For the reversible reaction ${\text{N}}_2(g)+3{\text{H}}_2(g)\rightleftharpoons 2{\text{NH}}_3(g)$, what is the expression for the equilibrium constant $K_c$?

Solution

1. The general reaction is $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$.
2. The equilibrium constant $K_c$ is given by $\frac{[\text{C}{]}^c[\text{D}{]}^d}{[\text{A}{]}^a[\text{B}{]}^b}$.
3. For the reaction ${\text{N}}_2+3{\text{H}}_2\rightleftharpoons 2{\text{NH}}_3$, the coefficients are $1, 3,$ and $2$.
4. Substituting these values, $K_c=\frac{[{\text{NH}}_3{]}^2}{[{\text{N}}_2][{\text{H}}_2{]}^3}$.
5. Hence the answer is (A).