1. Rotational heat capacity of Hydrogen molecule

2. Moment of inertia of hydrogen very small so rotational quanta too large
so classical treatment not used
rotational partition function calculated only by summation
But calculated value does not agree with experimental value

3. Difference is due to two nuclear spin isomers for hydrogen molecule
Nuclear spin of the two protons can be paired to get either symmetric combination (spin parallel) or antisymmetric combination (spin opposed)
Former is ortho and latter para hydrogen

4. Wave number ν= hJ/4π2Ic for rotational space of diatomic molecule
Spectra of ortho hydrogen has fairly intense lines corresponding to odd values of J
Para hydrogen has less intense lines corresponding to even value of J
Spectra of hydrogen has alternating intense and less intense rotational lines – shows that mixture contains ortho(75%) and para hydrogen

5. If system has non-interacting particles, wave function for over all system is product of wave function of all particles
For particles 1 and 2
a(1) or a(1) , b(2) or b(2)
For combination I =  a(1)b(2) = a(1)b(2)

6. If indistingushiable particles, same wave function can be available for both
So II = a(2) b(1) = a(2)b(1) is also valid
But by quantum mechanics this is not acceptable

7. To correct this – use in phase(+) and out phase(-) combinations of 1 and 2
for in phase = symmetric= s
Out phase = antisymmetric = a
s = c’(I +II) = c’[a(1)b(2) + a(2)b(1)
a = c’(I - II) = c’[a(1)b(2) – a(2)b(1)
C’ is normalising constant = 1/ 2

8. Interchanging of labels leaves s unchanged and a becomes - a
Etotal = Ee + Et + Ev + Er + En
total = e + t + v + r + n
If sum of numbers of proton , electron and neutron = P , is even
 must be symmetric and only s is allowed
If P is odd, antisymmetric- a is allowed

9. Symmetry of  on exchange will depend on whether exchange of nuclei 0r electrons or both
If only nuclei exchanged – effects of electron exchange not considered – but effect of nuclear exchange on e can be considered
r is characterised by quantum number J , each J value is associated with (2J + 1) function

10. r is symmetric if J is even
antisymmetric if J is odd
r is a function of angles  and π
Nuclear exchange means changing  to (-π) and  to (+π)
Hydrogen has odd number of particles so antisymmetric

11. n depends on inter nuclear separation so nuclear exchange has no effect
n is therefore symmetric
It hydrogen in electronic ground state e has symmetric g+ ,
+ means function is symmetric and symmetric w.r.t inversion as given by g
So over all wave function is symmetric

12. If H-H is assumed as A-B, each has spin+1/2 or -1/2
The possible combination is
A(+)B(+) + B(+)A(+) = A(+)B(+)
A(-)B(-) + B(-)A(-) = A(-)B(-)
A(+)B(-) + B(+)A(-)
A(-)B(+) + B(-)A(+) both 3 &4 are similar
These are symmetric nuclear spin function and has three different function.

13. A(+)B(-) - B(+)A(+) = 0
A(-)B(-) - B(-)A(-) = 0
A(+)B(-) - B(+)A(-)
A(-)B(+) - A(-)B(+) both 3 &4 are similar – these are antisymmetric and has only one function
1 & 2 are excluded

14. So number of symmetric wave function is 3 times more than antisymmetric wave function
In general – homo nuclear diatomic – of spin ‘i’ – ratio of symmetric to antisymmetric spin function is (i+1) : I
In general – one with greater value is ortho and the other is para

15. It is possible to find combinations of wave function which make total either symmetric or antisymmetric
e, v and t are symmetric – so overall symmetry determined by products of r x n
For hydrogen total wave function must be antisymmetric since ‘P’ is odd
total to be antisymmetric either r or n must be antisymmetric but not both

16. r symmetric J is even
n must be antisymmetric(s=0)
r antisymmetric J odd
n symmetric (s=1)
For pare J is even, s=0
Ortho J is odd , s= 1

17. In general – molecule in ortho state occupy odd rotational level- with pare occupy even rotational level
Єr =BJ(J+1) B= h2/8π2I
So fr(ortho) =
(2s+1)s=1 J=1,3,5..(2J+1)e-BJ(J+1)/kT …..(1)
Since each state has nuclear statistical weight (2s+1), s=0 for para and s=1 for ortho

18. So fr(para) =
(2s+1)s=0 J=0,2,4..(2J+1)e-BJ(J+1)/kT …..(2)
From (2)
as T→0 , all molecules are populated with J=0 , so no ortho form- not easy to convert ortho to para – without catalyst or other reagent

19. So fr(ortho) =
3J=1,3,5..(2J+1)e-BJ(J+1)/kT
So fr(para) =
1 J=0,2,4..(2J+1)e-BJ(J+1)/kT …..(1)
Relative numbers of ortho to para forms of hydrogen will be in their ratio of their partition function
At 300K, -B/kT = 0.3

20. Summation of ortho series is
3e e e-9 +….
For para
1+5e e-6 +….
For ortho series sum = 1.7
For para sum = 1.8

21. At higher temperatures two become similar
no/np = fo/fp =(2s+1)0/(2s+1)p = 3/1
ie 25% para and 75% ortho
To calculate Cv, evaluate f for ortho and para
Rotational heat capacity of ordinary H2 is thus 3/4Cv(ortho) + 1/4Cv(para) at a given temperature

22. At low temperatures , H2 molecules are in lowest possible rotational level
Єr = J(J+1)h2/8π2I
Since J=1 for ortho
fr(ortho) = 9e-h2/4π2I
For para J=o
fr(para) =1e-0h2/8π2I =1
Cv = d/dT[RT2dlnZ/dT]v

23. = d/dT[LkT2dlnf/dT]
Since lnf = constant
Cv(ortho) = 0
Cv(para) = 0
Rotational contribution to heat capacity is zero at low temperature
As temperature is raised rotational contribution becomes significance and specific heat increases.