You are using an enzyme catalyst, E, in your reactor that decomposes biomass X: \begin{array}{l} E \\ X \quad \rightarrow \quad \text { small molecules } \end{array} It is known

that the decomposition rate depends linearly on the amount of enzyme present in thereactor: \dot{r}_{\text {र,decomp }}=10 n_{R} \quad g / h r \text { (with } n_{R} \text { in mmoll } Unfortunately, given your current reactor conditions E has a tendency to degrade into aninactive form, I. E→IIrreversible enzyme deactivation Even worse, enzyme degradation is accelerated in the presence of the inactive form, such thatthe generation of the inactive form obeys the following empirical formula: \dot{r}_{I, G B N}=0.5+2 n_{I} \quad \text { mmolhr (with } n_{I} \text { in mmol }) Assume you initially supply the reactor with 100 mmol of enzyme E and1 kg of biomass X. (a) Write and solve the transient balance for the enzyme/inactive enzyme to determine how longbefore you run out of functional enzyme.(15 pts) (b) Write and solve the transient balance for the biomass to determine how much biomass is left undecomposed after the process stops?(10 pts)