Let's break the problem down step by step as per your image: --- ## **Given Data** ### **Reactions** - \( A \rightarrow B \) (rate constant \( k_1 \)) - \( A \rightarrow C \) (rate constant \( k_2 \)) ### **Rate Equation** - \( -r_A = (k_1 + k_2)C_A \) ### **Rate Constants (Arrhenius Equation)** - \( k_1 = 2.00 \times 10^5 e^{-700/RT} \) - \( k_2 = 5.00 \times 10^6 e^{-900/RT} \) ### **Initial Concentration** - \( C_{A} = 3. \) kmol/m³ --- ## **Required** 1. **Temperature for maximum selectivity to B** 2. **Reactor volume for \( X = .60 \) at optimal temperature** 3. **Yield of B** --- # **Step 1: Maximum Selectivity to B** **Selectivity (S) of B over C:** \[ S = \frac{\text{Rate of formation of B}}{\text{Rate of formation of C}} = \frac{k_1}{k_2} \] Substitute the Arrhenius expressions: \[ S = \frac{2. \times 10^5 e^{-700/RT}}{5. \times 10^6 e^{-900/RT}} \] \[ S = .04 \, e^{(900-700)/RT} = .04 e^{200/RT} \] To **maximize selectivity to B**, maximize \( S \). ### **Analysis:** - As \( T \) decreases, \( e^{200/RT} \) increases (since \( 1/T \) becomes larger). - **Therefore, selectivity is maximum at the lowest possible temperature.** But, you need to consider practical limits where the reaction is still reasonably fast. --- # **Step 2: Reactor Volume for \( X = .60 \) at Optimal Temperature** For a **CSTR** (assuming), the performance equation is: \[ V = \frac{F_{A} X}{-r_A} \] Where: - \( F_{A} = \text{inlet molar flow rate of A} \) - \( X = \text{conversion} = .60 \) - \( -r_A = (k_1 + k_2) C_A \) For a **batch reactor**: \[ \frac{dC_A}{dt} = -(k_1 + k_2)C_A \] \[ C_A = C_{A} e^{-(k_1 + k_2)t} \] \[ X = 1 - \frac{C_A}{C_{A}} = 1 - e^{-(k_1 + k_2)t} \] \[ t = -\frac{1}{k_1 + k_2} \ln(1-X) \] For a **PFR**: \[ V = F_{A} \int_{}^{X} \frac{dX}{-r_A} \] **But since flow rate is not given, let's assume batch mode unless specified.** --- ### **Calculate \( k_1 \) and \( k_2 \) at Optimal (Lowest) Temperature** Let’s assume a **reasonable temperature** (say, 300 K) for calculation. \[ R = 8.314\, \text{J/mol·K} \] \[ k_1 = 2. \times 10^5 \exp\left(-\frac{700}{8.314 \times 300}\right) \] \[ = 2. \times 10^5 \exp\left(-2.807\right) \] \[ = 2. \times 10^5 \times .0604 = 12080 \] \[ k_2 = 5. \times 10^6 \exp\left(-\frac{900}{8.314 \times 300}\right) \] \[ = 5. \times 10^6 \exp\left(-3.610\right) \] \[ = 5. \times 10^6 \times .0271 = 135500 \] Now, sum: \[ k_1 + k_2 = 12080 + 135500 = 147580\, \text{hr}^{-1} \] --- ### **Time for \( X = .60 \):** \[ t = -\frac{1}{k_1 + k_2} \ln(1-X) \] \[ t = -\frac{1}{147580} \ln(1-.60) \] \[ t = -\frac{1}{147580} \ln(.40) \] \[ t = -\frac{1}{147580} \times (-.9163) \] \[ t = 6.21 \times 10^{-6} \text{ hr} \] \[ t \approx .0224 \text{ s} \] --- # **Step 3: Yield of B** **Yield of B** is: \[ \text{Yield} = \text{Selectivity} \times \text{Conversion} \] At \( X = .60 \) and selectivity at 300 K: \[ S = .04 \exp\left(\frac{200}{8.314 \times 300}\right) \] \[ = .04 \exp(.802) \] \[ = .04 \times 2.23 = .089 \] \[ \text{Yield} = .60 \times .089 = .0534 \text{ or } 5.34\% \] --- # **Summary Table** | Step | Value/Expression | |------------------------------------|-----------------------------| | **Temp. for max selectivity** | Lowest possible (e.g., 300K)| | **Reactor time for X=.60** | \( t = .0224 \) s | | **Yield of B** | \( 5.34\% \) | --- ## **Final Answers** 1. **Temperature for maximum selectivity to B:** Lowest possible temperature (e.g., 300 K or lower if practical). 2. **Reactor time for \( X = .60 \) at optimal temperature:** \( t = .0224 \) s (for batch reactor, at 300 K). 3. **Yield of B:** \( 5.34\% \) at \( X = .60 \) and 300 K. --- **Note:** If you have a specific reactor type or different temperature, you can substitute those values accordingly!