The first excited vibrational energy level of ditomic chlorine (Cl2) is 558 cm1

Question

The first excited vibrational energy level of ditomic chlorine (Cl2) is 558 cm1 above the ground state. Wavenumbers, the units in which vibrational frequencies are usually recorded, are effectively units of energy, with 1cm1=1.986445×1023J. If every vibrational energy level is equally spaced, and has a degeneracy of 1, sum over the lowest 4 vibrational levels to obtain a vibrational partition function for chlorine.

Part A) Determine the average molar vibrational energy for chlorine at 298 K.

Part B) Determine the population of the zero level for chlorine at 298 K.

Part C) Determine the population of the first level for chlorine at 298 K.

Explanation / Answer

Part A) Energy = 1/2hv

with,

v = 558 cm-1 x 3 x 10^10 cm/s = 1.674 x 10^13 s-1

h = planck's constant

we get,

Average molar energy = 6.626 x 10^-34 x 1.674 x 10^13/6.023 x 10^23 = 1.84 x 10^-44 J/mol

Part B) the population n1/no,

n1/no = e^(-hv/KbT)

with,

/kb = Boltzman constant

T = 298 K

n1/mo = population ratio of v = 1 to v = 0 state

we get,

n1/no = e^(-6.626 x 10^-34 x 1.674 x 10^13/1.38 x 10^-23 x 298) = 0.068

Population of zero level = 1%

Part C) Population of first level = 6.8% of v = 0

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