Let's break the problem into clear steps. We're being asked to **estimate the tortuosity factor (τ)** for a catalyst, given: - The **effectiveness factor** (\(\eta\)) is **0.42**. - The reaction is **first-order**. - **Knudsen diffusion** is dominant. - **Cyclohexane feed rate**: \(2~\text{mol/s}\) - **Hydrogen feed rate**: \(8~\text{mol/s}\) (since 4:1 H2:C6H12) - **Cyclohexane conversion**: \(15.5\%\) - **Catalyst amount**: \(10.4~\text{g}\) - **Pellet diameter**: \(0.32~\text{cm}\) - **Pellet density**: \(1.332~\text{g/cm}^3\) - **Pellet porosity (\(\varepsilon_p\))**: \(0.59\) - **Pore volume**: \(0.48~\text{cm}^3/\text{g}\) - **Surface area**: \(240~\text{m}^2/\text{g}\) - **Temperature**: \(705~\text{K}\) - **Pressure**: \(1.480~\text{MPa}\) - **Reaction is first-order and isothermal** We need to **estimate the tortuosity (\(\tau\))**. ## Step 1: Write the expressions ### Effectiveness factor for first-order, isothermal, spherical catalyst: \[ \eta = \frac{3}{\phi^2}\left(\phi \coth \phi - 1\right) \] where \(\phi\) is the **Thiele modulus**. ### Thiele modulus for first-order, spherical pellet: \[ \phi = R \sqrt{\frac{k}{D_e}} \] - \(R\) = pellet radius - \(k\) = first-order rate constant (units \(1/\text{s}\)) - \(D_e\) = effective diffusivity of cyclohexane in pellet ### Effective diffusivity (\(D_e\)) for Knudsen diffusion: \[ D_e = D_K \frac{\varepsilon_p}{\tau} \] where \(D_K\) = Knudsen diffusivity. ## Step 2: Find the Thiele modulus (\(\phi\)) using the effectiveness factor \[ \eta = \frac{3}{\phi^2}\left(\phi \coth \phi - 1\right) \] Given \(\eta = 0.42\), solve for \(\phi\): This must be done numerically or by using a table. But for \(\eta = 0.42\), the value of \(\phi\) is typically around **2.6** (from effectiveness factor charts for 1st order reactions in spheres). ## Step 3: Collect and calculate all parameters ### Pellet radius, \(R\): \[ \text{Pellet diameter} = 0.32~\text{cm} \implies R = 0.16~\text{cm} = 1.6 \times 10^{-3}~\text{m} \] ### \(\phi = R \sqrt{\dfrac{k}{D_e}}\) So: \[ \phi^2 = R^2 \dfrac{k}{D_e} \implies \dfrac{k}{D_e} = \dfrac{\phi^2}{R^2} \] We need to get \(k\) and \(D_e\). --- ### Find the rate constant \(k\): Overall reaction (first-order in cyclohexane): \[ \text{Cyclohexane} \rightarrow \text{Benzene} + 3H_2 \] Rate law: \[ -\frac{dC_A}{dt} = k C_A \] But reactor is continuous flow, so let's relate rate to what is given. #### **Calculate the observed reaction rate per gram catalyst:** - Cyclohexane feed: \(2~\text{mol/s}\) - Conversion: \(15.5\%\) - Cyclohexane converted per second: \(2 \times 0.155 = 0.31~\text{mol/s}\) - Catalyst: \(10.4~\text{g}\) Observed rate per gram catalyst: \[ r_{obs} = \frac{0.31~\text{mol/s}}{10.4~\text{g}} = 0.0298~\text{mol/(g·s)} \] #### **Express observed rate in terms of \(k\):** For first-order, \[ \text{rate per gram catalyst} = k C_{A,\text{bulk}} \cdot \eta \] So, \[ k = \frac{r_{obs}}{\eta C_{A,\text{bulk}}} \] --- ### Calculate \(C_{A,\text{bulk}}\): - Total pressure: \(1.480~\text{MPa}\) = \(1.480 \times 10^6~\text{N/m}^2\) - Total feed: \(8 + 2 = 10~\text{mol/s}\) - Cyclohexane mole fraction: \(2/10 = 0.2\) Assume ideal gas: \[ C_{A,\text{bulk}} = y_A \frac{P}{RT} \] - \(y_A = 0.2\) - \(P = 1.480 \times 10^6~\text{N/m}^2\) - \(R = 8.314~\text{J/(mol·K)}\) - \(T = 705~\text{K}\) \[ C_{A,\text{bulk}} = 0.2 \times \frac{1.480 \times 10^6}{8.314 \times 705} \] \[ = 0.2 \times \frac{1.480 \times 10^6}{5,862.37} \] \[ = 0.2 \times 252.59 \] \[ = 50.52~\text{mol/m}^3 \] --- ### Now, calculate \(k\): \[ k = \frac{0.0298~\text{mol/(g·s)}}{0.42 \times 50.52~\text{mol/m}^3} \] But, for Thiele modulus, \(k\) must be per **m³ catalyst**. The rate above is per g catalyst. Let's convert 1 g catalyst to volume: - Pellet density: \(1.332~\text{g/cm}^3 = 1,332~\text{kg/m}^3\) - So, \(1~\text{g} = 0.00075~\text{cm}^3 = 7.5 \times 10^{-7}~\text{m}^3\) But let's check: \[ 1~\text{g} \times \frac{1~\text{cm}^3}{1.332~\text{g}} = 0.751~\text{cm}^3 = 7.51 \times 10^{-7}~\text{m}^3 \] So, \(0.0298~\text{mol/(g·s)}\) = \(0.0298 / 7.51 \times 10^{-7} = 39,700~\text{mol/(m}^3\text{·s)}\) Now, \[ k = \frac{39,700~\text{mol/(m}^3\text{·s)}}{0.42 \times 50.52~\text{mol/m}^3} \] \[ = \frac{39,700}{21.219} \] \[ = 1,871~\text{s}^{-1} \] --- ### Now, use Thiele modulus relation: \[ \phi^2 = R^2 \frac{k}{D_e} \implies D_e = \frac{R^2 k}{\phi^2} \] \[ R = 1.6 \times 10^{-3}~\text{m} \] \[ k = 1,871~\text{s}^{-1} \] \[ \phi^2 = (2.6)^2 = 6.76 \] \[ D_e = \frac{(1.6 \times 10^{-3})^2 \times 1,871}{6.76} \] \[ = \frac{2.56 \times 10^{-6} \times 1,871}{6.76} \] \[ = \frac{4.789 \times 10^{-3}}{6.76} \] \[ = 7.09 \times 10^{-4}~\text{m}^2/\text{s} \] --- ## Step 4: Calculate Knudsen diffusivity, \(D_K\): \[ D_K = \frac{2}{3} r_p \sqrt{\frac{8RT}{\pi M}} \] Where: - \(r_p\) = pore radius (in m) - \(R\) = 8.314 J/mol·K - \(T\) = 705 K - \(M\) = molecular weight of cyclohexane = 84 g/mol = 0.084 kg/mol We need \(r_p\). #### Estimate pore radius from pore volume and surface area: \[ \text{Pore volume per g} = 0.48~\text{cm}^3/\text{g} = 4.8 \times 10^{-4}~\text{m}^3/\text{g} \] \[ \text{Surface area per g} = 240~\text{m}^2/\text{g} \] Assume cylindrical pores: \[ \text{Pore volume} = \text{surface area} \times \text{average pore radius} \] \[ V_p = A \cdot r_p \implies r_p = \frac{V_p}{A} \] \[ V_p = 0.48~\text{cm}^3/\text{g} = 4.8 \times 10^{-4}~\text{cm}^3/\text{g} = 4.8 \times 10^{-7}~\text{m}^3/\text{g} \] \[ A = 240~\text{m}^2/\text{g} \] \[ r_p = \frac{4.8 \times 10^{-7}~\text{m}^3/\text{g}}{240~\text{m}^2/\text{g}} = 2.0 \times 10^{-9}~\text{m} \] So, \(r_p = 2.0~\text{nm}\) --- Now, plug into Knudsen diffusivity: \[ D_K = \frac{2}{3} r_p \sqrt{\frac{8RT}{\pi M}} \] \[ = \frac{2}{3} \times 2.0 \times 10^{-9}~\text{m} \times \sqrt{\frac{8 \times 8.314 \times 705}{\pi \times 0.084}} \] Calculate the sqrt term: \[ 8 \times 8.314 \times 705 = 8 \times 5,863 = 46,902 \] \[ \pi \times 0.084 = 0.2639 \] \[ \frac{46,902}{0.2639} = 177,706 \] \[ \sqrt{177,706} = 421.87 \] So, \[ D_K = \frac{2}{3} \times 2.0 \times 10^{-9} \times 421.87 \] \[ = 1.333 \times 10^{-9} \times 421.87 \] \[ = 5.63 \times 10^{-7}~\text{m}^2/\text{s} \] --- ## Step 5: Now relate \(D_e\) and \(D_K\) to find tortuosity (\(\tau\)): \[ D_e = D_K \frac{\varepsilon_p}{\tau} \] \[ \tau = D_K \frac{\varepsilon_p}{D_e} \implies \tau = \frac{D_K \cdot \varepsilon_p}{D_e} \] Plug in the numbers: - \(D_K = 5.63 \times 10^{-7}~\text{m}^2/\text{s}\) - \(\varepsilon_p = 0.59\) - \(D_e = 7.09 \times 10^{-4}~\text{m}^2/\text{s}\) Wait, but \(D_e\) is much larger than \(D_K\), which is inconsistent since \(D_e\) should be less than \(D_K\). Let's check the calculation for \(D_e\): Earlier, we calculated: \[ D_e = \frac{(1.6 \times 10^{-3})^2 \times 1,871}{6.76} \] \[ = \frac{2.56 \times 10^{-6} \times 1,871}{6.76} \] \[ = \frac{0.004789}{6.76} \] \[