The following elementary irreversible organic reaction takes place in a liquid phase reactor.

$A+B-$

It is carried out adiabatically in a flow reactor. An equal molar feed in A and $B$ enters at $27C,$ and the volumetric flow

rate is $2dm\frac{3}{sec}$ and $CA=0\mathrm{.}1$kmo$\frac{l}{m}3.$ Use the following information and answer the questions below:

$H^{0}A(273K)=-20$kca$\frac{l}{m}ol,H^{0}B(273K)=-15$kca$\frac{l}{m}ol,H^{0}C(273K)=-41$kca$\frac{l}{m}ol,C{P}_A=C{P}_B=15Ca\frac{l}{m}$olK;$C{P}_E=30$

$Ca\frac{l}{m}$olKk$=0\mathrm{.}01dm\frac{3}{?}($mol$.$sec$)$ at $27CE=10,000ca\frac{l}{m}ol$

a$)$ Plot and then analyse the conversion and temperature as a function of PFR volume up to where $x=0\mathrm{.}85.$ Discuss

the trends.

b$)$ What is the maximum inlet temperature one could have so that the boiling point of the liquid, $550K,$ would not be

exceeded even for complete conversion of $100\%.$

c$)$ Plot the heat that must be removed along the reactor $($Q vs$.$ V$)$ to maintain isothermal operation. Discuss the trend.

d$)$ Now use the following additional information and answer the question below:

$Ua=20ca\frac{l}{m}\frac{3}{m}i\frac{n}{K}$

Mass flow rate of coolant $=50\frac{g}{sec}$

$Ta0=450K,C_{\text{e}\text{q}\text{e}\text{e}\text{d}\text{a}\text{n}\text{t}}=1Ca\frac{l}{g}\frac{?}{K_{?}}$

Calculate the conversion if the reaction is carried out adiabatically in one $500dm3$ CSTR$,$ and then in two $250dm3$

CSTRs in series.$--------$Provide answer of D$.$