Question

1. Calculate the vibrational partition function of I2 (g) at 298.15K given the vibrational frequency for this molecule is 211.22 cm-1. What fraction of molecules are in the ground, first and second excited vibrational states?.

Answer 1

Apply

Wave number -->

Wave number(v) = 211.22 cm^-1........therefore

E = (6.636*10^-34)(3*10^8)(211.22*100)/(1.602*10^-19) = 0.02625 eV

Ground state energy

Eg = 0.02625 .

and nth state energy

En = 0.02625 /n^2.

Hence ratio of number of molecules in 1st excited to ground state

β ε =hcv /KT

β ε = (6.636*10^-34)(3*10^8)(211.22*100) / ((1.38*10^-23)(298.15))

β ε = 1.0219

substtiute in

q = 1/(1-exp(-x))

q = 1/(1-exp(-1.0219)) = 1.562

fracction --> 1/q = 1/1.562 = 0.6401