Let's solve the given problem step-by-step as requested: --- # (i) Governing Equations in Dimensionless Form Let’s define the dimensionless variables: - \(\psi = \frac{C_A}{C_{A\infty}}\) - \(\theta = \frac{T}{T_\infty}\) - \(\xi = \frac{r}{R}\) **Mass Balance:** \[ \frac{1}{r^2} \frac{d}{dr}\left(r^2 D_e \frac{dC_A}{dr}\right) - k(T)C_A = \] Substitute the dimensionless variables: \( r = \xi R \) \\ \( C_A = C_{A\infty}\psi \) Chain rule gives: \[ \frac{dC_A}{dr} = \frac{dC_A}{d\xi}\frac{d\xi}{dr} = \frac{1}{R}\frac{dC_A}{d\xi} \] and \[ \frac{d}{dr} \left( r^2 D_e \frac{dC_A}{dr} \right) = D_e\frac{d}{dr}\left(r^2\frac{1}{R}\frac{dC_A}{d\xi}\right) \] After substituting and simplifying: \[ \frac{1}{\xi^2} \frac{d}{d\xi}\left(\xi^2 \frac{d\psi}{d\xi}\right) - \phi^2 f(\theta)\psi = \] where \(\phi^2 = \frac{k_ R^2}{D_e}\) (Thiele modulus squared), and \(f(\theta)\) captures the temperature dependence of \(k(T)\). **Energy Balance:** \[ \frac{1}{r^2}\frac{d}{dr}\left(r^2 k_e \frac{dT}{dr}\right) + (-\Delta H_{Rx})k(T)C_A = \] Substitute the dimensionless variables: \( T = T_\infty \theta \) \[ \frac{1}{\xi^2}\frac{d}{d\xi}\left(\xi^2 \frac{d\theta}{d\xi}\right) + \beta \phi^2 f(\theta)\psi = \] where \(\beta = \frac{(-\Delta H_{Rx})C_{A\infty}}{T_\infty k_e}\) --- # (ii) Four Boundary Conditions At the center (\(\xi = \)), symmetry implies no flux: \[ \left.\frac{d\psi}{d\xi}\right|_{\xi=} = , \quad \left.\frac{d\theta}{d\xi}\right|_{\xi=} = \] At the surface (\(\xi = 1\)), account for external film resistance: \[ -\left.\frac{d\psi}{d\xi}\right|_{\xi=1} = \frac{k_c R}{D_e} (\psi_s - 1) \] \[ -\left.\frac{d\theta}{d\xi}\right|_{\xi=1} = \frac{h R}{k_e} (\theta_s - 1) \] --- # (iii) Linear Relationship Between \(\theta\) and \(\psi\) inside the Pellet Eliminate the reaction term: Divide the energy balance by the mass balance: \[ \frac{\frac{1}{\xi^2}\frac{d}{d\xi}(\xi^2 \frac{d\theta}{d\xi}) + \beta \phi^2 f(\theta)\psi}{\frac{1}{\xi^2} \frac{d}{d\xi}(\xi^2 \frac{d\psi}{d\xi}) - \phi^2 f(\theta)\psi} = 1 \] Inside the pellet, if \(\phi\) is large (high reaction rate), and assuming \(f(\theta)\) varies slowly: \[ \frac{d^2 \theta}{d\xi^2} \approx -\beta \frac{d^2 \psi}{d\xi^2} \] or, integrating (with appropriate constants absorbed into boundary conditions): \[ \theta = 1 + \beta (\psi - 1) \] This is a linear relation between temperature and concentration inside the pellet. --- # (iv) Sketch of Expected Profiles For both \(\psi(\xi)\) and \(\theta(\xi)\): - Both start at their center values (\(\xi=\)), with zero slope. - Both decrease monotonically toward the surface (\(\xi=1\)), reaching values determined by external mass and heat transfer. - There is a jump in slope at the surface due to finite film resistance. A sketch would show both \(\psi\) and \(\theta\) starting at higher values in the center, smoothly decreasing toward the bulk values at \(\xi=1\), but not reaching bulk conditions exactly due to external resistance. The profiles are qualitatively similar but not identical, as heat and mass transfer resistances may differ. --- **Summary of Final Answers:** (i) The dimensionless coupled ODEs are: \[ \frac{1}{\xi^2} \frac{d}{d\xi}(\xi^2 \frac{d\psi}{d\xi}) - \phi^2 f(\theta)\psi = \] \[ \frac{1}{\xi^2} \frac{d}{d\xi}(\xi^2 \frac{d\theta}{d\xi}) + \beta \phi^2 f(\theta)\psi = \] (ii) The four boundary conditions are: \[ \left.\frac{d\psi}{d\xi}\right|_{\xi=} = , \quad \left.\frac{d\theta}{d\xi}\right|_{\xi=} = \] \[ -\left.\frac{d\psi}{d\xi}\right|_{\xi=1} = \frac{k_c R}{D_e} (\psi_s - 1) \] \[ -\left.\frac{d\theta}{d\xi}\right|_{\xi=1} = \frac{h R}{k_e} (\theta_s - 1) \] (iii) The linear relationship is: \[ \theta = 1 + \beta (\psi - 1) \] (iv) Both \(\psi(\xi)\) and \(\theta(\xi)\) decrease smoothly from their center values (\(\xi=\)) to the surface (\(\xi=1\)), showing a gradient due to reaction and finite external film resistances, and do not reach the bulk value at the surface.