The Equilibrium State Chemical Equilibrium: The state reached when the concentrations of reactants...

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

=> dynamic state : rate forward = rate reverse

2NO2(g)N2O4(g)

c C + dDaA + bB

For a general reversible reaction:

Kc = [A]a[B]b

[C]c[D]d

Equilibrium constant expression

Equilibrium constantReactants

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 =

ExamplesWrite the equilibrium constant expression,

Kc, for each of the following reactions:

a. CH4(g) + H2O(g) CO(g) + 3 H2(g)

b. 2 NO(g) N2(g) + O2(g)

= 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

=1

Kc

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

ExamplesConsider the following equilibria

2CO2(g) + 2 CF4(g) 4COF2(g)

Write the equilibrium constant K’c expression

2NO2(g)N2O4(g)

Kp =N2O4

P

NO2 P

2

P is the partial pressure of that component

ExamplesIn 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?

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.

ExamplesConsider the following reaction and

corresponding value of Kc:

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

Kc = 6.2 x 102 at 25oC

What is the value of Kp at this temperature?

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

Examples

Consider the following unbalanced reaction(NH4)2S(s) NH3(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?

Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 13/19

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]t

a[B]tb

[C]tc[D]t

d

Reaction 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 = Kcno 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 from Initial ConcentrationsSteps to follow in calculating equilibirium

concentrations from initial concentrationWrite a balance equation for the reactionMake an ICE (Initial, Change, Equilibrium) table,

involves The initial concentrations The change in concentration on going to equilibrium, defined

as x The 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

Using the Equilibrium Constant

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

Quadratic equationax2 + bx + c = 0

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]

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)

ExamplesThe 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)

Examples

Consider the following reactionI2(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?

Chapter 13/30

Le 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.

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

Lecture NotesAlan D. Earhart

Southeast Community College • Lincoln, NE

Chapter 13Chemical Equilibrium

John E. McMurry • Robert C. Fay

C H E M I S T R YFifth Edition

• 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 Mixture: ConcentrationAdd reactant – denominator in Qc expression

becomes largerQc < Kc

To return to equilibrium, Qc must be increasesMore product must be made => reaction shifts

to the right

Altering an Equilibrium Mixture: ConcentrationRemove reactant – denominator in Qc

expression becomes smallerQc > Kc

To return to equilibrium, Qc must be decreasesLess product must be made => reaction shifts

to the left

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.

ExamplesConsider 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 mixture

a. COCl2 is added to the reaction mixture

b. Cl2 is added to the reaction mixture

c. COCl2 is removed from the reaction mixture: also account for the change using the reaction quotient Qc

ExamplesThe 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

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 expanding 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: Pressure and VolumeIf 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 gasNumerator will be larger

Qc > Kc

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

reactant)

Altering an Equilibrium Mixture: Pressure and VolumeReaction involves no change in the number moles

of gasNo effect on composition of equilibrium mixture

For heterogenous equilibrium mixtureEffect 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 productsNo effect on composition

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.

(1.00)(6.00)3

(3.96)2

= 0.0726 < 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 reducing the volume by a factor of 2.

ExamplesConsider the following reaction at chemical

equilibrium2 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?

ExamplesDoes 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)

2NH3(g)N2(g) + 3H2(g) H° = -2043 kJ

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

Copyright © 2008 Pearson Prentice Hall, Inc.

Chapter 13/48

• 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

ExamplesThe 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?

ExamplesIn 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 kJHow does the temperature amount of NO vary

with an increases in temperature?

The Effect of a Catalyst on EquilibriumCatalyst increases the rate of a chemical reaction

Provide a new, lower energy pathwayForward and reverse reactions pass through the

same transition stateRate for forward and reverse reactions increase by

the same factorDoes not affect the composition of the equilibrium

mixtureDoes not appear in the balance chemical equationCan influence choice of optimum condition for a

reaction