Le Chatelier's principle — predict shift when pressure increases on N₂O₄ ⇌ 2NO₂

Question

For the equilibrium: $N_2O_4 (g) \rightleftharpoons 2NO_2 (g)$

Predict the direction of shift when the pressure of the system is increased at constant temperature.

Solution — Step by Step

Le Chatelier’s Principle states: When a system at equilibrium is subjected to a change in conditions (concentration, pressure, temperature), the equilibrium shifts in the direction that opposes the change and re-establishes equilibrium.

For pressure changes, the key is counting moles of gas.

N_2O_4 (g) \rightleftharpoons 2NO_2 (g)

Left side (reactants): 1 mole of gas
Right side (products): 2 moles of gas

The forward reaction increases the number of moles. The reverse reaction decreases the number of moles.

When pressure increases, the system opposes this by reducing the number of gas moles (to reduce pressure). So the equilibrium shifts toward the side with fewer moles of gas — that is, the backward (reverse) direction.

N_2O_4 (g) \leftarrow 2NO_2 (g)

More $N_2O_4$ is formed; concentration of $NO_2$ decreases.

At constant temperature, $K_p$ remains unchanged. When total pressure increases, the partial pressures of all species increase. For this reaction, $Q_p > K_p$ momentarily (because $NO_2$ has squared term in denominator of $K_p = \frac{p_{N_2O_4}}{p_{NO_2}^2}$ ). The system reacts to decrease $Q_p$ back to $K_p$ — by consuming $NO_2$ and forming $N_2O_4$ . This confirms backward shift.

Conclusion: When pressure increases on the equilibrium $N_2O_4 \rightleftharpoons 2NO_2$ , the equilibrium shifts backward (to the left), favouring formation of $N_2O_4$ and reducing the amount of $NO_2$ .

Why This Works

The system “wants” to resist the increased pressure. The only way to reduce pressure (at constant volume and temperature) is to have fewer gas molecules. By shifting backward, 2 moles of $NO_2$ collapse into 1 mole of $N_2O_4$ , halving the number of gas particles and thus reducing the pressure. This is why Le Chatelier’s principle for pressure always points toward the side with fewer gas moles.

Notice the colour change: $N_2O_4$ is colourless, $NO_2$ is brown. Increased pressure causes the mixture to become lighter/less brown — a visible demonstration of the backward shift.

Alternative Method — Using Mole Fraction Argument

When pressure increases at constant volume, it means more gas is present in the same space (or same amount in smaller space). Le Chatelier’s analysis tells us the system will try to reduce its own pressure. The only chemical way to do this is to reduce total moles of gas. Backward reaction: $2 \text{ mol} \to 1 \text{ mol}$ — achieves this. Forward reaction: $1 \text{ mol} \to 2 \text{ mol}$ — would make things worse.

The $N_2O_4 \rightleftharpoons 2NO_2$ system is a favourite in JEE because it also shows a colour change (colourless to brown), making it easy to verify experimentally. In JEE problems, you may be asked about this using a piston — compressing the piston increases pressure, so expect backward shift and lighter brown colour.

Common Mistake

Students often confuse “increase in pressure by adding an inert gas at constant volume” with “increase in pressure by compressing the gas.” When an inert gas is added at constant volume, the partial pressures of $N_2O_4$ and $NO_2$ don’t change — only the total pressure increases. So there is no shift in equilibrium in that case. Le Chatelier’s pressure effect only applies when you change partial pressures of the reacting gases, not when you add a non-reacting gas.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Mole Fraction Argument

Common Mistake

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