Let's break down the problem and solve step by step: ## **Given Data** - **Reaction**: Cyclohexane → Benzene + H₂ (via Dehydrogenation, over Pt/Al₂O₃ catalyst) - **Temperature** (\(T\)): 705 K - **Pressure** (\(P\)): 1.480 MPa - **H₂ Feed Rate**: 8 mol/s - **Cyclohexane Feed Rate**: 2 mol/s - **Conversion of Cyclohexane (\(X\))**: 15.5% - **Catalyst amount (\(W\))**: 10.4 g **Catalyst Properties:** - Pellet diameter (\(d_p\)): 0.32 cm - **Pore volume (\(V_p\))**: 0.48 cm³/g - **Surface area (\(S\))**: 240 m²/g - **Pellet density (\(\rho_p\))**: 1.332 g/cm³ - **Pellet porosity (\(\varepsilon_p\))**: 0.59 cm³ voids/cm³ - **Effectiveness factor (\(\eta\))**: 0.42 **Assumptions:** - **First order kinetics** - **Knudsen diffusion is dominant** - **We are to estimate the tortuosity factor (\(\tau\))** --- # **Step 1: Calculate the Rate Constant from Reactor Data** For a plug-flow reactor: \[ \text{Moles of cyclohexane fed per second} = 2 \text{ mol/s} \] \[ \text{Conversion (\(X\))} = 0.155 \] \[ \text{Moles of cyclohexane reacted per second} = 2 \times 0.155 = 0.31 \text{ mol/s} \] \[ \text{Catalyst mass} = 10.4\ \text{g} \] \[ \text{Observed rate (per gram of catalyst)} = \frac{0.31}{10.4} = 0.0298\ \text{mol/s/g} \] --- # **Step 2: Calculate the Thiele Modulus (\(\phi\)) Using the Effectiveness Factor** For first-order reaction and spherical pellets, the effectiveness factor is: \[ \eta = \frac{3}{\phi^2} \left[\phi \coth(\phi) - 1\right] \] Given: \(\eta = 0.42\) We need to solve for \(\phi\). This equation is transcendental, so let's estimate \(\phi\): - For \(\phi = 2.5\), \[ \coth(2.5) = \frac{\cosh(2.5)}{\sinh(2.5)} \approx 1.013 \] \[ \eta = \frac{3}{2.5^2} [2.5 \times 1.013 - 1] = \frac{3}{6.25} [2.5325 - 1] = 0.48 \times 1.5325 = 0.735 \] - For \(\phi = 3.5\), \[ \coth(3.5) \approx 1.0005 \] \[ \eta = \frac{3}{12.25} [3.5 \times 1.0005 - 1] = 0.245 \times (3.5018 - 1) = 0.245 \times 2.5018 = 0.613 \] Try \(\phi = 5\): \[ \coth(5) \approx 1.00009 \] \[ \eta = \frac{3}{25} [5 \times 1.00009 - 1] = 0.12 \times (5.00045 - 1) = 0.12 \times 4.00045 = 0.480 \] Try \(\phi = 7\): \[ \coth(7) \approx 1.00001 \] \[ \eta = \frac{3}{49} [7 \times 1.00001 - 1] = 0.0612 \times (7.00007 - 1) = 0.0612 \times 6.00007 = 0.367 \] So, for \(\eta = 0.42\), interpolate between 5 (\(\eta=0.48\)) and 7 (\(\eta=0.367\)), or use a linear approximation: \[ \frac{0.48-0.42}{0.48-0.367} = \frac{0.06}{0.113} \approx 0.53 \] So, \(\phi \approx 5 + 0.53 \times (7-5) = 5 + 1.06 = 6.06\) **Thus, \(\boxed{\phi \approx 6.1}\)** --- # **Step 3: Thiele Modulus in Terms of Diffusivity and Rate Constant** For first-order, spherical pellet: \[ \phi = R \sqrt{\frac{k}{D_e}} \] Where: - \(R\): radius of pellet (cm) - \(k\): reaction rate constant (1/s) - \(D_e\): effective diffusivity (cm\(^2\)/s) --- # **Step 4: Get the Intrinsic Rate Constant, \(k\)** From effectiveness factor definition: \[ \text{Observed rate} = \eta \times \text{intrinsic rate} \] \[ \text{Intrinsic rate per gram catalyst} = \frac{\text{Observed rate}}{\eta} \] \[ = \frac{0.0298}{0.42} = 0.071 \text{ mol/s/g} \] But for a first-order reaction: \[ \text{rate} = k \cdot C_{A, s} \cdot \text{(catalyst amount)} \] But since we don't have the concentration at the surface, we need to use the Thiele modulus relationship. Alternatively, let's try to express everything in terms of the diffusivity and then the tortuosity. --- # **Step 5: Calculate Knudsen Diffusivity (\(D_K\))** Knudsen diffusivity for a single component: \[ D_K = \frac{9700 \cdot r_p}{\sqrt{M} \cdot \sqrt{T}} \] But more generally (SI units): \[ D_K = \frac{2}{3} r_p \left( \frac{8RT}{\pi M} \right)^{1/2} \] Where: - \(r_p\): pore radius (cm) - \(R\): gas constant (8.314 J/mol·K) - \(T\): temperature (K) - \(M\): molecular weight (kg/mol) First, let's estimate the pore radius. Given: - Pore volume per gram: \(V_p = 0.48\) cm³/g - Surface area per gram: \(S = 240\) m²/g = \(2.4 \times 10^5\) cm²/g For cylindrical pores: \[ \text{Pore volume per gram} = \text{surface area per gram} \times \frac{\text{pore radius}}{2} \] \[ V_p = S \times \frac{r_p}{2} \] \[ r_p = \frac{2 V_p}{S} \] Plug in: \[ r_p = \frac{2 \times 0.48}{2.4 \times 10^5} = \frac{0.96}{2.4 \times 10^5} = 4.0 \times 10^{-6} \text{ cm} \] --- # **Step 6: Calculate Knudsen Diffusivity** \[ D_K = \frac{2}{3} r_p \left(\frac{8RT}{\pi M}\right)^{1/2} \] - \(r_p = 4.0 \times 10^{-6}\) cm = \(4.0 \times 10^{-8}\) m - \(R = 8.314\) J/mol·K - \(T = 705\) K - For cyclohexane, \(M = 84\) g/mol = \(0.084\) kg/mol Calculate the square root term: \[ \frac{8RT}{\pi M} = \frac{8 \times 8.314 \times 705}{\pi \times 0.084} \] \[ 8 \times 8.314 = 66.512 \] \[ 66.512 \times 705 = 46,852.96 \] \[ \pi \times 0.084 = 0.2639 \] \[ \frac{46,852.96}{0.2639} \approx 177,575 \] \[ \sqrt{177,575} \approx 421.6 \] So, \[ D_K = \frac{2}{3} \times 4.0 \times 10^{-6} \times 421.6 \] \[ = \frac{8.0 \times 10^{-6}}{3} \times 421.6 = 2.6667 \times 10^{-6} \times 421.6 = 1.125 \times 10^{-3} \text{ cm}^2/\text{s} \] --- # **Step 7: Effective Diffusivity** \[ D_e = \frac{\varepsilon_p}{\tau} D_K \] Where: - \(\varepsilon_p = 0.59\) - \(\tau\) (tortuosity, to be found) - \(D_K = 1.125 \times 10^{-3}\) cm²/s --- # **Step 8: Rearranged Thiele Modulus Equation for \(\tau\)** \[ \phi = R \sqrt{\frac{k}{D_e}} \implies \phi^2 = R^2 \frac{k}{D_e} \implies D_e = R^2 \frac{k}{\phi^2} \] But also: \[ D_e = \frac{\varepsilon_p}{\tau} D_K \] Therefore, \[ \frac{\varepsilon_p}{\tau} D_K = R^2 \frac{k}{\phi^2} \implies \tau = \frac{\varepsilon_p D_K \phi^2}{R^2 k} \] --- # **Step 9: Plug in the Values** - \(\varepsilon_p = 0.59\) - \(D_K = 1.125 \times 10^{-3}\) cm²/s - \(\phi = 6.1\) - \(R = \frac{d_p}{2} = \frac{0.32}{2} = 0.16\) cm **We still need \(k\) (the first order rate constant).** Recall the "intrinsic" rate per gram: \[ \text{Intrinsic rate} = k \cdot C_{A, s} \cdot \text{mass} \] We need to estimate \(C_{A, s}\), the cyclohexane concentration at the surface (almost the same as bulk for low conversion). **Calculate molar concentration:** \[ C = \frac{P}{RT} \] - \(P = 1.48\) MPa \(= 1.48 \times 10^6\) Pa - \(R = 8.314\) J/mol·K - \(T = 705\) K \[ C = \frac{1.48 \times 10^6}{8.314 \times 705} = \frac{1.48 \times 10^6}{5,862} \approx 252.6 \text{ mol/m}^3 = 0.2526 \text{ mol/L} = 0.2526 \text{ mol/dm}^3 = 2.526 \times 10^{-4} \text{ mol/cm}^3 \] But this is total molar concentration. The cyclohexane partial pressure is: \[ \text{Total feed:} \quad \ce{H2}:\ce{C6H12} = 4:1 \text{ molar} \] So cyclohexane fraction = \(1/(4+1) = 0.2\) \[ P_A = 0.2 \times 1.48 \times 10^6 = 2.96 \times 10^5\ \text{Pa} \] So cyclohexane concentration: \[ C_{A} = \frac{2.96 \times 10^5}{8.314 \times 705} = \frac{2.96 \times 10^5}{5,862} \approx 50.5 \text{ mol/m}^3 = 0.0505 \text{ mol/L} = 5.05 \times 10^{-5} \text{ mol/cm}^3 \] --- **Intrinsic rate per gram:** \[ r = k \cdot C_A \cdot \text{mass} \] \[ \text{Intrinsic rate per gram} = 0.