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An amount of solid $	ext{NH}_4	ext{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 $	ext{NH}_3$ and $	ext{H}_2	ext{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 $	ext{NH}_4	ext{HS}$ decomposition at this temperature is A). $0.11$ B). $0.17$ C). $0.18$ D). $0.30$

Edvaya · Accepted Answer

The correct answer is $0.11$. Reaction: $	ext{NH}_4	ext{HS}(s) ightleftharpoons 	ext{NH}_3(g) + 	ext{H}_2	ext{S}(g)$.
Initial partial pressure of $	ext{NH}_3 = 0.50	ext{ atm}$.
Let the partial pressure of $	ext{H}_2	ext{S}$ at equilibrium be $P$. Due to stoichiometry, the partial pressure of $	ext{NH}_3$ will also increase by $P$.
At equilibrium:
$P_{	ext{NH}_3} = 0.50 + P$
$P_{	ext{H}_2	ext{S}} = P$
Total pressure = $P_{	ext{NH}_3} + P_{	ext{H}_2	ext{S}} = (0.50 + P) + P = 0.50 + 2P = 0.84	ext{ atm}$.
$2P = 0.34 \Rightarrow P = 0.17	ext{ atm}$.
Equilibrium partial pressures:
$P_{	ext{NH}_3} = 0.50 + 0.17 = 0.67	ext{ atm}$
$P_{	ext{H}_2	ext{S}} = 0.17	ext{ atm}$.
Equilibrium constant $K_p = P_{	ext{NH}_3} 	imes P_{	ext{H}_2	ext{S}} = 0.67 	imes 0.17 \approx 0.1139 \approx 0.11$.

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