Question

Calculate the vibrational partition function for CS₂ at 500 K given wavenumbers 658 cm^-1 (symmetric stretch), 397 cm^-1 (bend, two modes), and 1535 cm^-1 (asymmetric stretch).

Answer 1

Vibrational partition functional Z(vib),

Z(vib) = e^(-hv/2kt)/[1 - e^(-hv/kt)]

with,

h = planck's constant

v = vibrational frequency in Hz

k = Boltzmann constant

T = 500 K

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For CS2 (symmetric stretch) = 658 cm-1

c = vl [l = wavelength which is inverse of wavenumber]

v = (3 x 10^10 cm/s)(658 cm-1) = 1.974 x 10^13 Hz

hv/kT = 6.626 x 10^-34 x 1.974 x 10^13/1.38 x 10^-23 x 500 = 1.8956

Feeding this value in Vibrational partition function equation,

Z(vib) = e^(-1.8956/2)/[1 - e^(-1.8956)]

          = 0.456

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For CS2 (bend, two modes) = 397 cm-1

c = vl [l = wavelength which is inverse of wavenumber]

v = (3 x 10^10 cm/s)(397 cm-1) = 1.191 x 10^13 Hz

hv/kT = 6.626 x 10^-34 x 1.191 x 10^13/1.38 x 10^-23 x 500 = 1.144

Feeding this value in Vibrational partition function equation,

Z(vib) = e^(-1.144/2)/[1 - e^(-1.144)]

          = 2.60

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For CS2 (asymmetric stretch) = 1535 cm-1

c = vl [l = wavelength which is inverse of wavenumber]

v = (3 x 10^10 cm/s)(1535 cm-1) = 4.605 x 10^13 Hz

hv/kT = 6.626 x 10^-34 x 4.605 x 10^13/1.38 x 10^-23 x 500 = 4.422

Feeding this value in Vibrational partition function equation,

Z(vib) = e^(-4.422/2)/[1 - e^(-4.422)]

          = 0.111

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