chemistry: atoms first julia burdge & jason overby copyright (c) the mcgraw-hill companies, inc....

Chemistry: Atoms FirstJulia Burdge & Jason Overby

Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 15

Chemical Equilibrium

Kent L. McCorkle

Cosumnes River College

Sacramento, CA

Chemical EquilibriumChemical Equilibrium15

15.1 The Concept of Equilibrium15.2 The Equilibrium Constant

Calculating Equilibrium ConstantsMagnitude of the Equilibrium Constant

15.3 Equilibrium ExpressionsHeterogeneous EquilibriaManipulating Equilibrium ExpressionsGaseous Equilibria

15.4 Using Equilibrium Expressions to Solve ProblemsPredicting the Direction of a ReactionCalculating Equilibrium Concentrations

15.5 Factors that Affect Chemical EquilibriumAddition or Removal of a SubstanceChanges in Volume and PressureChanges in TemperatureCatalysis

The Concept of EquilibriumThe Concept of Equilibrium

The decomposition of N2O4 is a reversible process, meaning the products of the reaction can react to form reactants.

The system is in equilibrium when the rates of the forward reaction and the reverse reaction are the same.

rate forward = kf[N2O4] and rate reverse = kr[NO2]2

15.1

N2O4(g) ⇌ 2NO2(g)


Starting with N2O4


Starting with NO2


Some important things to remember about equilibrium are:

Equilibrium is a dynamic state—both the forward and reverse reactions continue to occur, although there is no net change in reactant and product concentration over time.

At equilibrium, the rates of the forward and reverse reactions are equal.

Equilibrium can be established starting with only reactants, with only products, or with any mixture of reactants and products.

The Equilibrium ConstantThe Equilibrium Constant

rate forward = rate reverse kf[N2O4] eq = kr[NO2]2eq

The subscript “eq” denotes a concentration at equilibrium.

Rearranging

The ratio of two constants (kf/kr) is also a constant:

15.2

equilibrium expression

N2O4(g) 2NO⇌ 2(g)

2

k

k

2 eqf

r 2 4 eq

NO

NO

2

K2 eq

c2 4 eq

NO

NOequilibrium constant


Note the relationship between the equilibrium constant and the balanced chemical equation:

2

K2 eq

c2 4 eq

NO

NON2O4(g) ⇌ 2NO2(g)


The reaction quotient (Qc ) is a fraction with product concentrations in the numerator and reactant concentrations in the denominator.

Each concentration is raised to a power equal to the corresponding stoichiometric coefficient in the balanced chemical equation.

aA + bB ⇌ cC + dD

This law of mass action applies to not only elementary reactions, but also to more complex reactions.

c d

a bQ Kc c

C D

A B(at equilibrium)

Worked Example 15.1

Strategy Use the law of mass action to write the equilibrium expression and plug in the equilibrium concentrations of all three species to evaluate Kc.

Carbonyl chloride (COCl2), also called phosgene, is a highly poisonous gas that

was used on the battlefield of World War I. It is produced by the reaction of carbon monoxide with chlorine gas:

CO(g) + Cl2(g) COCl⇌ 2(g)

In an experiment conducted at 74°C, the equilibrium concentrations of the

species involved in the reaction were as follows: [CO] = 1.2×10-2 M, [Cl2] =

0.054 M, and [COCl2] = 0.14 M. (a) Write the equilibrium expression, and (b)

determine the value of the equilibrium constant for this reaction at 74°C.

Solution (a) Kc =

(b) Kc = = 216 or 2.2×102

[COCl2][CO][Cl2]

(0.14)(1.2×10-2)(0.054)

Think About It When putting the equilibrium concentrations into the equilibrium expression, we leave out the units. It is common practice to express equilibrium constants without units.

Worked Example 15.2

Strategy Use the law of mass action to write reaction quotients.

Write reaction quotients for the following reactions:

(a) N2(g) + 3H2(g) 2NH⇌ 3(g)

(b) H2(g) + I2(g) 2HI(⇌ g)

(c) Ag+(aq) + 2NH3(aq) Ag(NH⇌ 3)2+(aq)

(d) 2O3(g) 3O⇌ 2(g)

(e) Cd2+(aq) + 4Br-(aq) CdBr⇌ 42-(aq)

(f) 2NO(g) + O2(g) 2NO⇌ 2(g)

Solution (a) Qc =

(d) Qc =

[NH3]2

[N2][H2]3(b) Qc =

[HI]2

[H2][I2](c) Qc =

[Ag(NH3)2+]

[Ag+][NH3]2

[O2]3

[O3]2(e) Qc =

[CdBr42-]

[Cd2+][Br-]4(f) Qc =

[NO2]2

[NO]2[O2]

Think About It With practice, writing reaction quotients becomes second nature. Without sufficient practice, it will seem inordinately difficult. It is important that you become proficient at this. It is very often the first step in solving equilibrium problems.


At any point during the progress of a reaction:

c d

a bQc

C D

A B

aA + bB ⇌ cC + dD


The value of the reaction quotient, Q, changes as the reaction progresses

N2O4(g) 2NO⇌ 2(g)


The equilibrium constant gives the extent a reaction will proceed at a particular temperature.

Three outcomes are possible:

1) The reaction will go essentially to completion and the equilibrium mixture will consist predominately of products.

Ag+(aq) + 2NH3(aq) Ag(NH⇌ 3)2+(g)Kc = 1.5 x 107 (at 25°C)

Large Kc, product favored


The equilibrium constant gives the extent a reaction will proceed at a particular temperature.

Three outcomes are possible:

2) The reaction will not occur to any significant degree, and the equilibrium mixture will consist predominantly of reactant.

N2(g) + O2(g) 2NO(⇌ g) Kc = 4.3 x 10–25 (at 25°C)

3) The reaction will proceed a significant degree but will not go to completion, and the equilibrium mixture will contain comparable amounts of both reactants and products.

Small Kc, reactant favored

Equilibrium ExpressionsEquilibrium Expressions

When the species in a reversible chemical reaction are not all in the same phase, the equilibrium is heterogeneous.

Only gaseous species and aqueous species appear in equilibrium expressions, pure solids and pure liquids do not.

15.3

CO2(g) + C(s) 2CO(⇌ g)

2Fe(s) + 3H2O(l) Fe⇌ 2O3(s) + 2H2(g)

2

Kc2

CO

CO

2Kc 2H

Worked Example 15.3

Strategy Use the law of mass action to write the equilibrium expressions for each reaction. Only gases and aqueous species appear in the expression.

Write equilibrium expressions for each of the following reactions:

(a) CaCO3(s) CaO(⇌ s) + CO2(g)

(b) Hg(l) + Hg2+(aq) Hg⇌ 22+(aq)

(c) 2Fe(s) + 3H2O(l) Fe⇌ 2O3(s) + 2H2(g)

(d) O2(g) + 2H2(g) 2H⇌ 2O(l)

Solution (a) Kc = [CO2] (b) Kc = [Hg2

2+][Hg2+]

(d) Kc = 1

[O2][H2]2(c) Kc = [H2]2

Think About It Like writing equilibrium expressions for homogeneous equilibria, writing equilibrium expressions for heterogeneous equilibria becomes second nature if you practice. The importance of developing this skill now cannot be overstated. Your ability to understand the principles and to solve many of the problems in and Chapters 16 to 19 depends on your ability to write equilibrium expressions correctly and easily.

Equilibrium ExpressionsEquilibrium Expressions

When a reversible chemical equation is manipulated, it is also necessary to make appropriate changes in the equilibrium expression and the equilibrium constant.

Worked Example 15.4

Strategy Begin by writing the equilibrium expressions for the reactions that are given. Then, determine the relationship of each equation’s equilibrium expression to the equilibrium expression of the original equations, and make the corresponding change to the equilibrium constant for each.

Kc = and Kc =

The following reactions have the indicated equilibrium constants at 100°C:

(1) 2NOBr(g) 2NO(⇌ g) + Br2(g) Kc = 0.014

(2) Br2(g) + Cl2(g) 2BrCl(⇌ g) Kc = 7.2

Determine the value of Kc for the following reactions at 100°C:

(a) 2NO(g) + Br2(g) 2NOBr(⇌ g) (d) 2NOBr(g) + Cl2(g) 2NO(⇌ g) + 2BrCl(g)

(b) 4NOBr(g) 4NO(⇌ g) + 2Br2(g) (e) NO(g) + BrCl(g) NOBr(⇌ g) + Cl2(g)

(c) NOBr(g) NO(⇌ g) + Br2(g)

12

12

[NO]2[Br2][NOBr]2

[BrCl]2

[Br2][Cl2]

Worked Example 15.4 (cont.)

Solution (a) This equation is the reverse of original equation 1. Its equilibrium expression is the reciprocal of that for the original equation:

Kc = = 1/0.014 = 71

(b) This is original equation 1 multiplied by a factor of 2. Its equilibrium expression is the original expression squared:

Kc = = (0.014)2 = 2.0×10-4

(c) This is the original equation 1 multiplied by . Its equilibrium expression is the square root of the original

Kc = or Kc = = (0.014)1/2 = 0.12

[NOBr]2

[NO]2[Br2]

[NO]2[Br2][NOBr]2

2

12

[NO]2[Br2][NOBr]2

1/2

2

22

][

][][

NOBr

BrNO


Solution (d) This is the sum of the original equations 1 and 2. Its equilibrium expression is the product of the two individual expressions:

Kc = = (0.014)(7.2) = 0.10

(e) Probably the simplest way to analyze this reaction is to recognize that it is the reverse of the reaction in part (d), multiplied by . Its equilibrium expression is the square root of the reciprocal of the expression in part (d):

Kc = = (1/0.10)1/2 = 3.2

[NO]2[BrCl]2

[NOBr]2[Cl2]

[NO]2[BrCl]2

[NOBr]2[Cl2]

1/2

12


Think About It The magnitude of an equilibrium constant reveals whether products or reactants are favored, so the reciprocal relationship between Kc values of forward and reverse reactions should make sense. A very large Kc value means that products are favored. In the reaction of hydrogen ion and hydroxide ion to form water, the value of Kc is very large, indicating that the product, water, is favored.

H+(aq) + OH-(aq) H⇌ 2O(l) Kc = 1.0×1014 (at 25°C)

Simply writing the equation backward doesn’t change the fact that water is the predominate species. In the reverse reaction, therefore, the favored species is on the reactant side:

H2O(l) H⇌ +(aq) + OH-(aq) Kc = 1.0×10-14 (at 25°C)

As a result, the magnitude of Kc should correspond to reactants being favored; that is, it should be very small.

Gaseous EquilibriaGaseous Equilibria

When an equilibrium expression contains only gases, we can write an alternate form of the expression in which the concentrations of gases are expressed as partial pressures (atm). Thus, for the equilibrium

we can write as

Kc = or KP =

The relationship between Kc and KP can be expressed as

where Δn = moles of gaseous products – moles of gaseous reactants.

N2O4(g) 2NO⇌ 2(g)

[NO2]2

[N2O4](PNO2)2

PN2O4

KP = Kc[(0.08206 L∙atm/K∙mol)×T]Δn

Worked Example 15.5

Strategy Write equilibrium expressions for each equation, expressing the concentrations of the gases in partial pressures.

Write KP expressions for (a) PCl3(g) + Cl2(g) PCl⇌ 5(g),

(b) O2(g) + 2H2(g) 2H⇌ 2O(l), and (c) F2(g) + H2(g) 2HF(⇌ g).

Solution (a) All the species in this equation are gases, so they will all appear in the KP expression.

KP =

(b) Only the reactants are gases. KP =

(c) All species are gases. KP =

(PPCl5)(PPCl3)(PCl2) 1

(PO2)(PH2)2

(PHF)2

(PF2)(PH2)

Think About It It isn’t necessary for every species in the reaction to be a gas–only those species that appear in the equilibrium expression.

Worked Example 15.6

Strategy Use KP = Kc[(0.08206 L∙atm/K∙mol)×T]Δn. Be sure to convert temperature in degrees Celsius to kelvins. Using Δn = moles of gaseous products – moles of gaseous reactants, Δn = 2(NO2) – 1(N2O4) = 1. T = 298K.

The equilibrium constant, Kc, for the reaction

N2O4(g) 2NO⇌ 2(g)

is 4.63×10-3 at 25°C. What is the value of KP at this temperature.

Solution

KP = Kc ×T

= (4.63×10-3)(0.08206 × 298)

= 0.113

0.08206 L∙atmK∙mol

Think About It Note that we have essentially disregarded the units of R and T so that the resulting equilibrium constant, KP, is unitless. Equilibrium constants commonly are treated as unitless quantities.

Using Equilibrium Expressions to Solve ProblemsUsing Equilibrium Expressions to Solve Problems

The equilibrium expression may be used to predict the direction of a reaction and to calculate equilibrium concentrations.

Predictions are made based on comparisons between Qc and Kc.

There are three possibilities:1) Q < K The ratio of initial concentrations of products to reactants is too small. To reach equilibrium, reactants must be converted to products. The system proceeds in the forward direction.

2) Q = K The initial concentrations are equilibrium concentrations. The system is at equilibrium.

3) Q > K The ratio of initial concentrations of products to reactants is too large. To reach equilibrium products must be converted to reactants. The system proceeds in the reverse direction.

15.4

Worked Example 15.7

Strategy Use the initial concentrations to calculate Qc, and then compare Qc with Kc.

Qc = = = 0.61

At 375°C, the equilibrium constant for the reaction

N2(g) + 3H2(g) 2NH⇌ 3(g)

is 1.2. At the state of a reaction, the concentrations of N2, H2, and NH3 are

0.071 M, 9.2×10-3 M, and 1.83×10-4 M, respectively. Determine whether the system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium.

[NH3]i2

[N2]i[H2]i3

(1.83×10-4)2

(0.071)(9.2×10-3)3

Strategy The calculated value of Qc is less than Kc. Therefore, the reaction is not at equilibrium and must proceed to the right to establish equilibrium.

Think About It In proceeding to the right, a reaction consumes reactants and produces more products. This increases the numerator in the reaction quotient and decreases the denominator. The result is an increase in Qc until it is equal to Kc, at which point equilibrium will be established.

Calculating Equilibrium ConcentrationsCalculating Equilibrium Concentrations

Equilibrium concentrations can be calculated from initial concentrations if the equilibrium constant is known.

C

H

C

H

C

H

C

HKc = 24.0 (200°C)

Initial concentration (M) 0.850 0

Change in concentration (M)

Equilibrium concentration (M)

–x

0.850 – x

+x

x

cis-Stilbene trans-Stilbene

cis-Stilbene trans-Stilbene ⇌





–x

0.850 – x

+x

x


Use the equilibrium concentrations, defined in terms of x, in the equilibrium expression:

trans

Kcisc

-stilbene

-stilbene

24.00.850

x

x

x = 0.816 M (Represents the change in initial concentrations)





–0.816

0.850 – 0.816

+0.816

0.816


Calculate the equilibrium concentrations of cis- and trans-stilbene:

[cis-stilbene] = (0.850 – x) M = (0.850 – 0.816) M = 0.034 M

[trans-stilbene] = x M = 0.816 M

Worked Example 15.8

Strategy Insert the starting concentrations that we know into the equilibrium table:

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide,H2(g) + I2(g) 2HI(⇌ g)

is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and I2?

Initial concentration (M) 0.240 0.240 0



H2 I2 2HI+ ⇌


Solution We define the change in concentration of one of the reactants as x. Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (2x). Combining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations.

Initial concentration (M) 0.240 0.240 0

Change in concentration (M) –x –x +2x

Equilibrium concentration (M) 0.240 – x 0.240 – x 2x

H2 I2 2HI+ ⇌


Solution Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x.

Kc =

54.3 = =

=

x = 0.189

Use the calculated value of x, we can determine the equilibrium concentration of each species as follows:

[H2] = (0.240 – x) M = 0.051 M[I2] = (0.240 – x) M = 0.051 M[HI] = 2x = 0.378 M

[HI]2

[H2][I2]

(2x)2

(0.240 – x)(0.240 – x)(2x)2

(0.240 – x)2

3.542x

0.240 – x

Think About It Always check your answer by inserting the calculated concentrations into the equilibrium expression:

The small difference between the calculated Kc and the one given in the problem statement is due to rounding.

= 54.9 ≈ Kc

[HI]2

[H2][I2](0.378)2

(0.051)2=

Worked Example 15.9

Strategy Using the initial concentrations, calculate the reaction quotient, Qc, and compare it to the value of Kc (given in the problem statement of Worked Example 15.8) to determine which direction the reaction will proceed to establish equilibrium. Then, construct an equilibrium table to determine the equilibrium concentrations.

For the same reaction and temperature as in Worked Example 15.8, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H2] = 0.00623 M, [I2] = 0.00414 M, [HI] = 0.0424 M.

[HI]2

[H2][I2](0.0424)2

(0.00623)(0.00414)= = 69.7


Strategy Therefore, Qc > Kc, so the system will proceed to the left (reverse) to reach equilibrium. The equilibrium table is

Initial concentration (M) 0.00623 0.00414 0.0424



H2 I2 2HI+ ⇌


Solution Because we know the reaction must proceed from right to left, we know that the concentration of HI will decrease and the concentrations of H2 and I2 will increase. Therefore, the table should be filled in as follows:

Next, we insert these expressions for the equilibrium concentration into the equilibrium expression and solve for x.

Initial concentration (M) 0.00623 0.00414 0.0424

Change in concentration (M) +x +x –2x

Equilibrium concentration (M) 0.00623 + x 0.00414 + x 0.0424 – 2x

H2 I2 2HI+ ⇌

[HI]2

[H2][I2]Kc =

(0.0424 – 2x)2

(0.00623 + x)(0.00414 + x)54.3 =


Solution It isn’t possible to solve this equation the way we did in Worked Example 15.8 (by taking the square root of both sides) because the concentrations of H2 and I2 are unequal. Instead, we have to carry out the multiplications.

54.3(2.58×10-5 + 1.04×10-2x + x2) = 1.80×10-3 – 1.70×10-1x + 4x2

Collecting terms we get50.3x2 + 0.735x – 4.00×10-4 = 0

This is a quadratic of the form ax2 + bx + c = 0. The solution for the quadratic equation [Appendix 1] is

x =

Here we have a = 50.3, b = 0.735, and c = -4.00×10-4, so

x =

a

acbb

2

42

)3.50(2

)1000.4)(3.50(4)735.0(735.0 42


Solution x = 5.25×10-4 or x = –0.0151

Only the first of these values, 5.25×10-4, makes sense because concentration cannot be a negative number. Using the calculated value of x, we can determine the equilibrium concentration of each species as follows:

[H2] = (0.00623 + x) M = 0.00676 M[I2] = (0.00414 + x) M = 0.00467 M[HI] = (0.0424 – 2x) M = 0.0414 M

Think About It Checking this result gives

[HI]2

[H2][I2](0.0414)2

(0.00676)(0.00467)= = 54.3Kc =

Worked Example 15.10

Strategy Construct an equilibrium table to determine the equilibrium partial pressures.

A mixture of 5.75 atm of H2 and 5.75 atm of I2 is contained in a 1.0-L vessel at 430°C. The equilibrium constant (KP) for the reaction

H2(g) + I2(g) 2HI(⇌ g)at this temperature is 54.3. Determine the equilibrium partial pressures of H2, I2, and HI.

Initial partial pressure (atm) 5.75 5.75 0

Change in partial pressure (atm) –x –x +2x

Equilibrium partial pressure (atm) 5.75 – x 5.75 – x 2x

H2 I2 2HI+ ⇌


Solution Setting the equilibrium expression equal to KP,

54.3 =

Taking the square root of both sides of the equation gives

=

The equilibrium partial pressures are PH2 = PI2 = 5.75 – 4.52 = 1.23 atm, and PHI = 9.04 atm.

(2x)2

(5.75 – x)2

3.542x

5.75 – x2x

5.75 – x7.369(5.75 – x) = 2x

42.37 – 7.369x = 2x

42.37 = 9.369x

x = 4.52

7.369 =

Think About It Plugging the calculated partial pressures into the equilibrium expression gives

The small difference between this result and the equilibrium constant given in the problem is due to rounding.

= 54.0(PHI)2

(PH2)(PI2)(9.04)2

(1.23)2=

Factors That Affect Chemical EquilibriumFactors That Affect Chemical Equilibrium

Le Châtelier’s principle states that when a stress is applied to a system at equilibrium, the system, will respond by shifting in the direction that minimizes the effect of the stress.

Stress refers to any of the following:

The addition of a reactant or product

The removal of a reactant or product

A change in volume of the system, resulting in a change in concentration or partial pressure of the reactants and products

A change in temperature

15.5


Consider the Haber process at 700 K:

N2(g) + 3H2(g) 2NH⇌ 3(g)

At equilibrium:

[N2] = 2.05 M [H2] = 1.56 M [NH3] = 1.52 M

Applying stress by the addition of N2 to give the following concentrations:

[N2] = 3.51 M [H2] = 1.56 M [NH3] = 1.52 M

3c

2

NH

N H

..

. .K

2 2

3 3

2

1 520 297

2 05 1 56

3c c

2

NH

N H

..

. .Q K

2 2

3 3

2

1 520 173

3 51 1 56

The reaction shifts to the right.

N2(g) + 3H2(g) 2NH⇌ 3(g)


N2(g) + 3H2(g) 2NH⇌ 3(g)


Addition of a reactant or removal of a product will cause an equilibrium to shift to the right.

Addition of a product or removal of a reactant will cause an equilibrium to shift to the left.


Strategy Use Le Châtelier’s principle to predict the direction of shift in each case. Remember that the position of the equilibrium is only changed by the addition or removal of a species that appears in the reaction quotient expression.

Because sulfur is a solid, it does not appear in the expression.

Hydrogen sulfide (H2S) is a contaminant commonly found in natural gas. It is removed by reaction with oxygen to produce elemental sulfur.

2H2S(g) + O2(g) 2S(⇌ s) + 2H2O(g)For each of the following scenarios, determine whether the equilibrium will shift to the right, shift to the left, or neither: (a) addition of O2(g), (b) removal of H2S(g), (c) removal of H2O(g), and (d) addition of S(s).

[H2O]2

[H2S][O2]Qc =


Solution

Changes in concentration of any of the other species will cause a change in the equilibrium position. Addition of a reactant or removal of a product that appears in the expression Qc will shift the equilibrium to the right:

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

Removal of a reactant or addition of a product that appears in the expression Qc will shift the equilibrium to the left:

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

(a) Shift to the right(d) No change

[H2O]2

[H2S][O2]Qc =

addition addition

removal

removal removal

addition

(b) Shift to the left (c) Shift to the right

Think About It In each case, analyze the effect the change will have on the value of Qc. In part (a), for example, O2 is added, so its concentration increases. Looking at the reaction quotient expression, we can see that a larger concentration of oxygen corresponds to a larger overall denominator – giving the overall fraction a smaller value. Thus, Q will temporarily be smaller that K and the reaction will have to shift to the right, consuming some of the added O2 (along with some of the H2S in the mixture) to reestablish equilibrium.


When volume is decreased, the equilibrium is driven toward the side with the smallest number of moles of gas.

N2O4(g) 2NO⇌ 2(g)

Equilibrium mixture:

[N2O4] = 0.643 M

[NO2] = 0.0547 M

2c

2 4

NO

NO

..

.K

2 2

30 05474 65 10

0 643

2c c

2 4

NO

NO

..

.Q K

2 2

30 10949 31 10

1 286

Volume decreases by half, concentrations are initially

doubled:

[N2O4] = 1.286 M

[NO2] = 0.1094 M

The reaction shifts to the left.N2O4(g) 2NO⇌ 2(g)


Strategy Determine which direction minimized the number of moles of gas in the reaction. Count only moles of gas.

For each reaction, predict in what direction the equilibrium will shift when the volume of the reaction vessel is decreased.

(a) PCl5(g) PCl⇌ 3(g) + Cl2(g)

(b) 2PbS(s) + 3O2(g) 2PbO(⇌ s) + 2SO2(g)

(c) H2(g) + I2(g) 2HI(⇌ g)

Solution (a) We have 1 mole of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the left.

(b) 3 moles of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the right.

(c) 2 moles of gas on each side, so no shift.

Think About It When there is a difference in the number of moles of gas, changing the volume of the reaction vessel will change the concentrations of reactant(s) and product(s)–but the system will remain at equilibrium. (Q will remain equal to K.)


Changes in volume and concentration do not change the value of the equilibrium constant.

A change in temperature can alter the value of the equilibrium constant.

Heat + N2O4(g) 2NO⇌ 2(g) ΔH° = 58.0 kJ/mol

Because the processes is endothermic, adding heat

shifts the equilibrium toward products


For any endothermic reaction, heat is a reactant:

Adding heat shifts the reaction towards products, Kc increases

Removing heat shifts the reaction towards reactants, Kc decreases.

For any exothermic reaction, heat is a product:

Adding heat shifts the reaction towards reactants, Kc decreases

Removing heat shifts the reaction towards products, Kc increases

heat + reactants products⇌ ΔH° > 0 kJ/mol

reactants products + heat⇌ ΔH° < 0 kJ/mol


CoCl42- + 6H2O Co(H⇌ 2O)6

2+ + 4Cl- + heat

blue pink


Catalysts act by reducing the activation energy of a reaction (Section 14.8) which occurs to the same extent for both the forward and reverse reactions.

As a result the addition of a catalyst reduces the time required to reach equilibrium but has no effect on equilibrium constants of the position of equilibrium.

Key PointsKey Points15

The Concept of EquilibriumThe Equilibrium Constant

Calculating Equilibrium ConstantsMagnitude of the Equilibrium Constant

Equilibrium ExpressionsHeterogeneous EquilibriaManipulating Equilibrium Expressions

Using Equilibrium Expressions to Solve ProblemsPredicting the Direction of a ReactionCalculating Equilibrium Concentrations

Factors that Affect Chemical EquilibriumAddition or Removal of a SubstanceChanges in Volume and PressureChanges in Temperature

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