Question

A container containing ammonia is allowed to decompose and the following equilibrium is established.
$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$.
Then, the equilibrium constant, $K_p$ is expressed as:
(Where $p_{N_2}$ = partial pressure of $N_2$ and $p_{N{H}_3}$ is the partial pressure of $N{H}_3$ )

• A. $K_p=\frac{3^3\times p_{N_2}^2}{p_{N{H}_3}}$
• B. $K_p=\frac{3^3\times p_{N_2}{p}_{H_2}^3}{p_{N{H}_3}^2}$
• C. $K_p=\frac{3^3\times p_{N_2}^4}{p_{N{H}_3}^2}$
• D. $K_p=\frac{3^{\left(3/2\right)}\times p_{N_2}^4}{p_{N{H}_3}^2}$

Answer 1

The correct option is C $K_p=\frac{3^3\times p_{N_2}^4}{p_{N{H}_3}^2}$

Given, equilibrium is,
$2N{H}_3\rightleftharpoons N_2+3H_2$
$\therefore K_p=\frac{p_{N_2}.p_{H_2}^3}{p_{N{H}_3}^2}$
Where, $p_{N_2},p_{H_2}\text{and}{p}_{N{H}_3}\text{are the partial pressure of}{N}_2,H_2\text{and}N{H}_3\text{respectively}$

Since initially only $N{H}_3$ is present hence, $3p_{N_2}=p_{H_2}$
$\therefore K_p=\frac{p_{N_2}\times (3p_{N_2}{)}^3}{p_{N{H}_3}^2}$

$=\frac{p_{N_2}\times 3^3.p_{N_2}^3}{p_{N{H}_3}^2}=\frac{3^3\times p_{N_2}^4}{p_{N{H}_3}^2}=\frac{3^3\times p_{N_2}^4}{p_{N{H}_3}^2}$