Consider the following consecutive first-order reaction: A -(K1)- B -(K2)- C. If K1 = 0.01 min^-1 and K1:K2 = 1:2, after what time from the start of reaction, the concentration of 'B' will be maximu

Question

Consider the following consecutive first-order reaction: A -(K1)-> B -(K2)-> C. If K1 = 0.01 min^-1 and K1:K2 = 1:2, after what time from the start of reaction, the concentration of 'B' will be maximum? (ln 2 = 0.7)

Prashant Kotian · Accepted Answer

Correct answer: A. Given K1 = 0.01 min^-1 and K1:K2 = 1:2 => K2 = 0.02 min^-1. The time at which the concentration of the intermediate B is maximum is t_max = ln(K2/K1) / (K2 - K1). Substituting the values: t_max = ln(0.02/0.01) / (0.02 - 0.01) = ln2 / 0.01 = 0.693/0.01 ≈ 70 min (using ln2 = 0.7).

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