For a two-step reaction: A ⇌(K1/K2) R + B ; R + C --(K3)-- Products (where R is a reactive intermediate whose concentration is maintained at some low steady-state throughout the reaction), the rate l

Question

For a two-step reaction: A ⇌(K1/K2) R + B ; R + C --(K3)--> Products (where R is a reactive intermediate whose concentration is maintained at some low steady-state throughout the reaction), the rate law expression will be

Prashant Kotian · Accepted Answer

Correct answer: A. Applying steady-state to R: d[R]/dt = k1[A] - k2[R][B] - k3[R][C] = 0. Solving for [R] gives [R] = k1[A] / (k2[B] + k3[C]). Rate = dx/dt = k3[R][C] = k3[C] * k1[A] / (k2[B] + k3[C]) = k1[A] / (1 + k2[B]/k3[C]). Therefore, correct answer is A.

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