chapter 13 chemical equilibrium. 13.1 describing chemical equilibrium reactants product reactants ...

Chapter 13Chapter 13

Chemical Equilibrium

13.1 Describing Chemical Equilibrium13.1 Describing Chemical Equilibrium

Reactants Product

Reactants Products

When substances react, they eventually form a mixture of reactants and products in dynamic equilibrium

forward

reverse

For a general reversible reaction:

c C + d Da A + b B

Double arrows show reversible reaction

The Equilibrium StateThe Equilibrium State

Many reactions do not go to completion instead of reachingChemical Equilibrium: -The state reached when the concentrations of reactants and products remain constant over time.-The rate forward and reverse have become equal

2NO2(g)N2O4(g)

BrownColorless

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

13.2 The Equilibrium Constant 13.2 The Equilibrium Constant KKcc

cC + dDaA + bB

For a homogeneous reaction

Kc = [A]a[B]b

[C]c[D]d

Equilibrium constant expressionwhen concentrations are used

Equilibrium constantReactants

ProductsEquilibrium equation:

Coefficient

=

Kc is temperature dependent

13.3The Equilibrium Constant 13.3The Equilibrium Constant KKpp

When writing an equilibrium expression for a gaseous reaction in terms of partially pressure, we call it equilibrium constant, Kp

2NO2(g)N2O4(g)

Kp =N2O4

P

NO2 P

2

P is the partial pressure of that component

ExamplesExamples Write the equilibrium constant, Kc and K’c for the

following reactions:

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

b. 2 SO2(g) + O2(g) 2 SO3(g)

ExampleaExampleaWrite Kp and K’p for the following reactions:CO(g)+ 2H2(g)   CH3OH(g)

The Equilibrium Constant The Equilibrium Constant KKcc

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) ´´ K c = ??????

Relating the Equilibrium Relating the Equilibrium Constant Constant KKp p and Kand Kcc

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.

ExampleExample Phosphorous pentachloride dissociate on heating

PCl5(g) PCl3(g) + Cl2(g)

If Kc equals 3.26 x 10-2 at 191oC, what is Kp at this temperature?

ExampleExample Consider the following reaction

 2NO (g) + O2 (g) 2NO2 (g)

If Kp = 1.48 x 104 at 184 C, what is Kc?

13.4Heterogeneous Equilibria13.4Heterogeneous 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

Heterogeneous EquilibriaHeterogeneous Equilibria

ExampleExample

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?

13.5 Using the Equilibrium 13.5 Using the Equilibrium ConstantConstant

When we know the numerical value of the equilibrium constant, we can make certain judgments about the extent of the chemical reaction

Predicting the Direction of Predicting the Direction of ReactionReaction

For a chemical system whether in equilibrium or not, we can calculate the value given by the law of mass action

We called it reaction quotient, Qc

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.

Using the Equilibrium Using the Equilibrium ConstantConstant

• If Qc = Kc

no net reaction occurs.

• If Qc < Kc

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

net reaction goes from right to left (products to reactants).

The equilibrium between reactant A and product B: A B

Finding Equilibrium concentrations Finding Equilibrium concentrations from Initial Concentrationsfrom Initial Concentrations

Steps to follow in calculating equilibrium 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

Calculating Equilibrium Calculating Equilibrium ConcentrationsConcentrations

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

Quadratic equation

ax2 + 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]

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)

Using the Equilibrium ConstantUsing the Equilibrium Constant

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)

ExamplesExamplesConsider the following reaction

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

A reaction mixture at 25oC initially 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?

ExampleExampleA flask initially contained hydrogen sulfide at

a pressure of 5.00 atm at 313 K. When the reaction reached equilibrium, the partial pressure of sulfur vapor was found to be 0.15 atm. What is the  Kp for the reaction?

2H2S(g) 2H2(g) + S2(g)

13.6 Le Châtelier’s 13.6 Le Châtelier’s PrinciplePrincipleLe 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.

13.7Altering an Equilibrium 13.7Altering an Equilibrium Mixture: ConcentrationMixture: Concentration

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

Altering an Equilibrium Mixture: Altering an Equilibrium Mixture: ConcentrationConcentration

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 Mixture: Altering an Equilibrium Mixture: ConcentrationConcentration

• 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

Altering an Equilibrium Altering an Equilibrium Mixture: ConcentrationMixture: Concentration

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.

Which direction will the net reaction shift to re-establish the equilibrium?

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

ExampleExample 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

ExampleExample 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 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

13.8 Altering an Equilibrium 13.8 Altering an Equilibrium Mixture: Changes in Pressure Mixture: Changes in Pressure and Volumeand Volume

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

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. Which direction will the net reaction shift?

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

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

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

• 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)

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

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

ExampleExample 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

• 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

13.9 Altering an Equilibrium 13.9 Altering an Equilibrium Mixture: TemperatureMixture: Temperature

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

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

ExampleExample 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

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

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