Question

An enzyme catalyzes the conversion of substrate A into product B. The rate equation for this reaction is \[ -r_A = \frac{C_A}{5 + C_A} \;\; \text{mol L}^{-1}\text{min}^{-1} \] 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.}

Hint:

For a CSTR with variable-rate kinetics, evaluate the rate at the exit concentration and use $F_{A0}X = -r_A V$ to get the reactor volume.

Answer 1

Correct Answer: 100

For an ideal MFR (CSTR) at steady state, the mole balance on A is
\[ F_{A0} - F_A = -r_A V, \] where $F_{A0}$ and $F_A$ are inlet and outlet molar flow rates of A, and $V$ is reactor volume.
Step 1: Express flows in terms of concentration and volumetric flow rate.
\[ F_{A0} = v_0 C_{A0}, \qquad F_A = v_0 C_A, \] with $v_0 = 10 \text{ L min}^{-1}$ and $C_{A0} = 10 \text{ mol L}^{-1}$.
Define conversion $X$ as
\[ X = \frac{F_{A0} - F_A}{F_{A0}} = \frac{C_{A0} - C_A}{C_{A0}}. \] Given $X = 0.5$, the outlet concentration is
\[ C_A = C_{A0}(1-X) = 10(1-0.5) = 5 \text{ mol L}^{-1}. \] Step 2: Evaluate the reaction rate at the exit concentration.
\[ -r_A = \frac{C_A}{5 + C_A} = \frac{5}{5+5} = \frac{5}{10} = 0.5 \text{ mol L}^{-1}\text{min}^{-1}. \] Step 3: Use the design equation $F_{A0X = -r_A V$.}
\[ F_{A0}X = v_0 C_{A0} X = 10 \times 10 \times 0.5 = 50 \text{ mol min}^{-1}. \] Thus,
\[ V = \frac{F_{A0}X}{-r_A} = \frac{50}{0.5} = 100 \text{ L}. \] Final Answer:
The required MFR volume is \[ \boxed{100 \text{ L}}. \]