5.1 g $N H_{4} H S$ is introduced in 3.0L evacuated flask at $327^{0} c$ , 30% of the solid $N H_{4} H S$ decomposed to $N H_{3}$ and $H_{2} S$ gases. The $K_{p}$ of the reaction at $327^{0} C$ is $(R = 0.082 a t m m o l e^{- 1} k^{- 1})$

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$0.242 \times 10^{- 4} a t m^{2}$

$4.9 \times 10^{- 3} a t m^{2}$

$2.42 \times 10^{- 1} a t m^{2}$

$2.42 \times 10^{- 1} a t m^{- 2}$

answer is C.

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Detailed Solution

$\underset{(S)}{N H_{4} H S} \overset{}{⇌} \underset{(g)}{N H_{3}} + \underset{(g)}{H_{2} S}$

No.of moles of $N H_{4} H S = \frac{5.1}{51} = 0.1 m o l e$

At t= 0 0.1 0 0
At t=eq 0.1 (1-0.03) 0.03 0.03
$K_{c} = [N H_{3}] [H_{2} S] = \frac{0.03}{3} \times \frac{0.03}{3} = 10^{- 4} {(m o l e . l t^{- 1})}^{2} K p = K c {(R T)}^{Δ n} K p = 10^{- 4} {(0.082 \times 600)}^{2} a t m^{2} K p = 0.242 a t m^{2} = 2.42 \times 10^{- 1} a t m^{2}$