Solid NH_4HS is placed in a flask containing NH_3 at 0.50 atm . It decomposes into NH_3 and H_2S . At equilibrium, total pressure is 0.84 atm . K_p is - Chemistry MCQ AIEEE 2026 | ExamTest.live

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Solid NH_{4} HS is placed in a flask containing NH_{3} at 0.50 atm . It decomposes into NH_{3} and H_{2} S . At equilibrium, total pressure is 0.84 atm . K_{p} is:

Solution & Step-by-step Explanation

NH_{4} HS (s) ⇌ NH_{3} (g) + H_{2} S (g) Initial pressure of NH_{3} = 0.50 .Let decomposition increase pressure by x .Equilibrium: P_{NH_{3}} = 0.50 + x; P_{H_{2} S} = x .Total pressure: 0.50 + 2 x = 0.84 ⟹ 2 x = 0.34 ⟹ x = 0.17 . K_{p} = (0.50 + 0.17) (0.17) = 0.67 \times 0.17 = 0.1139 .

Try it yourself before checking the explanation above.

Solid NH_{4} HS is placed in a flask containing NH_{3} at 0.50 atm . It decomposes into NH_{3} and H_{2} S . At equilibrium, total pressure is 0.84 atm . K_{p} is:

0.30

0.18

0.17

0.11

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