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?

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    $\begingroup$ 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 '19 at 9:10
  • $\begingroup$ @Karl there may be enough for an answer there. It seems the timescale is quite long, so long that there's probably not much chance of a re-freeze without extreme thermal insulation. $\endgroup$ – uhoh May 4 '19 at 12:58
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    $\begingroup$ 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 '19 at 20:48
  • $\begingroup$ 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 '19 at 21:05

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