A gaseous substance AB₂(g) converts to AB(g) in the presence of solid A as: AB₂(g) + A(s) ⇌ 2AB(g). The initial pressure and equilibrium pressure are 0.7 and 0.95 bar, respectively. Now, the equilibri

Question

A gaseous substance AB₂(g) converts to AB(g) in the presence of solid A as: AB₂(g) + A(s) ⇌ 2AB(g). The initial pressure and equilibrium pressure are 0.7 and 0.95 bar, respectively. Now, the equilibrium mixture is expanded reversibly and isothermally till the gas pressure falls to 0.4 bar. Then, which of the following statements is correct?

Prashant Kotian · Accepted Answer

Correct answer: B,C. Let initial pressure of AB₂ be 0.7. At equilibrium, P_AB₂ = 0.7 - x, P_AB = 2x. Total P = 0.7 + x = 0.95 => x = 0.25 bar. P_AB₂ = 0.45 bar, P_AB = 0.50 bar. K_p = (0.50)² / 0.45 = 5/9. (Option C). Upon expansion, P'_AB₂ + P'_AB = 0.4. Using K_p = (P'_AB)² / (0.4 - P'_AB) = 5/9, solving yields P'_AB ≈ 0.269 bar and P'_AB₂ ≈ 0.131 bar. Volume % of AB₂ = (0.131 / 0.4) × 100 ≈ 32.75% (nearly 32.5%). (Option B). Therefore, correct answer is B,C.

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