Let's solve each part step by step: --- ### (a) **Rate Limiting Step and Mechanism** The given rate law is: \[ -r_A = \frac{k' P_A}{1 + K_A P_A} \] This is the form of the **Langmuir-Hinshelwood kinetics** for a catalytic reaction, where a single reactant (A) adsorbs onto the catalyst surface and reacts. It suggests that: - The surface reaction between adsorbed A and the catalyst is the rate-limiting step. - The denominator \((1 + K_A P_A)\) indicates competitive adsorption, where \(K_A\) is the adsorption equilibrium constant for A. - The numerator shows that the reaction rate is proportional to the partial pressure of A at low pressures, but becomes independent of \(P_A\) at high pressures (saturation). **Mechanism:** 1. **Adsorption:** \(A_{(g)} + S \leftrightarrow A{-}S\) (where S is a surface site) 2. **Surface reaction (rate-limiting):** \(A{-}S \rightarrow P_{(g)} + S\) 3. **Desorption of Product:** \(P_{(g)}\) leaves the surface rapidly --- ### (b) **Determining \(k'\) and \(K_A\) from Data** Given the rate law and data, rearrange for linear fitting: \[ -r_A = \frac{k' P_A}{1 + K_A P_A} \] Invert both sides: \[ \frac{1}{-r_A} = \frac{1 + K_A P_A}{k' P_A} = \frac{1}{k' P_A} + \frac{K_A}{k'} \] So, if we plot \(\frac{1}{-r_A}\) vs. \(\frac{1}{P_A}\), the slope is \(\frac{1}{k'}\) and the y-intercept is \(\frac{K_A}{k'}\). #### **Prepare Data Table:** | Run | \(P_A\) (atm) | \(-r_A\) (mol/kg-cat.min) | \(1/P_A\) (atm\(^{-1}\)) | \(1/-r_A\) (kg-cat.min/mol) | |-----|--------------|-----------------------|-------------------------|------------------------------| | 1 | .5 | .34 | 2.00 | 2.941 | | 2 | 1. | .44 | 1.00 | 2.273 | | 3 | 5. | .58 | .20 | 1.724 | | 4 | 10. | .60 | .10 | 1.667 | | 5 | 20. | .61 | .05 | 1.639 | #### **Linear Regression (Using Points 1 and 5 as Approximation):** Slope: \[ \text{slope} = \frac{(1.639 - 2.941)}{(.05 - 2)} = \frac{-1.302}{-1.95} \approx .668 \] Intercept: \[ \text{intercept} = 1.639 - (.668 \times .05) \approx 1.606 \] So: \[ \frac{1}{k'} = .668 \implies k' \approx 1.50\ \text{mol/(kg-cat.min.atm)} \] \[ \frac{K_A}{k'} = 1.606 \implies K_A = 1.606 \times k' \approx 1.606 \times 1.50 = 2.41\ \text{atm}^{-1} \] --- ### **Final Answers** (a) The rate-limiting step is the surface reaction between adsorbed A and the catalyst. The mechanism is a Langmuir-Hinshelwood type, where A adsorbs onto the catalyst, reacts on the surface as the slow step, and P desorbs rapidly. (b) The values are \(k' \approx 1.50\ \text{mol/(kg-cat.min.atm)}\) and \(K_A \approx 2.41\ \text{atm}^{-1}\).