Chemistry 1011
TOPIC
Gaseous Chemical Equilibrium
TEXT REFERENCE
Masterton and Hurley Chapter 12
Chemistry 1011 Slot 5
1
12.2 The Equilibrium Constant
YOU ARE EXPECTED TO BE ABLE TO:
• Write an expression for the equilibrium constant,
K, for a gaseous reaction
• Recognize that the expression for K depends on
the form of the balanced chemical equation for the
reaction.
• Write an expression for the equilibrium constant,
K, for a gaseous reaction that includes a substance
in the solid or liquid phase.
Chemistry 1011 Slot 5
2
The Equilibrium Constant
• For a gaseous reaction, the equilibrium
constant can be written in terms of the
partial pressures (concentrations) of
reactants and products
• For aA(g) + bB (g)
cC (g) + dD (g)
• The equilibrium constant, Kp , is
(PC)c x (PD)d
K=
(PA)a x (PB)b
Chemistry 1011 Slot 5
3
The Equilibrium Constant
• Equilibrium partial pressures of the
products are in the numerator (top)
• Equilibrium partial pressures of the
reactants are in the denominator (bottom)
• Each partial pressure is raises to a power
equal to its coefficient in the balanced
equation
Chemistry 1011 Slot 5
4
Equilibrium Constant Example
• Ammonia is made industrially by the Haber
Process:
N2(g) + 3H2(g)
2NH3(g)
• The equilibrium constant, K, is
Kp = (PNH3)2
PN2 x (PH2)3
Chemistry 1011 Slot 5
5
Equilibrium Constant Example
• Sulfuric acid is a very important industrial
chemical. It is manufactured from sulfur
dioxide and oxygen
2SO2(g) + O2(g)
2SO3(g)
• The equilibrium constant, K, is
2
(P
)
SO3
Kp =
(PSO2 )2 x PO2
Chemistry 1011 Slot 5
6
Dependence of K on Equation
Stoichiometry
• The expression for K, and its value will depend on
how the equation is written
• For N2(g) + 3H2(g)
2NH3(g)
Kp = (PNH3)2
PN2 x (PH2)3
• For 1/2N2(g) + 3/2H2(g)
NH3(g)
Kp’ = (PNH3)
(PN2 )1/2 x (PH2)3/2
Chemistry 1011 Slot 5
7
Dependence of K on Equation Stoichiometry
• The Coefficient Rule:
– If coefficients in a balanced equation are
multiplied by a factor, n, then
– The equilibrium constant is raised to the nth
power
K’ = Kn
• The Reciprocal Rule:
– If the equation is written in reverse, then
K’’ = 1/K
Chemistry 1011 Slot 5
8
Adding Chemical Equations
• The Rule of Multiple Equilibria
• If a reaction can be expressed as the sum of
two or more reactions, K for the overall
reaction is equal to the PRODUCT of the
equilibrium constants for the individual
reactions
– If Reaction 3 = Reaction 1 + Reaction 2
– Then K(Reaction 3) = K(reaction 1) x K(Reaction 2)
Chemistry 1011 Slot 5
9
Multiple Equilibria Example
Reaction 1:
SO2(g) + 1/2O2(g)
Kp = 2.2 =
Reaction 2:
NO2(g)
Kp =
SO3(g)
(PSO3)
(PSO2 ) x (PO2 )1/2
NO(g) + 1/2O2(g)
1 /2
(P
)
x
(P
)
4.0 =
NO
O2
(PNO2 )
Adding the equations:
SO2(g) + NO2(g)
SO3(g) + NO(g)
Chemistry 1011 Slot 5
10
Multiple Equilibria Example
• The equilibrium constant expression for the
total reaction is
Kp = (PSO3) x (PNO)
(PSO2 ) x (PNO2 )
This is obtained by multiplying together the equilibrium
constant expressions for the two individual reactions
(PSO3)
x (PNO) x (PO2 )1/2
(PSO2 ) x (PO2 )1/2
(PNO2 )
Kp = 2.2 x 4.0 = 8.8
Chemistry 1011 Slot 5
11
Heterogeneous Equilibria
• Up to now, all of the reactions considered
have been homogeneous gas reactions
• In some cases, one or more of the
substances involved may be a liquid or solid
• This would be a heterogeneous system
• Example:
I2(s) I2(g)
Chemistry 1011 Slot 5
12
Heterogeneous Equilibria – the
Sublimation of Iodine
I2(s)
I2(g)
• The rate of the forward process depends only on
temperature. Sublimation will occur at constant
rate as long as some solid iodine remains
• The rate of the reverse reaction will depend on the
concentration (partial pressure) of iodine gas
• At equilibrium (in a closed system), this rate will
become constant
Kp = PI2(g)
Chemistry 1011 Slot 5
13
Heterogeneous Equilibria
• For heterogeneous equilibria, it is found
that:
– The position of equilibrium is independent of
the amount of solid or liquid component, as
long as some still remains
– Concentrations of solids and liquids are
constant
– Terms for solid or liquid components do not
appear in the expression for K
Chemistry 1011 Slot 5
14
Heterogeneous Equilibria – the
Decomposition of Calcium Carbonate
CaCO3(s)
CaO(s) + CO2(g)
• Written in terms of concentrations, the
equilibrium constant expression is:
K = [CaO(s)][CO2(g)]
[CaCO3(s)]
• But the concentrations of the solids are
constant, so
K = [CO2(g)], or Kp = PCO2
Chemistry 1011 Slot 5
15
Heterogeneous Equilibria with a
Liquid Component
• Example:
CO2(g) + H2(g)
CO(g) + H2O(l)
• In this case, the amount of water vapour in
the system is constant, since it will be in
equilibrium with the liquid water
• So,
Kp = (PCO)
(PCO3) (PH3)
Chemistry 1011 Slot 5
16