Q. An amount of solid $NH_{4}HS$ is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield $NH_{3}\ and\ H_{2}S$ gases 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
A) 0.11
B) 0.17
C) 0.18
D) 0.30
Explanation
(a)
$NH_{4}HS(s) \rightleftharpoons NH_{3}(g) + H_{2}S(g)$
Initially 1 0.5 0
At equilibrium $(1 - x)$ $(0.5 + x)$ $x$
Total pressure at equilibrium
$= p_{NH_{3}} + p_{H_{2}S}$
$= 0.5 + x + x = 0.84$
$\therefore$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 0.17$ atm
$\therefore$ $p_{NH_{3}} = 0.50 + 0.17 = 0.67$ atm
$p_{H_{2}S} = 0.17$ atm
$\therefore$ $K_{p} = p_{NH_{3}}.p_{H_{2}S}$
$= 0.67 \times 0.17 = 0.114$ atm
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.