Chapter 13

Chapter 13Chapter 13Chemical

Equilibrium

The Equilibrium StateThe Equilibrium StateChemical Equilibrium: The state reached when the concentrations of reactants and products remain constant over time.

2NO2(g)N2O4(g)BrownColorless

The Equilibrium StateThe Equilibrium State2NO2(g)N2O4(g)

The Equilibrium Constant The Equilibrium Constant KKcc


cC + dDaA + bB

For a general reversible reaction:

Kc = [A]a[B]b

[C]c[D]d

Equilibrium constant expressionEquilibrium constant

Reactants

ProductsEquilibrium equation:

Double arrows show reversible reaction


For the following reaction:

2NO2(g)N2O4(g)

[N2O4]

[NO2]2

= 4.64 x 10-3 (at 25 °C)Kc =

Learning CheckLearning Check

= 4.63 x 10-3

0.0429

(0.0141)2

[N2O4]

[NO2]2

Kc = = 4.64 x 10-3

0.0337

(0.0125)2

Experiment 2 Experiment 5

=


Kc =

The equilibrium constant and the equilibrium constant expression are for the chemical equation as written.

N2(g) + 3H2(g)2NH3(g)[NH3]2

[N2][H2]3

= 1Kc

2NH3(g)N2(g) + 3H2(g)[N2][H2]3

[NH3]2

Kc =

´

4NH3(g)2N2(g) + 6H2(g)[N2]2[H2]6

[NH3]4

= Kc´´K c =2

ExamplesExamplesWrite the equilibrium constant, Kc

and K’c for the following reactions:CH4(g) + H2O(g) CO(g) + 3 H2(g)

The Equilibrium Constant The Equilibrium Constant KKpp

Chapter 13/11

2NO2(g)N2O4(g)

Kp =N2O4

PNO2

P2

P is the partial pressure of that component

The Equilibrium Constant The Equilibrium Constant KKpp

n

Kp = Kc(RT)n

0.082058K mol

L atmR is the gas constant,

T is the absolute temperature (Kelvin).

is the number of moles of gaseous products minus the number of moles of gaseous reactants.

Heterogeneous EquilibriaHeterogeneous Equilibria

CaO(s) + CO2(g)CaCO3(s)LimeLimestone

(1)

(1)[CO2]= [CO2][CaCO3][CaO][CO2] =

Pure solids and pure liquids are not included.

Kc =

Kc = [CO2] Kp = PCO2

ExamplesExamples In the industrial synthesis of hydrogen, mixtures of CO and

H2 are enriched in H2 by allowing the CO to react with steam. The chemical equation for this so-called water-gas shift reaction is

CO(g) + H2O(g) CO2(g) + H2(g)What is the value of Kp at 700K if the partial pressures in an

equilibrium mixture at 700K are 1.31 atm of CO, 10.0 atm of H2O, 6.12 atm of CO2, and 20.3 atm of H2?

ExamplesExamples Consider the following unbalanced reaction

(NH4)2S(s) 2NH3(g) + H2S(g)An equilibrium mixture of this mixture at a certain

temperature was found to have [NH3] = 0.278 M and [H2S] = 0.355 M. What is the value of the equilibrium constant (Kc) at this temperature?

ExamplesExamples Consider the following reaction and corresponding

value of Kc:H2(g) + I2(s) 2HI(g)

Kc = 6.2 x 102 at 25oCWhat is the value of Kp at this temperature?

Heterogeneous EquilibriaHeterogeneous Equilibria

Using the Equilibrium Using the Equilibrium ConstantConstant

Kc = 4.2 x 10-48

2H2(g) + O2(g)2H2O(g)

(at 500 K)

Kc = 2.4 x 1047

2H2O(g)2H2(g) + O2(g)

(at 500 K)Kc = 57.0

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

(at 500 K)


cC + dDaA + bB

Qc =[A]ta[B]t

b

[C]tc[D]t

dReaction quotient:

The reaction quotient, Qc, is defined in the same way as the equilibrium constant, Kc, except that the concentrations in Qc are not necessarily equilibrium values.


• If Qc = Kc no net reaction occurs.

• If Qc < Kc net reaction goes from left to right (reactants to products).

• If Qc > Kcnet reaction goes from right to left (products to reactants).

Finding Equilibrium concentrations Finding Equilibrium concentrations from Initial Concentrationsfrom Initial Concentrations

Steps to follow in calculating equilibirium concentrations from initial concentrationWrite a balance equation for the reactionMake an ICE (Initial, Change, Equilibrium) table,

involvesThe initial concentrationsThe change in concentration on going to equilibrium,

defined as xThe equilibrium concentration

Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x

Calculate the equilibrium concentrations form the calculated value of x

Check your answers


Sometimes you must use quadratic equation to solve for x, choose the mathematical solution that makes chemical sense

Quadratic equationax2 + bx + c = 0

Finding Equilibrium Finding Equilibrium concentrations from Initial concentrations from Initial ConcentrationsConcentrations

For Homogeneous Mixture

aA(g) bB(g) + cC(g)

Initial concentration (M) [A]initial [B]initial [C]initial

Change (M) - ax + bx + cx

Equilibrium (M) [A]initial – ax [B]initial – bx [C]initial - cx

Kc = [products]

[reactants]

Finding Equilibrium Finding Equilibrium concentrations from Initial concentrations from Initial ConcentrationsConcentrations

For Heterogenous Mixture

aA(g) bB(g) + cC(s)

Initial concentration (M) [A]initial [B]initial ……..

Change (M) - ax + bx ……..

Equilibrium (M) [A]initial – ax [B]initial + bx ……….

Kc = [products]

[reactants]


At 700 K, 0.500 mol of HI is added to a 2.00 L container and allowed to come to equilibrium. Calculate the equilibrium concentrations of H2, I2, and HI . Kc is 57.0 at 700 K.

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

ExamplesExamples Consider the following reaction

I2(g) + Cl2(g) 2ICl(g) Kp = 81.9 (at 25oC)

A reaction mixture at 25oC intially contains PI2 = 0.100 atm, PCl2 = 0.100 atm, and PICl = 0.100 atm. Find the equilibrium pressure of I2, Cl2 and ICl at this temperature.

In which direction does the reaction favored?

ExamplesExamples The value of Kc for the reaction is 3.0 x 10-2.

Determine the equilibrium concentration if the initial concentration of water is 8.75 M

C(s) + H2O(g) CO(g) + H2(g)

Le Châtelier’s PrincipleLe Châtelier’s PrincipleLe Châtelier’s Principle: If a stress is applied to a reaction mixture at equilibrium, net reaction occurs in the direction that relieves the stress.• The concentration of reactants or

products can be changed.

• The pressure and volume can be changed.

• The temperature can be changed.

Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: ConcentrationConcentration

2NH3(g)N2(g) + 3H2(g)

Altering an Equilibrium Altering an Equilibrium Mixture: ConcentrationMixture: Concentration


• the concentration stress of an added reactant or product is relieved by net reaction in the direction that consumes the added substance.

• the concentration stress of a removed reactant or product is relieved by net reaction in the direction that replenishes the removed substance.

In general, when an equilibrium is disturbed by the addition or removal of any reactant or product, Le Châtelier’s principle predicts that

Altering an Equilibrium Altering an Equilibrium Mixture: ConcentrationMixture: Concentration Add reactant – denominator in Qc expression

becomes larger◦ Qc < Kc

◦ To return to equilibrium, Qc must be increases◦ More product must be made => reaction shifts to

the right Remove reactant – denominator in Qc expression

becomes smaller◦ Qc > Kc

◦ To return to equilibrium, Qc must be decreases◦ Less product must be made => reaction shifts to

the left

Altering an Equilibrium Altering an Equilibrium Mixture: ConcentrationMixture: Concentration


2NH3(g)N2(g) + 3H2(g) at 700 K, Kc = 0.291

Since Qc < Kc, more reactants will be consumed and the net reaction will be from left to right.

(1.50)(3.00)3

(1.98)2

= 0.0968 < KcQc =[N2][H2]3

[NH3]2

=

An equilibrium mixture of 0.50 M N2, 3.00 M H2, and 1.98 M NH3 is disturbed by increasing the N2 concentration to 1.50 M.

ExamplesExamples The reaction of iron (III) oxide with carbon monoxide

occurs in a blast furnace when iron ore is reduced to iron metal:Fe2O3(s) + 3 CO(g) 2 Fe(l) + 3CO2(g)

Use Le Chatellier’s principle to predict the direction of net reaction when an equilibrium mixture is disturbed by:a. adding Fe2O3

b. Removing CO2

c. Removing CO; also account for the change using the reaction quotient Qc

ExamplesExamples Consider the following reaction at equilibrium

CO(g) + Cl2(g) COCl2(g)Predict whether the reaction will shift left, shift right, or

remain unchanged upon each of the following reaction mixturea. COCl2 is added to the reaction mixtureb. Cl2 is added to the reaction mixturec. COCl2 is removed from the reaction mixture: also account for the change using the reaction quotient Qc

Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: Pressure and VolumePressure and Volume

2NH3(g)N2(g) + 3H2(g)


• an increase in pressure by reducing the volume will bring about net reaction in the direction that decreases the number of moles of gas.

• a decrease in pressure by enlarging the volume will bring about net reaction in the direction that increases the number of moles of gas.

In general, when an equilibrium is disturbed by a change in volume which results in a corresponding change in pressure, Le Châtelier’s principle predicts that

Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: Pressure and VolumePressure and Volume If reactant side has more moles of gas

◦ Denominator will be larger Qc < Kc

To return to equilibrium, Qc must be increased Reaction shifts toward fewer moles of gas (to the

product) If product side has more moles of gas

◦ Numerator will be larger Qc > Kc

To return to equilibrium, Qc must be decreases Reaction shifts toward fewer moles of gas (to the

reactant)


Reaction involves no change in the number moles of gas◦ No effect on composition of equilibrium mixture

For heterogenous equilibrium mixture◦ Effect of pressure changes on solids and liquids

can be ignored Volume is nearly independent of pressure

Change in pressure due to addition of inert gas ◦ No change in the molar concentration of

reactants or products◦ No effect on composition

ExamplesExamples Consider the following reaction at chemical equilibrium

2 KClO3(s) 2 KCl(s) + 3O2(g)a. What is the effect of decreasing the volume of the reaction mixture?b. Increasing the volume of the reaction mixture?c. Adding inert gas at constant volume?

ExamplesExamples Does the number moles of products increases,

decreases or remain the same when each of the following equilibria is subjected to a increase in pressure by decreasing the volume?◦ PCl5(g) PCl3(g) + Cl2(g)◦ CaO(s) + CO2(g) CaCO3(s)◦ 3 Fe(s) + 4H2O(g) Fe3O4(s) + 4 H2(g)

Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: TemperatureTemperature

2NH3(g)N2(g) + 3H2(g) H° = -2043 kJ (exothermic )

As the temperature increases, the equilibrium shifts from products to reactants.

Altering an Equilibrium Altering an Equilibrium Mixture: TemperatureMixture: Temperature

Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: TemperatureTemperature

• the equilibrium constant for an exothermic reaction (negative H°) decreases as the temperature increases.• Contains more reactant than product• Kc decreases with increasing

temperature• the equilibrium constant for an

endothermic reaction (positive H°) increases as the temperature increases.• Contains more product than reactant• Kc increases with increasing

temperature

In general, when an equilibrium is disturbed by a change in temperature, Le Châtelier’s principle predicts that

ExamplesExamples The following reaction is endothermic

CaCO3(s) CaO(s) + CO2(g)a. What is the effect of increasing the temperature of the reaction mixture?b. Decreasing the temperature?

ExamplesExamples In the first step of Ostwald process for the synthesis

of nitric acid, ammonia is oxidized to nitric oxide by the reaction:2 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l)ΔHo = -905 kJ

How does the temperature amount of NO vary with an increases in temperature?

The Effect of a Catalyst on The Effect of a Catalyst on EquilibriumEquilibrium

The Effect of a Catalyst on The Effect of a Catalyst on EquilibriumEquilibrium

Catalyst increases the rate of a chemical reaction◦ Provide a new, lower energy pathway◦ Forward and reverse reactions pass through the

same transition state◦ Rate for forward and reverse reactions increase by

the same factor◦ Does not affect the composition of the equilibrium

mixture◦ Does not appear in the balance chemical equation◦ Can influence choice of optimum condition for a

reaction

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Chapter 13