Question

Consider the following homogeneous isothermal liquid-phase parallel reactions carried out in three reactor configurations (having identical volumes and same operating temperatures), as shown in the figure.
\[ A + B \rightarrow D \qquad \text{Rate of formation of D: } r_D = k_1 C_A C_B \]\[ A + B \rightarrow U \qquad \text{Rate of formation of U: } r_U = k_2 C_A^2 C_B \]
where D is the desired product and U is the undesired product. The inlet concentrations of A and B are the same in all three configurations (\(C_{A0}=C_{B0}=1\ \text{mol L}^{-1}\)). The total molar feed flow rate of A (\(F_{A0}\)) and that of B (\(F_{B0}\)) are the same in all three configurations (\(F_{A0}=F_{B0}=10\ \text{mol min}^{-1}\)). In configuration I, \(F_{A0}\) is equally distributed among all the inlets. Similarly, in configuration III, \(F_{B0}\) is equally distributed among all the inlets. Assuming plug flow behavior, at steady state, which one of the following statements is CORRECT?

• A. Configuration I and Configuration III give the same selectivity of the desired product.
• B. Configuration I gives the highest selectivity of the desired product.
• C. Configuration II gives the highest selectivity of the desired product.
• D. Configuration III gives the highest selectivity of the desired product.

Hint:

The selectivity ratio rD/rU depends only on 1/CA (CB cancels out), so whichever configuration keeps the concentration of A lowest throughout the reactor gives the best selectivity of D.

Answer 1

The Correct Option is B

Step 1: Write the instantaneous selectivity of D.
\[ \frac{r_D}{r_U} = \frac{k_1}{k_2}\cdot\frac{1}{C_A} \]
C_B cancels; selectivity depends only on C_A, higher when C_A is lower.
Step 2: General rule.
Since the undesired reaction is second order in A, keeping C_A low throughout maximizes selectivity, achieved by feeding A gradually along the reactor length.
Step 3: Apply to the three configurations.
Configuration I feeds A via distributed side streams, keeping C_A low everywhere - highest selectivity. Configuration III does the opposite (A enters fully at inlet) - lowest selectivity. Configuration II is intermediate.
Step 4: Rule out option A.
The reactions are not symmetric in A and B (order 2 in A for the undesired reaction, order 1 in B), so distributing A versus B does not give equal selectivity.
Step 5: Conclusion.
\[ \boxed{\text{Configuration I gives the highest selectivity of the desired product.}} \]