Chapter 13, Problem 13.66QP

Consider the following mechanism for the enzyme-catalyzed reaction:

$\begin{array}{ll}\begin{array}{c}\text{E}+\text{S}\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons }}\text{ES}\\ \text{ES}\stackrel{k_2}{\to }\text{E}+\text{P}\end{array}\hfill & \begin{array}{l}(\text{fast equilibrium})\hfill \\ (\text{slow})\hfill \end{array}\hfill \end{array}$

Derive an expression for the rate law of the reaction in terms of the concentrations of E and S. (Hint: To solve for [ES], make use of the fact that, at equilibrium, the rate of forward reaction is equal to the rate of the reverse reaction.)