Question

For a diatomic gas phase molecule, the rotational energy is given by the expression:

E_j = B_e J(J+1)

where J is the rotational quantum number, and B_e is the rotational constant. The degeneracy for rotational states is given by:

g(J) = 2J+1

The normalization constant (total rotational partition function) for the Boltmann distribution for rotations is given by:

Q(J,T) = k_BT / B_e

Iodine, I_{2(g), has a rotational constant of} B_e = 0.037372 cm^-1. Determine the fraction of I₂ molecules in the J= 25 state at T=300.0K.

(1 J/mol = 0.083595 cm^-1).

Answer 1

total rotational partition function
Q(J,T) = kB*T(J)/Be(J) = (kB*T/hc)/Be(cm-1)
kBT/hc = 209.7 cm-1, at 300K
Q(J,300K) = 209.7/Be(cm-1)
Rotating the molecules by 180deg gives the same state.
So the rotation partition function for diatomic molecule is divided by symmetry number.
homonuclear diatomic molecule, symmetry number, σ = 2.

q(J,300K) = Q(J,300K)/2 = 104.85/Be(cm-1)
Iodine, I2(g), has a rotational constant of Be = 0.037372 cm-1.
q(J,300K) = 104.85/0.037372 = 2805.6

For a diatomic gas phase molecule, the rotational energy is given by the expression:
Ej = Be J(J+1)
Iodine, I2(g), has a rotational constant of Be = 0.037372 cm-1.
J= 25 state at T=300.0K.
rotational energy, E25(cm-1) = Be*25*(25+1) = 24.3 cm-1

Fraction of molecules   having the energy E25 at 300K, ni/n0 = exp(-E25/(kB*T*q)
kB = 1.38E-23 J  
h*c= 1.986E-25 J*m
kB(cm-1) = kB
ni/n0 = exp(-E25/(kB*T*q) = exp(-24.3/(kB*T*2805.6))

kB = 1.38E-23 J/K = 0.695 cm-1/K
kB*T = 0.695*300 = 208.5 cm-1
ni/n0 = = exp(-24.3/(208.5*2805.6)) = 1