An amount of solid ${NH}_{4} HS$ is placed in a flask already containing ammonia gas at a certain temperature and pressure of0.50 atm. Ammonium hydrogen sulphide decomposes to yield ${NH}_{3}$ and $H_{2} S$ gasses in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for ${NH}_{4} HS$ decomposition at this temperature is

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$0.17 {atm}^{2}$

$0.30 {atm}^{2}$

$0.18 {atm}^{2}$

$0.11 {atm}^{2}$

answer is B.

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Detailed Solution

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

$\begin{array}{l} t & 0.5 atm \\ t_{eq} & 0.50 atm + p p \end{array}$

We are given that $p_{eq} = 0.84 atm = (0.50 atm + p) + p$

This gives p = 0.17 atm

Hence

$K_{p} = (0.5 atm + 0.17 atm) (0.17 atm) = 0.1139 {atm}^{2}$

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