In a fast pre-equilibrium Step 1: A ⇌ I and Step 2: I → P (slow), what is the resulting rate law in terms of [A] and [B]?

• A. Rate = k2 [A]
• B. Rate = k2 K1 [A][B]
• C. Rate = k1 [A][B]
• D. Rate = k2 [A]^2

Answer: (B)

Fast pre-equilibrium: the first step A ⇌ I occurs rapidly and establishes equilibrium before the slow conversion to product. Write the forward and reverse rates as k1[A] and k−1[I], and define K1 = k1/k−1 = [I]/[A]. Because this step is at equilibrium, [I] = K1[A].

The overall rate is governed by the slow step I → P, so Rate = k2[I]. Substituting the expression for [I] gives Rate = k2 K1 [A]. This shows the reaction is first order in the initial reactant A, with an effective rate constant k2K1. There isn’t a second reactant involved in this mechanism, so a rate law that includes another concentration factor would not arise from this sequence.