Let's solve the problem step by step: --- ## **1. List Given Data** - **Molar flow rate:** \( F_{A} = 50 \) mol/min - **Initial concentration:** \( C_{A} = .2 \) mol/L - **Temperature:** \( T = 450 \) K - **Pressure:** \( P = 15 \) atm (not needed for this calculation) - **Catalyst weight:** \( W = 150 \) kg - **Bulk density:** \( \rho_b = 120 \) kg/m³ - **Rate constant at 450 K:** \( k = .85 \) L/(kg·min) - **Catalyst deactivation:** \( -\dfrac{da}{dt} = k_d a \) - **Arrhenius for \( k_d \):** \( k_{d} = .12 \) min⁻¹ \( E_d = 50.2 \) kJ/mol = 50200 J/mol - **Gas constant:** \( R = 8.314 \) J/(mol·K) - **Time for conversion calculation:** \( t = 60 \) min --- ## **2. Calculate Deactivation Rate Constant \( k_d \) at 450 K** \[ k_d = k_{d} \exp\left(-\frac{E_d}{RT}\right) \] \[ k_d = .12 \exp\left(-\frac{50200}{8.314 \times 450}\right) \] Calculate the exponent: \[ \frac{50200}{8.314 \times 450} = \frac{50200}{3741.3} \approx 13.42 \] So, \[ k_d = .12 \exp(-13.42) \] \[ \exp(-13.42) \approx 1.48 \times 10^{-6} \] \[ k_d = .12 \times 1.48 \times 10^{-6} \approx 1.78 \times 10^{-7} \text{ min}^{-1} \] --- ## **3. Catalyst Activity Decay** \[ -\frac{da}{dt} = k_d a \implies \frac{da}{a} = -k_d dt \] Integrate from \( a_ = 1 \) to \( a \) and \( t = \) to \( t \): \[ \ln a = -k_d t \implies a = \exp(-k_d t) \] --- ## **4. Plug Flow Reactor Conversion with Deactivation** **For a PFR with first-order, irreversible reaction and deactivation:** The rate at any time: \[ -r_A = k a C_A \] The mole balance (in terms of catalyst weight, \( W \)): \[ \frac{dF_A}{dW} = -r_A = -k a C_A \] For a differential catalyst slice (but since \( a \) is only a function of time, it's OK to treat as a batch decay for the activity): For the whole reactor, integrate over \( W \): \[ F_{A} X = \int_^W k a(t) C_A dW \] Since it's PFR and \( a \) is uniform throughout the reactor at a given time, use the design equation: \[ W = \frac{F_{A}}{k a} \ln\left(\frac{C_{A}}{C_A}\right) \] Or, in terms of conversion \( X \) (since \( C_A = C_{A}(1-X) \)): \[ W = \frac{F_{A}}{k a} \ln\left(\frac{1}{1-X}\right) \] But due to deactivation with time, after time \( t \), \( a = \exp(-k_d t) \). --- ### **At \( t = 60 \) min:** \[ a(60) = \exp(-k_d \times 60) \] \[ a(60) = \exp(-1.78 \times 10^{-7} \times 60) \approx \exp(-1.07 \times 10^{-5}) \approx .999989 \] So **deactivation is negligible** over 60 min (since \( k_d \) is extremely small). --- ### **Calculate Conversion:** \[ W = \frac{F_{A}}{k a} \ln\left(\frac{1}{1-X}\right) \] Solve for \( X \): \[ \ln\left(\frac{1}{1-X}\right) = \frac{W k a}{F_{A}} \] \[ 1-X = \exp\left(-\frac{W k a}{F_{A}}\right) \] \[ X = 1 - \exp\left(-\frac{W k a}{F_{A}}\right) \] Plug in values: - \( W = 150 \) kg - \( k = .85 \) L/(kg·min) - \( a \approx .999989 \) - \( F_{A} = 50 \) mol/min Calculate \( W k a / F_{A} \): \[ \frac{150 \times .85 \times .999989}{50} \approx \frac{127.498}{50} = 2.55 \] \[ X = 1 - \exp(-2.55) = 1 - .078 = .922 \] --- ## **5. Time to Reach \( a = .5 \) (50% Activity)** \[ a = \exp(-k_d t) \] \[ .5 = \exp(-k_d t) \] \[ \ln(.5) = -k_d t \] \[ t = -\frac{\ln(.5)}{k_d} = \frac{.6931}{1.78 \times 10^{-7}} \] \[ t = 3.89 \times 10^6 \text{ min} \] Convert to years for reference: \[ \frac{3.89 \times 10^6}{60 \times 24 \times 365} \approx 7.4 \text{ years} \] --- ## **6. **Final Answers in Boxes** - **Calculated exit conversion after 60 minutes:** **.92** - **Time for catalyst activity to decay to .5:** **3.9 × 10⁶ min** --- ### **Summary Table** | Quantity | Value | |------------------------------------------------|-------------------| | Exit Conversion after 60 min | .92 | | Time for activity to reach .5 | 3.9 × 10⁶ min | --- #### **Boxed Answers:** > The calculated exit conversion of A after 60 minutes is **.92**, and the time required for the catalyst activity to reach .5 is **3.9 × 10⁶ min**. --- If you need detailed step-by-step calculations or clarifications on any step, let me know!