5 .1 g NH _ 4 SH is introduced in 3 .0 L evacuated flask at 327 ^ o...

${NH}_{4}SH\left(s\right) ⇌ {NH}_{3}\left(g\right) + {H}_{2}S\left(g\right)$

Number of moles$= \dfrac{5 .1}{51} = 0 .1 mol$

	${NH}_{4}SH$	${NH}_{3}$	${H}_{2}S$
Initial concentration	$0 . 1mol$	$0$	$0$
Equilibrium concentration	$0 . 1\left(1 - \alpha\right)$	$0 . 1\alpha$	$0 . 1\alpha$

\left(\alpha = 30% = 0 .3\right) =$Degree of dissociation<br /><br />So, number of moles at equilibrium,</p><p>\left[{NH}_{3}\right] =$\left[{H}_{2}S\right] = 0 . 1 \times 0 . 3 = 0 . 03$</p><p>\left[{NH}_{4}SH\right]$ is not considered as it is a solid.

${K}_{c} = \left[{NH}_{3}\right]\left[{H}_{2}S\right]$

${K}_{c} = \left[\dfrac{0 . 03}{3}\right]\left[\dfrac{0 . 03}{3}\right] = {10}^{- 4}$

${K}_{p} = {10}^{- 4}{\left(0 . 082 \times 600\right)}^{2}$

{K}_{p} = 0 .242 {atm}^{2}