Let's work **step-by-step** through the derivation and explanation, as requested in the image. --- # 1. **Langmuir–Hinshelwood Mechanism: Surface Kinetics Formulation** ## **a. Surface Site Balance** For two species (A and I) competing for the same surface sites \( S \): \[ \theta_A + \theta_I + \theta_S = 1 \] Where: - \( \theta_A \): Fraction of sites occupied by A - \( \theta_I \): Fraction by I - \( \theta_S \): Fraction unoccupied ## **b. Adsorption Isotherms** Assume adsorption–desorption equilibrium: \[ A + S \rightleftharpoons A_{ads} \] \[ I + S \rightleftharpoons I_{ads} \] Langmuir isotherms: \[ \theta_A = \frac{K_A C_A \theta_S}{1} \] \[ \theta_I = \frac{K_I C_I \theta_S}{1} \] Solve for \( \theta_S \) using the site balance: \[ \theta_A + \theta_I + \theta_S = 1 \implies K_A C_A \theta_S + K_I C_I \theta_S + \theta_S = 1 \implies \theta_S (1 + K_A C_A + K_I C_I) = 1 \implies \theta_S = \frac{1}{1 + K_A C_A + K_I C_I} \] Now, \[ \theta_A = \frac{K_A C_A}{1 + K_A C_A + K_I C_I} \] \[ \theta_I = \frac{K_I C_I}{1 + K_A C_A + K_I C_I} \] ## **c. Surface Reaction Rate** If the **rate-limiting** step is the reaction of adsorbed A: \[ r_s = k(T) \theta_A \] where \( k(T) \) is the temperature-dependent rate constant. So, \[ r_s = k(T) \frac{K_A C_A}{1 + K_A C_A + K_I C_I} \] --- # 2. **Transient Mass Balance** For species \( A \) in a CSTR: \[ V \frac{dC_A}{dt} = F \left[C_{A,in}(t) - C_A \right] - a_s r_s \] where: - \( V \): reactor volume - \( F \): volumetric flow rate - \( C_A \): bulk concentration of A - \( C_{A,in}(t) \): inlet concentration (can be time-varying) - \( a_s \): catalyst surface area per reactor volume - \( r_s \): surface reaction rate (from above) Substitute \( r_s \): \[ V \frac{dC_A}{dt} = F \left[C_{A,in}(t) - C_A \right] - a_s k(T) \frac{K_A C_A}{1 + K_A C_A + K_I C_I} \] A similar equation can be written for \( C_I \) if there is consumption or reaction for I. --- # 3. **Energy (Thermal) Balance** Since the reaction is **exothermic**, include temperature: \[ \rho C_p V \frac{dT}{dt} = F \rho C_p (T_{in} - T) + (-\Delta H_r) a_s r_s - Q_{loss} \] - \( \rho C_p \): heat capacity per volume - \( T \): reactor temperature - \( T_{in} \): inlet temperature - \( \Delta H_r \): heat of reaction (negative for exothermic) - \( Q_{loss} \): heat loss term --- # 4. **Coupled Nonlinear ODE System** Together, you have a **coupled, nonlinear ODE system** for \( C_A(t) \), \( C_I(t) \), and \( T(t) \): \[ \boxed{ \begin{align*} V \frac{dC_A}{dt} &= F \left[C_{A,in}(t) - C_A \right] - a_s k(T) \frac{K_A C_A}{1 + K_A C_A + K_I C_I} \\ V \frac{dC_I}{dt} &= F \left[C_{I,in}(t) - C_I \right] \\ \rho C_p V \frac{dT}{dt} &= F \rho C_p (T_{in} - T) + (-\Delta H_r) a_s k(T) \frac{K_A C_A}{1 + K_A C_A + K_I C_I} - Q_{loss} \end{align*} } \] --- # 5. **Non-Dimensionalization** Define dimensionless variables (as in the image): \[ X = \frac{C_A}{C_{A1}}, \quad Q = \frac{T - T_{in}}{T_n}, \quad \tau = \frac{t F}{V} \] Re-write the ODEs using these variables to get a dimensionless dynamical system. --- # **Summary (Final Answer)** **The complete set of coupled, nonlinear ODEs** governing the transient behavior of the reactor are: \[ \boxed{ \begin{align*} V \frac{dC_A}{dt} &= F \left[C_{A,in}(t) - C_A \right] - a_s k(T) \frac{K_A C_A}{1 + K_A C_A + K_I C_I} \\ V \frac{dC_I}{dt} &= F \left[C_{I,in}(t) - C_I \right] \\ \rho C_p V \frac{dT}{dt} &= F \rho C_p (T_{in} - T) + (-\Delta H_r) a_s k(T) \frac{K_A C_A}{1 + K_A C_A + K_I C_I} - Q_{loss} \end{align*} } \] where all terms, reaction rates, and coverages are explicitly defined above. --- **Next steps:** You can non-dimensionalize these equations, analyze steady states, and study stability (thermal runaway, oscillations) as outlined in the image. If you want to see the non-dimensional form or steady-state analysis, let me know!