inorganic chemistry - Deuterium depleted water through fractional distilation

Could a packed fractional still of the same type used to distill ethanol to 95% be used to produce deuterium depleted drinking water? The still has a 1000 watt electric heating element and has a fractioning column that is 120cm long which is packed with stainless steel wool. If it is feasible what kind of performance might I expect?

Answer

Relying on the difference in vapor pressures of $\ce{HDO}$ and $\ce{H2O}$ is equivalent on relying on the equilibration of

$\ce{HDO(g) + H2O(l) <=> H2O(g) + HDO(l)}$

The fractionation factor $\alpha_{\ce{D}(l)/\ce{D}(g)}$ for this reaction is 1.015 at STP.

Deuterium ($\ce{^2H}$ or $\ce{D}$) is separated from protium ($\ce{^1H}$) commercially, usually via the Girdler sulfide process, which instead of relying on the above reaction, relies on

$\ce{ HDS + H2O <=> H2S + HDO}$

The fractionation factor $\alpha$ for this process is about 2.2.

Steam distillation for deuterium enrichment is certainly possible, but is not nearly as favorable as the industrial process. Given the very low concentration of deuterium in natural waters, the Rayleigh fractionation equation would be applicable:

$$R_{\ce{D}/\ce{H}} \approx R_{\ce{D}/\ce{H}_{init}} \times f^{\alpha-1}$$

In this equation:

• $R_{\ce{D}/\ce{H}}$ is the ratio of deuterium to hydrogen in the liquid water


• $R_{\ce{D}/\ce{H}_{init}}$ is the initial ratio of deuterium to hydrogen, or about 0.000156.


• $f$ is the fraction of the liquid water remaining after distillation.


• $\alpha$ is the fractionation factor mentioned above. However the value of 1.0015 is for the condensation of vapor phase water into liquid; performing the reverse reaction we should use the inverse, i.e. for distillation of existing liquid into vapor, $\alpha = 0.985$

With this equation and with mass balances (the total amount of deuterium and of protium doesn't change in the distillation process) you can find the amount of deuterium in the distillate.

If you distill off 100 grams of an initial 1000 grams of water, then $f=0.9$, $f^{\alpha-1} \approx 1.000156$, meaning the remaining 900 grams of liquid water will have a deuterium-to-hydrogen ratio of 0.0001562 instead of 0.000156.

The mass balance for deuterium would be:

$1000 \times 0.000156 = 100 R_{\ce{D}/{\ce{H}_{distillate}}} + 900 \times 0.0001562$

If I did all my math right, that means your distillate would have a deuterium-to-hydrogen ratio of 0.0001542, or about 1.2% less than the original water.

Hopefully this exercise convinces you that you would have to do a lot of distillation and burn through a ton of water to generate reasonable amounts of deuterium depleted water through simple steam distillation. There's a reason they don't do it that way commercially.