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Molecular Vibration Model 1: the Classical Harmonic Oscillator

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Molecular Vibration Model 1: the Classical Harmonic Oscillator Molecular Vibration Model 1: The Classical Harmonic Oscillator maximum maximum If the bond joining two atoms of masses, m and m , acts as a perfect V 1 2 compression elongaon spring, the two masses move back and forth as the bond compresses and expands with a characteristic frequency: ! ! !!× !! ν = where µ = !! ! !!! !! where µ is the reduced mass and k is the bond force constant. The force constant reflects how strongly the bond resists being stretched or compressed. Strong bonds have higher values for k and weak bonds have lower values for k. x < 0 0 x > 0 The figure above shows how the potential energy changes during the vibration. At every point, the total energy, E, is the sum of the potential, V, and kinetic, T, energy: E = T + V. The potential energy is at a maximum at the “turning points” where the kinetic energy is zero. Similarly, when the bond has its equilibrium length, x = 0, V = 0 and the atoms are moving with the maximum speed. Critical thinking questions 1. According to the equation above, what happens to the vibrational frequency if one of the atoms is replaced by a heavier mass (keeping k constant)? 2. According to the equation above, what happens to the vibrational frequency if the force constant is replaced by a larger value (keeping µ constant). 3. In general, what is the reduced mass of the homonuclear diatomic X2 if element X has atomic mass m? 4. What is the ratio of the reduced masses of H2 and D2? 5. What does the reduced mass of HX tend to as X becomes very heavy compared to H? Molecular Vibration Model 2: The Quantum Harmonic Oscillator The Schrödinger equation for the simple harmonic oscillator is given by: ℏ! !! Ĥψ(x) = (− + ½ kx2)ψ(x) = Eψ(x) !! !"! This gives the energy levels as ! εν = (ν + ½) ħ where ν = 0, 1, 2 .... ! 2 The figure opposite shows the form of ψ(x) for ν = 0, 1, 2 and 3. Critical thinking questions 1. The minimum energy for the vibrating molecule is not zero. (a) In terms of the wavefunction, why can’t the vibrating molecule have zero energy? (b) What is the minimum energy? 2. What is the energy difference between adjacent levels? 3. What is the average value of the bond length when the molecule has (a) ν = 0 and ν = 1? 4. If the figure above applies to H2, use your answers to Model 1 Q1 and Q4 to sketch on the levels for D2. 5. The potential energy curve for a real molecule is shown opposite together with that for a harmonic oscillator. (a) Explain the origin of the differences in the two functions. (b) Remembering your answer to question 3, how does the average bond length for a real molecule change as the vibrational energy increases? Molecular Vibration Model 3 The Anharmonic Oscillator For an anharmonic model, the vibrational energy levels (in cm-1) are given by: 2 G(v) = (v + ½)ωe - (v + ½) ωexe where ωe is the harmonic frequency and ωexe is the anharmonicity constant. v is the vibrational quantum number (v = 0, 1, 2, 3 ….) The dissociation energy from the bottom of the potential well, De, is given by: ! !! De = !!!!! This differs from the true dissociation energy, D0, as the minimum energy the molecule can have is the zero point energy, G(0). Hence: ! !! D0 = De – G(0) = – G(0) !!!!! Critical Thinking Questions 1. For the anharmonic case, what are the energies of the first 3 levels (v = 0, 1 and 2)? (a) G(0) = (b) G(1) = (c) G(2) = (d) G(3) = The fundamental is the strongest absorption in the infrared spectrum for the vibration. It corresponds to the transition from v = 0 to 1. Overtones correspond to transition from v = 0 to higher values of v. The intensity of overtones decreases rapidly as Δv gets larger. 2. By calculating the energy differences between the anharmonic energy levels below, work out expressions for the fundamental and the first and second overtones. (a) Fundamental: G(1) - G(0) = (b) First overtone: G(2) - G(0) = (c) Second overtone: G(3) - G(0) = 3. For HF, the fundamental appears at 4028.3 cm-1 and the first overtone appears at 7880.0 cm-1. Using your equations from Q2(a) and Q2(b) above, set up and solve the simultaneous equations for the two unknowns ωe and ωexe. 4. Using your values for ωe and ωexe, calculate the dissociation energy, D0. (Hint: do not forget to correct for the zero point energy. .
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