When 5 . 1g of solid NH _ 4 HS is introduced into a two litre...

Moles of ${NH}_{4}HS$ initially taken $= \dfrac{5 . 1g}{51g / mol} = 0 . 1mol$

volume of vessel $= 2L$

${NH}_{4}{HS}_{( s )} ⇌ {NH}_{3 ( g )} + {H}_{2}{S}_{( g )}$

$t = 00 . 1mol$

$\begin{cases}t = \infty & 0 . 1 ( 1 - 0 . 2 ) & 0 . 1 \times 0 . 2 & 0 . 1 \times 0 . 2\end{cases}$

Partial pressure of each gaseous component

$P = \dfrac{nRT}{V} = \dfrac{0 . 1 \times 0 . 2 \times 0 . 082 \times 300}{2}$

$= 0 . 246atm$

$⇒ {k}_{P} = {P}_{{NH}_{3}} \times {P}_{{H}_{2}S} = ( 0 . 246{)}^{2} = 0 . 060516$

$= 6 . 05 \times {10}^{- 2}$

$⇒ 6$