For the reaction $N H_{4} H S_{(} s) ⇌ N H_{3 (g)} + H_{2} S_{(g)}$ , the total pressure at equilibrium is $30 a t m$ . The value of $K_{p}$ ?

15 a t m^{2}

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225 a t m^{2}

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30 a t m^{2}

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15 a t m

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Solution

The correct option is C $225 a t m^{2}$
$N H_{4} H S (s) ⇌ N H_{3} (g) + H_{2} S (g)$
$K_{p} = P_{N H_{3}} \times P_{H_{2} S}$
Partial pressure of each gas ( $N H_{3}, H_{2} S$ ) = $\frac{30 a t m}{2} = 15 a t m$
$∴ K_{p} = (15 a t m)^{2} = 225 a t m^{2}$

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