Chapter 15Chapter 15

Overview

• Equilibrium Reactions

• Equilibrium Constants Kc & Kp

• Equilibrium Expression product/reactant

• Reaction Quotient Q

• Calculations

• Le Chatelier’s Principle

– disturbing the equilibrium

Overview, cont’d

• All reactions are reversible

• Dynamic Equilibrium

– When the rates of the forward and

– reverse reactions are equal

• Reactions do not “go to completion”

• Cannot use stoichiometric methods to calculate amount of products formed

A B

kf [A] = kr [B]

forward rate = reverse rate

[B ] = kf = constant

[A] kr

Equilibrium and Rates

• Ratio of Products to Reactant

– raised to each coefficient

• for example,

– N2 + 3H2 2NH3

– K = [NH3]2 [N2][H2]3

Equilibrium Constant Expression

N2 + 3H2 2NH3

Large K, more product

Product Favored

Small K, more reactant Reactant Favored

cont’d

aA + bB cC + dD

Kc = [C]c[D]d

[A]a[B]b

products

reactantsconcentration

In General:

Forms of Eq. Constant Expression

CO(g) + H2O(g) CO2(g) + H2(g)

Kc = [CO2][H2]

[CO][H2O]

4HCl(g) + O2(g) 2H2O(g) + 2Cl2(g)

Kc = [Cl2]2[H2O]2

[O2][HCl]4

Cont’d

2HI H2 + I2

Kf = [H2][I2]

[HI]2

H2 + I2 2HI

Kr = [HI]2 = 1

[H2][I2] Kf

Cont’d

2{ 2HI H2 + I2 }

Kf = [H2][I2]

[HI]2

4HI 2H2 + 2I2

K = Kf2 = [H2]2[I2]2

[HI]4

Example

N2O4 2NO2

Initial Initial Equilibrium Equilibrium KcN2O4 NO2 N2O4 NO2

0.0 0.02 0.0014 0.0172 0.211 0.0 0.03 0.0028 0.0243 0.211 0.0 0.04 0.0045 0.0310 0.213 0.02 0.0 0.0045 0.0310 0.213

Kc = [NO2]2 [N2O4]

generally unitless

N2 + 3H2 2NH3

Large K, more product

K > > 1 Product Favored

N2 + 3H2 2NH3

Small K, more reactant

Reactant Favored K < < 1

Review values of K:

Example:

N2 + 3H2 2NH3

Kc = 4.34 x 10-3 at 300°C = [NH3]2 [H2]2 [N2]

• What is K for reverse reaction?

• What is K for 2N2 + 6H2 4NH3 ?

• What is K for 4NH3 6H2 + 2N2 ?

Kc reverse = 230

Kc = 1.88 x 10-5

Kc = 5.31 x 104

Heterogeneous Equilibria:

• When pure solid or liquid is involved

– Pure solids & liquids do not appear in the equilibrium constant expression

• When H2O is a reactant or product and is the solvent

– H2O does not appear in the equilibrium constant expression

Examples:

• CaCO3(s) CaO(s) + CO2(g)

– conc. = mol = g/cm3 = density cm3 g/mol MM

• Kc = [CaO] [CO2] = (constant 1) [CO2] [CaCO3] (constant 2)

• Kc’ = Kc (constant 2) = [CO2] (constant 1)

both are constant

Multi-Step Equilibria

AgCl(s) Ag+(aq) + Cl-(aq) K1 = [Ag+][Cl-]

Ag+(aq) + 2NH3(aq) Ag(NH3)2

+(aq) K2 = [Ag(NH3)2

+] [Ag+][NH3]2

AgCl(s) + 2NH3(aq) Ag(NH3)2

+(aq) + Cl-(aq)

Ktot = K1 K2 = [Ag(NH3)2+][Cl-]

[NH3]2

Problem:1.00 mole of H2 & 1.00 mole of I2 are placed in a 1.0 L container at 520 K and allowed to come to equilibrium. Analysis reveals 0.12 mol of HI present at equilibrium. Calculate Kc. H2(g) + I2(g) 2HI(g)

initial 1.00 1.00 0 change -0.06 -0.06 +0.12

equil. 0.94 mol 0.94 mol +0.12

Kc = (0.12)2 = 1.6 x 10-2 (0.94)(0.94)

Conversion between Kp and Kc

• Kc

– Equilibrium constant using concentrations

• Kp

– Equilibrium constant using partial pressures

Kp = Kc (RT)n

P = n RT V

R = 0.0821 L atm/mol KT = Temperature in Kn = tot. mol product - tot. mol reactant

Problem:

For 2SO3(g) 2SO2(g) + O2(g)

Kc = 4.08 x 10-3 at 1000 K. Calculate Kp.

Kp = Kc(RT)n = 4.08 x 10-3 (0.0821 x 1000)1

Kp = 0.0335

Problem:

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

Kc = 0.105 at 472°C. Calculate Kp.

Kp = Kc(RT)n = 0.105 (0.0821 x 745)-2

Kp = 2.81 x 10-5

Applications of Eq. Constants

• Reaction Quotient

– Non-equilibrium concentrations used in the equilibrium constant expression

Q = K Reaction is at equilibrium

Q > K Reaction will shift left to equilibrium

Q < K Reaction will shift right to equilibrium

Problem:

Kc = 5.6 x 10-12 at 500 K for I2(g) 2I(g) [I2] = 0.020 M & [I] = 2.0 x 10-8 M. Is the reaction at equilibrium? Which direction will it shift to reach equilibrium?

Q = [I ]2 = (2.0 x 10-8)2 = 2.0 x 10-14 [I2] (0.020)

Q < K (2.0 x 10-14) < (5.6 x 10-12)

Reaction Shifts Right to get to equilibrium

not at equilibriumbecause Q K

Calculation of Eq. Concentrations

• use the stoichiometry of reaction

• initial concentration of all species

• change that occurs to all species

• equilibrium concentration of all species

• reaction will occur to reach the equilibrium point no matter the direction of reaction

Problem:

Cyclohexane, C6H12(g), can isomerize to form methylcyclopentane, C5H9CH3(g). The equilibrium constant at 25°C is 0.12. If the original amount was 0.045 mol cyclohexane in a 2.8 L flask, what are the concentrations at equilibrium?

C6H12 C5H9CH3

initial 0.045 mol 0

change -x +x

equil. 0.045 - x x

Cont’d

0.12 = (x) (0.045 - x)

Solve for x which is the equilibrium concentration of methylcyclopentane or the product

x = 4.8 x 10-3 mol C5H9CH3 in 2.8 L flask

[C5H9CH3] = 1.7 x 10-3 M

[C6H12] = 1.4 x 10-2 M

Problem:

For the reaction H2 + I2 2HI the Kc = 55.64. You start with 1.00 mol H2 and 1.00 mol I2 in a 0.500 L flask. Calculate the equilibrium concentrations of all species?

H2 + I2 2HI

initial 2.00 M 2.00 M 0

change -x -x +2x

equil. 2.00 - x 2.00 - x 2x

Cont’d

55.64 = (2x)2 (2.00 -x)2

(55.64)½ = (2x)2 (2.00 -x)2

x = 1.58 M

[HI] = 3.16 M [H2] = 0.42 M [I2] = 0.42 M

= 2x = 2.00 - x = 2.00 - x

½

perfect square

Problem:

For the reaction H2 + I2 2HI the Kc = 55.64. You start with 1.00 mol H2 and 0.50 mol I2 in a 0.500 L flask. Calculate the equilibrium concentrations of all species?

H2 + I2 2HI

initial 2.00 M 1.00 M 0

change -x -x +2x

equil. 2.00 - x 1.00 - x 2x

Cont’d

Kc = [HI]2

[H2][I2]

55.64 = (2x)2 (2.00 -x)(1.00 -x)

reduces to a quadratic equation:

x2 - 3.232 x + 2.155 = 0

not a perfect square

Quadratic Equation

x = - b ± b2 - 4ac 2a

for

ax2 + bx + c = 0

½

Cont’d

x = +3.232 ± 10.446 - 8.62 2

x = 1.616 ± 0.6755

[HI] = 1.88 M [H2] = 1.06 M [I2] = 0.060 M

= 2x = 2.00 - x = 1.00 - x

½

Le Chatelier’s Principle

• When a stress is applied to an equilibrium reaction, the equilibrium shifts to reduce the stress

• Types of Stress

– Addition or removal of reactant

– Addition or removal or product

– Increase or decrease of temperature

– Change in pressure or volume

2NOCl(g) Cl2(g) + 2NO(g) H = +77 kJ

temp.NOCl

reaction shifts

temp.NOCl

reaction shifts

pressure volu

me

reactionshifts

Cl2 NO

Cl2 NO

[C6H12] [C5H9CH3]

initial 1.4 x 10-2 + 1.0 x 10-2 M 1.7 x 10-3 M

change -x +x

equil. 2.4 x 10-2 - x 1.7 x 10-3 + x

0.12 = (1.7 x 10-3 + x) x = 1.05 x 10-3 M (2.4 x 10-2 - x)

[C6H12] = 0.023 M [C5H9CH3] = 0.0028 M

Addition of Reactant or Product

Changes in Temperature

• will change K

• for an endothermic reaction

– increasing T increases K

• for an exothermic reaction

– increasing T decreases K

A

B

En

erg

y

Reactions Path

withoutcatalyst

withcatalyst

rfrr

Ea (f)

Ea (r)

Effect of a Catalyst

2O3(g) 3O2(g) overall rxn

O3(g) O2(g) + O(g) fast

equil. rate1 = k1[O3]rate2 = k2[O2][O]

O(g) + O3(g) 2O2(g) slowrate3 = k3[O][O3]

rate 3 includes the conc. of an intermediate and the exptl. rate law will include only species that are present in measurable quantities

Reaction Mechanisms & Equilibria

k1

k3

k2

Substitution Method

at equilibrium k1[O3] = k2[O2][O]

rate3 =k3[O][O3] [O] = k1 [O3] k2 [O2]

rate3 = k3k1 [O3]2 or k2 [O2]

overall rate = k’ [O3]2 [O2]

substitute

Problem:

Derive the rate law for the following reaction given the mechanism step below:

OCl - (aq) + I -(aq) OI -

(aq) + Cl -(aq)

OCl - + H2O HOCl + OH - fast

I - + HOCl HOI + Cl - slow

HOI + OH - H2O + OI - fast

k1

k2 k3

k4

Cont’d

rate1 = k1 [OCl -][H2O] = rate 2 = k2 [HOCl][OH -]

[HOCl] = k1[OCl -][H2O] k2[OH -]

rate 3 = k3 [HOCl][I -]

rate 3 = k3k1[OCl -][H2O][I -]k2 [OH -]

overall rate = k’ [OCl -][I -] [OH -]

solvent