Freezing point of oxygen-18 water

Question

The freezing points of heavy water (3.8 °C) and tritiated water (4.49 °C) both seem to be well-known. I can't find anywhere that gives the freezing points of $\ce{H2^18O}$, $\ce{D2^18O}$, or $\ce{T2^18O}$, though. Are any of these values known? If not, is it at least known whether they'd be higher than with ordinary oxygen?

Answer 1

In this paper [1] from 1963, various properties of heavy-oxygen water are measured including the melting point of both $\ce{H2^{18}O}$ and $\ce{D2^{18}O}$.

The melting temperatures were measured at: $$\ce{H2^{18}O}:\ T_\mathrm{mp}=\pu{0.28\pm0.02^\circ C}$$ $$\ce{D2^{18}O}:\ T_\mathrm{mp}=\pu{4.02\pm0.02^\circ C}$$

So, as some commenters mentioned, the melting (or freezing) point of the heavy-oxygen water is slightly larger when it is protium on the heavy-oxygen, and quite a bit larger when deuterium is present. This basically means that the liquid has become more structured on average, which is to say that the average lifetime of a hydrogen bond is slightly longer mostly due to an increase in the average strength of a hydrogen bond.

I would be very surprised if there has been a measurement of $\ce{T2^{18}O}$ because tritium is very rare and not very easy to isolate. Additionally, heavy-oxygen is also quite uncommon, so purifying the $\ce{T2^{18}O}$ to make any reliable measurements would likely be very difficult. I would be very willing to guess, however, that the melting temperature is $\ce{T2^{18}O}:\ T_\mathrm{mp}\approx\pu{4.7^\circ C}$ by just noting that the isotopic substitiution effects seem to be roughly additive.

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Extra Credit: Trying to Interpret the Data:

As a note which is really not what you're asking but I think is really interesting, is why there is such a sharp increase in freezing temperature for deuterium (and hence why heavy-oxygen has these effects only on a smaller scale). Indeed, it is true that the $\ce{O-D}$ vibrational frequency in $\ce{D2O}$ is quite a bit smaller than that mode in $\ce{H2O}$, but it really isn't obvious that lowering this frequency should make $\ce{D2O}$ freeze at a higher temperature. For instance, all else being equal, a larger frequency in $\ce{H2O}$ would predict a shorter hydrogen-bond distance because the vibrationally averaged bond length will be longer (because the long bond lengths contribute more than the short ones because the potential is Morse-like). But, what I just said about the vibrationally averaged bond length did not allow the $\ce{O-O}$ distance to relax in $\ce{D2O}$, so if the $\ce{O-O}$ distance shrunk in $\ce{D2O}$, then we could say it's just the lower zero-point energy because then the shorter $\ce{O-O}$ distance would explain the stronger hydrogen-bonding we seem to be observing in $\ce{D2O}$. Well, very interestingly, the $\ce{O-O}$ distance in liquid $\ce{D2O}$ has been both theoretically and experimentally observed to increase slightly or stay exactly the same. See ref. [2] and experimental references therein. So both geometric parameters point towards water actually having a higher melting point even though this obviously isn't true.

What's the deal then? Well most of the answer seems to be that ordinarily we think of nuclei as being classical, i.e. the atom can be located precisely, but this type of reasoning is quite bad for hydrogen. The important point then is still that deuterium is heavier than hydrogen, though not because of the explicit contribution which the frequency lowering would have on the geometric parameters if everything else were the same. Rather, a hydrogen atom is much more spread out than a deuterium atom due to the fact that lighter atoms will essentially behave more quantum mechanically. In the sense of being more wavelike.

The fact that this delocalization effect for hydrogen is most important was demonstrated in ref. [3], where the authors find that the hydrogen-bond length cannot be directly related to the vibrational frequency in a one-to-one sort of way. This is due to a large sensitivity to the $\ce{OHO}$ angle. This then explains what we were missing up above because even though the average bond length in $\ce{H2O}$ is larger than the in $\ce{D2O}$, this sensitivity to the hydrogen-bonding angle greatly decreases the hydrogen-bonding strength in liquid $\ce{H2O}$ compared to $\ce{D2O}$ which does not delocalize as much and hence does not suffer from this angle problem.

As a final fun point, if one does a simulation of liquid $\ce{H2O}$ using classical nuclei, and then the same simulation is done using path integral methods to include nuclear quantum effects, the change in freezing point is a decrease of about $\pu{4^\circ C}$, which can only be due to this delocalization effect I describe.

I basically included all this to say that interpreting what this heavy-oxygen change really means can actually be pretty complicated. Whatever it is, there must be some structural change, and one would always expect isotopic substitution for a heavier isotope to make liquids more structured both from the frequency effects and the nuclear quantum delocalization.

Sorry if that was overkill :)

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[1] Steckel, F., & Szapiro, S. (1963). Physical properties of heavy oxygen water. Part 1.—Density and thermal expansion. Transactions of the Faraday Society, 59, 331-343.

[2] Chen, B., Ivanov, I., Klein, M. L., & Parrinello, M. (2003). Hydrogen bonding in water. Physical Review Letters, 91(21), 215503.

[3] Rey, R., Møller, K. B., & Hynes, J. T. (2002). Hydrogen bond dynamics in water and ultrafast infrared spectroscopy. The Journal of Physical Chemistry A, 106(50), 11993-11996.