Adding N₂ disturbs equilibrium. Reaction shifts right to consume added N₂, producing more NH₃.

What Happens When You Add More N₂ to Equilibrium? — Le Chatelier's

Question

The system $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is at equilibrium. What happens when we add more $N_2$ to the system at constant temperature and volume?

Solution — Step by Step

Adding $N_2$ increases its concentration. The system was balanced — now the forward reaction side suddenly has more reactant than the equilibrium demands.

Le Chatelier’s Principle says: the system will shift in the direction that reduces the disturbance. Since we’ve added $N_2$ (a reactant), the system shifts right — toward products — to consume the extra $N_2$ .

As the reaction shifts forward:

$[N_2]$ decreases (being consumed), but not back to its original value
$[H_2]$ decreases (also being consumed)
$[NH_3]$ increases (being produced)

The reaction shifts until a new equilibrium is reached. The value of $K_c$ doesn’t change — temperature hasn’t changed. But the equilibrium concentrations of all species are now different from before.

The equilibrium shifts to the right. More $NH_3$ is produced. The system reaches a new equilibrium position at which $Q_c = K_c$ once again.

Final Answer: Adding $N_2$ shifts the equilibrium to the right, producing more $NH_3$ .

Why This Works

When we add $N_2$ , the reaction quotient $Q_c$ temporarily becomes less than $K_c$ . The system isn’t at equilibrium anymore — the numerator of the equilibrium expression ( $[NH_3]^2$ ) is too small relative to the denominator ( $[N_2][H_2]^3$ ).

To restore balance, the forward reaction runs faster than the reverse until $Q_c$ climbs back up to $K_c$ . This is exactly what “shifting right” means — it’s not a vague statement, it’s the system adjusting rates to re-establish the equilibrium ratio.

The key insight: Le Chatelier is just a shortcut for the $Q$ vs $K$ logic. If you ever get confused, just calculate what adding a reactant does to $Q_c$ — it always drops below $K_c$ , so the forward reaction dominates.

Alternative Method — Using $Q_c$ vs $K_c$

Instead of stating Le Chatelier’s principle directly, we can reason through $Q_c$ :

K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}

At the original equilibrium, $Q_c = K_c$ . The moment we add $N_2$ , the denominator increases. So now:

Q_c = \frac{[NH_3]^2}{[N_2]_{\text{new}} \cdot [H_2]^3} < K_c

Since $Q_c < K_c$ , the reaction proceeds forward to produce more $NH_3$ and reduce $[H_2]$ and $[N_2]$ , until $Q_c = K_c$ again.

This method is more reliable on tricky questions where Le Chatelier’s “direction” isn’t immediately obvious (e.g., adding an inert gas, or changing volume with unequal moles).

In JEE/NEET MCQs, when they ask “what happens to equilibrium when X is added”, the fastest approach is: check whether X is a reactant or product, then apply $Q$ vs $K$ logic. For $N_2$ here — it’s a reactant, so $Q < K$ , so the reaction goes forward. Three seconds.

Common Mistake

Students often write: “Adding $N_2$ shifts equilibrium right, so $[N_2]$ returns to its original value.”

This is wrong. After the shift, $[N_2]$ is higher than the original equilibrium value — just not as high as the value immediately after addition. The system partially consumes the added $N_2$ , but cannot fully undo the change (that would violate $K_c$ conservation). This is a classic NCERT-based board exam trap — CBSE Class 11 papers have asked this exact follow-up.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using QcQ_cQc​ vs KcK_cKc​

Common Mistake

Try These Next

Alternative Method — Using $Q_c$ vs $K_c$