Equilibrium Constant Kp for Homogeneous Systems (Oxford AQA International A Level (IAL) Chemistry): Revision Note
Exam code: 9622
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
Kp, Equilibrium & Condition Changes
Le Chatelier's Principle can be applied to gaseous equilibria in the same way it is applied to aqueous systems
It states that if a change is made to a system in dynamic equilibrium, the position of the equilibrium moves to counteract this change
It can be used to predict the qualitative effects of changes in temperature and pressure on:
The position of equilibrium
The value of Kp
How temperature affects Kp
For a reaction that is exothermic in the forward direction:
Increasing the temperature pushes the equilibrium to the left
This means that there are more reactants and less products
The ratio of products to reactants decreases
Therefore, the value of Kp decreases
For a reaction that is endothermic in the forward direction:
Increasing the temperature pushes the equilibrium to the right
This means that there are less reactants and more products
The ratio of products to reactants increases
Therefore, the value of Kp increases
How pressure affects Kp
The value of Kp is not affected by any changes in pressure.
Changes in pressure cause a shift in the position of equilibrium to a new position which restores the value of Kp
This is similar to Kc when you change concentration in an aqueous equilibrium; a shift restores equilibrium to a new position maintaining Kc
How a catalyst affects Kp
If all other conditions stay the same, Kp is not affected by the presence of a catalyst
A catalyst speeds up both the forward and reverse reactions at the same rate
So, the ratio of products to reactants remains unchanged
Catalysts only cause a reaction to reach equilibrium faster
Catalysts therefore have no effect on the position of the equilibrium once this is reached
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