For the reaction: A + B → products, rate law is r = K[A]^2[B], where K = 0.5 M^-2 min^-1. The time in which the concentration of 'A' becomes half of its initial concentration if the initial concentrat

Question

For the reaction: A + B → products, rate law is r = K[A]^2[B], where K = 0.5 M^-2 min^-1. The time in which the concentration of 'A' becomes half of its initial concentration if the initial concentrations of 'A' and 'B' are 0.002 M and 2.0 M, respectively, is

Prashant Kotian · Accepted Answer

Correct answer: B. Since [B] >> [A], the reaction is pseudo-second order w.r.t A. k_pseudo = K[B] = 0.5 × 2.0 = 1.0 M^-1 min^-1. For a second-order reaction, t_1/2 = 1 / (k_pseudo × A0) = 1 / (1.0 × 0.002) = 500 min. Therefore, correct answer is B.

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