Question

The residence time distribution, \( E \), for a non-ideal flow reactor is given in the figure. A first-order liquid phase reaction with a rate constant \( 0.2 \, \text{min}^{-1} \) is carried out in the reactor. For an inlet reactant concentration of \( 2 \, \text{mol/L} \), the reactant concentration (in \( \text{mol/L} \)) in the exit stream is: \includegraphics[width=0.5\linewidth]{q46 CE.PNG}

• A. 0.905
• B. 0.452
• C. 1.902
• D. 0.502

Hint:

In residence time distribution problems, use the given \( E(t) \) function and calculate the outlet concentration by integrating over the effective residence time range.

Answer 1

The Correct Option is A

Step 1: Expression for outlet concentration in a non-ideal reactor. For a first-order reaction in a non-ideal reactor, the outlet concentration \( C_{\text{out}} \) is given by: \[ C_{\text{out}} = C_{\text{in}} \int_0^\infty E(t) e^{-kt} \, dt, \] where: - \( C_{\text{in}} = 2 \, \text{mol/L} \) (inlet concentration), - \( k = 0.2 \, \text{min}^{-1} \) (rate constant), - \( E(t) \) is the residence time distribution function. Step 2: Define \( E(t) \) from the figure. From the figure, \( E(t) \) is a rectangular function: \[ E(t) = \begin{cases} \frac{1}{2}, & \text{if } 3 \leq t \leq 5,
0, & \text{otherwise}. \end{cases} \] Step 3: Substitute \( E(t) \) into the integral. The integral becomes: \[ C_{\text{out}} = C_{\text{in}} \int_3^5 \frac{1}{2} e^{-0.2t} \, dt. \] Substitute \( C_{\text{in}} = 2 \): \[ C_{\text{out}} = 2 \cdot \frac{1}{2} \int_3^5 e^{-0.2t} \, dt = \int_3^5 e^{-0.2t} \, dt. \] Step 4: Evaluate the integral. The integral of \( e^{-0.2t} \) is: \[ \int e^{-0.2t} \, dt = \frac{e^{-0.2t}}{-0.2}. \] Evaluate the definite integral: \[ \int_3^5 e^{-0.2t} \, dt = \left[ \frac{e^{-0.2t}}{-0.2} \right]_3^5 = \frac{1}{0.2} \left( e^{-0.6} - e^{-1.0} \right). \] Step 5: Calculate the exponential terms. \[ e^{-0.6} \approx 0.5488, \quad e^{-1.0} \approx 0.3679. \] Substitute: \[ \int_3^5 e^{-0.2t} \, dt = \frac{1}{0.2} \left( 0.5488 - 0.3679 \right) = 5 \cdot 0.1809 = 0.905. \] Step 6: Conclusion. The reactant concentration in the exit stream is \( C_{\text{out}} = 0.905 \, \text{mol/L} \).