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(Interested in the general principle too, not just for water and its variants, but with particular emphasis on it.)

When you distill a weak ethanol-in-water solution, you get a condensate that's ethanol-enriched, but that only takes you asymptotically up to 95%, because ethanol and water form an azeotrope with that composition. Is there a similar "enrichment limit" one can achieve in enriching water with deuterium, using only distillation at a fixed pressure? (I know one can "break the azeotrope, but I don't want to get into this.) Or could one, given enough stages in a cascade, approach 100% enrichment arbitrarily closely?

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    $\begingroup$ What makes you think there might be one? $\endgroup$ – Gert Nov 26 '17 at 16:33
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    $\begingroup$ This is an interesting question, but you'd have to use deuterated ethanol as well. Mixing CH3CH2OH with D2O would yield a mixture of CH3CH2OH, CH3CH2OD, D2O, HDO, and H2O. $\endgroup$ – MaxW Nov 26 '17 at 16:38
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    $\begingroup$ Azeotrope forming or not : it is hard to achieve as the boiling points differ by about one degree Celsius. $\endgroup$ – Alchimista Nov 26 '17 at 17:01
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    $\begingroup$ @MaxW I'm not really interested in ethanol; I just used it as an example of an azeotrope-forming co-solvent with water. But yeah, the general form of my question stands: would ethanol and deuterated ethanol (any/all of them) form an azeotrope with each other?? $\endgroup$ – Bernd Jendrissek Nov 26 '17 at 21:54
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    $\begingroup$ @BerndJendrissek: there exists no algorithm that predicts the existence of an azeotrope. $\endgroup$ – Gert Nov 26 '17 at 22:02
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This is a good question, but no, there is no azeotrope in any isotopic mixture, as an azeotrope requires non-ideal mixing of two components to form a minimum or maximum critical point on the plot of total vapor pressure as a function of composition as shown in Figure 1 below. Isotopic Mixtures have a near-zero enthalpy of mixing, making isotopic mixtures the closest thing to an ideal liquid mixture. Because of this ideality of mixing the boiling of protium-deuterium oxide mixtures looks much like the phase diagram in Figure 2 which follows Raoult's law, though this is actually quite exaggerated compared to semi-heavy water mixtures. Even if there was a theoretical point at which an azeotrope could exist for the negligible enthalpy of mixing it would occur at extreme concentrations of $\ce{D2O}$ or $\ce{H2O}$ and be unobservable due to the "doped" species existing as semi-heavy water ($\ce{DHO}$). This is further evidenced by the fact that other physical properties of isotopic mixtures follow a linear curve when measured as a function of concentration.

VP(c)
Figure 1. Generic Plot of vapor pressure as a function of composition

Boiling Phase Diagram
Figure 2. Generic Boiling phase diagram for ideal solutions

To answer your question about distillation only enrichment, it is possible though impractical. Commercial enrichment of $\ce{D2O}$ first uses hydrogen sulfide in the Gridler-Sulfide process, once enriched to around 30%, distillation is used to finish the process to the desired isotopic purity.

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    $\begingroup$ +1 You might add that a mixture has to deviate quite a bit from ideality before the vapour pressure curve can get a minimum or maximum. No chance for isotopomeres. $\endgroup$ – Karl Nov 29 '17 at 21:49
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    $\begingroup$ +1 Can you expand a bit on "Even if there was a theoretical point at which an azeotrope could exist for the negligible enthalpy of mixing it would occur at extreme concentrations of D2O or H2O" (and assume that the pairs are H2O/HDO and HDO/D2O)? While the non-ideality of mixing is certainly very small (and mind you, the H/D chemical difference is much bigger than U-235/U-238), the difference in boiling points is also very small, so you wouldn't need much non-ideality for an azeotrope to form. $\endgroup$ – Bernd Jendrissek Nov 30 '17 at 10:45
  • $\begingroup$ @BerndJendrissek I put that in there to preempt that one critic that always asks "what about at 99.99995876%". In order for there to be an enthalpy of mixing there has to be some sort of interaction of the mixed species. For $\ce{H2O}$/$\ce{D2O}$ this is the formation of $\ce{HDO}$. If you have only $\ce{HDO}$ and one other species of water then there is no reaction to be had from mixing, thus no enthalpy, thus no azeotrope. $\endgroup$ – A.K. Nov 30 '17 at 23:45
  • $\begingroup$ @A.K. I guess I'm that one critic :p I disagree with any "obviousness" of H2O/HDO mixing producing no reaction - while any chemical difference between H and D is necessarily small, it's larger than between any other isotopes. I'm happy to accept an empirical claim like "we know that from 0.000014% to 99.99978% there's no azeotrope", but I'm unfortunately not convinced by any of these "molecular arguments" that none can exist. $\endgroup$ – Bernd Jendrissek Dec 2 '17 at 3:06
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Distillation alone has been used on an industrial scale to produce heavy water, but it requires a multi-stage process and uses a lot of energy due to the small difference in boiling points. That would seem to confirm that water and deuterium oxide do not form an azeotrope.

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    $\begingroup$ Do you have a reference to back this up? I thought that generally in systems of isotopically substituted molecules an infinite set of azeotropic points is possible. And "distillation" refers to the process of low-temperature rectification of liquid hydrogen, followed by combustion of isolated $\ce{D2}$ with $\ce{O2}$. $\endgroup$ – andselisk Nov 28 '17 at 12:51
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    $\begingroup$ @andselisk Where did you get the idea of azeotropic points and isotopes? And by far the cheapest processes to enrich deuterium involve water not hydrogen (one process uses water and hydrogen sulphide followed by distillation of the enriched deuterium water, the other just uses distillation of water). $\endgroup$ – matt_black Nov 28 '17 at 13:52
  • $\begingroup$ @matt_black From the back of my head. The difference in b.p. between D2O and H2O is only 1.44 °C, and max. separation factor of about 1.05 (whereas it can be >10 for electrolysis). Separation via distillation only is extremely inefficient; it can lead to enriched heavy water content in combination with the methods you've mentioned, but distillation only is a dead end. $\endgroup$ – andselisk Nov 28 '17 at 15:51
  • $\begingroup$ It does seem to indicate that there is no light/heavy water azeotrope, but can we exclude the azeotrope being at, say, 99.9993% heavy water? $\endgroup$ – Bernd Jendrissek Nov 29 '17 at 13:48
  • $\begingroup$ Aceotropes form between substances of different polarity. If heptane and hexane don't, I can't see it happen for H2O, HDO and D2O. The vapour pressure of the ideal mixture has a constant slope with composition, you have to deviate quite a bit from ideality to get a curve with a maximum/minimum. $\endgroup$ – Karl Nov 29 '17 at 21:23

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