1. can the stability of ortho-hydrogen greater than para-hydrogen can be stated as this?

    The intrinsic energy of ortho is greater than para-hydrogen and also with rise in temperature %of ortho form increase while para decreases and ultimately at room temperature it's 75% ortho and 25% para-hydrogen. Plus as we know that the everything in nature tries to attain lowest possible energy level, and ortho form with opposite spins counteracts each others magnetic moments, making it energetically stable.

  2. Why is it that pure para can be obtained at lower temperature such as 20 K but pure ortho form is difficult to obtain?

    I mean when ortho is more stable it should be ortho which should be available more easily than para form?

  • 3
    $\begingroup$ Ortho is not more stable. $\endgroup$ – Ivan Neretin Sep 22 '15 at 18:26
  • $\begingroup$ I have read every where that ortho is the more stable form. $\endgroup$ – shaistha Sep 22 '15 at 18:34
  • 2
    $\begingroup$ OK, let's abandon the word "stable" altogether, for it is confusing. Para is the one with lower energy. $\endgroup$ – Ivan Neretin Sep 22 '15 at 18:36

From "Statistical Mechanics: A Concise Introduction for Chemists" by Benjamin Widom (pp. 35-36):

For the diatom of ordinary hydrogen ($\ce{{}^{1}_{1} H}$), where each nucleus is just a proton of spin quantum number 1/2, there are three possible symmetric states of the pair of spins (nuclear spin triplet) and only one possible antisymmetric nuclear spin state (spin singlet). [...]

At high temperatures, which generally means $T \gg > \theta_{\mathrm{rot}}$ (where $\theta_{\mathrm{rot}}$ is the characteristic rotational temperature) and so for $\ce{H2}$ means $T$ well above 88 K, the three ortho nuclear-spin states and one para state are equally populated, so equilibrium hydrogen at high temperatures is a mixture of ortho- and para-hydrogen in the ratio 3:1. In pure hydrogen the magnetic interactions between molecules that could cause transitions between nuclear spin states are so weak that such hydrogen may be cooled to arbitrarily low temperatures with the ortho:para ratio remaining 3:1. That is normal hydrogen. Paramagnetic impurities such as $\ce{O2}$, or a catalyst such as $\ce{Pt}$ on the surface of which $\ce{H2}$ molecules may dissociate, allowing the atoms to recombine with their nuclear spins in a different relative orientation, catalyze the ortho-para conversion. Cooled in the presence of such a paramagnetic impurity or such a catalyst, then, the hydrogen may have its ortho:para ratio continually adjust to that which would correspond to thermal equilibrium at the given low temperature. [...]

Cooled in such a way, then, hydrogen becomes pure para-$\ce{H2}$ at low temperatures, in contrast to "normal" hydrogen, which remains a mixture of ortho and para in the ratio 3:1. If the hydrogen is cooled in the presence of a catalyst that catalyzes the ortho-para conversion and thus allows it to become pure para-$\ce{H2}$ at low T, and if the catalyst is then removed and the hydrogen warmed back up to 298 K, say, the result is a sample of pure para-$\ce{H2}$ at normal laboratory temperatures. Pure ortho-$\ce{H2}$ cannot be prepared in this way.


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