That's a really interesting thought. I'll give a rough answer, based on the approximation that each D or H in the ice or water has the same energy (enthalpy, to be precise) regardless of whether the nearby sites have D's or H's. From the 3.8°C freezing T of D
2O together with the latent heat of freezing of H
2O (and the assumption that D
2O has roughly the same thermodynamic parameters) you can calculate how much the free-energy of moving a single D to ice from liquid is below that of the same move for H. I get about 7*10
-23 J. Now the thermal energy scale (absolute T* Boltzmann's constant) at around 0°C (or 3.8°C, all the same at this level of approximation) is around 4*10
-21 J. So the free-energy preference for freezing the D's is, in thermal energy units, under 2%. The relative enrichment of the D concentration in the ice in a partially frozen solution is under 2%.
Now you might wonder if, by holding the temperature at say 1°C, you could get some ice to form that only contains D
2O or DHO, since H2O doesn't freeze at all at that T. The loss of entropy in separating the D from the H would prevent any ice from forming in ordinary low-D water at that point. It takes a lot of energy to unmix things.
Mike W.
(published on 04/07/11)
