Chapter 13, Problem 13.6P

The elementary irreversible liquid-phase reaction

$\text{A}\text{+}2\text{B}\underset{}{\to }\text{C}$

is to be carried out in a semibatch reactor in which B is fed to A. The volume of A in the reactor is 10 dm³, the initial concentration of A in the reactor is 5 mol/dm³, and the initial temperature in the reactor is 27°C. Species B is fed at a temperature of 52°C and a concentration of 4 M. It is desired to obtain at least 80% conversion of A in as short a time as possible, but at the same time the temperature of the reactor must not rise above 130°C. You should try to make approximately 120 mol of C in a 24-hour day, allowing for 30 minutes to empty and fill the reactor between each batch. The coolant flow rate through the reactor is 2000 mol/min. There is a heat exchanger in the reactor.

1. (a) What volumetric feed rate (dm³/min) do you recommend?
2. (b) How would your answer or strategy change if the maximum coolant rate dropped to 200 mol/min? To 20 mol/min?

Additional information:

$\Delta H_{\text{Rx}}^{°}\text{=}-55.000\text{cal/mol}\text{A}$

${\text{C}}_{{\text{P}}_{\text{A}}}=35\text{cal/mol}\cdot \text{K,}{\text{C}}_{{\text{P}}_{\text{B}}}=20\text{cal/mol}\cdot \text{K,}{\text{C}}_{{\text{P}}_{\text{C}}}=75\text{cal/mol}\cdot \text{K}$

$k=0.0005\frac{{\text{dm}}^6}{{\text{mol}}^{\text{2}}\cdot \min }\text{at}\text{27}°\text{C}\text{}\text{with}\text{}E=8000\text{cal/mol}$

$UA=2500\frac{\text{cal}}{\min \cdot \text{K}}\text{with}{T}_{\text{a}}=17°\text{C}$

C_P(coolant) = 18 cal/mol · K