[chico-03]

Hello,

Currently working with deuterium oxide (D2O), I have some questions about the behaviour of this substance.

• I was trying to calculate the vapour pressure corresponding to a given temperature for D2O. That's possible for water thanks to the Antoine equation. When I looked for the parameters A, B and C for D2O, they were equal to zero. Why, knowing that the behaviour of D2O is not so far from the one of water? And what is the meaning of parameters of Antoine equation equal to zero?
• In a vacuum system, when diffusing D2O through a film and measuring the current on the other side with a quadrupole mass spectrometer, the mass 19 is always higher than the deuterium oxide, regardless of the time of pumping before. Any  idea of why this happens?

Any suggestion is welcome!
Thank you

[Borek]

That's possible for water thanks to the Antoine equation. When I looked for the parameters A, B and C for D2O, they were equal to zero.

Where did you found them?

In many databases zero means "we have no idea what the real value is, but we have to put something here".

[chico-03]

Thank you for your answer, that's good to know.

It was a physical property data bank found on Elsevier.

[Enthalpy]

https://webbook.nist.gov/cgi/cbook.cgi?ID=C7789200&Mask=4
https://webbook.nist.gov/cgi/inchi?ID=C7732185&Mask=4

[chico-03]

Thank you a lot for the links Enthalpy.
I forgot to mention that I'm working between 20 and 50°C and Liu and Lindsay start only at 106°C.

[Corribus]

Do you need to calculate or are you just looking for values? Values for heavy water vapor pressure in your temperature range of interest exist:

https://www.nist.gov/system/files/documents/srd/jpcrd612.pdf

In a vacuum system, when diffusing D2O through a film and measuring the current on the other side with a quadrupole mass spectrometer, the mass 19 is always higher than the deuterium oxide, regardless of the time of pumping before. Any  idea of why this happens?

Could you elaborate on this point, particularly the bolded part? I don't understand what you mean. Maybe describing your experiment setup and what you're trying to do would be helpful.

[chico-03]

Hi Corribus,
I could find values and they are in accordance with the paper you gave me, thanks! I was trying to find a way to calculate them, but I think that I'll content myself with the values.

You're right, that was too short.
I have several polymer films, whose I want to evaluate the permeability to water (m/z 18), thanks to a mass spectrometer. In order to differentiate  easily the residual water present in the room and the water used for the measure, I use D2O (m/z 20).
On the graph, m/z 19 is clearly higher than D2O, even if the diffusion is made with pure D2O. Do you have an idea why? (% openening correpsonds to the opening of a butterfly valve that reduce or not the pumping speed in the measuring room)

[Corribus]

My off the cuff guess would be exchange reaction to form semi-heavy water, since you have not taken steps to remove regular (not heavy) water:

H₂O + D₂O 2HDO

This reaction is thermodynamically favorable (ΔG ~ -4.2 kJ/mol in gas phase at 298 K based on standard values published at NIST), mostly due to entropy contribution.

This sounds like a very complicated and possibly flawed way to do what is a fairly standard assessment of water vapor permeability. Aside from the fact that keeping water and heavy water separate is impossible, you are assuming that water and heavy water have the same permeation properties. This may be true in some cases but particularly in polymers that interact strongly with water, I'm not sure this is a good assumption. At the least, it should be justified. I guess it depends on what kind of error you would be willing to accept. I feel heavy water is unnecessary any way, because you can just baseline the background water vapor concentration, which should be as close to zero as possible anyway, since any water vapor on the other end of the polymer will affect the diffusion/permeation rate.

There are instruments solely designed to make a water vapor transmission rate measurement. Of course, not everyone has one of these. The typical way to do it "poor man style" is to find a container with a circular opening of known cross-sectional area, fill it with a known weight of water, press your film across the opening, and monitor the weight of water in the container as a function of time. As the water passes through the film, the water weight in the container drops. From this you can calculate the transmission rate and permeability of the film. This doesn't work great with films of very low permeability, but it can give you a reasonable measurement of permeability in most cases. It is best to do it in a low humidity, well-ventilated external environment so that you have very little water vapor pressure on the distal side of the film.

(You can also do this the opposite way - put a moisture absorber like calcium chloride in the container, seal with your polymer, and then place the whole assembly in a humidity chamber, measure the weight gain of the absorber over time. I seem to recall the two methods differ in whether you are interested in earlier or later time points in the diffusion curve based on the accuracy of measuring weight gain versus weight loss.)

Edit:

Here's a method that uses tritiated water to measure WVTR and directly compares to gravimetric method. This may be a better way to go since any exchange of a tritiated water and regular water will result in same species, in most cases you won't be forming something new.
 
https://www.sciencedirect.com/science/article/pii/S0022354915472041

I guess the assumption that heavy and regular water permeation kinetics are identical is probably OK for hydrophobic polymers like HDPE used in the linked study. I'd be more careful with the assumption in cases where the polymer is hydrophilic, since binding capacity/kinetics may be different for heavy water compared to regular water. I didn't find anything to support or refute that based on quick search, but it might be something to keep in mind. This is also discussed a little at the end of the article.

[chico-03]

Thank you so much Corribus, this is such a great exaplation! That is very helpful.

We actually did remove as much water as possible, by heating up the whole system to 80°C and having a pressure of 10^-8 mbar into the measurement chamber thanks to a turbo pump. But I guess this is still not enough.

The polymer parylene is actually rather hydrophobic. Which properties should I look at to determine if the binding capacity of the D2O is similar to water?

The study you mentionned is very interesting. If I understood it correctly, it works the same as the measure of total body water with diluted tritiated water. So is it correct when I say that, opposite to the behaviour of D2O, the amount of diluated tritiated water doesn't diminish when it is together with water?

[Corribus]

The tritiated water used in the paper is only singly tritiated, not doubly as in your case. Although the paper doesn't say this explicitly, it is probably heavily diluted with water, such that the concentration of HTO is probably quite low. If you have a large amount of water and a small fraction is singly tritiated water, you will still have exchange reactions. There are four possibilities here:
(1)   H2O + H2O = H2O + H2O
(2)   H2O + HTO = HTO + H2O
(3)   HTO + HTO = HTO + HTO
(4)   HTO + HTO = H2O + T2O
Process (1) is trivial and does not affect the concentration of H2O or HTO.
Process (2) and (3) likewise is trivial for the same reason.
Process (4) is not trivial as it affects the concentration of H2O and HTO, but if the concentration of HTO is very low, the effect will be insignificant because the probability of HTO encountering another HTO will be very low.

In your experimental setup, the effect is evidently not insignificant because your fully deuterated and regular water concentrations are both high enough that the probability of exchange events is high. (Although see far below for another idea here.)

The method in that paper detects permeated water vapor using a scintillation counter. An implicit assumption seems to be that the mole fraction of HTO on both sides of the film is the same at all times. Using this assumption they can calculate the total mass of water vapor that crosses through the polymer per time at steady state conditions by scaling with the scintillation rate of known pressures of water vapor. This assumption would fail if:
(1) There is considerable frequency of exchange reactions between different water species (which would occur if the concentration of tritiated water is high).
(2) The diffusion coefficients of HTO and H2O within the polymer are significantly different.
(3) If the gas sorption coefficient (partition coefficient) of the two species are different. I.e., if the solubility of HTO in the polymer is different from that of H2O.
(4) Diffusion is so slow that it is competitive with the radiation half-life of tritium. Which would be impressive since the half-life of tritium is a decade.
(5) If the polymer interacts with water so strongly that it can facilitate proton exchange reactions with HTO.

I think these would only really become a potential problem for a very hydrophilic polymer that has significant swelling in water. Your polymer doesn't seem like it would fit in this category (although you can do a simple swelling experiment - go you have a TGA? - to rule it out).

I wonder - instead of using pure heavy water, have you considered using diluted heavy water (with enough equilibration time to allow must species to be HDO)? This would essentially replicate what they're doing with tritiated water, except you'd be using mass spec for HDO instead of scintillation counting with HTO. You'd just have make some calibration curves to determine the relationship between HDO and total water vapor concentration. The only downside I would see is sensitivity. I'm not sure what concentrations of HDO would be detectable with your equipment. Also, you'd probably have a non-negligible background because there's a sizeable concentration of semiheavy water naturally.

On the other hand, I wonder whether the exchange to form semiheavy water really matters here. If the exchange to form HDO from D2O and water is happening in your source mixture, you can probably still get a good measurement of permeability by scaling your measured concentrations of D2O accordingly. I.e., as long as you know the "constant" concentration/pressure of D2O on one side of the film, and the evolving concentration on the other side, the concentration of other species doesn't matter, as long as it is constant. I would just want to check and see if the mole fractions of HDO and D2O are the same on both sides of the film.

Also, I could imagine scenarios where water may not be the only source of proton exchange. Exchange can happen if protons in the polymer or any polymer additives are labile. (For example, a lot of polymer UV blockers or antioxidants are fully of hydroxyl groups. These could easily undergo exchange reactions with D2O.) Some things to think about.

[chico-03]

Dear Corribus,
Once again, thank you so much for your answer!

Unfortunately, I think that dilution of D2O is not a possibility.  We already have problems with sensitivity (the QMS doesn't have a multiplier) and even at a pressure of several tens of mbar for the D2O, the detection is extremely low.

The penultimate paragraph is very interesting also. The level of pressure before and after the film are actually very different (1*10^-1 to 30 mbar before the film, and after around 1*10^-7 mbar), so I guess that the mole fractions are quite different. However, do you think there is still a way to calculate which amount of D2O from the begining actually contributes to the signal of HDO and D2O?

After your explanation, considering that our polymer is rather hydrophobic and it doesn't contain any hydroxyl group, I will assume that water and D2O have the same diffusive properties.

Something is still unclear from one of your previous answer:

H₂O + D₂O 2HDO

This reaction is thermodynamically favorable (ΔG ~ -4.2 kJ/mol in gas phase at 298 K based on standard values published at NIST), mostly due to entropy contribution.

I understand what it means, but I couldn't find the information myself.
This is the data I found about free energy reactions for water at NIST. I'm afraid to ask, but where did you find the info about its reaction with D2O?
https://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Mask=1EBF

[Corribus]

It's a little hard to be real specific because I am still not exactly understanding your experiment (for instance, the figure you supplied above). But just doing a thought experiment here, supposing you start with a single species A on one side of your film, and you end up with two species A + B on the other side of the film, it stands to reason that something is happening while A is permeating through the film. Since both A and B come from A, you might be able to use a mass balance approach and calculate your permeability metrics as the sum of A + B. In principle whatever causes the conversion of A to B would be subject to laws of chemical kinetics so I'd have to think a bit about the assumptions involved. But, it may OK.

Just curious, is there a reason you don't ditch heavy water altogether and just do regular water? If you are pumping your system down to remove atmospheric (background) water, I don't see why you can't just use the residual as a baseline for your measurement. It's the concentration (pressure) difference across the film that matters. If your baseline is the same on both sides of the film, then the baseline can just be subtracted out.

Regarding the Gibbs energy calculation, you can just look up standard heats of formation and standard entropy values for each species (make sure they're at the right temperature and phase). From these you can determine the enthalpy and entropy changes for the reaction, and then the Gibbs energy change.

[chico-03]

Just curious, is there a reason you don't ditch heavy water altogether and just do regular water?

Because that was the proposal of my work and now I need to explain why it doesn't work 
But I'll surely try with normal water to have a comparison.

Regarding the Gibbs energy, I couldn't get if it was a calculation or a given value, so thank you for the hint!

For the same work, I'll probably need to compare D2O/water diffusion to helium diffusion, both measured with a QMS. Would you say that this is comaprable, or the nature of the molecule and the element are too different?

[Corribus]

Permeation of gas through polymer depends on two basic factors: diffusion coefficient of the gas within the polymer and solubility of gas within the polymer. You might crudely think of the diffusion coefficient as being related to how fast individual gas molecules are able to move through transient pore spaces in the polymer (although, to be clear, the diffusion coefficient is not a rate). The solubility coefficient may be thought of as representing how many gas molecules can fit within a certain volume of polymer. To take an analogy, if a farmer is trying to shuttle her goats across a river using a raft, the diffusion coefficient represents how fast the raft can move through the water, and the solubility coefficient is how many goats can fit on the raft. So, even if she can row the raft really fast, if she can only fit one goat on the raft, it will still take a long time to get her goats across the river.

In a Fickian model, the diffusion coefficient is most related to how big each gas molecule is compared to the average pore space of the polymer. Smaller molecules have tend to have larger diffusion coefficients - they can physically move through the polymer more quickly than larger molecules, which tend to collide with polymer strands more frequently. The solubility coefficient is most related to how polar the gas molecule is compared to how polar the polymer is, although size does also play a role here. Solubility coefficients are a little harder to guess at, particularly for molecules that have similar polarity. But in any case, all things being equal: small, nonpolar molecules will permeate through nonpolar polymers faster than large, polar molecules will. And the pore space size of the polymer - which depends not only on the polymer structure but also how it is processed - also makes a big difference.

Consider polyethylene as a typical example. You can see some gas permeabilities, diffusion coefficients, and solubility coefficients here:

https://ogst.ifpenergiesnouvelles.fr/articles/ogst/pdf/2001/03/flaconneche2_v56n3.pdf

Notice, for instance that helium has a much larger diffusion coefficient than argon. Both are nonpolar but the argon atom is much larger than a helium atom, therefore has a smaller diffusion coefficient. A nitrogen molecule is slightly larger than an argon atom and has both a slightly lower diffusion coefficient and a slightly lower solubility. And so on. You can also see differences between low, medium, and high density polyethylene. The higher the polymer density, the less space there is for gas molecules to fit, with consequences to both the diffusion coefficient and (less so) solubility.

On its face, then, you would guess that water, being small and highly polar, would have a large diffusion coefficient but low solubility. Indeed the solubility of water in a nonpolar polymer like LDPE is quite low, which is why polyethylene is such a great moisture barrier (go back to the small but fast raft analogy of getting goats across a river). The interesting thing is that while the diffusion coefficient of water is pretty large (compared to large molecules, say), it is still much smaller than comparably sized nonpolar molecules. For instance, from data I found, diffusion coefficient of nitrogen (size ~300 pm) in LDPE is 2-3 orders of magnitude larger than water (size ~275 pm) at the same temperature. What gives here?

Water is a funky molecule. Whereas nonpolar permeants like helium or nitrogen gas tend to diffuse as individual molecules, it is thought that the strong hydrogen bonding of water combined with the highly hydrophobic LDPE environment means that when water partitions into LDPE, it does so as tightly bound water clusters. So, the effective size of the permeant is much larger than that of individual water molecules, with a correspondingly smaller diffusion coefficient - the diffusion coefficient decreases exponentially as a function of the permeant diameter. Moreover those water clusters may change size as they permeate through the polymer. In this situation diffusion isn't really Fickian any more so a lot of those elementary considerations about diffusion coefficients and permeant size kind of go out the window. You may read more about this here (dx.doi.org/10.1021/ie102221a | Wang et al. Ind. Eng. Chem. Res. 2011, 50, 6447–6454, https://pubs.acs.org/doi/pdf/10.1021/ie102221a).

Maybe that helps you think about diffusion and permeation of helium and water through your polymer. Hopefully you are learning that permeation is complicated and more difficult to measure than many people realize.