Question

Ammonia under a pressure of $20atm$ at ${27}^{o}C$ is heated to ${327}^{o}C$ in a closed vessel. Under these conditions ${NH}_3$ is partially decomposed to $N_2$ and $H_2$ according to the equation $2{NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$.
After decomposition at constant volume in a vessel, the pressure increases to $45atm$. What is the percentage of ammonia dissociated?

Answer 1

$2{NH}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$
$t=0$ $a$ $0$ $0$
$t_{eq.}$ $a-a\alpha$ $a\alpha /\alpha 2$ $a+a\alpha$
$n$ $P$ $V$
$a$ $20$ $V$
$a+a\alpha$ $40$ $V$
$V=\frac{nRT}{P}=\frac{a\times R\times 400}{20}...\left(1\right)$
$V=\frac{nRT}{P}=\frac{a(1+\alpha )\times R\times 600}{45}...\left(2\right)$
$\alpha =0.5$
Percentage dissociation of ${NH}_3=\alpha \times 100=50\%$