An amount of solid $\mathrm{NH}_{4} \mathrm{HS}$ is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm . Pressure. Ammonium hydrogen sulphide decomposes to yield $\mathrm{NH}_{3}$ and $\mathrm{H}_{2} \mathrm{~S}$ 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 $\mathrm{NH}_{4} \mathrm{HS}$ decomposition at this temperature is

JEE Main · 2005 · 08 May 2005 · Offline · Equilibrium

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)0.30
(B)0.18
(C)0.17
(D)0.11✓ Correct

Answer: (D)

Solution

The solid decomposes as

NH X_{4} HS (s) NH X_{3} (g) + H X_{2} S (g)

, but the flask already holds

NH X_{3}

0.50

atm. If decomposition produces

x

atm of

H X_{2} S

, it produces an equal

x

NH X_{3}

, so the total pressure becomes

(0.50 + x) + x = 0.84

, giving

x = 0.17

atm. Then

P_{NH X_{3}} = 0.67

atm and

P_{H X_{2} S} = 0.17

atm, and since the pure solid does not enter the expression,

K_{p} = P_{NH X_{3}} P_{H X_{2} S} = 0.67 \times 0.17 \approx 0.11

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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JEE Main · 2005 · 08 May 2005 · Offline · Equilibrium

Solution

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