# Step-by-Step Solution Let's solve each part of the question systematically for the first-order reaction in a Plug Flow Reactor (PFR). ## **Given Data** - Reaction: \( A \rightarrow B \) - Rate law: \( -r_A = kC_A \) - Rate constant: \( k = .25\, \text{min}^{-1} \) - Inlet concentration: \( C_{A} = 2\, \text{mol/L} \) - Volumetric flow rate: \( v_ = 5\, \text{L/min} \) - Reactor volume: \( V = 20\, \text{L} \) --- ## **1. Calculate the Conversion of A (\(X\))** For a first-order reaction in a PFR, the design equation is: \[ \int_{C_{A}}^{C_A} \frac{dC_A}{-r_A} = \frac{V}{v_} \] Substitute the rate law \( -r_A = kC_A \): \[ \int_{C_{A}}^{C_A} \frac{dC_A}{kC_A} = \frac{V}{v_} \] \[ \frac{1}{k}\int_{C_{A}}^{C_A} \frac{dC_A}{C_A} = \frac{V}{v_} \] \[ \frac{1}{k} \ln\left(\frac{C_A}{C_{A}}\right) = \frac{V}{v_} \] \[ \ln\left(\frac{C_A}{C_{A}}\right) = k \frac{V}{v_} \] \[ \frac{C_A}{C_{A}} = \exp\left(-k \frac{V}{v_}\right) \] \[ X = 1 - \frac{C_A}{C_{A}} \] Plug in the values: \[ k = .25\, \text{min}^{-1}, \quad V = 20\, \text{L}, \quad v_ = 5\, \text{L/min} \] \[ k \frac{V}{v_} = .25 \times \frac{20}{5} = .25 \times 4 = 1 \] \[ \frac{C_A}{C_{A}} = e^{-1} = .3679 \] \[ X = 1 - .3679 = .6321 \] **Final Answer (1):** The conversion of A is **.632** (or **63.2%**). --- ## **2. Determine the Exit Concentration of A (\(C_A\))** \[ C_A = C_{A} \cdot (1 - X) \] But from above, \[ C_A = C_{A} \cdot e^{-kV/v_} \] \[ C_A = 2 \cdot .3679 = .7358\, \text{mol/L} \] **Final Answer (2):** The exit concentration of A is **.736 mol/L**. --- ## **3. Calculate the Space Time (\(\tau\))** Space time is defined as: \[ \tau = \frac{V}{v_} \] Given: \[ V = 20\, \text{L}, \quad v_ = 5\, \text{L/min} \] \[ \tau = \frac{20}{5} = 4\, \text{min} \] **Final Answer (3):** The space time (\(\tau\)) is **4 min**. --- ## **4. Reactor Volume Required for 80% Conversion (\(X = .80\))** Let \( X = .80 \), so \( \frac{C_A}{C_{A}} = 1 - X = .20 \). For a first-order PFR: \[ \frac{C_A}{C_{A}} = e^{-kV/v_} \] \[ .20 = e^{-kV/v_} \] Take the natural log on both sides: \[ \ln(.20) = -kV/v_ \] \[ -1.6094 = -kV/v_ \] \[ kV/v_ = 1.6094 \] \[ V = \frac{1.6094 \cdot v_}{k} \] \[ V = \frac{1.6094 \cdot 5}{.25} = \frac{8.047}{.25} = 32.19\, \text{L} \] **Final Answer (4):** The reactor volume required for 80% conversion is **32.2 L**. --- ## **Summary Table** | Part | Quantity | Value | |------|---------------------------------|-----------------| | 1 | Conversion of A (\(X\)) | .632 (63.2%) | | 2 | Exit concentration (\(C_A\)) | .736 mol/L | | 3 | Space time (\(\tau\)) | 4 min | | 4 | Volume for 80% conversion (\(V\))| 32.2 L |