This relation may be derived by applying the law of mass action to the two reactions in (5.13.3-1). under the simplifying assumption that the reversible reaction between the enzyme and substrate reaches steady state much faster than the subsequent forward reaction forming the product. If the first and second reactions are denoted by subscripts 1 and 2, respectively, the law of mass action for the first (reversible) and second (forwar) reaction produces $$\begin{equation} \begin{aligned}\hat{\zeta}_{1} & =k_{F1}c^{e}c^{s}-k_{R1}c^{es},\quad\hat{\zeta}_{2}=k_{F2}c^{es}\,,\\ \hat{c}^{s} & =-\hat{\zeta}_{1},\quad\hat{c}^{p}=\hat{\zeta}_{2},\quad\hat{c}^{es}=\hat{\zeta}_{1}-\hat{\zeta}_{2}\,. \end{aligned} \label{eq529} \end{equation}$$ The total enzyme concentration remains constant at $c_{0}^{e}=c^{e}+c^{es}$ , so that$\hat{\zeta}_{1}=k_{F1}c_{0}^{e}c^{s}-\left(k_{F1}c^{s}+k_{R1}\right)c^{es}$ . Assuming that the first reaction equilibrates much faster than the second is equivalent to letting $\hat{\zeta}_{1}\approx0$ , in which case $$\begin{equation} c^{es}\approx\frac{c_{0}^{e}c^{s}}{c^{s}+K_{m}}\,,\label{eq530} \end{equation}$$ where $K_{m}=k_{R1}/k_{F1}$ . Then, $$\hat{\zeta}_{2}=\frac{V_{\max}c^{s}}{c^{s}+K_{m}}\,,$$ where $V_{\max}=k_{F2}c_{0}^{e}$ represents the maximum value of $\hat{\zeta}_{2}$ , when $K_{m}\ll c^{s}$ . In practice, choosing $k_{F1}\gg k_{F2}$ can produce the desired effect.