Chemical Equilibrium: Numerical Problems Set (3)

Question

For the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ at $500$ K, $K_c = 0.5$ mol$^{-2}$L$^2$. If $1$ mole of $N_2$ and $3$ moles of $H_2$ are placed in a $1$ L container, find the equilibrium concentrations.

Solution — Step by Step

Why This Works

The ICE (Initial-Change-Equilibrium) table organises the algebra so you don’t lose track of stoichiometry. The change in each species is governed by the stoichiometric coefficients in the balanced equation: for every $1$ mole of $N_2$ that reacts, $3$ moles of $H_2$ react and $2$ moles of $NH_3$ form.

Substituting equilibrium concentrations into $K_c$ gives a polynomial in $x$, which we solve.

Alternative Method

For approximate solutions when $K_c$ is small or large, assume small $x$ (or $1-x \approx 1$). Here $K_c$ isn’t extreme, so the approximation isn’t accurate — we need numerical solution.

For homework problems with small $K$ values, just write $1 - x \approx 1$, simplify, and solve linearly — much faster.

Common Mistake

Students often forget to cube $(3 - 3x)$. The reaction has stoichiometry $3$ for $H_2$, so the equilibrium expression has $[H_2]^3$. Missing the cube gives wildly wrong answers.

Another trap: writing $K_c = [NH_3]/([N_2][H_2])$, which would be the expression if the reaction were $N_2 + H_2 \rightleftharpoons NH_3$ — wrong stoichiometry. Always copy the balanced equation carefully when writing $K_c$.