# 7.8.2 Effect of Pressure Change

A pressure change obtained by changing the volume can affect the yield of products in case of a gaseous reaction where the total number of moles of gaseous reactants and total number of moles of gaseous products are different.

In applying Le Chatelier’s principle to a heterogeneous equilibrium the effect of pressure changes on solids and liquids can be ignored because the volume (and concentration) of a solution/liquid is nearly independent of pressure.

Consider the reaction,

CO(g) + 3H2(g) ⇔  CH4(g) + H2O(g)

Here, 4 mol of gaseous reactants (CO + 3H2) become 2 mol of gaseous products (CH4 + H2O). Suppose equilibrium mixture (for above reaction) kept in a cylinder fitted with a piston at constant temperature is compressed to one half of its original volume.

Then, total pressure will be doubled (according to pV = constant). The partial pressure and therefore, concentration of reactants and products have changed and the mixture is no longer at equilibrium.

The direction in which the reaction goes to re-establish equilibrium can be predicted by applying the Le Chatelier’s principle. Since pressure has doubled, the equilibrium now shifts in the forward direction, a direction in which the number of moles of the gas or pressure decreases (we know pressure is proportional to moles of the gas).

This can also be understood by using reaction quotient, Qc. Let [CO], [H2], [CH4] and [H2O] be the molar concentrations at equilibrium for methanation reaction.

When volume of the reaction mixture is halved, the partial pressure and the concentration are doubled. We obtain the reaction quotient by replacing each equilibrium concentration by double its value.

QC = [CH4(g)][ H2O(g)]/[ CO(g)][H2(g)]3

As Qc < Kc , the reaction proceeds in the forward direction

In the reaction C(s)+ CO(g) ⇔ 2CO(g) when pressure is increased, the reaction goes in the reverse direction because the number of moles of gas increases in the forward direction.