12.2 The Equilibrium Constant
Chemistry 1011
YOU ARE EXPECTED TO BE ABLE TO:
TOPIC
• 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.
Gaseous Chemical Equilibrium
TEXT REFERENCE
Masterton and Hurley Chapter 12
Chemistry 1011 Slot 5
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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
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Equilibrium Constant Example
Chemistry 1011 Slot 5
• 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
Kp = (PSO3)
2
(PSO ) x PO
2
2
Chemistry 1011 Slot 5
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Equilibrium Constant Example
• Ammonia is made industrially by the Haber
Process:
N2(g) + 3H2(g)
2NH3(g)
• The equilibrium constant, K, is
Kp = (PNH )2
3
PN x (PH )3
2
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
(P )c x (PD)d
K= Ca
(PA) x (PB)b
Chemistry 1011 Slot 5
Chemistry 1011 Slot 5
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Chemistry 1011 Slot 5
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6
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Dependence of K on Equation Stoichiometry
Dependence of K on Equation
Stoichiometry
• The Coefficient Rule:
• 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
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Adding Chemical Equations
Reaction 1:
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Chemistry 1011 Slot 5
SO2(g) + 1/2O2(g)
Kp = 2.2 =
SO3(g)
(PSO )
3
(PSO ) x (PO )1/2
2
Reaction 2:
NO2(g)
Kp =
2
NO(g) + 1/2O2(g)
1 /2
4.0 = (PNO) x (PO2 )
(PNO )
2
Adding the equations:
SO2(g) + NO2(g)
– If Reaction 3 = Reaction 1 + Reaction 2
– Then K(Reaction 3) = K(reaction 1) x K(Reaction 2)
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Multiple Equilibria Example
SO3(g) + NO(g)
Chemistry 1011 Slot 5
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Heterogeneous Equilibria
• The equilibrium constant expression for the
total reaction is
Kp = (PSO3) x (PNO)
(PSO ) x (PNO )
2
2
This is obtained by multiplying together the equilibrium
constant expressions for the two individual reactions
(PSO )
x (PNO) x (PO )1/2
3
2
(PSO ) x (PO )1/2
(PNO )
2
Kp = 2.2 x 4.0 = 8.8
Chemistry 1011 Slot 5
Chemistry 1011 Slot 5
Multiple Equilibria Example
• 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
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• The Reciprocal Rule:
– If the equation is written in reverse, then
K’’ = 1/K
Chemistry 1011 Slot 5
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– If coefficients in a balanced equation are
multiplied by a factor, n, then
– The equilibrium constant is raised to the nth
power
K’ = Kn
11
• 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
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2
Heterogeneous Equilibria – the
Sublimation of Iodine
Heterogeneous Equilibria
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
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– 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
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Heterogeneous Equilibria with a
Liquid Component
Heterogeneous Equilibria – the
Decomposition of Calcium Carbonate
• Example:
CO(g) + H2O(l)
CO2(g) + H2(g)
• In this case, the amount of water vapour in
the system is constant, since it will be in
equilibrium with the liquid water
• So,
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 = PCO
Kp = (PCO)
(PCO3) (PH3)
2
Chemistry 1011 Slot 5
• For heterogeneous equilibria, it is found
that:
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Chemistry 1011 Slot 5
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