If the slow step in a mechanism involves A and B reacting to form AB, the rate law is typically:

If the slow step in a mechanism involves A and B reacting to form AB, the rate law is typically:

• A. rate = k[A][B]
• B. rate = k[A]
• C. rate = k[B]
• D. rate = k[AB]

Answer: (A)

The rate of the overall reaction is governed by the slowest step in the mechanism. If the slow step is A + B -> AB, then this step is an elementary bimolecular process, and its rate law follows the stoichiometry of the reactants in that step. That means the rate is proportional to the frequency of collisions between A and B, giving rate = k[A][B]. This makes the reaction second order overall, first order in each reactant. AB is the product of this slow step, so its concentration does not appear in the rate law for the slow step. If the slow step involved only one reactant, you’d get rate = k[A] or rate = k[B]; if AB were the reacting species, you’d expect a rate depending on [AB], which isn’t the case here.

The rate of the overall reaction is governed by the slowest step in the mechanism. If the slow step is A + B -> AB, then this step is an elementary bimolecular process, and its rate law follows the stoichiometry of the reactants in that step. That means the rate is proportional to the frequency of collisions between A and B, giving rate = k[A][B]. This makes the reaction second order overall, first order in each reactant. AB is the product of this slow step, so its concentration does not appear in the rate law for the slow step. If the slow step involved only one reactant, you’d get rate = k[A] or rate = k[B]; if AB were the reacting species, you’d expect a rate depending on [AB], which isn’t the case here.