$5.1 g {\mathrm{NH}}_4\mathrm{SH}$ is introduced in $3.0 \mathrm{L}$ evacuated flask at $327°\mathrm{C}$. $30\%$ of the solid ${\mathrm{NH}}_4\mathrm{SH}$ decomposed to ${\mathrm{NH}}_3$ and ${\mathrm{H}}_2 \mathrm{S}$ as gases. The ${\mathrm{K}}_{\mathrm{p}}$ of the reaction at $327°\mathrm{C}$ is $\mathrm{R}=0.082 \mathrm{L}$ atm ${\mathrm{mol}}^{-1}{ \mathrm{K}}^{-1}$, Molar mass of$\mathrm{S}=32 \mathrm{g}{ \mathrm{mol}}^{-1}$, molar mass of $\mathrm{N}=14 \mathrm{g}{ \mathrm{mol}}^{-\left(A\right)}$

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

$\mathrm{n}=\frac{5.1}{51}=.1$ mole $0 0$

$.1(-1-\alpha ) .1\alpha .1\alpha$

$\alpha =30\%=.3$

So, the number of moles equilibrium

$.1(1-.3).1\times .3.1\times .3$

=.07=.03=03

Now use $\mathrm{PV}=\mathrm{nRT}$ at equilibrium

${\mathrm{P}}_{\text{total }}\times 3\text{ lit }=(.03+.03)\times .082\times 600$

${\mathrm{P}}_{\text{total }}=.984 \mathrm{atm}$

${\mathrm{P}}_{{\mathrm{NH}}_3}={\mathrm{P}}_{{\mathrm{H}}_2 \mathrm{S}}=\frac{{\mathrm{P}}_{\text{total }}}{2}=0.492$

${\mathrm{Sok}}_{\mathrm{p}}={\mathrm{P}}_{{\mathrm{NH}}_3},{\mathrm{PH}}_2 \mathrm{S}=(0.492)(0.492)$

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

Hence the correct answer is (D) $0.242{ \mathrm{atm}}^2$.