51 minutes ago, exchemist said:

Hmm, I see what you mean. There are no extra degrees of freedom, though. Diatomic molecules all have 2 rotational degrees of freedom. But ortho can only populate odd numbered rotational energy levels while para can populate only even levels. I had to look this up (it's badly explained or not explained in Wiki) but it appears the issue is that ortho hydrogen is a triplet state, in which the total nuclear spin of 1 can be orientated +1, 0 or -1 with respect to the axis of rotation, thereby multiplying the numbers of rotational states available by 3, i.e. each rotational level has 3-fold degeneracy, whereas the para states do not. So at RTP, with kT>> ε for rotation, you end up with a  3:1 ratio, just because there are more ways for ortho to have a certain amount of rotational energy.

I think that's it, at least. 

     

I'll have to sleep on that one. When I wake up may be I'll have figured out how what you said was different to what I said. Thanks nevertheless!

25 minutes ago, sethoflagos said:

I'll have to sleep on that one. When I wake up may be I'll have figured out how what you said was different to what I said. Thanks nevertheless!

Ah the light dawns - maybe. I didn't realise that by "extra degrees of freedom" you might have in mind the 3-fold degeneracy of the triplet rotational states. If so then, er, yes, we're saying the same thing! 

1 minute ago, exchemist said:

Ah the light dawns - maybe. I didn't realise that by "extra degrees of freedom" you might have in mind the 3-fold degeneracy of the triplet rotational states. If so then, er, yes, we're saying the same thing! 

Now I can sleep with an untroubled mind!

8 hours ago, sethoflagos said:

Now I can sleep with an untroubled mind!

I suppose there are also in theory spin isomers of other diatomic gases. But probably the spacing between rotational levels is too small for this to produce discernable effects. It’s something I’d never previously thought about.

7 minutes ago, exchemist said:

I suppose there are also in theory spin isomers of other diatomic gases. But probably the spacing between rotational levels is too small for this to produce discernable effects. It’s something I’d never previously thought about.

Check entropy of mixing. The slightest difference in particle properties produces a significant step change in entropy irrespective of the degree of difference. It's this aspect that got me thinking about these spin isomers in the first place. Again: Gibbs' Paradox.

Edited March 3, 20232 yr by sethoflagos

1 hour ago, sethoflagos said:

Check entropy of mixing. The slightest difference in particle properties produces a significant step change in entropy irrespective of the degree of difference. It's this aspect that got me thinking about these spin isomers in the first place. Again: Gibbs' Paradox.

Well then, since the spin of ¹⁴N is +1, that of ortho N2 will be +2, with potential components +2, +1, 0 -1, -2, so 5-fold degeneracy, vs single states for the para version. I suppose this should show up in entropy calculations for nitrogen in some way.   

17 hours ago, exchemist said:

Well then, since the spin of ¹⁴N is +1, that of ortho N2 will be +2, with potential components +2, +1, 0 -1, -2, so 5-fold degeneracy, vs single states for the para version. I suppose this should show up in entropy calculations for nitrogen in some way.   

I'm not aware of there being such a marked anomaly in nitrogen specific heats as for hydrogen. Perhaps it's something to do with the relative atomic masses. Without knowing the dU and transition temperature values, it's hard to evaluate. But it should appear much like a gradual phase change over some specific temperature band. My guess is that the effect is rather small otherwise it would be cropping up in the literature quite frequently. 

 

When you get cold enough for spin energy to be significant, most things (including nitrogen) have already frozen solid.
So you don't get the same effects that you see with hydrogen and deuterium (and, I guess, tritium if you have any).

38 minutes ago, John Cuthber said:

When you get cold enough for spin energy to be significant, most things (including nitrogen) have already frozen solid.
So you don't get the same effects that you see with hydrogen and deuterium (and, I guess, tritium if you have any).

Indeed. However, if there is 5-fold degeneracy in the rotational levels (in the ortho spin isomer), would that not be expected to affect the gas phase molar entropy? 

On 3/5/2023 at 12:13 PM, exchemist said:

Indeed. However, if there is 5-fold degeneracy in the rotational levels (in the ortho spin isomer), would that not be expected to affect the gas phase molar entropy? 

In the absence of a response maybe I can add my further thoughts.

If this effect gives rise to six distinct forms of hitrogen molecule (5 para, 1 ortho) and they distribute evenly at equilibrium, then surely this must contribute ln(6) R = 14.9 JK^-1mol^-1 to the standard entropy? 

However when I check references like Chem.libretexts  the standard entropy for nitrogen (S⁰ = 191.6 JK^-1mol^-1) doesn't appear anomalously high. Compare with O₂ (S⁰ = 205.1 JK^-1mol^-1) & CO (S⁰ = 197.7 JK^-1mol^-1). 

Puzzling.

atomic nuclei are able to sense temperature because of the thermal energy that is present in the environment. This thermal energy is composed of the kinetic energy of the particles in the environment, which is determined by the temperature. As the temperature increases, the kinetic energy of the particles increases, and the protons can move around more, allowing them to absorb or release energy. This energy can be used to change the spin of the protons, which is what happens in the case of ortho and para hydrogen.