Let's break down and solve **Question 8** step by step as requested. --- # Question 8: Effect of Slow Reaction Rate on Diffusion **Given:** - Gas A diffuses from point 1 to a catalyst surface at point 2 and reacts: \( 2A \rightarrow B \). - Gas B diffuses back a distance δ to point 1. - \( D_{AB} = 0.2 \times 10^{-4} \ \text{m}^2/\text{s} \) - \( x_{A1} = 0.97 \) - \( P = 101.21 \ \text{kPa} \) - \( \delta = 3.0 \ \text{mm} = 3.0 \times 10^{-3} \ \text{m} \) - \( T = 298 \ \text{K} \) --- ## **a. Derive the equation for \( N_A \) for a very fast reaction** **For a very fast reaction:** - All A that reaches the surface reacts. - \( x_{A2} \approx 0 \) (concentration of A at the surface is zero). **Fick's Law for steady-state diffusion:** \[ N_A = - D_{AB} \frac{dc_A}{dz} \] In terms of mole fractions and total pressure: \[ N_A = \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) \] Since \( x_{A2} \approx 0 \): \[ N_A = \frac{D_{AB} P}{RT\delta} x_{A1} \] --- ## **b. Calculate \( N_A \) using the provided values** Plug in the values: - \( D_{AB} = 0.2 \times 10^{-4} \ \text{m}^2/\text{s} \) - \( P = 101.21 \times 10^3 \ \text{Pa} \) - \( x_{A1} = 0.97 \) - \( \delta = 3.0 \times 10^{-3} \ \text{m} \) - \( R = 8.314 \ \text{J/(mol·K)} \) - \( T = 298 \ \text{K} \) \[ N_A = \frac{(0.2 \times 10^{-4}) \times (101.21 \times 10^3)}{8.314 \times 298 \times 3.0 \times 10^{-3}} \times 0.97 \] Calculate denominator: \[ 8.314 \times 298 \times 3.0 \times 10^{-3} = 8.314 \times 298 = 2477.572 \] \[ 2477.572 \times 3.0 \times 10^{-3} = 7.4327 \] Calculate numerator: \[ 0.2 \times 10^{-4} = 2 \times 10^{-5} \] \[ 2 \times 10^{-5} \times 101210 = 2.0242 \] Plug in: \[ N_A = \frac{2.0242}{7.4327} \times 0.97 = 0.263 \times 0.97 = 0.255 \ \text{mol/m}^2\text{s} \] **Final Answer for part (b):** \[ \boxed{N_A = 0.255 \ \text{mol/m}^2\text{s}} \] --- ## **c. Slow first-order reaction (\( k_1 \) is the reaction constant)** **At steady state, rate of diffusion = rate of reaction at the surface.** For first-order reaction: \[ \text{Rate of reaction at surface} = k_1 c_{A2} \] \[ N_A = \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) = k_1 c_{A2} \] \[ c_{A2} = \frac{P x_{A2}}{RT} \] \[ N_A = k_1 \frac{P x_{A2}}{RT} \] Combine the two equations: \[ N_A = \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) = k_1 \frac{P x_{A2}}{RT} \] \[ \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) = k_1 \frac{P x_{A2}}{RT} \] \[ \frac{D_{AB}}{\delta} (x_{A1} - x_{A2}) = k_1 x_{A2} \] \[ \frac{D_{AB}}{\delta} x_{A1} = \left( \frac{D_{AB}}{\delta} + k_1 \right)x_{A2} \] \[ x_{A2} = \frac{ \frac{D_{AB}}{\delta} x_{A1} }{ \frac{D_{AB}}{\delta} + k_1 } \] \[ N_A = k_1 \frac{P x_{A2}}{RT} \] --- ## **d. Calculate \( N_A \) and \( x_{A2} \) for \( k_1 = 0.52 \times 10^{-2} \ \text{m/s} \)** First, calculate \( \frac{D_{AB}}{\delta} \): \[ \frac{D_{AB}}{\delta} = \frac{0.2 \times 10^{-4}}{3.0 \times 10^{-3}} = \frac{2 \times 10^{-5}}{3 \times 10^{-3}} = \frac{2}{3} \times 10^{-2} = 0.6667 \times 10^{-2} \ \text{m/s} \] Now plug into the formula for \( x_{A2} \): \[ x_{A2} = \frac{ 0.6667 \times 10^{-2} \times 0.97 }{ 0.6667 \times 10^{-2} + 0.52 \times 10^{-2} } \] \[ = \frac{ 0.6465 \times 10^{-2} }{ 1.1867 \times 10^{-2} } \] \[ = 0.545 \] Now, calculate \( N_A \): \[ N_A = k_1 \frac{P x_{A2}}{RT} \] \[ = (0.52 \times 10^{-2}) \times \frac{101210 \times 0.545}{8.314 \times 298} \] First, calculate denominator: \[ 8.314 \times 298 = 2477.572 \] Numerator: \[ 101210 \times 0.545 = 55159.45 \] \[ \frac{55159.45}{2477.572} = 22.27 \] \[ N_A = 0.0052 \times 22.27 = 0.116 \ \text{mol/m}^2\text{s} \] --- ## **Summary of Answers** ### **(b) For very fast reaction:** \[ \boxed{N_A = 0.255 \ \text{mol/m}^2\text{s}} \] ### **(d) For slow first-order reaction:** \[ \boxed{x_{A2} = 0.545} \] \[ \boxed{N_A = 0.116 \ \text{mol/m}^2\text{s}} \] --- ## **List of Knowns and Unknowns** **Knowns:** - \( D_{AB}, x_{A1}, P, \delta, T, R \) - \( k_1 \) (for part d) **Unknowns:** - \( N_A \) (flux of A) - \( x_{A2} \) (mole fraction of A at the surface, for slow reaction) --- ## **Diagram Representation** ``` Point 1 (bulk) Distance δ Point 2 (catalyst surface) |----------------------|-----------------|--------------------------| x_A1 ≠ 0 ← Diffusion x_A2 (unknown or ≈ 0) → Reaction A diffuses from left (x_A1) to right (x_A2) and reacts on the surface. ``` --- Let me know if you need a hand-drawn or labeled diagram!

# Question 8: Effect of Slow Reaction Rate on Diffusion **Given:** - Gas A diffuses from point 1 to a catalyst surface at point 2 and reacts: \( 2A \rightarrow B \). - Gas B diffuses back a distance \(\delta\) to point 1. - \( D_{AB} = 0.2 \times 10^{-4} \ \text{m}^2/\text{s} \) - \( x_{A1} = 0.97 \) - \( P = 101.21 \ \text{kPa} \) - \( \delta = 1.30 \ \text{mm} = 1.30 \times 10^{-3} \ \text{m} \) - \( T = 298 \ \text{K} \) --- ## **a. Derive the equation for \( N_A \) for a very fast reaction** **For a very fast reaction:** - All A that reaches the surface reacts. - \( x_{A2} \approx 0 \) (concentration of A at the surface is zero). **Fick's Law for steady-state diffusion:** \[ N_A = - D_{AB} \frac{dc_A}{dz} \] In terms of mole fractions and total pressure: \[ N_A = \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) \] Since \( x_{A2} \approx 0 \): \[ N_A = \frac{D_{AB} P}{RT\delta} x_{A1} \] --- ## **b. Calculate \( N_A \) using the provided values** Plug in the values: - \( D_{AB} = 0.2 \times 10^{-4} \ \text{m}^2/\text{s} \) - \( P = 101.21 \times 10^3 \ \text{Pa} \) - \( x_{A1} = 0.97 \) - \( \delta = 1.30 \times 10^{-3} \ \text{m} \) - \( R = 8.314 \ \text{J/(mol·K)} \) - \( T = 298 \ \text{K} \) \[ N_A = \frac{(0.2 \times 10^{-4}) \times (101.21 \times 10^3)}{8.314 \times 298 \times 1.30 \times 10^{-3}} \times 0.97 \] ### **Calculate Denominator:** \[ 8.314 \times 298 \times 1.30 \times 10^{-3} = 8.314 \times 298 = 2477.572 \] \[ 2477.572 \times 1.30 \times 10^{-3} = 3.2218 \] ### **Calculate Numerator:** \[ 0.2 \times 10^{-4} = 2 \times 10^{-5} \] \[ 2 \times 10^{-5} \times 101210 = 2.0242 \] ### **Final Calculation:** \[ N_A = \frac{2.0242}{3.2218} \times 0.97 = 0.629 \times 0.97 = 0.611 \ \text{mol/m}^2\text{s} \] **Final Answer for part (b):** \[ \boxed{N_A = 0.611 \ \text{mol/m}^2\text{s}} \] --- ## **c. Slow first-order reaction (\( k_1 \) is the reaction constant)** **At steady state, rate of diffusion = rate of reaction at the surface.** For first-order reaction: \[ \text{Rate of reaction at surface} = k_1 c_{A2} \] \[ N_A = \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) = k_1 c_{A2} \] \[ c_{A2} = \frac{P x_{A2}}{RT} \] \[ N_A = k_1 \frac{P x_{A2}}{RT} \] Combine the two equations: \[ \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) = k_1 \frac{P x_{A2}}{RT} \] \[ \frac{D_{AB}}{\delta} (x_{A1} - x_{A2}) = k_1 x_{A2} \] \[ \frac{D_{AB}}{\delta} x_{A1} = \left( \frac{D_{AB}}{\delta} + k_1 \right)x_{A2} \] \[ x_{A2} = \frac{ \frac{D_{AB}}{\delta} x_{A1} }{ \frac{D_{AB}}{\delta} + k_1 } \] \[ N_A = k_1 \frac{P x_{A2}}{RT} \] --- ## **d. Calculate \( N_A \) and \( x_{A2} \) for \( k_1 = 0.52 \times 10^{-2} \ \text{m/s} \)** ### **Calculate \( \frac{D_{AB}}{\delta} \):** \[ \frac{D_{AB}}{\delta} = \frac{0.2 \times 10^{-4}}{1.30 \times 10^{-3}} = \frac{2 \times 10^{-5}}{1.30 \times 10^{-3}} \approx 0.1538 \times 10^{-2} \ \text{m/s} \] ### **Calculate \( x_{A2} \):** \[ x_{A2} = \frac{ 0.1538 \times 10^{-2} \times 0.97 }{ 0.1538 \times 10^{-2} + 0.52 \times 10^{-2} } \] \[ = \frac{ 0.14918 \times 10^{-2} }{ 0.6738 \times 10^{-2} } = 0.221 \] ### **Calculate \( N_A \):** \[ N_A = k_1 \frac{P x_{A2}}{RT} \] \[ = (0.52 \times 10^{-2}) \times \frac{101210 \times 0.221}{8.314 \times 298} \] #### **Calculate Denominator:** \[ 8.314 \times 298 = 2477.572 \] #### **Calculate Numerator:** \[ 101210 \times 0.221 = 22327.41 \] #### **Final Calculation:** \[ \frac{22327.41}{2477.572} \approx 9.00 \] \[ N_A = 0.0052 \times 9.00 = 0.0468 \ \text{mol/m}^2\text{s} \] --- ## **Summary of Answers** ### **(b) For very fast reaction:** \[ \boxed{N_A = 0.611 \ \text{mol/m}^2\text{s}} \] ### **(d) For slow first-order reaction:** \[ \boxed{x_{A2} = 0.221} \] \[ \boxed{N_A = 0.0468 \ \text{mol/m}^2\text{s}} \] --- ## **List of Knowns and Unknowns** **Knowns:** - \( D_{AB}, x_{A1}, P, \delta, T, R \) - \( k_1 \) (for part d) **Unknowns:** - \( N_A \) (flux of A) - \( x_{A2} \) (mole fraction of A at the surface, for slow reaction)

# Question 8: Effect of Slow Reaction Rate on Diffusion **Given:** - Gas A diffuses from point 1 to a catalyst surface at point 2 and reacts: \( 2A \rightarrow B \). - Gas B diffuses back a distance \(\delta\) to point 1. - \( D_{AB} = 0.2 \times 10^{-4} \ \text{m}^2/\text{s} \) - \( x_{A1} = 0.97 \) - \( P = 101.21 \ \text{kPa} \) - \( \delta = 1.30 \ \text{mm} = 1.30 \times 10^{-3} \ \text{m} \) - \( T = 298 \ \text{K} \) --- ## a. Derive the equation for \( N_A \) for a very fast reaction **For a very fast reaction:** - All A that reaches the surface reacts. - \( x_{A2} \approx 0 \) (concentration of A at the surface is zero). **Fick's Law for steady-state diffusion:** \[ N_A = - D_{AB} \frac{dc_A}{dz} \] In terms of mole fractions and total pressure: \[ N_A = \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) \] Since \( x_{A2} \approx 0 \): \[ N_A = \frac{D_{AB} P}{RT\delta} x_{A1} \] --- ## b. Calculate \( N_A \) using the provided values Plug in the values: - \( D_{AB} = 0.2 \times 10^{-4} \ \text{m}^2/\text{s} \) - \( P = 101.21 \times 10^3 \ \text{Pa} \) - \( x_{A1} = 0.97 \) - \( \delta = 1.30 \times 10^{-3} \ \text{m} \) - \( R = 8.314 \ \text{J/(mol·K)} \) - \( T = 298 \ \text{K} \) ### Calculate Denominator: \[ 8.314 \times 298 \times 1.30 \times 10^{-3} = 8.314 \times 298 = 2477.572 \] \[ 2477.572 \times 1.30 \times 10^{-3} = 3.2218 \] ### Calculate Numerator: \[ 0.2 \times 10^{-4} = 2 \times 10^{-5} \] \[ 2 \times 10^{-5} \times 101210 = 2.0242 \] ### Final Calculation: \[ N_A = \frac{2.0242}{3.2218} \times 0.97 = 0.629 \times 0.97 = 0.611 \ \text{mol/m}^2\text{s} \] **Final Answer for part (b):** \[ \boxed{N_A = 0.611 \ \text{mol/m}^2\text{s}} \] --- ## c. Slow first-order reaction (\( k_1 \) is the reaction constant) **At steady state, rate of diffusion = rate of reaction at the surface.** For first-order reaction: \[ \text{Rate of reaction at surface} = k_1 c_{A2} \] \[ N_A = \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) = k_1 c_{A2} \] \[ c_{A2} = \frac{P x_{A2}}{RT} \] \[ N_A = k_1 \frac{P x_{A2}}{RT} \] Combine the two equations: \[ \frac{D_{AB} P}{RT\delta} (x_{A1} - x_{A2}) = k_1 \frac{P x_{A2}}{RT} \] \[ \frac{D_{AB}}{\delta} (x_{A1} - x_{A2}) = k_1 x_{A2} \] \[ \frac{D_{AB}}{\delta} x_{A1} = \left( \frac{D_{AB}}{\delta} + k_1 \right)x_{A2} \] \[ x_{A2} = \frac{ \frac{D_{AB}}{\delta} x_{A1} }{ \frac{D_{AB}}{\delta} + k_1 } \] \[ N_A = k_1 \frac{P x_{A2}}{RT} \] --- ## d. Calculate \( N_A \) and \( x_{A2} \) for \( k_1 = 0.52 \times 10^{-2} \ \text{m/s} \) ### Calculate \( \frac{D_{AB}}{\delta} \): \[ \frac{D_{AB}}{\delta} = \frac{0.2 \times 10^{-4}}{1.30 \times 10^{-3}} = \frac{2 \times 10^{-5}}{1.30 \times 10^{-3}} \approx 0.1538 \times 10^{-2} \ \text{m/s} \] ### Calculate \( x_{A2} \): \[ x_{A2} = \frac{ 0.1538 \times 10^{-2} \times 0.97 }{ 0.1538 \times 10^{-2} + 0.52 \times 10^{-2} } \] \[ = \frac{ 0.14918 \times 10^{-2} }{ 0.6738 \times 10^{-2} } = 0.221 \] ### Calculate \( N_A \): \[ N_A = k_1 \frac{P x_{A2}}{RT} \] \[ = (0.52 \times 10^{-2}) \times \frac{101210 \times 0.221}{8.314 \times 298} \] #### Calculate Denominator: \[ 8.314 \times 298 = 2477.572 \] #### Calculate Numerator: \[ 101210 \times 0.221 = 22327.41 \] #### Final Calculation: \[ \frac{22327.41}{2477.572} \approx 9.00 \] \[ N_A = 0.0052 \times 9.00 = 0.0468 \ \text{mol/m}^2\text{s} \] --- ## Summary of Answers ### (b) For very fast reaction: \[ \boxed{N_A = 0.611 \ \text{mol/m}^2\text{s}} \] ### (d) For slow first-order reaction: \[ \boxed{x_{A2} = 0.221} \] \[ \boxed{N_A = 0.0468 \ \text{mol/m}^2\text{s}} \] --- ## List of Knowns and Unknowns **Knowns:** - \( D_{AB}, x_{A1}, P, \delta, T, R \) - \( k_1 \) (for part d) **Unknowns:** - \( N_A \) (flux of A) - \( x_{A2} \) (mole fraction of A at the surface, for slow reaction)