In previous lesson, we have obtained molar balance equations for different reactors. Here we are going to apply that molar balance equation for a CSTR reactor to get familiar with that equation.

CSTR means **continuous stirred tank reactor**

Problem

There is a CSTR which has inflow and outflow operates at steady state. A and B are reactants which enters to the
reactor at a rate of 10.01 mol/min. A and B is react and produce C and D.

A + B → 2C + D.

at steady state A is reacted at a rate of 8.94 mol/min

Calculate followings.

a) producing rate of products

b) rate of A and B in inflow

Answer

a)

According to the reaction, one A molecule and one B molecule react together and produce **two** C molecules and
one D molecule. Therfore producing rate of C is twice as reducing rate of A.

producing rate of C = 8.94 mol/min * 2 = 17.88 mol/min

producing rate of D = 8.94 mol/min

b)

This reaction occurs in steady state. Therefore no accumulation in this reactor. Here we are going to apply our quantities in
moles. Our time basis is per **minute**.

A is reacting and reducing in the reactor. So generation is negative.

flow rate of A in inflow, F_{A}|_{in} = 10.01 mol/min

generation of A, r_{A}= -8.94 mol/min

accumulation = 0

substitute those values to the molar balance equation,

10.01 + ( - 8.94 ) = F_{A}|_{out} + 0

F_{A}|_{out} = 1.07 mol/min

Flow rates of A and B in inflow is same. Also A and B reacts 1:1. therefore reaction rate of A and B is same and also flow rate of B in outflow is same as flow rate of A in outflow.

F