Problem $4(25/100)$: Two liquid phase, elementary reactions A $$ D and A $$ U take place

simultaneously in a continuous stirred tank reactor with a heat exchanger under steady state

conditions. The reactor volume is $100L.$ Diluted A is fed to the reactor. The total feed rate is $10$

$\frac{L}{m}in,$ the feed concentration of $A$ is $0\mathrm{.}1mo\frac{l}{L},$ and the inlet temperature is $250K.$ The constant

heat capacity of the solution is $0\mathrm{.}5k\frac{J}{\text{m}\text{o}\text{l}\text{K}}.$ The gas constant is $8\mathrm{.}314\frac{J}{\text{m}\text{o}\text{l}\text{K}}.$ Parameters are

listed below:

Heat exchange rate and coolant temperature: $UA=0\mathrm{.}01k\frac{J}{\text{m}\text{i}\text{n}\text{K}},T_a=250K.$

The reaction heat, reference rate constant, and activation energy for the reaction $AD$ :

$H_{Rrn,l}=-120k\frac{J}{m}ol,k_1=0\mathrm{.}01mi{n}^{-1}$ at $300K,E_1=41\mathrm{.}6k\frac{J}{m}ol.$

The reaction heat, reference rate constant, and activation energy for the reaction $AU$ :

$H_{Rxn,2}=50k\frac{J}{m}ol,k_2=0\mathrm{.}001mi{n}^{-1}$ at $300K,E_2=49\mathrm{.}9k\frac{J}{m}ol.$

a$)$ What would be the conversion of $A$ if the reactor is operated isothermally at $405K?$

b$)$ Derive the expression for the heat generation term $G(T)$ solely as a function of the residence

time $(),$ reaction rate constants $(k_1$ and $k_2),$ and reaction enthalpies and $H_{Rmn,2}).$

c$)$ Is it possible to operate the reactor with heat exchange in such a way that the steady state reactor

temperature is $405K?$ Justify your answer using numerical or$/$and graphical solution.