Question

In a constant volume batch reactor, enzyme E catalyzes the transformation of reactant A to product R as follows:

A -----> R

The rate of consumption of A is given as rA= 200CACE/(2 + CA) (mol/l min). If we introduce enzyme with an initial concentration of 0.001 mol/l and reactant with an initial

concentration of CA0=10 mol/l into a constant volume batch reactor and let the reaction proceed, find the time needed for the concentration of reactant to drop to 0.025 mol/l. Note that the concentration of enzyme remains unchanged during the reaction.

Answer 1

The rate of reaction is given by:

- rA= 200 CA CE / (2 + CA)

Initial concentration of enzyme ; CEo = 0.001 mol / L

The concentration of enzyme remains unchanged during the reaction.

Therefore;

CE = CEo = 0.001 mol / L

Therefore; the rate of reaction expression becomes:

- rA= 200 X 0.001 CA / (2 + CA)

which gives;

- rA = 0.2 CA / (2 + CA)

The unsteady state mole balance for a batch reactor gives:

dCA / dt = rA

- dCA / dt =  0.2 CA / (2 + CA)

which is the governing equation for CA.

The above equation can be integrated to get the concentration as a function of time.

Therefore;

The time needed for the concentration of reactant to drop to 0.025 mol/l = 110 min