Q. For $NH_{4}HS\left(\right.s\left.\right)\rightleftharpoonsNH_{3}\left(\right.g\left.\right)+H_{2}S\left(\right.g\left.\right)$ , if $K_{P}=64atm^{2}$ , equilibrium pressure of mixture is

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Solution:

$P_{N H_{3}}=P_{H_{2} S}$
So $K_{P}=\left(\right.P_{N H_{3}}\left(\left.\right)^{2}$
$64=\left(\right.P_{N H_{3}}\left(\left.\right)^{2}$
$P_{N H_{3}}=8$ atm
Total pressure $=P_{N H_{3}}+P_{H_{2} S}$
$=8+8=16$ atm