solve steady state of simple product reaction

consider these biochemical reactions (simpler than usual michaelis-menten setup):


$E + S \rightarrow^{k\_{f}} ES$


$ES \rightarrow^{k\_{r}} E + S$


$E$ and $S$ reversibly make $ES$. the forward/reverse rate constants are equal: $k\_{f} = k\_{r} = k$. how can you solve for steady state concentration of $[ES]$ given initial amounts $[E]\_{0}, [S]\_{0}$ and assuming $[ES]\_0 = 0$?


$d[ES]/dt = k[E][S] - k[ES] = 0$


rate constant cancels:


$[ES] = [E][S]$


then $[E] = E\_{t} - [ES]$ where $E\_{t}$ is total enzyme concentration. by substitution:


$[ES] = [S]([E]\_{t} - [ES])$


$[ES] = [S][E]\_{t} - [ES][S]$


$[ES] + [ES][S] = [S][E]\_{t}$


$[ES] = \frac{[S][E]\_{t}}{1 + [S]}$


but if $E\_{0} = 10, S\_{0} = 5$ then $[ES] = (5 \* 10) / (5 + 1) \approx 8.3$ but the right answer is closer to 4. what's wrong with this?




There are $3$ equations for $3$ unknowns $[E],[S],[ES]$ given by


$$\left\{\begin{array}{l}
k\_f[E][S]=k\_r[ES],\\
[E]+[ES]=E\_0,\\
[S]+[ES]=S\_0.
\end{array}\right.$$


We can use the last two equations to cancel $[E]=E\_0-[ES]$ and $[S]=S\_0-[ES]$ and plug them into the first equation to get a quadratic equation for $[ES]$ given by


$$k\_f(E\_0-[ES])(S\_0-[ES])=k\_r[ES].$$


Then we plug in numbers $k\_r/k\_f=1$, $E\_0=10$, $S\_0=5$ to obtain $[ES]=8\pm\sqrt{14}$. Since we have $[ES]\leq\min\{E\_0,S\_0\}=5$, the solution is $[ES]=8-\sqrt{14}\approx 4.26$.


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