Chemical Equilibrium

1
Chapter 15

• Chemical Equilibrium

2
Overview

• Equilibrium Reactions
• Equilibrium Constants Kc Kp
• Equilibrium Expression product/reactant
• Reaction Quotient Q
• Calculations
• Le Chateliers Principle
• disturbing the equilibrium

3
Overview, contd

• 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

4
Equilibrium and Rates

• A B
• kf A kr B
• forward rate reverse rate
• B kf constant
• A kr

5
Equilibrium Constant Expression

• Ratio of Products to Reactant
• raised to each coefficient
• for example,
• N2 3H2 2NH3
• K NH32 N2H23

6
contd

• N2 3H2 2NH3
• Large K, more product
• Product Favored
• Small K, more reactant
• Reactant Favored

7
In General

• aA bB cC dD
• Kc CcDd AaBb

products
reactants
concentration
8
Forms of Eq. Constant Expression

• CO(g) H2O(g) CO2(g)
  H2(g)
• Kc CO2H2
• COH2O
• 4HCl(g) O2(g) 2H2O(g)
  2Cl2(g)
• Kc Cl22H2O2
• O2HCl4

9
Contd

• 2HI H2 I2
• Kf H2I2
• HI2
• H2 I2 2HI
• Kr HI2 1
• H2I2 Kf

10
Contd

• 2 2HI H2 I2 Kf
  H2I2
• HI2
• 4HI 2H2 2I2
• K Kf2 H22I22
• HI4

11
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 NO22 N2O4

generally unitless
12
Review values of K

• N2 3H2 2NH3
• Large K, more product
• K gt gt 1 Product Favored
• N2 3H2 2NH3
• Small K, more reactant
• Reactant Favored K lt lt 1

13
Example

• N2 3H2 2NH3
• Kc 4.34 x 10-3 at 300C NH32
  H22 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
14
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

15
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)

16
Multi-Step Equilibria

• AgCl(s) Ag(aq) Cl-(aq) K1
  AgCl-
• Ag(aq) 2NH3(aq) Ag(NH3)2(aq)
  K2 Ag(NH3)2 AgNH32
• AgCl(s) 2NH3(aq)
  Ag(NH3)2(aq) Cl-(aq)
• Ktot K1 K2 Ag(NH3)2Cl-
  NH32

17
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)

18
Conversion between Kp and Kc

• Kc
• Equilibrium constant using concentrations
• Kp
• Equilibrium constant using partial pressures
• Kp Kc (RT)Dn P n RT
  V
• R 0.0821 L atm/mol K T Temperature
  in K Dn tot. mol product - tot. mol reactant

19
Problem

• For 2SO3(g) 2SO2(g) O2(g)
• Kc 4.08 x 10-3 at 1000 K. Calculate Kp.
• Kp Kc(RT)Dn 4.08 x 10-3 (0.0821 x 1000)1
  
• Kp 0.0335

20
Problem

• For 3H2(g) N2(g) 2NH3(g)
• Kc 0.105 at 472C. Calculate Kp.
• Kp Kc(RT)Dn 0.105 (0.0821 x 745)-2
• Kp 2.81 x 10-5

21
Applications of Eq. Constants

• Reaction Quotient
• Non-equilibrium concentrations used in the
  equilibrium constant expression
• Q K Reaction is at equilibrium
• Q gt K Reaction will shift left to equilibrium
• Q lt K Reaction will shift right to equilibrium

22
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 lt K (2.0 x 10-14) lt (5.6
  x 10-12)
• Reaction Shifts Right to get to equilibrium

not at equilibriumbecause Q ? K
23
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

24
Problem

• Cyclohexane, C6H12(g), can isomerize to form
  methylcyclopentane, C5H9CH3(g). The equilibrium
  constant at 25C 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

25
Contd

• 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

26
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

27
Contd
perfect square

• 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
28
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

29
Contd

• Kc HI2 H2I2
• 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
30
Quadratic Equation

• x - b b2 - 4ac 2a
• for
• ax2 bx c 0

½
31
Contd
½

• 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

32
Le Chateliers 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

33
2NOCl(g) Cl2(g) 2NO(g)
DH 77 kJ
reaction shifts
temp.NOCl
Cl2 NO
temp.NOCl
reaction shifts
Cl2 NO
pressure
reactionshifts
volume
34
Addition of Reactant or Product

• 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

35
Changes in Temperature

• will change K
• for an endothermic reaction
• increasing T increases K
• for an exothermic reaction
• increasing T decreases K

36
Effect of a Catalyst
withoutcatalyst
rf
rr
Ea (f)
Ea (r)
Energy
A
B
Reactions Path
37
Reaction Mechanisms Equilibria

• 2O3(g) 3O2(g) overall rxn
• O3(g) O2(g) O(g) fast
  equil. rate1 k1O3 rate2
  k2O2O
• O(g) O3(g) 2O2(g)
  slow rate3 k3OO3
• rate 3 includes the conc. of an intermediate and
  the exptl. rate law will include only species
  that are present in measurable quantities

k1
k2
k3
38
Substitution Method

• at equilibrium k1O3 k2O2O
• rate3 k3OO3 O k1 O3
  k2 O2
• rate3 k3k1 O32 or k2 O2
• overall rate k O32 O2

substitute
39
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
40
Contd

• rate1 k1 OCl -H2O rate 2 k2
  HOClOH -
• HOCl k1OCl -H2O k2OH
  -
• rate 3 k3 HOClI -
• rate 3 k3k1OCl -H2OI - k2 OH
  -
• overall rate k OCl -I - OH -

solvent