In the mechanism A ⇌ I (fast equilibrium) followed by I -> P (slow), which step is rate-determining?

• A. The fast pre-equilibrium A ⇌ I
• B. The product formation is instantaneous
• C. The slow step I → P
• D. The overall rate is controlled by A

Answer: (C)

The key idea is that in a sequence of steps, the slowest step controls the overall rate—the rate-determining step. Here, the first step A ⇌ I is fast and reaches equilibrium quickly, so it sets up a relationship between A and I without bottlenecking the process. The production of P depends on the slow step I → P, so the pace of the reaction is limited by how fast that step occurs.

The rate of product formation comes from the slow step: rate = k2[I]. Because the first step is fast and at equilibrium, the intermediate concentration is [I] = K_eq[A]. Substituting gives rate = k2 K_eq [A], meaning the overall rate is determined by the slow step but appears first-order in A due to the fast pre-equilibrium.

So, the step that truly governs how fast P is formed is the slow one, I → P. The fast pre-equilibrium cannot limit the rate, and product formation is not instantaneous, and the overall rate is not controlled by A in isolation—the bottleneck is the slow step.