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give answer in 3 step with explanation at the end of each step and final answer at the endHere we use a statistical approach to calculate how the vibrational internal energy and heat capacity change when NO binds to a surface. Assume the vibrational motion of 14N80 can be treated as a quantum mechanical linear harmonic oscillation. In the gas phase the vibrational temperature of NO is 8,;,=2719K. v PartA Assuming T=300K, calculate the vibrational partition function of NO in the gas phase. 0 “eos Quit - rT] Request Answer v PartB Calculate the molar vibrational internal energy of NO in the gas phase at T=300K. EB oo Uas= | Value Units Request Answer v PartC Using your results from parts A and B, calculate the molar vibrational heat capacity for CO in the gas phase at T=300K. vi] RR Cvuiv=| Value Units v PartB Calculate the molar vibrational internal energy of NO in the gas phase at T=300K. EB oo Uas= | Value Units Request Answer v PartC Using your results from parts A and B, calculate the molar vibrational heat capacity for CO in the gas phase at T=300K. vi] RR Cvuiv=| Value Units ~ PartD Suppose NO binds to a surface with the oxygen immobilized on the protein surface leaving the nitrogen atom is free to continue a vibrational motion. Assuming the NO bond spring constant remains unchanged, calculate the vibrational temperture for ¥N'60 bound to the surface at T=300K. Hint: Calculate the reduced masses of free NO and NO bound with the O immobilized. Elo - Buis,surace= | Value Units v PartE Calculate the change in the molar vibrational internal energy when NO binds to the surface at T=300K. EB os" AUw= | Value Units wv PartF Calculate the change in the molar vibrational heat capacity when NO binds to a surface at T=300K. ams » 0 ? Request Answer

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give answer in 3 step with explanation at the end of each step and final answer at the endUploaded ImageUploaded ImageUploaded ImageUploaded ImageUploaded ImageHere we use a statistical approach to calculate how the vibrational internal energy and heat capacity change when NO binds to a surface. Assume the vibrational motion of 14N80 can be treated as a quantum mechanical linear harmonic oscillation. In the gas phase the vibrational temperature of NO is 8,;,=2719K. v PartA Assuming T=300K, calculate the vibrational partition function of NO in the gas phase. 0 “eos Quit - rT] Request Answer v PartB Calculate the molar vibrational internal energy of NO in the gas phase at T=300K. EB oo Uas= | Value Units Request Answer v PartC Using your results from parts A and B, calculate the molar vibrational heat capacity for CO in the gas phase at T=300K. vi] RR Cvuiv=| Value Units v PartB Calculate the molar vibrational internal energy of NO in the gas phase at T=300K. EB oo Uas= | Value Units Request Answer v PartC Using your results from parts A and B, calculate the molar vibrational heat capacity for CO in the gas phase at T=300K. vi] RR Cvuiv=| Value Units ~ PartD Suppose NO binds to a surface with the oxygen immobilized on the protein surface leaving the nitrogen atom is free to continue a vibrational motion. Assuming the NO bond spring constant remains unchanged, calculate the vibrational temperture for ¥N'60 bound to the surface at T=300K. Hint: Calculate the reduced masses of free NO and NO bound with the O immobilized. Elo - Buis,surace= | Value Units v PartE Calculate the change in the molar vibrational internal energy when NO binds to the surface at T=300K. EB os" AUw= | Value Units wv PartF Calculate the change in the molar vibrational heat capacity when NO binds to a surface at T=300K. ams » 0 ? Request Answer

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Asked by: amu
Created at: 2025-07-27 04:27:58

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Let's solve **Part A** step by step: --- ### **Step 1: Write the formula for the vibrational partition function** For a diatomic molecule, the vibrational partition function is: \[ q_{vib} = \frac{1}{1 - \exp\left(-\frac{\theta_{vib}}{T}\right)} \] where: - \(\theta_{vib}\) = vibrational temperature - \(T\) = temperature in Kelvin **Explanation:** This formula comes from the quantum harmonic oscillator model for vibrational energy levels. --- ### **Step 2: Plug in the given values** Given: - \(\theta_{vib} = 2719 \ \text{K}\) - \(T = 300 \ \text{K}\) So, \[ q_{vib} = \frac{1}{1 - \exp\left(-\frac{2719}{300}\right)} \] **Explanation:** We substitute the values directly into the formula. --- ### **Step 3: Calculate the value** Calculate the exponent: \[ \frac{2719}{300} \approx 9.063 \] So, \[ q_{vib} = \frac{1}{1 - \exp(-9.063)} \] \[ \exp(-9.063) \approx 1.16 \times 10^{-4} \] Therefore, \[ q_{vib} \approx \frac{1}{1 - 1.16 \times 10^{-4}} \approx \frac{1}{0.999884} \approx 1.00012 \] **Explanation:** The exponent is large and negative, so the exponential term is very small. The partition function is thus just slightly above 1. --- ## **Final Answer (Part A):** \[ \boxed{q_{vib} = 1.00012} \] --- Let me know if you'd like to proceed to Part B!

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