Chemical Equilibrium: Real-World Scenarios (2)

Question

In the Haber process for ammonia synthesis: $N_2 + 3H_2 \rightleftharpoons 2NH_3$. At a certain temperature, $K_c = 0.5 \, \text{M}^{-2}$. If we start with $[N_2] = 1 \, \text{M}$, $[H_2] = 1 \, \text{M}$, $[NH_3] = 1 \, \text{M}$, predict the direction of the reaction.

Solution — Step by Step

Why This Works

The Q vs K comparison is the universal direction-prediction tool for any equilibrium problem. Three cases:

• $Q < K$: forward reaction dominates (system makes more products to reach equilibrium).
• $Q = K$: system is already at equilibrium.
• $Q > K$: backward reaction dominates (system breaks down products to reach equilibrium).

This works because $Q$ and $K$ have the same form — $Q$ at equilibrium just IS $K$. So comparing them tells us how far we are from equilibrium and which way to go.

Alternative Method — Le Chatelier’s Principle Reasoning

Think about it physically: we have 1 M of each species. The equilibrium constant $K_c = 0.5$ tells us at equilibrium, the products’ concentrations would be modest compared to reactants (since $K < 1$).

Currently, $[NH_3]$ is at the same level as reactants — too high. So the reaction breaks down ammonia until the ratio matches $K_c$.

Le Chatelier gives the same conclusion as Q vs K, but the latter is more rigorous and works for any starting concentrations.

Common Mistake

Students often forget the cube on $[H_2]$ and write $Q = [NH_3]^2/([N_2][H_2])$. The exponents in $Q$ (and $K$) match the stoichiometric coefficients in the balanced equation.

For Class 11 boards, the Q vs K logic earns full marks on any direction-of-reaction question.