Kc = [NH₃]²/([N₂][H₂]³). Products over reactants, raise to stoichiometric coefficients.

Write the Expression for Kc of N₂ + 3H₂ ⇌ 2NH₃

Question

Write the expression for the equilibrium constant $K_c$ for the reaction:

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

Solution — Step by Step

Products are on the right side of the equilibrium arrow: $NH_3$ . Reactants are on the left: $N_2$ and $H_2$ . The $K_c$ expression always puts products in the numerator and reactants in the denominator.

We represent the equilibrium concentration of each species using square brackets. So we write $[NH_3]$ , $[N_2]$ , and $[H_2]$ for ammonia, nitrogen, and hydrogen respectively.

This is where the balancing comes in. The coefficient of $NH_3$ is 2, so it becomes $[NH_3]^2$ . The coefficient of $N_2$ is 1 (write nothing or just the bracket). The coefficient of $H_2$ is 3, so it becomes $[H_2]^3$ .

Put it all together — products over reactants, each raised to its coefficient:

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

This is the final answer.

Why This Works

The $K_c$ expression comes directly from the law of mass action, which says that at equilibrium, the ratio of product concentrations to reactant concentrations (each raised to stoichiometric coefficients) is a constant at a given temperature.

The stoichiometric coefficients become exponents because equilibrium is a result of rate balance — rate expressions involve concentration raised to the order of the reaction, and for elementary steps those orders equal the coefficients. The exponents encode how many moles participate in each direction.

Pure solids and pure liquids are not included in $K_c$ expressions because their concentrations don’t change during a reaction. Only dissolved species and gases appear in the expression.

Alternative Method

Some students find it easier to use the general formula directly. For any equilibrium:

aA + bB \rightleftharpoons cC + dD

K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Map our reaction onto this: $a = 1$ (N₂), $b = 3$ (H₂), $c = 2$ (NH₃). Substitute straight in:

K_c = \frac{[NH_3]^2}{[N_2]^1 [H_2]^3}

Same result. This template approach is faster in exams once you’ve drilled it a few times.

In NEET and JEE Main MCQs, they sometimes reverse the reaction and ask for $K_c$ of $2NH_3 \rightleftharpoons N_2 + 3H_2$ . The answer for that would be the reciprocal — $K_c' = \frac{[N_2][H_2]^3}{[NH_3]^2}$ . If you also multiply the equation by a factor $n$ , the new $K_c = (K_c')^n$ .

Common Mistake

The most common error is forgetting to apply the coefficients as exponents. Students write $K_c = \frac{[NH_3]}{[N_2][H_2]}$ — ignoring the 2 and the 3 entirely. This costs a guaranteed mark in board exams. The stoichiometric coefficient always becomes the power of the concentration term. No exceptions.

A related slip: writing $[H_2]^3$ in the numerator instead of the denominator. Always check — $H_2$ is a reactant, so it goes below the fraction line.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next