The following equilibrium exists in a closed system at $ 25^{\circ} \mathrm{C} \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $. When a sample of pure $ \mathrm{NH}_{4} \mathrm{HS} $ (s) is placed in an evacuated vessel and allowed to reach equilibrium at $ 25^{\circ} \mathrm{C} $, the total pressure is $ 0.66 $ atm. Find the value of $ \mathrm{K}_{\mathrm{p}} $. (Answer by rounding off up to $ 2 $ decimal places.)

Question

The following equilibrium exists in a closed system at $ 25^{\circ} \mathrm{C} \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $. When a sample of pure $ \mathrm{NH}_{4} \mathrm{HS} $ (s) is placed in an evacuated vessel and allowed to reach equilibrium at $ 25^{\circ} \mathrm{C} $, the total pressure is $ 0.66 $ atm. Find the value of $ \mathrm{K}_{\mathrm{p}} $. (Answer by rounding off up to $ 2 $ decimal places.),

MARKS App · Accepted Answer

At equilibrium same moles of NH 3 and H 2 S is formed. NH 4 HS ( s ) ⇌ NH 3 ( g ) + H 2 S ( g ) Initially : x 0 0 At equilibrium - P P Given: Total pressure = 0 . 66 atm P + P = 0 . 66 atm P = 0 . 33 atm K P = P NH 3 × P H 2 S For pure solid and pure liquid, active mass is taken as unity. K P = 0 . 33 × 0 . 33 = 0 . 11

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