Consider the following mechanism for the enzymecatalyzed reaction:
$$\mathrm{E}+\mathrm{S} \frac{k_{1}}{k_{-1}} \mathrm{ES} \quad \text { (fast equilibrium) }$$
$$\mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} \quad \text { (slow) }$$
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.)