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Chemical Equilibrium
Chemical Equilibrium
Rate of forward and reverse reactions are
equal
Reactant and product concentrations no
longer change
Represent a reaction at dynamic
equilibrium with a double arrow
A B
Haber Process:
N2(g) + 3H2(g)  2NH3(g)
Equilibrium
Equilibrium
Chemical Clock Demo
http://lecturedemo.ph.unimelb.edu.au/Heat
-Thermodynamics/Non-equilibriumThermodynamics/Hg-1-Chemical-Clock-AB
Equilibrium Constant (Keq)
Law of mass action expresses relationship
between concentrations of reactants and
products at equilibrium
Equilibrium Constant Expressions
aA + bB  cC + dD
Keq = (PC)c(PD)d / (PA)a(PB)b for gasses
Keq = [C]c[D]d / [A]a[B]b
for solutions
Exercise 1
 Write equilibrium constant expressions
for the following reversible reactions:
a) H2(g) + I2(g)  2 HI(g)
b) 2Cl2(g) + 2H2O(g)  4HCl(g) + O2(g)
a) Kc = [HI]2 / [H2][I2] or Kp = P2HI / PH2PI2
b) Kc= [HCl]4[O2] / [Cl2]2[H2O]2 or
Kp = P4HClPO2 / P2Cl2P2H2O
Equilibrium Constant (Keq)
 Equilibrium constant expression depends only
on stoichiometry of reaction, not the mechanism
 Value of Keq varies only with temperature
Keq > 1: reaction favors products
Keq < 1: reaction favors reactants
 Need to specify temperature as well as direction
when writing an equilibrium constant
 Keq is a unitless value
Summary Points
 Equilibrium constant of a reaction in the reverse
direction is the inverse of the equilibrium
constant of the forward reaction
 The equilibrium constant of a reaction that has
been multiplied by a number is the equilibrium
constant raised to a power equal to that number
 The equilibrium constant for a net reaction of two
or more steps is the product of the constants of
the individual steps
Heterogeneous Equilibria
Contains substances at equilibrium in
different phases
Solids and liquids do not appear in the
equilibrium expression because they do
not have varying concentrations
Calculating Equilibrium Constants
 Tabulate the known initial and equilibrium
concentrations of all species in the equilibrium
constant expression
 For the species where both initial and
equilibrium concentrations are known,
calculate the change in concentration that
occurs
 Use the stoichiometry of the reaction to
calculate the changes in all species
 Calculate equilibrium concentration from initial
and changes in concentration
Summary of Steps to Calculate
Equilibrium Concentrations
1. Write a balanced equation for the reaction
2. Identify the knowns and unknowns
3. Assign the variable X to the unknown (usually
a change in concentration)
4. Prepare a table of concentrations
1. Row 1: chemicals involved
Row 2: initial concentrations
Row 3 Changes in concentrations (X appears)
Row 4: Equilibrium concentrations (rows 2 + 3)
Summary of Steps to Calculate
Equilibrium Concentrations
5. Write equilibrium expression
6. Solve for unknown variable using K and
data from row 4. Determine of
approximations can be made.
7. Answer the question
8. Check for validity of approximations. For
expressions of C+x or C-x, x can be
omitted if (x/C)100 < 5%
Example:
Solution Equilibrium
NH3(aq) + H2O(l)  NH4+(aq) + OH- (aq)
Enough ammonia is dissolved in 50 liters of
water at 25oC to produce a solution that is
0.0124 M in ammonia. The solution is
then allowed to come to equilibrium.
Analysis of the equilibrium mixture shows
that the concentration of OH- is
4.64 x 10-4 M. Calculate Keq at 25oC for
the reaction.
Initial
NH3(aq) M H2O(l)
NH4+(aq) M OH-(aq) M
.0124
0
0
Change
Equilibrium
4.64 x 10-4
After stoichiometry
NH3(aq) M
H2O(l)
NH4+(aq) M
OH-(aq) M
Initial
.0124
0
0
Change
- 4.64 x 10-4
+ 4.64 x 10-4
+ 4.64 x 10-4
Equilibrium
0.0119
4.64 x 10-4
4.64 x 10-4
Answer
Keq = [NH4+][OH-] / [NH3]

= (4.64 x 10-4)(4.64 x 10-4) / 0.0119

= 1.81 x 10-5
Applications of Equilibrium Constants
Predicting Direction of Reaction
 Compare the starting position to the equilibrium
position in order to determine how the reaction
must progress in order to achieve equilibrium
 When we substitute reactant and product
concentration or partial pressure into a Keq
expression, we achieve the reaction quotient (Q)
If Q = Keq then the system is at equilibrium
If Q > Keq then reaction moves to the left
If Q < Keq then reaction moves to the right
Applications of Equilibrium Constants
Calculating Equilibrium Concentrations
Use the ICE process described earlier
Example:
A 1.0 L container initially holds 0.015 mol
of H2 and 0.020 mol of I2 at 721 K. What
are the concentrations of H2, I2 and HI
after the system has achieved a state of
equilibrium? The value of Keq is 50.5 for
the reaction:
H2(g) + I2(g)  2HI(g)
Gas Equilibrium ICE
H2(g)
I2(g)
HI(g)
Initial (M)
.015
.020
0
Change (M)
-x
-x
2x
Equilibrium (M)
0.015 –x
0.020 –x
2x
Gas Equilibrium
Keq = [HI]2 / [H2][I2]
= (2x)2 / (0.015 –x)(0.020-x) = 50.5
46.5x2 – 1.77x + 0.0152 = 0
Solve quadratic equation
X = 0.025 M and 0.0131M
[H2] = 0.002 M, [I2] = 0.007 M, [HI] = 0.026 M
Le Chatelier’s Principle
If a system at equilibrium is disturbed by a
change in temperature, pressure, or
concentration the system will shift its
equilibrium to counteract the disturbance.
Le Chatelier : Change in Reactant or
Product Concentration
Shifts away from an increase in a
substance and toward a decrease in a
substance.
N2(g) + 3H2(g)  2NH3(g)
Addition of N2 shifts to right
Addition of NH3 shifts to left
Haber process removes NH3 as it is
produced, shifts right to continually
produce more NH3
Le Chatelier: Effect of Volume or
Pressure
A change in volume, shifts to reduce the
pressure (fewer moles of gas)
N2O4(g)  2NO2(g)
Volume decreases, shifts to left
Pressure decreases, shifts to right
Does not change the value of Keq at
constant temperature
Le Chatelier: Effect of Temperature
Changes
Keq changes as temperature changes
Treat temperature as a reactant or product
and determine shift accordingly
Increase temperature of endothermic 
increase Keq
Increase temperature of exothermic 
decrease in Keq
Le Chatelier: Effect of Catalysts
Catalyst lowers activation energy of both
the forward and reverse reactions by the
same amount
The rates of both forward and reverse
reactions increases by the same amount
Value of Keq does not change, but
equilibrium is reached sooner
Exercise 8
 What changes in the equilibrium composition
of the reaction N2(g) + O2(g)  2NO(g) will
occur at constant temperature if
a) The partial pressure of N2(g) is increase
b) The pressure of the NO(g) is decreased
c) the total pressure of the system is increased
d) The total volume of the system is increased
Exercise 8 answer
a)
b)
c)
d)
Shifts to right
Shifts to right
Unchanged
No change