Q.59
An enzyme catalyzes the conversion of substrate A into product B.
The rate equation for this reaction is

−r_A = C_A / (5 + C_A) mol L⁻¹ min⁻¹

Substrate A at an initial concentration of 10 mol L⁻¹ enters an ideal mixed flow reactor (MFR)
at a flow rate of 10 L min⁻¹.
The volume of the MFR required for 50% conversion of substrate to product is
___________ L.

MFR Volume Calculation for 50% Conversion Using Nonlinear Enzyme Kinetics

Introduction

Reactor sizing problems frequently appear in GATE Biotechnology, Chemical Engineering, and Bioprocess exams. This question considers an enzyme-catalyzed reaction happening in a Mixed Flow Reactor (MFR) using a nonlinear rate law.

The rate equation is:

−rA = CA / (5 + CA) mol L⁻¹ min⁻¹

Given feed concentration and flow rate, we calculate the reactor volume required to achieve 50% conversion.

Problem Data

• Rate law: −rA = CA / (5 + CA)
• Feed concentration: CA0 = 10 mol L⁻¹
• Flow rate: Q = 10 L min⁻¹
• Desired conversion: X = 0.50

Step-by-Step Solution

1. Outlet Concentration

CA = CA0(1 − X) = 10(1 − 0.5) = 5 mol L⁻¹

2. Reaction Rate at Reactor Exit

−rA = 5 / (5 + 5) = 5/10 = 0.5 mol L⁻¹ min⁻¹

3. MFR Design Equation

For an ideal MFR:

V = (F_A0 × X) / (−rA)

Where FA0 = Q × CA0 = 10 × 10 = 100 mol min⁻¹

4. Substitute Values

V = (100 × 0.5) / 0.5 = 100 L

Final Answer

Required MFR Volume = 100 L

Conclusion

Applying standard reactor material balances and substituting the given nonlinear rate expression, the volume of an ideal MFR required to achieve 50% conversion of substrate A to product B is:

⭐ 100 L