A definite amount of solid $N H_{4} H S$ is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. $N H_{4} H S$ decomposes to give $N H_{3}$ and $H_{2} S$ and at equilibrium total pressure in flask is 0.84 atm. The equilibrium constant for the reaction is $(a t m^{2})$ :

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0.18

0.11

0.30

0.17

answer is D.

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Detailed Solution

$\begin{array}{l} N H_{4} H S ⇌ N H_{3} + H_{3} S (g) \\ a 0.5 \\ a-x 0.5+x x \\ P_{T} = 0.5 + 2 x = 0.84 \\ X = 0.17 \\ K_{P} = (0.67) (0.17) \Rightarrow 0.1139 \end{array}$

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