Consider a mechanism in which A ⇌ I (fast equilibrium) is followed by I → P (slow). What is the overall rate law expressed in terms of [A]?

• A. Rate = k2 K1 [A]
• B. Rate = k2 [A]
• C. Rate = k1 [A]
• D. Rate = k-1 [A]

Answer: (A)

In this scenario, a fast equilibrium A ⇌ I sets up the relationship between the concentrations of A and the intermediate I, and a slower step I → P controls the overall rate. Because the first step is fast and reversible, it reaches equilibrium quickly, so the concentrations satisfy k1[A] = k−1[I], giving [I] = K1[A], where K1 = k1/k−1.

The overall rate is governed by the slow step, so Rate = rate of I → P = k2[I]. Substituting [I] from the equilibrium gives Rate = k2(K1[A]) = k2 K1 [A]. Since K1 = k1/k−1, this can also be written as Rate = (k2 k1 / k−1)[A].

So the rate is proportional to [A] with an effective constant k2K1, reflecting the fast pre-equilibrium that channels A into I before the slow transformation to product.