Let's solve each part step-by-step: --- ## **1. Gas-Solid System: Derivation of Global Reaction Rate** ### **Given:** - Molar flux in gas film: \( N_A = k_g(C_{Ab} - C_{As}) \) - Surface reaction rate: \( -r_A = k_s C_{As} \) ### **Step 1: Set Molar Flux = Surface Reaction Rate** At steady state, the rate at which A arrives at the surface = rate at which A reacts: \[ N_A = -r_A \] \[ k_g(C_{Ab} - C_{As}) = k_s C_{As} \] ### **Step 2: Solve for Surface Concentration \( C_{As} \)** \[ k_g C_{Ab} - k_g C_{As} = k_s C_{As} \] \[ k_g C_{Ab} = (k_g + k_s) C_{As} \] \[ C_{As} = \frac{k_g}{k_g + k_s} C_{Ab} \] ### **Step 3: Expression for Overall Rate Constant \( K_{oa} \)** Overall (observable) reaction rate: \[ N_A = k_g (C_{Ab} - C_{As}) = k_g \left(C_{Ab} - \frac{k_g}{k_g + k_s} C_{Ab}\right) \] \[ = k_g C_{Ab} \left(1 - \frac{k_g}{k_g + k_s}\right) = k_g C_{Ab} \frac{k_s}{k_g + k_s} \] \[ N_A = K_{oa} C_{Ab} \] where \[ K_{oa} = \frac{k_g k_s}{k_g + k_s} \] Or, in terms of resistances: \[ \frac{1}{K_{oa}} = \frac{1}{k_g} + \frac{1}{k_s} \] --- ## **2. Enzyme-Catalyzed Reaction in a CSTR** ### **Given:** - \( V_{max} = .5 \) mmol·L\(^{-1}\)·s\(^{-1}\) - \( K_m = 2. \) mmol/L - Feed concentration, \( C_{A} = 10 \) mmol/L - Reactor volume, \( V = 5 \) L - Desired conversion, \( X = .95 \) - Michaelis-Menten rate: \( -r_S = \frac{V_{max}C_S}{K_m + C_S} \) ### **Step 1: Material Balance for CSTR** At steady state, \[ \text{Input} - \text{Output} + \text{Generation} = \] \[ F C_{A} - F C_{A} - r_S V = \] \[ F(C_{A} - C_{A}) = r_S V \] Let \( v_ = F \) (feed flow rate). ### **Step 2: Express \( C_{A} \) in Terms of Conversion** \[ X = \frac{C_{A} - C_{A}}{C_{A}} \implies C_{A} = C_{A}(1 - X) \] Given \( X = .95 \), \( C_{A} = 10 \times .05 = .5 \) mmol/L ### **Step 3: Substitute into the Balance** \[ v_ (C_{A} - C_A) = r_S V \] \[ v_ (10 - .5) = \frac{V_{max} \cdot .5}{K_m + .5} \cdot 5 \] \[ v_ \cdot 9.5 = \frac{.5 \times .5}{2. + .5} \cdot 5 \] \[ = \frac{.25}{2.5} \cdot 5 = .1 \cdot 5 = .5 \text{ mmol/s} \] \[ v_ = \frac{.5}{9.5} \approx .0526 \text{ L/s} \] **Final answer:** \[ \boxed{v_ \approx .0526~\text{L/s}} \] **Competitive inhibition:** Competitive inhibition increases the apparent \( K_m \), reducing the rate at a given substrate concentration, which **increases the required residence time** for the same conversion. --- ## **3. Plotting Residence Time Distribution and Variance** ### **Given:** - \( E(t) = \frac{n^n}{(n-1)! \tau} t^{n-1} e^{-nt/\tau} \) - Residence time \( \tau \) - Variance for \( n \) tanks: \( \sigma^2 = \frac{\tau^2}{n} \) ### **Step 1: Variance for 4-Tank System with \(\tau = 20\) min** \[ \sigma^2 = \frac{\tau^2}{n} = \frac{20^2}{4} = \frac{400}{4} = 100~\text{min}^2 \] **Final answer:** \[ \boxed{\sigma^2 = 100~\text{min}^2} \] --- ## **Summary Table** | Part | Answer | |------|--------| | Gas-solid system overall rate | \( \frac{1}{K_{oa}} = \frac{1}{k_g} + \frac{1}{k_s} \) | | CSTR feed flow for 95% conversion | \( v_ \approx .0526~\text{L/s} \) | | Variance, 4 tanks, 20 min total | \( \sigma^2 = 100~\text{min}^2 \) | --- If you need the RTD plots or a more detailed discussion on competitive inhibition, let me know!