For the reaction ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$ at 400 °C, the equilibrium constant $K_c=0.42$. If the initial concentrations are $[{\text{N}}_2]=1.00;\text{M}$, $[{\text{H}}_2]=3.00;\text{M}$ and $[{\text{NH}}_3]=0.200;\text{M}$, calculate the equilibrium concentration of ${\text{NH}}_3$.

Step-by-step solution

1. Let $y$ be the increase in ${\text{NH}}_3$ (since $Q<K$, the forward reaction proceeds).
2. At equilibrium: $[{\text{N}}_2]=1-0.5y,;[{\text{H}}_2]=3-1.5y,;[{\text{NH}}_3]=0.200+y$.
3. Write the equilibrium expression $K_c=\frac{[{\text{NH}}_3{]}^2}{[{\text{N}}_2][{\text{H}}_2{]}^3}=0.42$.
4. Substitute the concentrations: $\frac{(0.200+y{)}^2}{(1-0.5y)(3-1.5y{)}^3}=0.42$.
5. Solve the resulting equation numerically; the physically relevant root is $yapprox0.87$.
6. Hence $[{\text{NH}}_3{]}_{\text{eq}}=0.200+0.87\approx 1.07;\text{M}$.

Concept tested: Calculating equilibrium concentrations using $K_c$

Published 16 October 2026 · How this solution was prepared and checked