Chemical Equilibrium: Exam-Pattern Drill (10)

Question

For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $K_c = 1.6 \times 10^{-2}$ at $400°\text{C}$. If $[N_2] = 1.0\ \text{M}$ and $[H_2] = 2.0\ \text{M}$ at equilibrium, find $[NH_3]$.

Solution — Step by Step

Final answer: $[NH_3] \approx 0.358\ \text{M}$.

Why This Works

The equilibrium constant expression uses stoichiometric coefficients as exponents. Get the powers right and the rest is just substitution. JEE Main routinely tests this pattern with two- and three-component equilibria.

The key insight: $K_c$ is a number tied to the reaction as written. Reverse the reaction, and $K$ becomes $1/K$. Multiply coefficients by $n$, and $K$ becomes $K^n$.

Alternative Method

Use $K_p$ if pressures are given instead of molarities. The relation is $K_p = K_c (RT)^{\Delta n}$, where $\Delta n$ is the change in moles of gas. For this reaction, $\Delta n = 2 - 4 = -2$.

Common Mistake