In a closed rigid vessel, the following equilibrium partial pressures are measured: N₂ = 100 mm, H₂ = 400 mm and NH₃ = 1000 mm. Now, nitrogen is removed from the vessel until the pressure of hydrogen

Question

In a closed rigid vessel, the following equilibrium partial pressures are measured: N₂ = 100 mm, H₂ = 400 mm and NH₃ = 1000 mm. Now, nitrogen is removed from the vessel until the pressure of hydrogen at equilibrium is equal to 700 mm. The new equilibrium partial pressure of N₂ is

Prashant Kotian · Accepted Answer

Correct answer: A. For N₂ + 3H₂ ⇌ 2NH₃, K_p = (1000)² / (100 × 400³) = 1 / 6400 mm⁻². H₂ increases from 400 to 700 mm, meaning 300 mm of H₂ formed. This implies the reaction shifted backward, creating 100 mm of N₂ and consuming 200 mm of NH₃. New P_NH₃ = 1000 - 200 = 800 mm. Let new P_N₂ be x. K_p = (800)² / (x × 700³) = 1 / 6400. Solving for x: x = (800² × 6400) / 700³ = (640000 × 6400) / 343000000 = 4096 / 343 ≈ 11.94 mm. Therefore, correct answer is A.

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