The classical vibrational frequency of HF is 1.240 Times 10^14Hz. Compute the sp

Question

The classical vibrational frequency of HF is 1.240 Times 10^14Hz. Compute the spacing between the HF vibrational levels, assuming that the vibrational eigenstates are accurately described by a harmonic oscillator potential. What is the energy in units of kJ/mol? Is this large or small relative to the energy of a C-H bond (415 kJ/mol)? If the masses of hydrogen and fluorine were 1 g and 19 g, respectively, what would the vibrational energy spacing be in units of kJ/mol? Would this system behave classically? Provide a complete explanation for your answer.

Explanation / Answer

a) The ground state vibrational energy is therefore

E0 = 1/2 h0

E0 = 6.62×1034 X 1.240 X 10^14 / 2 = 4.1 X 10^-20 Joules

The first excited state vibrational energy is

E1 = 3 /2 h0 = 12.3 X 10^-20 Joules

So difference = 8.2 X 10^-20 Joules

For 1 mole = 8.2 X 10^-20 Joules X 6.023 X 10^23 = 49.38 X 10^3 Joules = 49.38 KJ

So it is small relative to C-H bond

b) we know that

force constant = k = 42 02 µ .

µ . = reduced mass = mH X mF / mH + mF = 1x19 / 1+19 = 0.95 amu = 0.95 X 1.66 ×1024 g = 1.577 X 1024g

k = 4 X 9.8 X (1.240 X 10^14)2 X 1.577 X 1024g = 95.05 X 10^4 gs2

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