For dissociation of a gas N₂O₅ as: N₂O₅(g) ⇌ 2NO₂(g) + 1/2 O₂(g). The reaction is performed at constant temperature and volume. If D is the vapour density of equilibrium mixture, P_o is initial pressu

Question

For dissociation of a gas N₂O₅ as: N₂O₅(g) ⇌ 2NO₂(g) + 1/2 O₂(g). The reaction is performed at constant temperature and volume. If D is the vapour density of equilibrium mixture, P_o is initial pressure of N₂O₅(g) and M is molecular mass of N₂O₅, then the correct information(s) at the equilibrium is/are

Prashant Kotian · Accepted Answer

Correct answer: A,B,C. For N₂O₅ ⇌ 2NO₂ + 1/2 O₂, Δn = 1.5. Let α be the degree of dissociation. The average molecular mass of the mixture is M_avg = 2D. The degree of dissociation α = (M - M_avg) / (Δn × M_avg) = (M - 2D) / (1.5 × 2D) = (M - 2D) / 3D (Option B is correct). Total equilibrium moles / Initial moles = (1 + 1.5α) = M / 2D. So, total pressure P = P_o(M / 2D) (Option A is correct). Partial pressure of N₂O₅ = P_o(1 - α) = P_o[1 - (M - 2D)/3D] = P_o(5D - M)/3D (Option C is correct). Partial pressure of O₂ = P_o(0.5α) = P_o(M - 2D)/6D, which conflicts with Option D. Therefore, correct answer is A,B,C.

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