Equilibrium constant Kp

Let’s start with some terms…

Mole fraction

The mole fraction is the ratio of moles of a gas in a mixture to the total moles of gas.

Say you had a mixture of two gases, 3 moles of gas A and 4 moles of gas B. Altogether, there are 7 moles in total. The mole fraction of A is 3/7 and for B, 4/7.

Partial pressure & mole fractions

A mixture of gases will exert a total pressure that is the sum of all the individual gas pressures (the partial pressures). The partial pressure can be considered the pressure that a gas, if not in a mixture, would alone exert.

The relationship between the two can be expressed as:

partial pressure = mole fraction x total pressure

Calculating Kp Using Partial Pressures

An equilibrium is affected by the partial pressures of gaseous reactants and products. An equilibrium expression can be written in terms of the partial pressures of the gases involved for the equilibrium constant, Kp.

Consider the following reaction:

Then…

Therefore, at equilibrium:

Remember, only gases are of concern for AQA - solids and liquids are ignored in Kp equilibrium expressions. Also, the equilibrium constant is only affected by changes in temperature.

Write expressions for the following equilibria

Temperature and Kp

In an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant and vice-versa for endothermic ones. Not only that, but equilibrium also changes with temperature.

Recall, from Le Chatelier's Principle, that increasing temperature favours the reaction which absorbs heat. In an exothermic reaction in the forward direction, the back reaction will be favoured when temperature increases – equilibrium will move to the left.

Since reactants are favoured over products, Kp will decrease. This is demonstrated by the following, stripped-down equilibrium expression…

The denominator increases which causes a reduction in Kp

Pressure and the equilibrium position

Increasing pressure, according to Le Chatelier, means that equilibrium will move in a way to reduce the effect of increasing pressure. Since pressure is the result of particle collisions, it makes sense that increased pressure can be countered by equilibrium shifting towards the side producing fewer molecules.

The reverse argument is true for a decrease in pressure. Note that if the number of molecules are the same on either side, equilibrium will not shift.

Examples (with answers)

1

The following equilibrium is reached at 700 K

The partial pressures are:

What is the Kp for this reaction?

2

Nitrogen dioxide gas fades from dark brown to colourless dinitrogen tetroxide gas in the following reaction:

When equilibrium is reached…

What is Kp for the reaction?

3

The partial pressures, at equilibrium, for the reaction above are as follows:

What is the Kp for the reaction?

4

For the following reaction at 375K...

… the partial pressures were recorded:

Calculate Kp

Answers

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