If t_1/2 of a second-order reaction is 1.0 h. After what time, the amount will be 25% of the initial — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

For a second-order reaction: 1/[A]_t - 1/[A]_0 = Kt. (1) For half-life ([A]_t = 0.5 [A]_0): t_1/2 = 1/(K[A]_0) = 1.0 h. (2) For the concentration to be 25% of the initial amount ([A]_t = 0.25 [A]_0): 4/[A]_0 - 1/[A]_0 = Kt => t = 3/(K[A]_0). Since t_1/2 = 1/(K[A]_0) = 1.0 h, we get t = 3 × 1.0 h = 3.0 h.