Question

$N_2\left(g\right)+{3H}_2\left(g\right)\rightleftharpoons {2NH}_3\left(g\right)$

For the reaction initially the mole was 1 : 3 of $N_2$ and $H_2$. At equilibrium 50% of each has reacted. If the equilibrium pressure is p, the partial pressure of ${NH}_3$ at equilibrium is :

• A. $\frac{p}{3}$
• B. $\frac{p}{4}$
• C. $\frac{p}{6}$
• D. $\frac{p}{8}$

Answer 1

The correct option is A $\frac{p}{3}$

$N_2+3H_2\rightleftharpoons 2N{H}_3$

1 3 0

(1-$x$) ($3-3x$) $2x$

$1-x=\frac{1}{2}$

$\therefore x=\frac{1}{2}$

Mole of $H_2=$ 1.5 mol

Mole of $N_2=$ 0.5 mol

Mole of $N{H}_3=$ 1.0 mol

Now,

Mole fraction of $N{H}_3$ =$\frac{1}{3}$

partial pressure = $p\ast \frac{1}{3}=\frac{p}{3}$