Let's address the **first problem** step by step: ## Problem Restatement For a non-isothermal PFR (Plug Flow Reactor) with the reversible exothermic reaction \(A \rightleftharpoons B\), as temperature increases, the equilibrium conversion decreases but the forward reaction rate constant increases. We are to: - Derive the equation for the **Locus of Maximum Rates** on the temperature-conversion phase plane. - Explain how **adiabatic beds with interstage cooling** can be used to approximate this optimal path, and calculate the required heat duty for the intercoolers to restore the feed temperature to the optimal inlet condition for each stage. --- ## Step 1: **Reaction and Rate Law** For a reversible reaction: \[ A \rightleftharpoons B \] The rate expression is: \[ r = k_f C_A - k_r C_B \] where \(k_f\) and \(k_r\) are the forward and reverse rate constants. The equilibrium constant: \[ K_{eq} = \frac{k_f}{k_r} = \frac{C_{B,eq}}{C_{A,eq}} \] --- ## Step 2: **Non-isothermal PFR Material and Energy Balances** Mole balance for A: \[ \frac{dF_A}{dV} = -r \] Energy balance for adiabatic PFR: \[ \frac{dT}{dX} = - \frac{\Delta H_{rxn}}{C_p} \] where \( X \) is conversion, \( \Delta H_{rxn} \) is the heat of reaction (negative for exothermic), and \( C_p \) is the heat capacity. --- ## Step 3: **Rate as a Function of Conversion and Temperature** Express concentrations in terms of conversion (\(X\)): \[ C_A = C_{A}(1 - X) \] \[ C_B = C_{A} X \] So, the rate: \[ r = k_f C_{A}(1 - X) - k_r C_{A} X \] \[ r = C_{A} \left[ k_f (1 - X) - k_r X \right] \] But, \( k_r = \frac{k_f}{K_{eq}} \), so: \[ r = C_{A} k_f \left[ (1 - X) - \frac{X}{K_{eq}} \right] \] --- ## Step 4: **Temperature Dependence** The forward rate constant follows Arrhenius law: \[ k_f = k_{f} \exp\left(-\frac{E}{RT}\right) \] The equilibrium constant depends on temperature (van't Hoff equation): \[ \ln K_{eq} = -\frac{\Delta H_{rxn}}{RT} + \text{const} \] So, \[ K_{eq} = K_ \exp\left(-\frac{\Delta H_{rxn}}{RT}\right) \] --- ## Step 5: **Locus of Maximum Rates** To find the **locus of maximum rates** in the \(T\)–\(X\) plane, set \(\frac{\partial r}{\partial T} = \) at constant \(X\): \[ r = C_{A} k_f \left[ (1 - X) - \frac{X}{K_{eq}} \right] \] Take derivative with respect to \(T\): \[ \frac{\partial r}{\partial T} = C_{A} \left\{ \frac{d k_f}{dT} \left[ (1 - X) - \frac{X}{K_{eq}} \right] + k_f \left( \frac{X}{K_{eq}^2} \frac{d K_{eq}}{dT} \right) \right\} \] Set \(\frac{\partial r}{\partial T} = \): \[ \frac{d k_f}{dT} \left[ (1 - X) - \frac{X}{K_{eq}} \right] + k_f \left( \frac{X}{K_{eq}^2} \frac{d K_{eq}}{dT} \right) = \] Now, \[ \frac{d k_f}{dT} = k_f \left( \frac{E}{RT^2} \right) \] \[ \frac{d K_{eq}}{dT} = K_{eq} \left( \frac{\Delta H_{rxn}}{RT^2} \right) \] Substitute: \[ k_f \left( \frac{E}{RT^2} \right) \left[ (1 - X) - \frac{X}{K_{eq}} \right] + k_f \left( \frac{X}{K_{eq}^2} K_{eq} \frac{\Delta H_{rxn}}{RT^2} \right) = \] \[ k_f \left\{ \frac{E}{RT^2} \left[ (1 - X) - \frac{X}{K_{eq}} \right] + \frac{X}{K_{eq}} \frac{\Delta H_{rxn}}{RT^2} \right\} = \] \[ \frac{E}{RT^2} (1 - X) - \frac{E}{RT^2} \frac{X}{K_{eq}} + \frac{\Delta H_{rxn}}{RT^2} \frac{X}{K_{eq}} = \] \[ \frac{E}{RT^2} (1 - X) + \frac{X}{K_{eq}} \frac{(\Delta H_{rxn} - E)}{RT^2} = \] Divide by \(\frac{1}{RT^2}\): \[ E (1 - X) + \frac{X}{K_{eq}} (\Delta H_{rxn} - E) = \] \[ E (1 - X) = (E - \Delta H_{rxn}) \frac{X}{K_{eq}} \] \[ K_{eq} = \frac{(E - \Delta H_{rxn}) X}{E(1 - X)} \] This is the **Locus of Maximum Rates** equation in the temperature-conversion phase plane. Recall \(K_{eq}\) is a function of \(T\), so this equation defines a curve in the \(T\)-\(X\) plane. --- ## Step 6: **Adiabatic Beds with Interstage Cooling** Each adiabatic bed follows a trajectory in the \(T\)-\(X\) plane, dictated by: \[ dT = - \frac{\Delta H_{rxn}}{C_p} dX \] To approximate the locus, use several adiabatic beds in series, with intercoolers between beds to reset the temperature to the optimal inlet condition. The heat duty for each intercooler is: \[ Q_{cool} = n_{total} C_p (T_{exit} - T_{inlet}) \] where \(T_{exit}\) is the exit temperature from the previous adiabatic section, and \(T_{inlet}\) is the desired optimal inlet temperature for the next section (from the maximum rate locus). By restoring the feed temperature to the locus at each stage, you maximize the rate at each bed. --- ## Final Answers **Locus of Maximum Rates:** \[ K_{eq}(T) = \frac{(E - \Delta H_{rxn}) X}{E(1 - X)} \] with \(K_{eq}(T)\) given by \(K_ \exp\left(-\frac{\Delta H_{rxn}}{RT}\right)\). **Interstage Cooling Trajectory:** Use a series of adiabatic beds with intercoolers to reset the feed temperature to the locus, with heat duties per stage given by: \[ Q_{cool} = n_{total} C_p (T_{exit} - T_{inlet}) \] where \(T_{inlet}\) is found from the locus equation for the next stage. --- **Summary:** You derived the locus of maximum rates in the temperature-conversion phase plane as an explicit relation between \(K_{eq}(T)\) and \(X\). To operate near this optimal path, use a sequence of adiabatic beds with intercooling to repeatedly bring the feed temperature back to the locus at each stage, calculating the necessary heat removal with the given energy balance.