Question

You conducted a batch experiment in the lab for 10 minutes to degrade a toxic compound, which follows first-order kinetics. The compound degrades from \( 2 \times 10^{-3} \, {M} \) to \( 2 \times 10^{-4} \, {M} \). The information from the lab experiment will be used to design a plug flow reactor in field conditions.
Given field conditions:
• Flow rate of contaminated water to be treated: 1 m\(^3\)/hour
• Concentration of toxic compound in contaminated water: \( 5 \times 10^{-1} \, {M} \)
• Target concentration of toxic compound in treated water: \( 1 \times 10^{-4} \, {M} \)
• Temperature is the same in lab and field conditions.
The required volume of the plug flow reactor is ________ m\(^3\). (rounded off to two decimal places)

Hint:

For first-order reactions in plug flow reactors, the volume is directly related to the flow rate, rate constant, and concentration change. Make sure the rate constant is determined under the same conditions as the field operation.

Answer 1

Step 1: First-order reaction rate constant from the batch experiment.
The reaction follows first-order kinetics, so the rate law is given by: \[ \ln \left( \frac{C_0}{C_t} \right) = k \cdot t \] where:
\( C_0 = 2 \times 10^{-3} \, {M} \) is the initial concentration
\( C_t = 2 \times 10^{-4} \, {M} \) is the concentration after time \( t \)
\( k \) is the first-order rate constant
\( t = 10 \, {minutes} = \frac{10}{60} \, {hours} = \frac{1}{6} \, {hours} \)
Rearranging the equation to solve for \( k \): \[ k = \frac{1}{t} \ln \left( \frac{C_0}{C_t} \right) \] Substitute the values: \[ k = \frac{1}{\frac{1}{6}} \ln \left( \frac{2 \times 10^{-3}}{2 \times 10^{-4}} \right) = 6 \ln \left( 10 \right) = 6 \times 2.3026 = 13.8156 \, {hour}^{-1} \] Step 2: Use the rate constant to calculate the reactor volume.
For a plug flow reactor, the volume \( V \) can be determined from the equation for first-order kinetics: \[ V = \frac{Q}{k} \ln \left( \frac{C_0}{C_t} \right) \] where:
\( Q = 1 \, {m}^3/{hour} \) is the flow rate
\( C_0 = 5 \times 10^{-1} \, {M} \) is the initial concentration in the field
\( C_t = 1 \times 10^{-4} \, {M} \) is the target concentration in treated water
\( k = 13.8156 \, {hour}^{-1} \) is the rate constant Substitute the values: \[ V = \frac{1}{13.8156} \ln \left( \frac{5 \times 10^{-1}}{1 \times 10^{-4}} \right) = \frac{1}{13.8156} \ln (5000) \] \[ V = \frac{1}{13.8156} \times 8.5172 = 0.616 \, {m}^3 \] Step 3: Conclusion.
The required volume of the plug flow reactor is approximately: \[ \boxed{0.62} \, {m}^3 \]