Let's solve the problem step by step: --- ## **Step 1: Understanding the Problem** We are given the rate constants (\( k \)) at different temperatures (\( T \)) for a bimolecular gas-phase reaction. We are asked to **calculate the Gibbs free energy of activation (\( \Delta G^\ddagger \)) at 298 K**. --- ## **Step 2: Key Equation - Eyring Equation** The Eyring equation relates the rate constant to the Gibbs free energy of activation: \[ k = \frac{k_B T}{h} e^{-\Delta G^\ddagger / RT} \] Where: - \( k \) = rate constant - \( k_B \) = Boltzmann constant (\(1.381 \times 10^{-23}\) J/K) - \( h \) = Planck constant (\(6.626 \times 10^{-34}\) J·s) - \( T \) = temperature (in K) - \( R \) = gas constant (\(8.314\) J/(mol·K)) - \( \Delta G^\ddagger \) = Gibbs free energy of activation --- ## **Step 3: Rearranging for \(\Delta G^\ddagger\)** Take the natural logarithm: \[ \ln k = \ln\left(\frac{k_B T}{h}\right) - \frac{\Delta G^\ddagger}{RT} \] So, \[ \Delta G^\ddagger = -RT \left[ \ln k - \ln\left(\frac{k_B T}{h}\right) \right] \] Or, \[ \Delta G^\ddagger = RT \ln\left(\frac{k_B T}{h k}\right) \] --- ## **Step 4: Plug in the Values** Given: - \( T = 298\, \text{K} \) - \( k = 3.55 \times 10^6 \, \text{dm}^3\, \text{mol}^{-1}\,\text{s}^{-1} \) - Convert \( k \) to SI units (\( \text{m}^3\,\text{mol}^{-1}\,\text{s}^{-1} \)): \( 1\,\text{dm}^3 = 10^{-3}\,\text{m}^3 \), so \( k = 3.55 \times 10^3\,\text{m}^3\,\text{mol}^{-1}\,\text{s}^{-1} \) Constants: - \( k_B = 1.381 \times 10^{-23}\,\text{J/K} \) - \( h = 6.626 \times 10^{-34}\,\text{J·s} \) - \( R = 8.314\,\text{J/(mol·K)} \) --- ## **Step 5: Calculation** \[ \Delta G^\ddagger = RT \ln\left(\frac{k_B T}{h k}\right) \] Plug in values: 1. **Compute \(\frac{k_B T}{h k}\):** \[ \frac{k_B \cdot T}{h \cdot k} = \frac{1.381 \times 10^{-23} \times 298}{6.626 \times 10^{-34} \times 3.55 \times 10^3} \] Numerator: \(1.381 \times 10^{-23} \times 298 = 4.11238 \times 10^{-21}\) Denominator: \(6.626 \times 10^{-34} \times 3.55 \times 10^3 = 2.35223 \times 10^{-30}\) So: \[ \frac{4.11238 \times 10^{-21}}{2.35223 \times 10^{-30}} \approx 1.749 \times 10^9 \] 2. **Compute \(\ln(1.749 \times 10^9)\):** \[ \ln(1.749 \times 10^9) = \ln(1.749) + \ln(10^9) \approx .560 + 20.723 = 21.283 \] 3. **Compute \(RT\):** \[ RT = 8.314 \times 298 = 2477.572 \text{ J/mol} \] 4. **Calculate \(\Delta G^\ddagger\):** \[ \Delta G^\ddagger = 2477.572 \times 21.283 \approx 52744 \text{ J/mol} = 52.7 \text{ kJ/mol} \] --- ## **Final Answer** \[ \boxed{ \Delta G^\ddagger = 52.7 \ \text{ kJ/mol} \text{ at 298 K} } \] --- ### **Summary Table** | Quantity | Value | |-------------------------|------------------------| | \( T \) | 298 K | | \( k \) | \(3.55 \times 10^3\,\text{m}^3\,\text{mol}^{-1}\,\text{s}^{-1}\) | | \( R \) | 8.314 J/(mol·K) | | \( k_B \) | \(1.381 \times 10^{-23}\) J/K | | \( h \) | \(6.626 \times 10^{-34}\) J·s | | \( \Delta G^\ddagger \) | \(\mathbf{52.7~kJ/mol}\) | --- **Let me know if you need further clarification or calculations!**