the equilibrium concentrations of all species.

Plan: In this case we are not given any of the equilibrium concentrations. We must develop some relationships

that relate the initial concentrations to those at equilibrium. The procedure is similar in many regards to that

outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial concentrations.

Solve: First, we note the initial concentrations of H2 and I2 in the 1.000-L flask:

Second, we construct a table in which we tabulate the initial concentrations:

Equilibrium

55.
SAMPLE EXERCISE 15.12 continued

Third, we use the stoichiometry of the reaction to determine the changes in concentration that occur as the

reaction proceeds to equilibrium. The concentrations of H2 and I2 will decrease as equilibrium is established,

and that of HI will increase. Let’s represent the change in concentration of H2 by the variable x. The balanced

chemical equation tells us the relationship between the changes in the concentrations of the three gases:

Fourth, we use the initial concentrations and the changes in concentrations, as dictated by stoichiometry, to

express the equilibrium concentrations. With all our entries, our table now looks like this:

Fifth, we substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the

single unknown, x:

Equilibrium

Third, we use the stoichiometry of the reaction to determine the changes in concentration that occur as the

reaction proceeds to equilibrium. The concentrations of H2 and I2 will decrease as equilibrium is established,

and that of HI will increase. Let’s represent the change in concentration of H2 by the variable x. The balanced

chemical equation tells us the relationship between the changes in the concentrations of the three gases:

Fourth, we use the initial concentrations and the changes in concentrations, as dictated by stoichiometry, to

express the equilibrium concentrations. With all our entries, our table now looks like this:

Fifth, we substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the

single unknown, x:

Equilibrium

56.
SAMPLE EXERCISE 15.12 continued

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this

expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative

concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this

solution. We then use x = 0.935 to find the equilibrium concentrations:

Check: We can check our solution by putting these numbers into the equilibrium-constant expression:

Equilibrium

If you have an equation-solving calculator, you can solve this equation directly for x. If not, expand this

expression to obtain a quadratic equation in x:

Solving the quadratic equation (Appendix A.3) leads to two solutions for x:

When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative

concentrations of H2 and I2. Because a negative concentration is not chemically meaningful, we reject this

solution. We then use x = 0.935 to find the equilibrium concentrations:

Check: We can check our solution by putting these numbers into the equilibrium-constant expression:

Equilibrium

57.
SAMPLE EXERCISE 15.12 continued

Comment: Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions will

not be chemically meaningful and should be rejected.

PRACTICE EXERCISE

For the equilibrium the equilibrium constant Kp has the value 0.497 at 500 K.

A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium

pressures of PCl5 , PCl3 , and Cl2 at this temperature?

Equilibrium

Comment: Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions will

not be chemically meaningful and should be rejected.

PRACTICE EXERCISE

For the equilibrium the equilibrium constant Kp has the value 0.497 at 500 K.

A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium

pressures of PCl5 , PCl3 , and Cl2 at this temperature?

Equilibrium

58.
Le Châtelier’s

Principle

Equilibrium

Principle

Equilibrium

59.
Le Châtelier’s Principle

“If a system at equilibrium is disturbed by

a change in temperature, pressure, or the

concentration of one of the components,

the system will shift its equilibrium

position so as to counteract the effect of

the disturbance.”

Equilibrium

“If a system at equilibrium is disturbed by

a change in temperature, pressure, or the

concentration of one of the components,

the system will shift its equilibrium

position so as to counteract the effect of

the disturbance.”

Equilibrium

60.
What Happens When More of a

Reactant Is Added to a System?

Play insert CD

Equilibrium

Reactant Is Added to a System?

Play insert CD

Equilibrium

61.
The Haber Process

The transformation of nitrogen and hydrogen into

ammonia (NH3) is of tremendous significance in

agriculture, where ammonia-based fertilizers are of

utmost importance.

Equilibrium

The transformation of nitrogen and hydrogen into

ammonia (NH3) is of tremendous significance in

agriculture, where ammonia-based fertilizers are of

utmost importance.

Equilibrium

62.
The Haber Process

If H2 is added to the

system, N2 will be

consumed and the

two reagents will

form more NH3.

Equilibrium

If H2 is added to the

system, N2 will be

consumed and the

two reagents will

form more NH3.

Equilibrium

63.
The Haber Process

This apparatus

helps push the

equilibrium to the

right by removing

the ammonia (NH3)

from the system as

a liquid.

Equilibrium

This apparatus

helps push the

equilibrium to the

right by removing

the ammonia (NH3)

from the system as

a liquid.

Equilibrium

64.
The Effect of Changes in Pressure

Equilibrium

Equilibrium

65.
The Effect of Changes in

Temperature

Co(H2O)62+(aq) + 4 Cl(aq) CoCl4 (aq) + 6 H2O (l)

Equilibrium

Temperature

Co(H2O)62+(aq) + 4 Cl(aq) CoCl4 (aq) + 6 H2O (l)

Equilibrium

66.
The Effect of Changes in

Temperature

Play insert CD

Equilibrium

Temperature

Play insert CD

Equilibrium

67.
Catalysts increase the rate of both the

forward and reverse reactions.

Equilibrium

forward and reverse reactions.

Equilibrium

68.
Equilibrium is achieved faster, but the

equilibrium composition remains unaltered.

Equilibrium

equilibrium composition remains unaltered.

Equilibrium

69.
SAMPLE EXERCISE 15.13 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium

Consider the equilibrium

In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure

is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?

Analyze: We are given a series of changes to be made to a system at equilibrium and are asked to predict

what effect each change will have on the position of the equilibrium.

Plan: Le Châtelier’s principle can be used to determine the effects of each of these changes.

Solve: (a) The system will adjust to decrease the concentration of the added N2O4, so the equilibrium shifts to

the right, in the direction of products.

(b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the

equilibrium shifts to the right.

(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial

pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the position of the equilibrium.

(d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas

molecules); thus, the equilibrium shifts to the right. (This is the opposite of the effect observed in Figure 15.13,

where the volume was decreased.)

(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation.

Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium

shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of the

equilibrium constant, K. Equilibrium

Consider the equilibrium

In which direction will the equilibrium shift when (a) N2O4 is added, (b) NO2 is removed, (c) the total pressure

is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased?

Analyze: We are given a series of changes to be made to a system at equilibrium and are asked to predict

what effect each change will have on the position of the equilibrium.

Plan: Le Châtelier’s principle can be used to determine the effects of each of these changes.

Solve: (a) The system will adjust to decrease the concentration of the added N2O4, so the equilibrium shifts to

the right, in the direction of products.

(b) The system will adjust to the removal of NO2 by shifting to the side that produces more NO2; thus, the

equilibrium shifts to the right.

(c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial

pressures of NO2 and N2O4 are therefore unchanged, and there is no shift in the position of the equilibrium.

(d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas

molecules); thus, the equilibrium shifts to the right. (This is the opposite of the effect observed in Figure 15.13,

where the volume was decreased.)

(e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation.

Decreasing the temperature will shift the equilibrium in the direction that produces heat, so the equilibrium

shifts to the left, toward the formation of more N2O4. Note that only this last change also affects the value of the

equilibrium constant, K. Equilibrium

70.
SAMPLE EXERCISE 15.13 continued

PRACTICE EXERCISE

For the reaction

in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the temperature is decreased, (c)

the volume of the reaction system is increased, (d) PCl3(g) is added?

Answers: (a) right, (b) left, (c) right, (d) left

Equilibrium

PRACTICE EXERCISE

For the reaction

in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the temperature is decreased, (c)

the volume of the reaction system is increased, (d) PCl3(g) is added?

Answers: (a) right, (b) left, (c) right, (d) left

Equilibrium

71.
SAMPLE EXERCISE 15.14 Predicting the Effect of Temperature on K

(a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the

(b) Determine how the equilibrium constant for this reaction should change with temperature.

Analyze: We are asked to determine the standard enthalpy change of a reaction and how the equilibrium

constant for the reaction varies with temperature.

Plan: (a) We can use standard enthalpies of formation to calculate H° for the reaction. (b) We can then use

Le Châtelier’s principle to determine what effect temperature will have on the equilibrium constant.

Solve: (a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar

enthalpies of formation of the products, each multiplied by its coefficient in the balanced chemical equation,

less the same quantities for the reactants. At 25°C, for NH3(g) is The values for

H2(g) and N2(g) are zero by definition because the enthalpies of formation of the elements in their normal states

at 25°C are defined as zero (Section 5.7). Because 2 mol of NH3 is formed, the total enthalpy change is

Equilibrium

(a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy change for the

(b) Determine how the equilibrium constant for this reaction should change with temperature.

Analyze: We are asked to determine the standard enthalpy change of a reaction and how the equilibrium

constant for the reaction varies with temperature.

Plan: (a) We can use standard enthalpies of formation to calculate H° for the reaction. (b) We can then use

Le Châtelier’s principle to determine what effect temperature will have on the equilibrium constant.

Solve: (a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar

enthalpies of formation of the products, each multiplied by its coefficient in the balanced chemical equation,

less the same quantities for the reactants. At 25°C, for NH3(g) is The values for

H2(g) and N2(g) are zero by definition because the enthalpies of formation of the elements in their normal states

at 25°C are defined as zero (Section 5.7). Because 2 mol of NH3 is formed, the total enthalpy change is

Equilibrium

72.
SAMPLE EXERCISE 15.14 continued

(b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction.

An increase in temperature causes the reaction to shift in the direction of less NH 3 and more N2 and H2. This

effect is seen in the values for Kp presented in Table 15.2. Notice that Kp changes markedly with changes in

temperature and that it is larger at lower temperatures.

Comment: The fact that Kp for the formation of NH3 from N2 and H2 decreases with increasing temperature

is a matter of great practical importance. To form NH3 at a reasonable rate requires higher temperatures. At

higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to NH 3 is

smaller. To compensate for this, higher pressures are needed because high pressure favors NH 3 formation.

Equilibrium

(b) Because the reaction in the forward direction is exothermic, we can consider heat a product of the reaction.

An increase in temperature causes the reaction to shift in the direction of less NH 3 and more N2 and H2. This

effect is seen in the values for Kp presented in Table 15.2. Notice that Kp changes markedly with changes in

temperature and that it is larger at lower temperatures.

Comment: The fact that Kp for the formation of NH3 from N2 and H2 decreases with increasing temperature

is a matter of great practical importance. To form NH3 at a reasonable rate requires higher temperatures. At

higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to NH 3 is

smaller. To compensate for this, higher pressures are needed because high pressure favors NH 3 formation.

Equilibrium

73.
SAMPLE EXERCISE 15.14 continued

PRACTICE EXERCISE

Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction

Use this result to determine how the equilibrium constant for the reaction should change with temperature.

the equilibrium constant will increase with increasing temperature

Equilibrium

PRACTICE EXERCISE

Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction

Use this result to determine how the equilibrium constant for the reaction should change with temperature.

the equilibrium constant will increase with increasing temperature

Equilibrium

74.
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

At temperatures near 800°C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form

CO and H2:

The mixture of gases that results is an important industrial fuel called water gas. (a) At 800°C the equilibrium

constant for this reaction is Kp = 14.1. What are the equilibrium partial pressures of H2O, CO, and H2 in the

equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H 2O in a 1.00-L vessel?

(b) What is the minimum amount of carbon required to achieve equilibrium under these conditions? (c) What is

the total pressure in the vessel at equilibrium? (d) At 25°C the value of Kp for this reaction is 1.7 10–21. Is the

reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H2 at equilibrium, should

the pressure of the system be increased or decreased?

Solution (a) To determine the equilibrium partial pressures, we use the ideal gas equation, first determining

the starting partial pressure of hydrogen.

We then construct a table of starting partial pressures and their changes as equilibrium is achieved:

Equilibrium

At temperatures near 800°C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form

CO and H2:

The mixture of gases that results is an important industrial fuel called water gas. (a) At 800°C the equilibrium

constant for this reaction is Kp = 14.1. What are the equilibrium partial pressures of H2O, CO, and H2 in the

equilibrium mixture at this temperature if we start with solid carbon and 0.100 mol of H 2O in a 1.00-L vessel?

(b) What is the minimum amount of carbon required to achieve equilibrium under these conditions? (c) What is

the total pressure in the vessel at equilibrium? (d) At 25°C the value of Kp for this reaction is 1.7 10–21. Is the

reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H2 at equilibrium, should

the pressure of the system be increased or decreased?

Solution (a) To determine the equilibrium partial pressures, we use the ideal gas equation, first determining

the starting partial pressure of hydrogen.

We then construct a table of starting partial pressures and their changes as equilibrium is achieved:

Equilibrium

75.
SAMPLE INTEGRATIVE EXERCISE continued

There are no entries in the table under C(s) because the reactant, being a solid, does not appear in the

equilibrium-constant expression. Substituting the equilibrium partial pressures of the other species into the

equilibrium-constant expression for the reaction gives

Multiplying through by the denominator gives a quadratic equation in x:

Solving this equation for x using the quadratic formula yields x = 6.14 atm. Hence, the equilibrium partial

pressures are and

(b) Part (a) shows that x = 6.14atm of H2O must react in order for the system to achieve equilibrium. We can

use the ideal-gas equation to convert this partial pressure into a mole amount.

Thus, 0.0697 mol of H2O and the same amount of C must react to achieve equilibrium. As a result, there

must be at least 0.0697 mol of C (0.836 g C) present among the reactants at the start of the reaction.

(c) The total pressure in the vessel at equilibrium is simply the sum of the equilibrium partial pressures:

Equilibrium

There are no entries in the table under C(s) because the reactant, being a solid, does not appear in the

equilibrium-constant expression. Substituting the equilibrium partial pressures of the other species into the

equilibrium-constant expression for the reaction gives

Multiplying through by the denominator gives a quadratic equation in x:

Solving this equation for x using the quadratic formula yields x = 6.14 atm. Hence, the equilibrium partial

pressures are and

(b) Part (a) shows that x = 6.14atm of H2O must react in order for the system to achieve equilibrium. We can

use the ideal-gas equation to convert this partial pressure into a mole amount.

Thus, 0.0697 mol of H2O and the same amount of C must react to achieve equilibrium. As a result, there

must be at least 0.0697 mol of C (0.836 g C) present among the reactants at the start of the reaction.

(c) The total pressure in the vessel at equilibrium is simply the sum of the equilibrium partial pressures:

Equilibrium

76.
SAMPLE INTEGRATIVE EXERCISE continued

(d) In discussing Le Châtelier’s principle, we saw that endothermic reactions exhibit an increase in Kp with

increasing temperature. Because the equilibrium constant for this reaction increases as temperature increases,

the reaction must be endothermic. From the enthalpies of formation given in Appendix C, we can verify our

prediction by calculating the enthalpy change for the reaction,

The positive sign for H° indicates that the reaction is endothermic.

(e) According to Le Châtelier’s principle, a decrease in the pressure causes a gaseous equilibrium to shift

toward the side of the equation with the greater number of moles of gas. In this case there are two moles of gas

on the product side and only one on the reactant side. Therefore, the pressure should be reduced to maximize

the yield of the CO and H2.

Equilibrium

(d) In discussing Le Châtelier’s principle, we saw that endothermic reactions exhibit an increase in Kp with

increasing temperature. Because the equilibrium constant for this reaction increases as temperature increases,

the reaction must be endothermic. From the enthalpies of formation given in Appendix C, we can verify our

prediction by calculating the enthalpy change for the reaction,

The positive sign for H° indicates that the reaction is endothermic.

(e) According to Le Châtelier’s principle, a decrease in the pressure causes a gaseous equilibrium to shift

toward the side of the equation with the greater number of moles of gas. In this case there are two moles of gas

on the product side and only one on the reactant side. Therefore, the pressure should be reduced to maximize

the yield of the CO and H2.

Equilibrium