Kc & Kp (OCR A Level Chemistry A): Revision Note
Kc & Kp Expressions
As previously discussed, 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 the general reaction
aA + bB ⇌ cC + dD
Kc = 
Gaseous Equilibria
This section covers how we manage gases in calculating the equilibrium constant and how an equilibrium yield is affected by the partial pressures of reactants and products
Many industrial process involve reactions between gases so this application has important consequences for controlling reaction conditions
In the generic example above, if all the substances are gases, we can show the equation with that state symbol
aA (g) + bB (g) ⇌ cC (g) + dD (g)
We can write a different equilibrium expression in terms of the partial pressure of the gases
This equilibrium constant is called Kp and is defined as follows
Equilibrium expression linking the partial pressures of reactants and products at equilibrium
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)
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
Answer 1
Answer 2
Kc & Kp Calculations
Kp calculations
Kp calculations are a step-by-step process in which you need to find
the mole fractions of the gases present
their partial pressures
the Kp expression
the value of Kp
Worked Example
Finding Kp
Hydrogen and bromine were mixed in a flask in a 1: 1 ratio and allowed to reach equilibrium at 450 K. When equilibrium had been achieved the total pressure in the flask was 140 kPa and the mole fraction of bromine was 0.35.
The equation for the reaction is
H2 (g) + Br2 (g) ⇌ 2HBr (g)
Determine the partial pressures for each gas at equilibrium and the value of Kp
Another style of Kp calculation involves being given the value of Kp and working backwards to deduce the partial pressure of one of the gases
Worked Example
Finding Partial Pressure
Kp for the dissociation equilibrium reaction of nitrogen monoxide is 0.0021.
2NO (g) ⇌ N2 (g) + O2 (g)
If pure NO is introduced into a reaction flask at an initial pressure of 100 kPa, what is the equilibrium partial pressure of nitrogen?
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