Effect of Pressure Change

Effect of Pressure Change on Equilibrium

When we talk about changing pressure in a gaseous reaction, we almost always mean changing the volume of the container. A pressure change obtained by changing the volume can affect the yield of products only in those gaseous reactions where the total number of moles of gaseous reactants differs from the total number of moles of gaseous products. If the number of moles of gas is the same on both sides, changing the volume (and hence the pressure) has no effect on the equilibrium position.

For heterogeneous equilibria involving solids or liquids, the effect of pressure changes on these condensed phases can be ignored. The volume (and therefore the concentration) of a solid or a liquid is nearly independent of pressure. Only the gaseous components matter.

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The Core Idea: Le Chatelier’s Principle

Consider the methanation reaction:

Here, 4 moles of gaseous reactants (1 CO + 3 H₂) become 2 moles of gaseous products (1 CH₄ + 1 H₂O). The total number of gas moles decreases in the forward direction.

Now suppose the equilibrium mixture is kept in a cylinder fitted with a piston at constant temperature. If we compress the mixture to one half of its original volume, the total pressure will double (according to Boyle’s law, $pV = \text{constant}$ at constant $T$). The partial pressures — and therefore the concentrations — of all reactants and products have changed. The mixture is no longer at equilibrium.

Le Chatelier’s principle tells us the direction in which the reaction will shift to re-establish equilibrium. Since the pressure has doubled, the equilibrium will shift in the direction that reduces the pressure. Pressure is proportional to the total number of moles of gas (at constant $T$ and $V$). The forward direction produces fewer moles of gas (2 moles) than the reverse direction (4 moles). Therefore, the equilibrium shifts in the forward direction to reduce the total number of gas moles and hence reduce the pressure.

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Proving the Shift Using the Reaction Quotient

We can also understand this shift quantitatively using the reaction quotient, $Q_c$.

Let $[\text{CO}]$, $[\text{H}_2]$, $[\text{CH}_4]$, and $[\text{H}_2\text{O}]$ be the molar concentrations at equilibrium for the methanation reaction. At equilibrium, we have:

When the volume of the reaction mixture is halved, the concentration of each gas doubles (because concentration = moles/volume, and volume is halved while moles remain the same initially). So the new concentrations are $2[\text{CO}]$, $2[\text{H}_2]$, $2[\text{CH}_4]$, and $2[\text{H}_2\text{O}]$.

We obtain the reaction quotient $Q_c$ by substituting these doubled concentrations into the equilibrium expression:

Simplify the numerator and denominator:

But $\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}$ is exactly $K_c$. Therefore:

Since $Q_c < K_c$, the reaction proceeds in the forward direction to re-establish equilibrium. This matches the prediction from Le Chatelier’s principle.

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The Opposite Case: When Forward Direction Increases Moles of Gas

Consider the reaction: …