Question

A first-order liquid phase reaction \( \text{A} \rightarrow \text{B} \) is carried out in two isothermal plug flow reactors (PFRs) of volume \( 1 \, \text{m}^3 \) each, connected in series. The feed flow rate and concentration of \( \text{A} \) to the first reactor are \( 10 \, \text{m}^3/\text{h} \) and \( 1 \, \text{kmol}/\text{m}^3 \), respectively. At steady-state, the concentration of \( \text{A} \) at the exit of the second reactor is \( 0.2 \, \text{kmol}/\text{m}^3 \). If the two PFRs are replaced by two equal-volume continuously stirred tank reactors (CSTRs) to achieve the same overall steady-state conversion, the volume of each CSTR, in \( \text{m}^3 \), is:

• A. 1.54
• B. 3.84
• C. 7.28
• D. 1.98

Hint:

For first-order liquid phase reactions, there is conversion to PFR and then in CSTRs.

Answer 1

The Correct Option is A

Overall Conversion: \[ X = 1 - \frac{C_{A2}}{C_{A1}} = 0.8. \] For PFR: \[ C_A = C_{A0} e^{-k \tau_1}, \] where \[ \tau_1 = \frac{V_1}{v} = 0.1. \] For Overall: \[ C_A = C_{A0} e^{-k (\tau_1 + \tau_2)}, \] where \[ \tau_2 = \frac{V_2}{v} = 0.1. \] Substitute the given values: \[ 0.2 = 1 e^{-k \cdot 0.2}. \] Taking the natural logarithm: \[ \ln(0.2) = -0.2k. \] \[ k = \frac{5 \ln(5)}{1} = 8.047. \] Replacing Both PFR with Two CSTRs: Both CSTRs have equal volume. \[ \frac{C_{A2}}{C_{A1}} = \frac{1}{1 + k \tau}. \] For the first CSTR: \[ \frac{C_{A1}}{C_{A0}} = \frac{1}{1 + k \tau}. \] For the second CSTR: \[ \frac{C_{A2}}{C_{A1}} = \frac{1}{1 + k \tau}. \] Overall: \[ \frac{C_{A2}}{C_{A0}} = \left(\frac{1}{1 + k \tau}\right)^2. \] Substitute the values: \[ \frac{C_{A2}}{C_{A0}} = 0.2 = \left(\frac{1}{1 + 8.047 \cdot \tau}\right)^2. \] Solving for \( \tau \): \[ \sqrt{0.2} = \frac{1}{1 + 8.047 \cdot \tau}. \] \[ 1 + 8.047 \cdot \tau = \frac{1}{\sqrt{0.2}}. \] \[ \tau = \frac{0.15316}{2} = 0.15316 \, \text{s}. \] Volume of Each CSTR: \[ V = \tau v = 1.5316 \, \text{m}^3. \] Final Answer: The volume of each CSTR is: \[ V = 1.5316 \, \text{m}^3. \]