# Solution Let's solve each part step by step. --- ## **Given Data** - **Bulk concentration of A, \( C_{A,bulk} = 0.002 \) kmol/m³** - **Rate of reaction per catalyst area, \( r_A = 0.08 \) kmol/(h·m²)** - **Total pressure, \( P = 1.5 \) bar** - **Temperature, \( T = 250^\circ \mathrm{C} = 523 \) K** - **Diffusivity at 1 bar, 100°C: \( D_{AB} = 0.16 \) cm²/s** - **Reaction is instantaneous** - **All resistance in stagnant film** - **Units: 1 kmol = 1000 mol; 1 bar = 10⁵ Pa; 0.16 cm²/s = \( 1.6 \times 10^{-5} \) m²/s** --- ## **Step 1: Convert Rate to SI Units** Given: \( r_A = 0.08 \) kmol/(h·m²) - \( 1 \) kmol = 1000 mol - \( 1 \) hr = 3600 s So, \[ r_A = 0.08 \ \text{kmol/(h·m}^2) = 0.08 \times 1000 \ \text{mol/(3600 s·m}^2) \] \[ = \frac{80}{3600} = 0.0222 \ \text{mol/(s·m}^2) \] --- ## **Step 2: Calculate Diffusivity at 1.5 bar, 250°C** ### **2.1. Use Diffusivity Temperature and Pressure Correction** \[ D_{AB} \propto \frac{T^{3/2}}{P} \] Given: - \( D_{AB,ref} = 0.16 \) cm²/s at \( T_{ref} = 373 \) K (\( 100^\circ \mathrm{C} \)), \( P_{ref} = 1 \) bar - \( T = 523 \) K, \( P = 1.5 \) bar \[ D_{AB,new} = D_{AB,ref} \left( \frac{T}{T_{ref}} \right)^{3/2} \left( \frac{P_{ref}}{P} \right) \] Plug in values: \[ D_{AB,new} = 0.16 \left( \frac{523}{373} \right)^{3/2} \left( \frac{1}{1.5} \right) \] Calculate \( \left( \frac{523}{373} \right)^{3/2} \): - \( \frac{523}{373} = 1.402 \) - \( (1.402)^{3/2} = 1.402^{1.5} \) - \( \sqrt{1.402} = 1.185 \) - \( 1.185 \times 1.402 = 1.661 \) So, \[ D_{AB,new} = 0.16 \times 1.661 \times \frac{1}{1.5} \] \[ = 0.16 \times 1.661 \times 0.6667 \] \[ = 0.16 \times 1.107 = 0.177 \ \text{cm}^2/\text{s} \] Convert to m²/s: \[ 0.177 \ \text{cm}^2/\text{s} = 0.177 \times 10^{-4} \ \text{m}^2/\text{s} = 1.77 \times 10^{-5} \ \text{m}^2/\text{s} \] --- ## **Step 3: Relationship for Flux and Mass Transfer Coefficient** For an **instantaneous reaction**, the concentration of A at the catalyst surface is zero (\( C_{A,surface} = 0 \)). The flux is determined by diffusion through the film. \[ \text{Flux}, N_A = \frac{D_{AB}}{\delta} (C_{A,bulk} - C_{A,surface}) \] But \( C_{A,surface} = 0 \): \[ N_A = \frac{D_{AB}}{\delta} C_{A,bulk} \] The **rate/flux per unit area** is given as \( r_A \): \[ r_A = N_A = \frac{D_{AB}}{\delta} C_{A,bulk} \] --- ## **Step 4: Calculate Film Thickness (\( \delta \))** \[ \delta = \frac{D_{AB} C_{A,bulk}}{r_A} \] Plug in values: - \( D_{AB} = 1.77 \times 10^{-5} \) m²/s - \( C_{A,bulk} = 0.002 \) kmol/m³ = 2 mol/m³ - \( r_A = 0.0222 \) mol/(s·m²) \[ \delta = \frac{1.77 \times 10^{-5} \times 2}{0.0222} \] \[ = \frac{3.54 \times 10^{-5}}{0.0222} \] \[ = 1.595 \times 10^{-3} \ \text{m} \] \[ = 1.60 \ \text{mm} \] --- ## **Step 5: Calculate Mass Transfer Coefficient \( k_G \) (m/s)** \[ k_G = \frac{D_{AB}}{\delta} \] \[ k_G = \frac{1.77 \times 10^{-5}}{1.595 \times 10^{-3}} \] \[ = 1.11 \times 10^{-2} \ \text{m/s} \] --- ## **Step 6: Calculate Mass Transfer Coefficient in Partial Pressure Units \( k_y \)** \[ k_y = k_G \cdot \frac{C_{Tot}}{P_{Tot}} \] Where \( C_{Tot} \) is the total molar concentration (ideal gas law): \[ C_{Tot} = \frac{P}{RT} \] - \( P = 1.5 \) bar = \( 1.5 \times 10^5 \) Pa - \( R = 8.314 \) J/(mol·K) - \( T = 523 \) K \[ C_{Tot} = \frac{1.5 \times 10^5}{8.314 \times 523} \] \[ = \frac{1.5 \times 10^5}{4344} \] \[ = 34.54 \ \text{mol/m}^3 \] Now, \[ k_y = k_G \cdot \frac{C_{Tot}}{P} \] \[ k_y = 1.11 \times 10^{-2} \cdot \frac{34.54}{1.5 \times 10^5} \] \[ = 1.11 \times 10^{-2} \cdot 2.3027 \times 10^{-4} \] \[ = 2.56 \times 10^{-6} \ \text{mol/(s·m}^2\cdot\text{Pa)} \] Alternatively, if asked for units of m/s in y-units (mol fraction): \[ k_y' = k_G \cdot C_{Tot} \] \[ = 1.11 \times 10^{-2} \times 34.54 = 0.384 \ \text{mol/(s·m}^2\cdot\text{y}_A) \] --- ## **Step 7: Concentration of A Halfway in the Film** The concentration profile in a stagnant film (steady state, 1D) is linear. \[ C_A(x) = C_{A,bulk} - (C_{A,bulk} - C_{A,surface}) \frac{x}{\delta} \] Since \( C_{A,surface} = 0 \): \[ C_A(x) = C_{A,bulk} \left(1 - \frac{x}{\delta}\right) \] At \( x = \delta/2 \): \[ C_A(\delta/2) = C_{A,bulk} \left(1 - \frac{1}{2}\right) = 0.5 \, C_{A,bulk} \] \[ = 0.5 \times 2 = 1.0 \ \text{mol/m}^3 \] Or, in kmol/m³: \( 0.001 \) kmol/m³ --- ## **Step 8: Summary Table of Answers** | Quantity | Value | |--------------------------------------|--------------------------------------| | Film Thickness (\( \delta \)) | 1.60 mm ( \(1.60 \times 10^{-3}\) m )| | Mass Transfer Coefficient (\( k_G \))| \(1.11 \times 10^{-2}\) m/s | | Mass Transfer Coefficient (\( k_y \))| \(2.56 \times 10^{-6}\) mol/(s·m²·Pa)| | \( C_A \) halfway through film | 1.0 mol/m³ (0.001 kmol/m³) | --- ## **(b) Individual and Overall Mass Transfer Coefficient** ### **Individual Film Mass Transfer Coefficient** - Represents the resistance to mass transfer **in a single phase** (e.g., gas side or liquid side). - Defined as \( k_G \) for gas phase, \( k_L \) for liquid phase. - Used when mass transfer is controlled primarily by one phase. ### **Overall Mass Transfer Coefficient** - Accounts for **all resistances** in series (e.g., gas and liquid films, interfacial resistance). - Defined as \( K_G \) or \( K_L \). - Used when both phases contribute significant resistance. - Related to individual coefficients by: \[ \frac{1}{K_G} = \frac{1}{k_G} + \frac{m}{k_L} \] Where \( m \) is the slope of the equilibrium line. ### **When to Use Which?** - **Individual coefficient**: When only one phase dominates the resistance (as in this problem, with only a gas film). - **Overall coefficient**: When both phases are important, or when you want to relate total driving force (difference in bulk concentrations) to the flux. --- ## **Final Answers** ### **1. Film thickness (\( \delta \)):** \[ \boxed{1.60 \ \text{mm} \ (1.60 \times 10^{-3}\ \text{m})} \] ### **2. Mass transfer coefficients:** - \( k_G = \boxed{1.11 \times 10^{-2}\ \text{m/s}} \) - \( k_y = \boxed{2.56 \times 10^{-6}\ \text{mol/(s·m}^2\cdot\text{Pa})} \) ### **3. Concentration of A halfway through the film:** \[ \boxed{1.0 \ \text{mol/m}^3\ (0.001 \ \text{kmol/m}^3)} \] --- **(b) Individual film mass transfer coefficient refers to the rate of mass transfer in a single phase (gas or liquid), reflecting the resistance offered by that phase alone. Overall mass transfer coefficient sums up all resistances in series (gas film, liquid film, interface), and is used when both phases matter.**