Increase pressure → shifts towards fewer moles (products). More NH₃ formed.

Le Chatelier's Principle — Effect of Pressure on N₂ + 3H₂ ⇌ 2NH₃

Question

The following equilibrium exists in a closed container:

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

What happens to the equilibrium position when the pressure on the system is increased at constant temperature? Explain using Le Chatelier’s Principle.

Solution — Step by Step

Reactant side: 1 mole N₂ + 3 moles H₂ = 4 moles of gas

Product side: 2 moles of NH₃

We have 4 moles on the left and 2 moles on the right.

When pressure increases, the system is under stress. Le Chatelier’s Principle says the equilibrium will shift in the direction that relieves this stress.

Fewer moles of gas means lower pressure. So the system shifts towards the side with fewer moles — the product side (right).

The equilibrium shifts forward (towards NH₃ formation).

More NH₃ is produced, and the concentrations of N₂ and H₂ decrease until a new equilibrium is established.

Here’s where students often confuse themselves: Kp does not change.

Temperature is constant, so the equilibrium constant stays the same. Only the position of equilibrium shifts — the ratio of products to reactants adjusts, but Kp remains fixed.

\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}

If $\Delta n_g < 0$ : Increased pressure → forward shift
If $\Delta n_g > 0$ : Increased pressure → backward shift
If $\Delta n_g = 0$ : Pressure has no effect on equilibrium position

For this reaction: $\Delta n_g = 2 - 4 = -2$ → forward shift on increasing pressure

Why This Works

Pressure and concentration are linked for gases — increasing pressure at constant volume increases the concentration of all species. The system responds by consuming more reactant molecules (4 moles) and producing fewer product molecules (2 moles), which brings the pressure back down.

This is exactly the principle behind the Haber Process for industrial ammonia synthesis. Engineers run the reaction at 200–300 atm precisely because high pressure favours NH₃ formation. The yield improves significantly compared to running at atmospheric pressure.

Notice that this works because the reaction involves gases with unequal moles on both sides. If both sides had the same total moles of gas, changing pressure would shift concentrations equally and the equilibrium position would not change at all.

Alternative Method — Using Kp Expression

Write the expression for $K_p$ :

K_p = \frac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3}

When total pressure increases, all partial pressures increase. The denominator has 4 pressure terms multiplied together ( $P_{N_2} \times P_{H_2}^3$ ) while the numerator has only 2 ( $P_{NH_3}^2$ ). So the denominator grows faster than the numerator.

This means $Q_p < K_p$ momentarily. The reaction must proceed forward to restore $K_p$ — producing more NH₃ until equilibrium is re-established. Same answer, different path.

The Kp method is especially useful when $\Delta n_g = 0$ — like in $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ . Here you can quickly verify that changing pressure leaves $Q_p = K_p$ unchanged, so equilibrium position is unaffected.

Common Mistake

“Increasing pressure always favours the product side.”

This is wrong. Pressure shift depends entirely on $\Delta n_g$ , not on which side is “products.”

For the reverse of our reaction — decomposition of NH₃:

2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)

Here, increasing pressure would shift equilibrium backward (towards NH₃, which is now the reactant side) because the reactant side (2 moles) has fewer moles than the product side (4 moles).

Always count moles first. Never assume.

Final Answer: Increasing pressure shifts the equilibrium forward, producing more NH₃, because the forward reaction reduces the total number of moles of gas ( $\Delta n_g = -2$ ), thereby relieving the increased pressure. Kp remains unchanged.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Kp Expression

Common Mistake

Try These Next