# Kinetics of converting a water + heavy water mixture to HDO

Imagine an initial 50%-50% mixture of normal ($$\rm H_2 \rm O$$) and heavy water ($$\rm D_2 \rm O$$).

I think, simply mixing them, the result will exponentially converge to an equilibric state, where about a half of the solution is $$\rm{HDO}$$.

What is the time constant with what the system is going to the equilibric state?

Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

• Chemistry SE policy, in contrary to e.g. Physics SE, is to avoid MathJax/LaTeX in titles unless absolutely necessary, for indexing/searching reasons. Jan 8 '20 at 2:02
• I'm pretty sure that proton/deuteron transfer is so fast that the limiting step here is diffusion/mixing. That is, what you're looking for is the time it takes for the protons or deuterons to be randomly distributed throughout the volume. That will be a function of the mixing process used, so it will vary with conditions. Jan 8 '20 at 2:15
• This may help with the kinetics: en.wikipedia.org/wiki/Self-ionization_of_water#Mechanism Jan 8 '20 at 2:18
• As @Poutnik states, it depends on ionization. At room temperature, only ~1 in 10,000,000 molecules are ionized, so the chances of an OD- encountering an H+, or of an OH- meeting up with a D+, must be calculated, and the number of collisions/second, as well. (Actually, the species shown are oversimplified, but you get the drift.) Jan 8 '20 at 2:28
• @Karnsten Theis Sure. Jan 9 '20 at 21:34

[Comment by Poutnik] Important is also en.wikipedia.org/wiki/Grotthuss_mechanism for the proton interchange. Mobility of H3O+ and OH- gives a hint it must be fast.

If you compare the diffusion coefficient of hydroxide ($$\pu{5.270e9 m^2/s}$$) to that of fluoride ($$\pu{1.460e9 m^2/s}$$), you might be surprised to see such a difference despite their comparable size. The mechanism for the higher diffusion rate are acid/base reactions with neighboring waters. Instead of swapping positions, a water donates a proton to hydroxide, turning into hydroxide.

$$\ce{H-O- + H-O-H <=> H-O-H + O-H-}$$

With one deuterated water, it leads to formation of the desired product:

$$\ce{H-O- + D-O-D <=> H-O-D + O-D-}$$

If the acid/base reaction were slow compared to diffusion, there would be little effect on the diffusion constant of hydroxide, and it would be similar to that of fluoride. However, because the main mechanism of hydroxide transport is by acid/base (i.e. the protons move and the oxygen atoms don't have to), you get a more than four-fold difference. This means acid/base reactions are on a time scale faster than diffusion.

This is also reflected in the high molar conductivities, as stated in the comment by Poutnik.

So as stated in other answers, the exchange will be fast compared to the time required to mix the two liquids.

The analogous mechanism would work for $$\ce{H3O+}$$ but it is harder to compare to other ions, so I started with hydroxide. You can imagine that the rates are heavily dependent on temperature and pH. However, even at neutral pH the rate is fast.

Does some "easy" way exist to estimate the reaction speed in such systems, or it has to be determined experimentally?

Here is a paper where they used computational chemistry to look at this question: http://omh.umeche.maine.edu/pdfs/JChemPhys_135_124505.01pdf.pdf. I would not consider this "easy", though.

The key word is "mixing". If you take a glass cylinder and fill it halfway with D2O (the heavier water, so put it on the bottom) and fill the cylinder ever so gently with H2O, the mixing will be determined by diffusion (NO mechanical mixing). But the experimental techniques required to analyze and determine the rates would be expensive and time-consuming.

Now if in a practical situation you need to make HOD, you could fill a beaker halfway with one and the rest of the way with the other (doesn't matter which way), and stir with a glass rod or a high speed mixer - or shake the mix in a bottle. The time to make HOD will be as fast as you can mix by shaking or stirring.

However, the idea of watching a diffusion experiment sounds like a fun thing. A cylinder could be filled halfway with dyed water, then covered with pure water, and the color equalization could be observed. It should probably be done with pure water on the bottom and dyed water on top, too, for comparison. And probably with more than one dye (in separate experiments) to see if the dye is important in the diffusion rate, or whether the dye is just carried by larger bunches of water.

The exchange of protons in amines or between H2O and other hydroxylic compounds has been found to be measurably slow (e.g., https://pubs.acs.org/doi/abs/10.1021/ja01652a030, which dates to 1953). A longer article in Wikipedia (https://en.wikipedia.org/wiki/Hydrogen–deuterium_exchange) discusses proteins and other compounds that are separated from the H2O/D2O environment and analyzed separately. What I am suggesting (and may be on shaky ground) is that H2O and D2O in solution together are not as independent as molecules of proteins or alcohols. In H2O, all the hydrogens are equivalent and exchange on a time scale that might be calculable; same for D2O. But when you put bulk H2O and bulk D2O together, the hydrogen bonded clusters will re-form, rearrange, simply by being there, so that if you try to analyze for HOD, you are selecting one molecule (say, by evaporation) from a cluster that is primarily determined by how much mixing you have done. If you add a drop of H2O to a liter of D2O, you may be quick enough to recapture most of that drop. But if you mix the drop (takes time: maybe one or two seconds), those H2O molecules are distributed throughout the liter, forced into new hydrogen bonding by the mixing, and not identifiable as H2O because they are in clusters of molecules. Pull out an oxygen, and it might have an H and a D, or 2D, but not likely 2H, because of the randomness induced by the mixing.

• I am not sure in that. The water molecules are very stable, and only about $10^{-7}$ of them is dissociated. I think, a $D_2O+H_2O\rightarrow 2HDO$ requires at least some quantum tunneling of the protons/deuterons (low probability), or some more complex reaction involving $H_3O^+$ / $HO^-$ ions (which are relative rare). Can you please make clear in your answer, why would it be so fast? Jan 9 '20 at 12:11
• A single water molecule is very stable, as you correctly state. If you put a few together, there will be hydrogen bonds and clumps of water which are not just H2O. The few protons are not just H+, not even just H3O+, but more like H5O2+ and H7O3+, and so on until the electrical attraction of the proton is diminished. Perhaps I'm being too practical, not theoretical enough. Measuring the formation of HOD from H2O + D2O requires observing HOD: difficult. Perhaps H2OD+ + D2O --> D2OH+ + HOD could be observed by NMR line broadening. I'm editing my original answer to include more. Jan 9 '20 at 15:45
• Do not forget the fast proton exchanging effect, mentioned in the other comment. H3O+ + H2O <-> H2O + H3O+, resp OH- + H2O <-> H2O + OH-, standing behind very high molar conductivities of both ions. Jan 9 '20 at 15:56