Background (hydrogen)

In the case of recently liquified hydrogen (which is quite cold of course) it must be re-equilibrated before loading on to a rocket as fuel to avoid a sudden exothermic equilibration of the ratio of the ortho- and para- forms. This is because the nuclear spin degree of freedom (singlet vs triplet) initially remains hot even when the other degrees of freedom of the molecule are cold.

The question What is the ortho/para issue with LH2 as a fuel? quotes Hydrogen Fundamentals on a hydrogen safety website:

Liquid hydrogen (LH2) has the advantage of extreme cleanliness and the more economic type of storage, however, on the expense of a significant energy consumption of about one third of its heat of combustion. Another drawback is the unavoidable loss by boil off which is typical to maintain the cold temperature in the tank. The evaporation rate is even enhanced when ortho hydrogen is stored. The heat liberated during the ortho-para conversion at 20 K is huge with 670 kJ/kg compared to a figure of 446 kJ/kg for the latent heat of vaporization at the same temperature. This represents a safety issue requiring a design of the hydrogen loop which is able to remove the heat of conversion in a safe manner.


If instead I started with extremely cold water ice (we've now switched from $\ce{H2}$ to $\ce{H2O}$) say at liquid helium temperature, and add a enough heat to quickly raise it to 0°C and just melt it to liquid, it would be mostly para- water. This will be out of equilibrium, and the nuclear spins will still want to warm up as well.

What would happen next? Would the water quickly refreeze in a fraction of a second? Or would it take hours or days to quietly re-equilibrate?

  • 10
    $\begingroup$ Tikhonov and Volkov science.sciencemag.org/content/296/5577/2363 say it takes quite some time for the spin states to re-equillibrate, months in solid, minutes in liquid state. $\endgroup$
    – Karl
    Apr 27, 2019 at 9:10
  • 2
    $\begingroup$ Good question! I found this link interesting. www1.lsbu.ac.uk/water/ortho_para_water.html One thing I don't understand is how, if in liquid water there is extremely fast hydrogen atom exchange between water molecules, the equilibration between ortho and para can take so long. Acid-base chemistry happens in water at (very fast) diffusion-limited rates because H exchange is so facile. So why doesn't para / ortho equilibration take the same amount of time? $\endgroup$
    – Curt F.
    Jun 3, 2019 at 20:48
  • 1
    $\begingroup$ The Volkov and Tikhonov paper (sort of) addresses my question. They say: *[the rates in liquid water] are 10$^6$ greater than what could be expected from the rate of proton exchange between $\ce{H2O}$ molecules in liquid water (5). We conclude that fast proton exchange obviously does not lead to the fast OP conversion. The exchange without OP transitions, i.e., without change of energy (of resonance character), distinctly dominates. $\endgroup$
    – Curt F.
    Jun 3, 2019 at 21:05
  • 1
    $\begingroup$ Is this a good article to address your question? $\endgroup$
    – DialFrost
    Sep 19, 2022 at 10:23
  • 1
    $\begingroup$ @uhoh I'm not sure but, a bunch of sources online state that the OPR (ortho-para ratio) is usually 3:1 at room temp. and above, and 1:1 at the temp. of liquification of air (-196 degrees C), so at 0 degrees you should about 3:1 - so more ortho instead of para right? $\endgroup$
    – DialFrost
    Sep 19, 2022 at 12:05

1 Answer 1


Note that this answer is not complete. Note that wherever a "*" is placed in the answer, it means it requires figures or explanations to support it. If anyone found them or anything related, please share them! :)

Now unfortunately I do not have a complete and reliable answer to this, so I strongly welcome any comments objecting or adding on to my answer.


It should take a few hours to re-equilibrate*.

Longer answer

... and just melt it to liquid, it would be mostly para-water

To those who don't understand why this is so I'll add a short explanation. At lower temperatures, especially at around 0K-50K, para-water is more stable than ortho-water as shown by Statistical ortho-to-para ratio of water desorbed from ice at 10 kelvin:

As shown with Fig. 1, curve A, $T_{spin}$ may be used as a probe for low-temperature regions because para-$\ce{H_2O}$ ($J_{K_a,K_c=0_{00}}$) is more stable than ortho-H2O ($J_{K_a,K_c=1_{01}}$) in the gas phase owing to the $\ce{23.8 cm^{−1}}$ (34.2 K) rotational energy difference between them.

OPR of H2O as function of temperature

Now back to the main question. We know that it takes energy (likely a lot of it*) for the ortho-para transition to occur based on A Mechanism for the para-ortho Conversion of Hydrogen by Diamagnetic Substances:

According to the Wigner theory (I) non-dissociative conversion of the hydrogen molecule between pm-u and ortho states can occur only in the presence of an external inhomogeneous magnetic field.

We also know that all that energy cannot be obtained and used in the ortho-para transition this quickly, and many sources suggest it would take hours to re-equilibrate*, two of such sources are here on the Natural Ortho-Para Conversion rate in liquid and gaseous hydrogen and Separation of Water into Its Ortho and Para Isomers.

Note that from Wikipedia - Spin Isomers of Hydrogen:

This is the T = 0 intercept seen in the molar energy of orthohydrogen. Since "normal" room-temperature hydrogen is a 3:1 ortho:para mixture, its molar residual rotational energy at low temperature is (3/4) × 2Rθrot ≈ 1091 J/mol,[citation needed] which is somewhat larger than the enthalpy of vaporization of normal hydrogen, 904 J/mol at the boiling point, Tb ≈ 20.369 K.[10] Notably, the boiling points of parahydrogen and normal (3:1) hydrogen are nearly equal; for parahydrogen ∆Hvap ≈ 898 J/mol at Tb ≈ 20.277 K, and it follows that nearly all the residual rotational energy of orthohydrogen is retained in the liquid state.

One source states from Iopscience - The Ortho-to-para Ratio of Water Molecules Desorbed from Ice Made from Para-water Monomers at 11 K:

Ortho-Para conversion and graph


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.