Let's solve each task step by step as requested: --- # 1. **Derive Equilibrium Conversion Expression** For the reaction A ⇌ B, the equilibrium constant is \( K = \frac{C_B}{C_A} \). If \( X_{eq} \) is the equilibrium conversion, then: - \( C_A = C_{A} (1 - X_{eq}) \) - \( C_B = C_{A} X_{eq} \) So, \[ K = \frac{C_B}{C_A} = \frac{C_{A} X_{eq}}{C_{A}(1 - X_{eq})} = \frac{X_{eq}}{1 - X_{eq}} \] Thus, \[ X_{eq} = \frac{K}{1+K} \] --- # 2. **Write PFR Mole Balance (Reversible Reaction)** For a PFR (Plug Flow Reactor), the mole balance is: \[ \frac{dF_A}{dV} = -r_A \] For a reversible reaction, \[ r_A = k C_A - k_{-1} C_B \] But since \( k_{-1} = \frac{k}{K} \), \[ r_A = k C_A - \frac{k}{K} C_B \] Substitute \( C_A = C_{A}(1-X) \) and \( C_B = C_{A} X \): \[ r_A = k C_{A} (1-X) - \frac{k}{K} C_{A} X = k C_{A} \left( 1-X - \frac{X}{K} \right ) \] \[ r_A = k C_{A} \left( 1-X \left(1 + \frac{1}{K}\right) \right ) \] --- # 3. **Temperature-Conversion Relation (Adiabatic)** For an adiabatic reactor, The energy balance gives: \[ T = T_ + \frac{ -\Delta H }{ \rho C_p } C_{A} X \] where \( \Delta H \) is the heat of reaction (per mole A), \( \rho \) is density, \( C_p \) is heat capacity. --- # 4. **Calculate Equilibrium Conversion** Given \( K = 3 \): From step 1, \[ X_{eq} = \frac{K}{1+K} = \frac{3}{1+3} = \frac{3}{4} = .75 \] --- # 5. **Reactor Volume for 90% of Equilibrium Conversion** 90% of equilibrium conversion is: \[ X_{90\%} = .9 \times .75 = .675 \] For a PFR, \[ V = \int_^{X_{90\%}} \frac{F_{A}}{-r_A} dX \] Given: \( F_{A} = C_{A} v_ = 30 \times .5 = 15 \) mol/min \( k = 1. \) min\(^{-1} \) From step 2, \[ -r_A = k C_{A} \left(1 - X \left(1 + \frac{1}{K}\right)\right) \] \[ -r_A = 1 \times 30 \left(1 - X \left(1 + \frac{1}{3}\right)\right) = 30 \left(1 - X \cdot 1.333\right) \] \[ V = \int_^{.675} \frac{15}{30(1 - 1.333X)} dX = \int_^{.675} \frac{.5}{1 - 1.333X} dX \] \[ V = .5 \int_^{.675} \frac{dX}{1 - 1.333X} \] Let \( a = 1.333 \): \[ V = .5 \left[ -\frac{1}{a} \ln|1 - aX| \right]_^{.675} \] \[ = .5 \left[ -\frac{1}{1.333} \left( \ln|1 - 1.333 \times .675| - \ln|1| \right) \right] \] \[ 1 - 1.333 \times .675 = 1 - .9 = .1 \] \[ V = .5 \left( -\frac{1}{1.333} (\ln .1 - ) \right ) \] \[ = .5 \times \frac{1}{1.333} \times 2.3026 \] \[ = .5 \times .750 \times 2.3026 \approx .863\, \text{m}^3 \] --- # 6. **Effect of Increasing Feed Temperature** Increasing feed temperature increases the initial reaction rate due to increased \( k \), but for an exothermic reaction (since most reactions are exothermic unless specified), the equilibrium constant \( K \) decreases with increasing temperature, resulting in a lower equilibrium conversion. Thus, while rate improves, the maximum achievable conversion decreases at higher temperature. --- # **Summary of Answers** 1. \( X_{eq} = \frac{K}{1+K} \) 2. PFR mole balance: \( \frac{dF_A}{dV} = -r_A = kC_{A}[1 - X(1+\frac{1}{K})] \) 3. Adiabatic relation: \( T = T_ + \frac{ -\Delta H }{ \rho C_p } C_{A} X \) 4. \( X_{eq} = .75 \) 5. Required reactor volume for 90% of equilibrium conversion is approximately \( .863\, \text{m}^3 \) 6. Increasing feed temperature decreases equilibrium conversion for exothermic reactions due to a decrease in \( K \).