At a certain temperature, the equilibrium constant of the reaction: N₂(g) + O₂(g) ⇌ 2NO(g) is 0.0004. Assuming air to be a mixture of four volumes of nitrogen with one volume of oxygen, the percentage

Question

At a certain temperature, the equilibrium constant of the reaction: N₂(g) + O₂(g) ⇌ 2NO(g) is 0.0004. Assuming air to be a mixture of four volumes of nitrogen with one volume of oxygen, the percentage of nitric oxide, by volume, in the gas produced by allowing air to reach equilibrium at this temperature, is

Prashant Kotian · Accepted Answer

Correct answer: B. Let total pressure be P. Initial pressures: P_N₂ ≈ 0.8P, P_O₂ ≈ 0.2P. Assuming small conversion, P_N₂ and P_O₂ stay roughly the same. K_p = P_NO² / (0.8P × 0.2P) = 0.0004. P_NO² = 0.0004 × 0.16P² = 64 × 10⁻⁶ P². Thus, P_NO = 8 × 10⁻³ P = 0.008P. Percentage by volume = (P_NO / P) × 100 = 0.8%. Therefore, correct answer is B.

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