Let's break down the **adiabatic reactor** problem step-by-step. --- ## **Given Data** ### **Reaction** \[ \text{Cyclopentadiene (C)} + \text{Benzoquinone (B)} \rightarrow \text{Product (M)} \] ### **Kinetic Data** - \( k = 6.853 \times 10^{11} e^{-E/RT} \) (m³/mole·ksec) - \( E = 76.480 \) kJ/mole - \( C_{A} = .204 \) kmole/m³ (Cyclopentadiene) - \( C_{B} = .408 \) kmole/m³ (Benzoquinone) - **Feed rate**: \( 142 \) cm³/sec \( = 1.42 \times 10^{-4} \) m³/sec - **Reactor volume**: \( 279 \) cm³ \( = 2.79 \) L \( = 2.79 \times 10^{-3} \) m³ - **Inlet temperature**: \( 25^\circ C = 298 \) K ### **Thermodynamics** - \( \Delta H_R^\circ = -548 \) kJ/mole at 25°C - \( C_p = 4.2 \) J/cm³·K \( = 420 \) J/L·K \( = 420 \) kJ/m³·K --- ## **To Find** - **(a)** Final (effluent) temperature, \( T \) - **(b)** Fractional conversion, \( X \) --- ## **Step 1: Moles of Reactants per Second** **Volumetric flow rate**: \( Q = 1.42 \times 10^{-4} \) m³/sec \[ \text{Moles of C/sec} = C_{A} \cdot Q = .204 \times 1.42 \times 10^{-4} = 2.90 \times 10^{-5} \text{ kmol/sec} \] \[ \text{Moles of B/sec} = C_{B} \cdot Q = .408 \times 1.42 \times 10^{-4} = 5.79 \times 10^{-5} \text{ kmol/sec} \] --- ## **Step 2: Adiabatic Energy Balance** For **adiabatic** reactors: \[ \text{Enthalpy change} = \text{Heat of reaction} \times \text{moles reacted} \] \[ \Delta T = \frac{-\Delta H_R \cdot n_{reacted}}{C_p \cdot Q} \] Let \( X \) = fractional conversion of limiting reactant (Cyclopentadiene). \[ n_{reacted}/\text{sec} = X \cdot \text{moles of C entering/sec} \] So, \[ \Delta T = \frac{-\Delta H_R \cdot (C_{A} Q X)}{C_p Q} = \frac{-\Delta H_R \cdot C_{A} X}{C_p} \] \[ T = T_{inlet} + \Delta T = 298 + \frac{-(-548,000) \cdot .204 \cdot X}{4,200,000} \] \[ T = 298 + \frac{111,792 X}{4,200} \] \[ T = 298 + 26.62 X \] --- ## **Step 3: Reactor Design Equation (CSTR)** For **CSTR**: \[ \text{Mole balance: } F_{A} - F_{A} + r V = \] \[ F_{A} - F_{A} = \text{rate of disappearance of A} \] \[ F_{A} = F_{A}(1 - X) \] \[ r = k(T) C_{A} C_{B} \] \[ C_{A} = C_{A}(1 - X) \] \[ C_{B} = C_{B} - C_{A} X \] \[ X = \frac{\text{rate} \times V}{F_{A}} = \frac{k(T) C_{A} C_{B} V}{C_{A} Q} \] --- ## **Step 4: Arrhenius Equation for \( k(T) \)** \[ k(T) = 6.853 \times 10^{11} \exp\left(-\frac{76,480}{8.314 T}\right) \] --- ## **Step 5: Solve Simultaneously** 1. **Expression for \( T \):** \[ T = 298 + 26.62 X \] 2. **Expression for \( X \):** \[ X = \frac{k(T) C_{A} C_{B} V}{C_{A} Q} \] where - \( C_{A} = C_{A}(1 - X) \) - \( C_{B} = C_{B} - C_{A} X \) - \( V = 2.79 \times 10^{-3} \) m³ - \( Q = 1.42 \times 10^{-4} \) m³/sec --- ## **Step 6: Substitute Values** \[ X = \frac{k(T) \cdot .204 (1 - X) \cdot (.408 - .204 X) \cdot 2.79 \times 10^{-3}}{.204 \cdot 1.42 \times 10^{-4}} \] \[ X = \frac{k(T) \cdot (1 - X) \cdot (.408 - .204 X) \cdot 2.79 \times 10^{-3}}{1.42 \times 10^{-4}} \] --- ## **Step 7: Iterative Solution** ### **Guess values of \( X \) and solve for \( T \), then use \( T \) to solve for \( k(T) \) and substitute back until convergence.** #### **Try \( X = .5 \) (50% conversion):** \[ T = 298 + 26.62 \times .5 = 311.3 \text{ K} \] \[ k(311.3) = 6.853 \times 10^{11} \exp\left(-\frac{76,480}{8.314 \times 311.3}\right) \] \[ \frac{76,480}{8.314 \times 311.3} = \frac{76,480}{2,588.3} = 29.55 \] \[ k(311.3) = 6.853 \times 10^{11} \exp(-29.55) = 6.853 \times 10^{11} \times 1.45 \times 10^{-13} = .099 \text{ m}^3/(\text{mol·ksec}) \] Now, \[ C_A = .204(1-.5) = .102 \] \[ C_B = .408 - .204 \times .5 = .306 \] \[ X = \frac{.099 \times .102 \times .306 \times 2.79 \times 10^{-3}}{1.42 \times 10^{-4}} \] \[ = \frac{.099 \times .102 \times .306 \times 2.79 \times 10^{-3}}{1.42 \times 10^{-4}} \] Calculate numerator: - \( .099 \times .102 = .010098 \) - \( .010098 \times .306 = .0030899 \) - \( .0030899 \times 2.79 \times 10^{-3} = 8.6248 \times 10^{-6} \) \( X = \frac{8.6248 \times 10^{-6}}{1.42 \times 10^{-4}} = .0608 \) This value is much less than the assumed .5. Let's try a smaller \( X \). --- #### **Try \( X = .1 \):** \[ T = 298 + 26.62 \times .1 = 300.66 \text{ K} \] \[ k(300.66) = 6.853 \times 10^{11} \exp\left(-\frac{76,480}{8.314 \times 300.66}\right) \] \[ 8.314 \times 300.66 = 2,501.5 \] \[ \frac{76,480}{2,501.5} = 30.59 \] \[ \exp(-30.59) = 5.20 \times 10^{-14} \] \[ k = 6.853 \times 10^{11} \times 5.20 \times 10^{-14} = .0356 \] Now: \[ C_A = .204 \times .9 = .1836 \] \[ C_B = .408 - .204 \times .1 = .3876 \] \[ X = \frac{.0356 \times .1836 \times .3876 \times 2.79 \times 10^{-3}}{1.42 \times 10^{-4}} \] Calculate numerator: - \( .0356 \times .1836 = .00654 \) - \( .00654 \times .3876 = .002537 \) - \( .002537 \times 2.79 \times 10^{-3} = 7.083 \times 10^{-6} \) \[ X = \frac{7.083 \times 10^{-6}}{1.42 \times 10^{-4}} = .0499 \] Still too low. Let's try \( X = .05 \): --- #### **Try \( X = .05 \):** \[ T = 298 + 26.62 \times .05 = 299.33 \text{ K} \] \[ 8.314 \times 299.33 = 2,488.6 \] \[ \frac{76,480}{2,488.6} = 30.74 \] \[ \exp(-30.74) = 4.72 \times 10^{-14} \] \[ k = 6.853 \times 10^{11} \times 4.72 \times 10^{-14} = .0323 \] \[ C_A = .204 \times .95 = .1938 \] \[ C_B = .408 - .204 \times .05 = .3978 \] \[ X = \frac{.0323 \times .1938 \times .3978 \times 2.79 \times 10^{-3}}{1.42 \times 10^{-4}} \] Numerator: - \( .0323 \times .1938 = .006263 \) - \( .006263 \times .3978 = .002492 \) - \( .002492 \times 2.79 \times 10^{-3} = 6.954 \times 10^{-6} \) \[ X = \frac{6.954 \times 10^{-6}}{1.42 \times 10^{-4}} = .049 \] So, at \( X \approx .05 \), the calculation is consistent. --- ## **Step 8: Final Answers** ### **(a) Temperature of effluent:** \[ T = 298 + 26.62 \times .05 = 299.33~\text{K} = 26.2^\circ C \] ### **(b) Fraction conversion:** \[ X = .05 = 5\% \] --- ## **Summary Table** | Quantity | Value | |----------|-------| | **Fractional Conversion (X)** | **.05** or **5%** | | **Effluent Temperature (T)** | **299.3 K** or **26.2°C** | --- ### **Conclusion** - **Fractional conversion** is **5%**. - **Effluent temperature** is **26.2°C**. **Note:** For higher conversion, you would need either higher inlet temperature or a larger reactor volume. --- Let me know if you need the calculation for the **fractional conversion in a batch reactor** or **for the mixed-feed (adiabatic operation)**!