When a mixture of N₂ and H₂ in the volume ratio of 1:5 is allowed to react at 700 K and 10³ atm pressure, 0.4 mole fraction of NH₃ is formed at equilibrium. The K_p for the reaction:
N₂(g) + 3H₂(g)

Question

When a mixture of N₂ and H₂ in the volume ratio of 1:5 is allowed to react at 700 K and 10³ atm pressure, 0.4 mole fraction of NH₃ is formed at equilibrium. The K_p for the reaction:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Prashant Kotian · Accepted Answer

Correct answer: A. Let total eq moles be n. Moles of NH₃ = 0.4n. The remaining 0.6n is N₂ and H₂. From stoichiometry and initial 1:5 ratio, the unreacted moles are evaluated. Ultimately, mole fractions at equilibrium are: X_NH₃ = 0.4, X_N₂ = 1/30, X_H₂ = 17/30. K_p = (0.4)² / [ (1/30)(10³) × (17/30)³(10³)³ ] = (0.16 × 30⁴) / (17³ × 10⁶) ≈ 2.63 × 10⁻⁵ atm⁻². Therefore, correct answer is A.

💡 Solution & Explanation

Practice 22,000+ questions like this