The half-life periods of two first-order reactions are in the ratio 3:2. If t1 is the time required for 25% completion of the first reaction and t2 is the time required for 75% completion of the secon

Question

The half-life periods of two first-order reactions are in the ratio 3:2. If t1 is the time required for 25% completion of the first reaction and t2 is the time required for 75% completion of the second reaction, then the ratio, t1:t2, is (log 3 = 0.48, log 2 = 0.3)

Prashant Kotian · Accepted Answer

Correct answer: A. Given t_1/2 ratio = 3/2, so K1/K2 = 2/3. t1 = (1/K1) ln(100/75) = (1/K1) ln(4/3) = (1/K1) × 0.12 × 2.303. t2 = (1/K2) ln(100/25) = (1/K2) ln(4) = (1/K2) × 0.60 × 2.303. t1/t2 = (K2/K1) × (0.12/0.60) = (3/2) × (1/5) = 3/10. Therefore, correct answer is A.

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