What is the rate constant of forward reaction? — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

At equilibrium, K_f[A]_e = K_b[B]_e => K_f(0.10) = K_b(0.05) => K_b = 2K_f. Integrating rate law: ln(([A]0 - [A]_e)/([A]_t - [A]_e)) = (K_f + K_b)t. At t=10, ln((0.15-0.10)/(0.125-0.10)) = ln(0.05/0.025) = ln(2) = 0.693. So, 10(3K_f) = 0.693 => K_f = 0.0231 min^-1. Therefore, correct answer is A.