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Characteristics of Equilibrium Constant

A definite am...

Question

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

A

0.30

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B

0.18

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C

0.17

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D

0.11

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Solution

The correct option is D $0.11$
For given reaction
$N H_{4} H S ⇌ N H_{3} (g) + H_{2} S (g) a 0.5 a t m a - x 0.5 + x x$
Hence, the total pressure $= 0.5 + 2 x = 0.84$
Therefore,
$x = 0.17$
$K_{p} = p_{N H_{3}} . p_{H_{2} S}$
$= (0.67) (0.17)$
$= 0.1139$

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