For a mechanism in which Step 1: A + B → C is slow and Step 2: C → P is fast, what is the rate law?

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

Answer: (A)

The key idea is that the overall rate is governed by the slowest step in a mechanism, the rate-determining step. Here, the first step A + B → C is slow, so it sets how fast the reaction proceeds. Since this step is bimolecular, its rate depends on both reactants A and B. That gives the rate law as Rate = k1[A][B].

The second step C → P is fast and just converts the intermediate C into product. Because it happens quickly, it doesn’t limit the overall rate, and the intermediate C does not appear in the rate law.

So the expression Rate = k1[A][B] is the one that matches a slow first step with a bimolecular collision between A and B.