Condition for (t_1/2)_A = (t_1/2)_B is — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

For B (a first-order decay with rate constant 2k), (t_1/2)_B = (ln 2) / 2k. Setting this equal to (t_1/2)_A from the previous question: (ln 2) / 2k = (1/2k) ln(b / (b - a)). Thus, ln 2 = ln(b / (b - a)) => 2 = b / (b - a) => 2b - 2a = b => b = 2a. Therefore, correct answer is A.