Pressure and Volume

8 min read Updated Mar 26, 2026

Pressure and volume changes only matter for reactions involving gases. Liquids and solids are essentially incompressible, so changing the pressure of the container has no effect on their concentrations.

The Core Rule

When pressure increases (or volume decreases), the system shifts toward the side with fewer moles of gas. When pressure decreases (or volume increases), the system shifts toward the side with more moles of gas.

Why? The system is trying to relieve the stress. If you squeeze the container (increase pressure), the system responds by reducing the total number of gas molecules - which reduces the pressure.

Example: The Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Reactant side: 1 + 3 = 4 moles of gas
Product side: 2 moles of gas

Increase pressure: Shifts right (toward 2 moles, fewer gas molecules)
Decrease pressure: Shifts left (toward 4 moles, more gas molecules)

This is why the Haber process uses high pressure (200 atm) - it pushes the equilibrium toward ammonia production.

Photograph of a flask of dark carbonated liquid with CO2 molecular models and upward arrows overlaid, showing dissolved CO2 molecules escaping from solution into the gas phase when pressure is reduced by opening the container — When you open a bottle of soda, the pressure drops and dissolved CO₂ escapes as bubbles. This is Le Chatelier's principle in action: decreasing the pressure above the liquid shifts the equilibrium CO₂(aq) ⇌ CO₂(g) toward the gas phase (the side with more volume). *Credit: OpenStax Chemistry 2e, CC BY 4.0*

When Moles of Gas Are Equal

If delta-n = 0 (same number of gas moles on both sides), pressure changes have no effect on the equilibrium.

Example: H₂(g) + I₂(g) ⇌ 2HI(g) has 2 moles of gas on each side. Changing pressure does not shift this equilibrium.

This distinction - inert gas at constant volume vs. constant pressure - is a classic MCAT trap question. Remember: partial pressures determine Q, not total pressure.

Why Does This Work? The Q Explanation

When you decrease the volume of a container, all gas concentrations increase (same moles, smaller volume). But they do not all increase equally in the Q expression - the side with more moles of gas gets “raised to higher powers” in the expression. This changes Q in a predictable direction:

If more moles of gas are in the numerator (products), Q increases above K, so the reaction shifts left
If more moles of gas are in the denominator (reactants), Q decreases below K, so the reaction shifts right

The net result: the shift always goes toward the side with fewer moles of gas.

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), what happens if the volume of the container is halved?

Click to reveal answer

Equilibrium shifts to the right (toward SO₃). Halving the volume doubles the pressure. Reactant side has 3 moles of gas (2 + 1), product side has 2 moles. The system shifts toward the side with fewer moles of gas (products) to reduce pressure.

Argon gas is added to a sealed rigid container holding N₂O₄(g) ⇌ 2NO₂(g) at equilibrium. Does the equilibrium shift?

Click to reveal answer

No shift occurs. Adding an inert gas at constant volume increases total pressure but does NOT change the partial pressures of N₂O₄ or NO₂. Since Q depends on partial pressures, Q is unchanged and equals K. No shift.