$[50]$ As you see in the right Figure, liquid phase reactions are being carried out in the glass

tube, in which light irradiation can trigger the reactions. Outside the glass tube, no reactions

are observed. In the left$-$hand side, there is a tank, in which only reactant $A$ is present at

initial time. By using the pump installed beside the tank, mixture from tank is drawn out at

pre$-$determined fixed volumetric rate $v$ and the mixture flows to the reactor inlet. The

reacted mixture goes back to the original tank. The volumes of the tank and the glass tube

are $V_T$ and $V_R,$ respectively. Be sure to use ${}_T(=\frac{V_T}{v}),$ and ${}_R(=\frac{V_R}{v})$ in all your answers.

Inside the reactor, chemical reactions are taking place as follow. The reaction rate constants

for the first and the second reactions are $k_1$ and $k_2,$ respectively. The orders of the reactions

are $0\mathrm{.}5$ and $1,$ respectively.

Reactions:

$($a$)[15]$ Derive material balance equations for A and $R$ in the glass tube reactor. If variables are functions of time, you must

designate them as such $($eg$.f(t)).$ Otherwise, you must not.

$($b$)[15]$ Derive material balance equations for A and $R$ in the tank. Same rules are applied as in the $($a$).$

$($c$)[10]$ If you want to get $R$ as much as you can, what do you suggest without changing this configuration?

$($d$)[10]$ If the glass tube reactor is replaced with a MFR$,$ what change do you expect in terms of the concentrations of A$,$ R and

$S$ in the tank compared with the original configuration? Be sure to explain why.