Let's solve the problem step by step: --- ## **Given Data** - \( x_{AL} = \) (Liquid phase bulk mole fraction of A) - \( y_{AG} = .02 \) (Gas phase bulk mole fraction of A) - **Overall mass transfer coefficient:** \( K_y = 5 \) kmol A/(h·m²·atm) - **Relationship between individual coefficients:** \( k'_x = 2.5 k'_y \) - **System conditions:** \( T = 273.15 \) K, \( P = 2 \) atm - **Equilibrium (Henry's Law):** \( y = 5x \) - **Equal molar counter-diffusion** --- ## **a) Interface Concentrations (\( x_{Ai} \), \( y_{Ai} \))** ### **Mass Transfer Across Gas Film** \[ N_A = K_y (y_{AG} - y_{Ai}) = K_x (x_{Ai} - x_{AL}) \] Using Henry's Law at the interface: \[ y_{Ai} = 5 x_{Ai} \] --- ### **Set up the two-film theory:** \[ \frac{1}{K_y} = \frac{1}{k'_y} + \frac{m}{k'_x} \] where \( m \) is the slope from Henry's Law: \( m = 5 \). Given: \[ k'_x = 2.5 k'_y \implies k'_y = \frac{k'_x}{2.5} \] --- #### **Graphical Solution Method:** 1. **Draw the lines:** - Bulk values: \( x_{AL} = \), \( y_{AG} = .02 \) - Equilibrium: \( y = 5x \) 2. **Plot operating line:** - From \( (x_{AL}, y_{AG}) = (, .02) \) to \( (x_{Ai}, y_{Ai}) \), and at the interface, \( y_{Ai} = 5 x_{Ai} \). 3. **Set up the flux equations for both films:** \[ N_A = k'_y (y_{AG} - y_{Ai}) = k'_x (x_{Ai} - x_{AL}) \] \[ k'_y (.02 - y_{Ai}) = k'_x (x_{Ai} - ) \] Since \( y_{Ai} = 5 x_{Ai} \) and \( k'_x = 2.5 k'_y \): \[ k'_y (.02 - 5 x_{Ai}) = 2.5 k'_y x_{Ai} \] \[ .02 - 5 x_{Ai} = 2.5 x_{Ai} \] \[ .02 = 7.5 x_{Ai} \] \[ x_{Ai} = \frac{.02}{7.5} = .00267 \] \[ y_{Ai} = 5 x_{Ai} = .0133 \] --- ### **Interface Concentrations:** - \( \boxed{x_{Ai} = .00267} \) - \( \boxed{y_{Ai} = .0133} \) --- ## **b) Liquid and Gas Film Mass Transfer Coefficients (\( k'_x, k'_y \))** Given: - \( K_y = 5 \) kmol/(h·m²·atm) - \( k'_x = 2.5 k'_y \) - Use the resistance-in-series equation: \[ \frac{1}{K_y} = \frac{1}{k'_y} + \frac{m}{k'_x} \] Substitute \( k'_x = 2.5 k'_y \), \( m = 5 \): \[ \frac{1}{K_y} = \frac{1}{k'_y} + \frac{5}{2.5 k'_y} \] \[ \frac{1}{K_y} = \frac{1}{k'_y} + \frac{2}{k'_y} \] \[ \frac{1}{K_y} = \frac{3}{k'_y} \] \[ k'_y = 3 K_y = 3 \times 5 = 15\ \text{kmol}/(\text{h} \cdot \text{m}^2 \cdot \text{atm}) \] \[ k'_x = 2.5 k'_y = 2.5 \times 15 = 37.5\ \text{kmol}/(\text{h} \cdot \text{m}^2 \cdot \text{atm}) \] --- ### **Film Mass Transfer Coefficients:** - \( \boxed{k'_y = 15\ \text{kmol}/(\text{h} \cdot \text{m}^2 \cdot \text{atm})} \) - \( \boxed{k'_x = 37.5\ \text{kmol}/(\text{h} \cdot \text{m}^2 \cdot \text{atm})} \) --- ## **c) Flux (\( N_A \))** Use: \[ N_A = K_y (y_{AG} - y_{Ai}) = 5 \times (.02 - .0133) = 5 \times .0067 = .0335\ \text{kmol}/(\text{h} \cdot \text{m}^2) \] --- ### **Final Answers** 1. **Interface concentrations:** - \( x_{Ai} = .00267 \) - \( y_{Ai} = .0133 \) 2. **Film coefficients:** - \( k'_x = 37.5\ \text{kmol}/(\text{h} \cdot \text{m}^2 \cdot \text{atm}) \) - \( k'_y = 15\ \text{kmol}/(\text{h} \cdot \text{m}^2 \cdot \text{atm}) \) 3. **Flux:** - \( N_A = .0335\ \text{kmol}/(\text{h} \cdot \text{m}^2) \) --- **Let me know if you need a diagram or more details on any step!**