For a fast pre-equilibrium where Step 1 is A ⇌ I and Step 2 I → P is slow, the rate law can often be expressed in terms of the equilibrium constant K1 and a rate constant. Which expression is correct?

For a fast pre-equilibrium where Step 1 is A ⇌ I and Step 2 I → P is slow, the rate law can often be expressed in terms of the equilibrium constant K1 and a rate constant. Which expression is correct?

• A. Rate = k2 K1 [A]
• B. Rate = k1 [A][I]
• C. Rate = k3 [I]
• D. Rate = k2 [A][B]

Answer: (A)

Fast pre-equilibrium means the first step A ⇌ I reaches equilibrium quickly, and the second step I → P is the slow, rate-determining step. The overall rate is governed by the slow step, so Rate = k2[I]. At equilibrium, [I] = K1[A], where K1 is the equilibrium constant for A ⇌ I. Substituting gives Rate = k2 K1 [A]. This directly matches the expression in the option. The other forms don’t fit because they either imply the rate depends on the intermediate concentration [I] without relating it to [A], or involve a species not present in the mechanism, or assume the wrong step controls the rate.

Fast pre-equilibrium means the first step A ⇌ I reaches equilibrium quickly, and the second step I → P is the slow, rate-determining step. The overall rate is governed by the slow step, so Rate = k2[I]. At equilibrium, [I] = K1[A], where K1 is the equilibrium constant for A ⇌ I. Substituting gives Rate = k2 K1 [A]. This directly matches the expression in the option. The other forms don’t fit because they either imply the rate depends on the intermediate concentration [I] without relating it to [A], or involve a species not present in the mechanism, or assume the wrong step controls the rate.