Kp Expressions (AQA A Level Chemistry): Revision Note
Exam code: 7405
Kp Expressions
The equilibrium expression links the equilibrium constant, Kc, to the concentrations of reactants and products at equilibrium taking the stoichiometry of the equation into account
So, for a given reaction:
aA + bB ⇌ cC + dD
Gaseous Equilibria
If all the substances in the general equation above are gases:
aA (g) + bB (g) ⇌ cC (g) + dD (g)
The equilibrium constant Kp is deduced from a gaseous reversible reaction equation
Kp is defined as:
Where:
a, b, c and d are the respective number of moles of each reactant and product
paA (g) and pbB (g) are the equilibrium partial pressures of A and B, kPa
pcC (g) and pdD (g) are the equilibrium partial pressures of C and D, in kPa
Equilibrium partial pressure is calculated from the:
The total number of moles
The mole fraction
The total pressure
Solids and liquids are ignored in Kp equilibrium expressions
The Kp of a reaction is constant and only changes if the temperature of the reaction changes
Examiner Tips and Tricks
There are a variety of ways to represent the partial pressure terms in a Kp expression.
The only key point is do not use square brackets as these represent concentration and, therefore, imply a Kc expression.
Worked Example
Write a Kp expression for the following equilibria and deduce the units of Kp:
N2 (g) + 3H2 (g)
2NH3 (g)
N2O4 (g)
2NO2 (g)
2SO2 (g) + O2 (g)
2SO3 (g)
Answers:
N2 (g) + 3H2 (g)
2NH3 (g)
Kp =
Units =
= kPa-2
N2O4 (g)
2NO2 (g)
Kp =
Units =
= kPa
2SO2 (g) + O2 (g)
2SO3 (g)
Kp =
Units =
= kPa-1
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