For the reversible reaction: A(aq) ⇌ B(aq), which is first order in both directions, K_f = 1.38/300 min^-1. The variation in concentrations with time shows [A] dropping from 0.3 to 0.2 and [B] rising

Question

For the reversible reaction: A(aq) ⇌ B(aq), which is first order in both directions, K_f = 1.38/300 min^-1. The variation in concentrations with time shows [A] dropping from 0.3 to 0.2 and [B] rising from 0 to 0.1 at equilibrium. The time (in min) in which 25% of 'A' would be converted into 'B', is (ln 2 = 0.69)

Prashant Kotian · Accepted Answer

Correct answer: 0100. K_eq = [B]eq / [A]eq = 0.1 / 0.2 = 0.5. K_b = K_f / 0.5 = 2K_f. Total rate constant K_tot = K_f + K_b = 3K_f = 3(1.38/300) = 0.0138 min^-1. At 25% conversion, [A]t = 0.3 - 0.25(0.3) = 0.225. K_tot × t = ln(([A]0 - [A]eq) / ([A]t - [A]eq)) = ln((0.3 - 0.2) / (0.225 - 0.2)) = ln(0.1 / 0.025) = ln 4 = 2(0.69) = 1.38. t = 1.38 / 0.0138 = 100 min. Therefore, correct answer is 0100.

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