JEE Main 2019 — Chemical Equilibrium Question with Solution

${\mathrm{NH}}_4\mathrm{SH} \left(\mathrm{s}\right)\rightleftharpoons {\mathrm{NH}}_3 \left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{S} \left(\mathrm{g}\right)$

Number of moles$=\frac{5.1}{51}=0.1\mathrm{ }\mathrm{mol}$

	${\mathrm{NH}}_4\mathrm{SH}$	${\mathrm{NH}}_3$	${\mathrm{H}}_2\mathrm{S}$
Initial concentration	$0.1 \mathrm{mol}$	$0$	$0$
Equilibrium concentration	$0.1\left(1-\mathrm{\alpha }\right)$	$0.1\mathrm{\alpha }$	$0.1\mathrm{\alpha }$

$\left(\mathrm{\alpha }=30\mathrm{\%}=0.3\right)=$ Degree of dissociation

So, number of moles at equilibrium,

$\left[{\mathrm{NH}}_3\right]=\left[{\mathrm{H}}_2\mathrm{S}\right]=0.1\times 0.3=0.03$

$\left[{\mathrm{NH}}_4\mathrm{SH}\right]$ is not considered as it is a solid.

${\mathrm{K}}_{\mathrm{c}}=\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2\mathrm{S}\right]$

${\mathrm{K}}_{\mathrm{c}}=\left[\frac{0.03}{3}\right]\left[\frac{0.03}{3}\right]={10}^{-4}$

${\mathrm{K}}_{\mathrm{p}}={10}^{-4} {\left(0.082\times 600\right)}^2$

${\mathrm{K}}_{\mathrm{p}}=0.242\mathrm{ }{\mathrm{atm}}^2$