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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