Effect of Pressure on Equilibrium (Higher Tier)

Pressure and Equilibrium

Pressure changes affect equilibria involving gases. This is because gases are compressible, so changing the pressure alters the concentration of gaseous molecules.

However, pressure only affects equilibria where the number of gas molecules differs between the reactants and products. If the total moles of gas are the same on both sides, changing pressure has no effect on the position of equilibrium.

When pressure is increased, the equilibrium shifts towards the side with fewer gas molecules. This reduces the total pressure by favouring the side with less volume of gas.

Conversely, when pressure is decreased, the equilibrium shifts towards the side with more gas molecules, increasing the total pressure.

For example, consider the equilibrium:

$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$

On the left, there are 4 moles of gas (1 mole $\mathrm{N_2}$ + 3 moles $\mathrm{H_2}$), and on the right, 2 moles of $\mathrm{NH_3}$.

Increasing pressure shifts the equilibrium to the right (towards ammonia), because this side has fewer gas molecules.

If the reaction involved equal moles of gas on both sides, changing pressure would not affect the equilibrium position.

Le Chatelier6s Principle Application

Le Chatelier6s Principle states that if a system at equilibrium is subjected to a change, the system will respond to oppose that change.

Applying this to pressure changes:

• If pressure increases, the system shifts to reduce pressure by favouring the side with fewer gas molecules.
• If pressure decreases, the system shifts to increase pressure by favouring the side with more gas molecules.

This shift changes the equilibrium position but does not affect the equilibrium constant, which depends only on temperature. The equilibrium constant remains unchanged because pressure changes alter concentrations but do not change the fundamental energy balance of the reaction.

For example, in the reaction:

$\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$

There are 3 moles of gas on the left and 2 moles on the right. Increasing pressure shifts equilibrium to the right, producing more $\mathrm{SO_3}$.

This principle helps predict how equilibrium will shift when pressure changes, which is important in industrial processes.

For instance, if a reaction has:

• More moles of gas on the reactant side 1 increasing pressure shifts equilibrium to products.
• More moles of gas on the product side 1 increasing pressure shifts equilibrium to reactants.
• Equal moles of gas on both sides 1 pressure changes have no effect.

For example, if a reaction is:

$\mathrm{A(g) + B(g) \rightleftharpoons C(g)}$

There are 2 moles of gas on the left and 1 mole on the right. Increasing pressure shifts equilibrium to the right.

For instance, if a reaction is:

$\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$

There are 2 moles of gas on the left and 1 mole on the right. Increasing pressure shifts equilibrium to the right, producing more $\mathrm{N_2O_4}$.

Industrial Relevance

Pressure is an important factor in industry to optimise the yield of products in reversible reactions involving gases.

A key example is the Haber process for ammonia production:

$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$

This reaction produces ammonia, which is vital for fertilisers.

Because there are 4 moles of gas on the left and 2 moles on the right, increasing pressure shifts equilibrium to the right, increasing ammonia yield.

However, very high pressures require stronger, more expensive equipment and increase costs and safety risks.

Therefore, industry balances pressure to maximise yield while keeping costs and safety manageable.

For example, the Haber process typically uses pressures around 200 atmospheres (about 20,000 kPa) to increase ammonia production efficiently without excessive cost.

Lower pressures reduce yield but are cheaper and safer, so a compromise pressure is chosen.