Let's go step-by-step through each part. --- ### **Part D:** **Calculate the vibrational temperature for \( ^{14}N^{16}O \) bound to the surface at \( T=300K \) (with O immobilized).** #### **Step 1: Reduced Mass Calculation** - **For free NO:** Reduced mass, \[ \mu_{\text{free}} = \frac{m_N m_O}{m_N + m_O} \] where \( m_N = 14 \) u, \( m_O = 16 \) u. \[ \mu_{\text{free}} = \frac{14 \times 16}{14+16} = \frac{224}{30} = 7.47 \text{ u} \] - **For NO with O immobilized:** Reduced mass (since O is fixed, only N moves): \[ \mu_{\text{surface}} = m_N = 14 \text{ u} \] #### **Step 2: Vibrational Temperature (\( \theta_{\text{vib}} \))** The vibrational temperature is: \[ \theta_{\text{vib}} = \frac{h\nu}{k_B} \] And, \[ \nu = \frac{1}{2\pi} \sqrt{\frac{k}{\mu}} \] Since spring constant \( k \) is unchanged, the ratio of vibrational temperatures is: \[ \frac{\theta_{\text{vib,surface}}}{\theta_{\text{vib,free}}} = \sqrt{\frac{\mu_{\text{free}}}{\mu_{\text{surface}}}} \] \[ \theta_{\text{vib,surface}} = \theta_{\text{vib,free}} \sqrt{\frac{\mu_{\text{free}}}{\mu_{\text{surface}}}} \] - For NO, \( \theta_{\text{vib,free}} \approx 2740 \) K (you can look this up in a constants table). \[ \theta_{\text{vib,surface}} = 2740 \times \sqrt{\frac{7.47}{14}} \] \[ \sqrt{\frac{7.47}{14}} = \sqrt{0.534} = 0.731 \] \[ \theta_{\text{vib,surface}} = 2740 \times 0.731 \approx 2002 \text{ K} \] **Explanation:** We first found the new reduced mass (since O is immobilized, only N vibrates). Then, knowing vibrational temperature depends inversely on the square root of reduced mass, we scaled the standard value for free NO by the ratio of reduced masses. **Final Answer (Part D):** \[ \boxed{2002 \ \text{K}} \] --- ### **Part E:** **Calculate the change in the molar vibrational internal energy when NO binds to the surface at T=300K.** - Vibrational energy per mole: \[ U_{\text{vib}} = R \frac{\theta_{\text{vib}}}{e^{\theta_{\text{vib}}/T}-1} \] So, \[ \Delta U_{\text{vib}} = U_{\text{vib,surface}} - U_{\text{vib,free}} \] Let's calculate: #### For free NO: \[ U_{\text{vib,free}} = R \frac{2740}{e^{2740/300}-1} \] \[ \frac{2740}{300} = 9.13 \] \[ e^{9.13} \approx 9233 \] \[ U_{\text{vib,free}} = 8.314 \times \frac{2740}{9233-1} = 8.314 \times \frac{2740}{9232} = 8.314 \times 0.2966 = 2.465 \text{ J/mol} \] #### For surface-bound NO: \[ U_{\text{vib,surface}} = R \frac{2002}{e^{2002/300}-1} \] \[ \frac{2002}{300} = 6.673 \] \[ e^{6.673} \approx 790.6 \] \[ U_{\text{vib,surface}} = 8.314 \times \frac{2002}{790.6-1} = 8.314 \times \frac{2002}{789.6} = 8.314 \times 2.536 = 21.10 \text{ J/mol} \] #### **Change:** \[ \Delta U_{\text{vib}} = 21.10 - 2.465 = 18.64 \text{ J/mol} \] **Explanation:** We used the formula for vibrational energy, plugged in the vibrational temperatures for both free and surface-bound NO, and subtracted to find the change. **Final Answer (Part E):** \[ \boxed{18.6 \ \text{J/mol}} \] --- ### **Part F:** **Calculate the change in the molar vibrational heat capacity at T=300 K.** - Vibrational heat capacity, \[ C_{V,\text{vib}} = R \left( \frac{\theta_{\text{vib}}}{T} \right)^2 \frac{e^{\theta_{\text{vib}}/T}}{(e^{\theta_{\text{vib}}/T} - 1)^2} \] #### For free NO: \[ \left( \frac{2740}{300} \right)^2 \frac{e^{9.13}}{(e^{9.13}-1)^2} \] \[ (9.13)^2 = 83.39 \] \[ \frac{e^{9.13}}{(e^{9.13}-1)^2} = \frac{9233}{(9232)^2} = \frac{9233}{85230824} = 1.08 \times 10^{-4} \] \[ C_{V,\text{vib,free}} = 8.314 \times 83.39 \times 1.08 \times 10^{-4} = 8.314 \times 0.00901 = 0.0749 \text{ J/mol K} \] #### For surface-bound NO: \[ \left( \frac{2002}{300} \right)^2 \frac{e^{6.673}}{(e^{6.673}-1)^2} \] \[ (6.673)^2 = 44.53 \] \[ \frac{e^{6.673}}{(e^{6.673}-1)^2} = \frac{790.6}{(789.6)^2} = \frac{790.6}{623507} = 0.001268 \] \[ C_{V,\text{vib,surface}} = 8.314 \times 44.53 \times 0.001268 = 8.314 \times 0.0565 = 0.470 \text{ J/mol K} \] #### **Change:** \[ \Delta C_{V,\text{vib}} = 0.470 - 0.0749 = 0.395 \text{ J/mol K} \] **Explanation:** We calculated the vibrational heat capacity for both free and surface-bound NO, then subtracted to find the change. **Final Answer (Part F):** \[ \boxed{0.40 \ \text{J/mol K}} \] --- ## **Summary of Final Answers:** - **Part D:** \( \boxed{2002 \ \text{K}} \) - **Part E:** \( \boxed{18.6 \ \text{J/mol}} \) - **Part F:** \( \boxed{0.40 \ \text{J/mol K}} \) Let me know if you need any clarification or further explanation!