Certainly! Let’s tackle each part step by step, with explanations after each, and summarize at the end. --- ### **Step 1: Expression for the Overall Rate Constant (R)** **a) Provide an expression for the overall rate constant R of this first-order reaction.** **Explanation:** For a first-order surface reaction (like H₂ dissociating over Cu(111)), and using **Transition State Theory (TST)**, the rate constant is given by: \[ R = k = \frac{k_B T}{h} \exp\left(-\frac{E_a}{k_B T}\right) \] Where: - \( k_B \) = Boltzmann constant (\(1.38 \times 10^{-23}\) J/K) - \( h \) = Planck’s constant (\(6.626 \times 10^{-34}\) J·s) - \( T \) = temperature in Kelvin - \( E_a \) = activation energy (from the potential energy map) **Summary of Step 1:** The overall rate constant for the first-order reaction is: \[ \boxed{k = \frac{k_B T}{h} \exp\left(-\frac{E_a}{k_B T}\right)} \] --- ### **Step 2: Predict the Rate Constant at 298 K** **b) Predict the rate constant for dissociation of a mole of H₂ at 298K. (Hint: use pre-exponential term \(k_B T/h\))** **Explanation:** Let’s plug in the values for \( T = 298 \) K. Since the exact \( E_a \) is not specified from your question, let's call it \( E_a \) (in J/mol). First, compute the pre-exponential factor: \[ \frac{k_B T}{h} = \frac{1.38 \times 10^{-23} \times 298}{6.626 \times 10^{-34}} = \frac{4.1124 \times 10^{-21}}{6.626 \times 10^{-34}} \approx 6.2 \times 10^{12} \ \text{s}^{-1} \] Now, the exponential factor: \[ \exp\left(-\frac{E_a}{k_B T}\right) \] If \( E_a \) is given in kJ/mol, convert to J (and use \( R \), the gas constant, for per mole): \[ \exp\left(-\frac{E_a}{R T}\right) \] So, the general answer is: \[ k_{298} = 6.2 \times 10^{12} \cdot \exp\left(-\frac{E_a}{R \times 298}\right) \] **Summary of Step 2:** The rate constant at 298 K is: \[ \boxed{ k_{298} = 6.2 \times 10^{12} \exp\left(-\frac{E_a}{R \times 298}\right) \ \text{s}^{-1} } \] where \(E_a\) is in J/mol, and \(R = 8.314\) J/mol·K. --- ### **Step 3: Effect of Increasing Temperature to 398 K** **c) By how much does the rate constant increase if the temperature is increased to 398K? Explain this result.** **Explanation:** Let’s write the rate constant at 398K: \[ k_{398} = \frac{k_B \times 398}{h} \exp\left(-\frac{E_a}{R \times 398}\right) \] \[ \frac{k_B \times 398}{h} \approx \frac{1.38 \times 10^{-23} \times 398}{6.626 \times 10^{-34}} \approx 8.3 \times 10^{12} \ \text{s}^{-1} \] So, \[ k_{398} = 8.3 \times 10^{12} \exp\left(-\frac{E_a}{R \times 398}\right) \] Now, the **ratio** of rate constants is: \[ \frac{k_{398}}{k_{298}} = \frac{8.3 \times 10^{12}}{6.2 \times 10^{12}} \cdot \frac{\exp\left(-\frac{E_a}{R \times 398}\right)}{\exp\left(-\frac{E_a}{R \times 298}\right)} \] \[ = 1.34 \cdot \exp\left(-\frac{E_a}{R} \left[\frac{1}{398} - \frac{1}{298}\right]\right) \] \[ = 1.34 \cdot \exp\left(\frac{E_a}{R} \left[\frac{1}{298} - \frac{1}{398}\right]\right) \] This shows the **rate constant increases** both due to a higher pre-factor and because the exponential term becomes less negative (activation barrier easier to overcome at higher T). **Summary of Step 3:** The increase is: \[ \boxed{ \frac{k_{398}}{k_{298}} = 1.34 \cdot \exp\left(\frac{E_a}{R} \left[\frac{1}{298} - \frac{1}{398}\right]\right) } \] This means the rate constant increases significantly with temperature, especially for reactions with high activation energies. --- ## **Final Answers and Explanation** 1. **Expression for overall rate constant:** \[ k = \frac{k_B T}{h} \exp\left(-\frac{E_a}{k_B T}\right) \] 2. **Rate constant at 298 K:** \[ k_{298} = 6.2 \times 10^{12} \exp\left(-\frac{E_a}{R \times 298}\right) \ \text{s}^{-1} \] 3. **Increase in rate constant at 398 K:** \[ \frac{k_{398}}{k_{298}} = 1.34 \cdot \exp\left(\frac{E_a}{R} \left[\frac{1}{298} - \frac{1}{398}\right]\right) \] **Explanation:** - The rate constant increases with temperature due to both the pre-exponential (linear in T) and an exponential factor which becomes much less negative as T increases. - For reactions with high activation energies, the exponential term dominates, so the rate increases sharply with temperature. This explains the dramatic increase in rate constant for even small temperature increases in surface reactions like H₂ dissociation. --- **If you provide a value for \(E_a\), I can plug in numbers for you!**