EXAMPLE $1$

An enzyme is produced for manufacturing a sun protection lotion. Given kinetic data for the enzyme reaction with $y_{\text{m}\text{a}\text{x}}=2,5$mmo$\frac{l}{m}3s,K_m=8\mathrm{.}9mM,$ and initial substrate $S_i=12mM,$ what would be the time required for $95\%$ conversion in a batch bioreactor?

$v=-\frac{dS}{dt}=\frac{v_{\text{m}\text{a}\text{x}}S}{K_M+S}$

$\frac{dS}{dt}=\frac{r_{\text{m}\text{a}\text{x}}S}{K_{m+1}+S}$

$-{\displaystyle \underset{0}{\overset{1}{}}}dt={\displaystyle \underset{s}{\overset{s}{}}}\frac{K_m+S}{v_{m1}S}ds$

Solution

$V_{\text{m}\text{a}\text{x}}=(2\mathrm{.}5\frac{\text{m}\text{m}\text{o}\text{l}}{m^3(s)})(\frac{3600(s)}{(h)})(\frac{1m^3}{1000(L)})=9\frac{\text{m}\text{m}\text{o}\text{l}}{hL}$

$t_{\text{h}\text{o}\text{d}\text{i}}=\frac{8\mathrm{.}9mM}{9m\frac{M}{h}}ln\frac{12}{0\mathrm{.}6}+\frac{(0\mathrm{.}95)(12)}{9}=4\mathrm{.}23h$