[rpadoan]

Dear all,

I am having troubles in answering 3 small questions, does anyone of you have a clue about it? Thanks for your help


1- How can I calculate the speed in which water from air at a fixed RH is absorbed in an anhydrous solution of Ethanol?

2-   How will the increasing presence of water in the solution influence this speed? 

3-   Will this absorption process reach an equilibrium state?


Any suggestion/comment is more than welcome

Greetings
Roberto

[Honclbrif]

What are your own thoughts on these questions so far?

[rpadoan]

Thanks for the quick reply Honclbrif!

My thoughts about this questions so far are the followings:

Water contained in air is in its gas form (if there is no saturation) so it bound with ethanol that is in a liquid phase creating an heterogeneous equilibrium.

However this is only partly true because water (gas) when in contact with ethanol (liquid) is bound to it in its liquid form and not anymore as a gas so the equilibrium is not between two heterogeneous substances.

The concentration of ethanol is very high so I think the solution will never reach an equilibrium, ethanol will keep absorbing water from air until it is completely evaporated. Than in this case what I need is the evaporation speed of ethanol and the absorption speed of water (gas) in ethanol so I know how much time I have to leave the pot open and keep it under a certain water concentration.   

It looks to me that two different reactions occur : One between two heterogenic substances and one between two homogeneous substances. Which formula I have to apply and in which way to calculate the equilibrium is the key point for me because from that it would than be possible to obtain the speed of the reaction.

Another consideration is that during this reaction ethanol tends to evaporate too so the amount of ethanol in the starting solution reduce enhancing the concentration of water in the solution. In an extreme situation what I imagine is a pot of anhydrous ethanol that will start absorbing water from the air if left open and at the same time it will start evaporating. So different scenarios could occur:

1-   The anhydrous ethanol absorbs enough water from the air to become an azeotrope and then evaporate together with water living nothing in the pot
2-   The anhydrous ethanol absorbs more water than its azeotrope mixture so the ethanol will evaporate and water will residues will remain in the pot

A third option in which the anhydrous ethanol absorbs less water than its azeotrope seems to me impossible because anyway due to the evaporation of ethanol the solution will reach an azeotrope condition so water and ethanol will evaporate together living nothing in the pot.

Comments are VERY welcome!

Greetings
Roberto

[rpadoan]

Maybe I should specify something..

I am not a chemist, I am a restorer/conservator but as you can immagine we have to face chemistry very often in our field. This may explain the presence of erroneous terminology and/or difficulties in expressing my toughts with correct chemical formulations.

Again, any help and/or comment is more than welcome.

Greetings
Roberto

[orgopete]

As presented, I don't think one can arrive at an answer. This is what I think we or someone could determine. If you have  closed container and if you knew the amounts of water and ethanol in it, they could tell you the concentration of the liquid and vapor at a given temperature and pressure.

If you have an open container, then the concentrations will depend more on the surface area, the volume of the container, the temperature, the pressure, the mixing or rate of diffusion, and the humidity. At equilibrium, no ethanol will be in the pot because the concentration of ethanol in the vapor phase is very low. There is a lot more water to be in equilibrium than ethanol.

It might be helpful if you told us what you are trying to accomplish or why you are asking the question.

[rpadoan]

Very well,

I think it is better that I re-formulate my questions than and I explain you the background of the problem.

We have an adhesive  that is prepared by using ethanol, it is a cellulose Ether. This adhesive is used for conservation treatments on water sensible media. We want to conduct some tests to verify whether the 4% water content of the standard ethanol generally used to prepare this adhesive might have a negative influence on the media or not. Before starting this investigation I want to know  how fast a solution of anhydrous alcohol will absorb water from air before reaching the 4% concentration of water because if my adhesive pot (prepared with anhydrous alcohol) is reaching this water concentration in less than 8 hrs there is no use in starting a complete research project on the use of anhydrous alcohol. Why? Because in a normal working environment this pot will stay open for few hours, it will be mixed by the brush used to apply the adhesive and of course it will be spread on the surface of the treated object. If, at the end of a working day, the pot of adhesive produced with anhydrous alcohol will have the same water content of the one prepared with standard alcohol (4% water) there is no need in using an anhydrous alcohol. Moreover, even if standard alcohol is  used it would be nice to know how much water will be absorb from this open pot (starting already with a 4% water content) until the end of the working day (8 hrs), will this water concentration increase? If yes how much? This is essential for us to prevent damages on the media.

`N.B. I do not need a direct answer to the problem (although it would be helpful)  I need calculations, formulas and laws I need to apply so that if I change the parameters of the problem I can obtain results that could best fit our situation. An experimental method to verify the water content of the pot is also very welcome (we can eventually make use of some facilities at the University).
 
Now to resume the data of the problem we can start form an ideal condition:
 
Environmental condition:

Temperature 20° C
Relative humidity 50%
Pressure  Average sea level pressure ( 101.325 kPa)

Pot:
Volume 250 ml
Exposed surface of the adhesive circa 40 cm2  (diameter 7cm)

Starting solutions:
-   Anhydrous Ethanol
-   4% water 5% Methanol 91%Ethanol

Exposure time
8 hrs

Mixing
The pot is not mixed but the brash is inserted in the pot……..but let’s start with an ideal situation where the pot is left open and untouched.


Questions:
-Water and ethanol content of the two solutions after 8 hrs exposure.

I hope it is more clear now what we need to achieve.

Thanks
Roberto

[rpadoan]

Still not clear what the question is?

[opsomath]

It will be very hard to estimate the water content from first principles. If you put a gun to my head and made me guess, I would guess that after 8h you would have perhaps 0.5% water in an open container of absolute alcohol.

The easiest way will be to find a gas chromatograph with a TCD or other detector that can pick up water, and inject your solution. The experiment should only take a few minutes.

[orgopete]

Let me constrain this problem slightly (greatly). Let us say that with the given conditions, the pot has a total volume of 500 mL and no air is present. The partial pressures then should equilibrate to the bp and concentration of the liquid. Now, let us assume you increase the water content of the vapor phase in the container. Then it will increase the water content of the pot. Any equilibrium will increase the water content as water is much more abundant.

However, I don't think this is going to be an equilibrium question. My instinct is that evaporation is going to be much faster than condensation into the liquid, within reason. Given the nature of this problem, this is what I think more likely to result. Since the solvent is quite volatile, some cooling of the pot will occur. This can increase the condensation somewhat. Brushing the solvent onto a surface will significantly increase the evaporation rate. This will result in a cooling of that surface. In the presence of water vapor, you will greatly increase the amount of condensation that can occur. If water is the problem, then I think the humidity of the air and absorption of water at the applied surface will be a greater variable in the performance than the condensation into the pot on your application day.

I think on low humidity days, absorption into the pot and condensation on the brushed surface will be low. On high humidity days, the absorption into the pot may he higher, but the absorption onto the brushed surface may be a greater problem. I think this may still be a problem that you can model. How much solvent in evaporating, how large is the heat sink of the applied surface, how much could the temperature change, what is the dew point? Obviously, I am just guessing here.

[Arkcon]

As was mentioned before, an analytical experiment is probably best.  You don't have to rely on GC, many analytical labs will perform a Karl Fisher titration.  Prepare your solution, and use it for a day, in a mock trial.  At various time points, seal a sample from the bucket in an airtight vial for moisture analysis.  You'll have the answer, including any other effects, like stirring the vapor layer by dipping the brush, and you won't have to account for non-ideal real world results deviating from the theoretical.

[rpadoan]

Dear all, first of all many thanks for all your answers they all are very good reflection points for me and my problem!

Opsomath:  I am sure it will be hard but not impossible, as  Otopede said it is a problem that can be modeled and that is exactly what I want and need to do. If I may ask you, with or without the shooting gun, how did you reach the conclusion of the  0.5 % value?

Orgopete: If I understand correctly your sentence you say that a closed pot with 250ml ethanol and 250 ml ethanol in the vapor phase will reach an equilibrium and if I increase the water content in the vapor phase this will eventually be absorbed by the liquid phase and reach an equilibrium. So now the question is again how do I calculate this and over all, how fast will this happen? 

Arkcon: I will surely perform some analytical tests but creating a model that can give us an average idea of the quantity of water absorbed by the adhesive before it is completely dry based on the amount of adhesive, surface treated and the variable factors like temperature and RH, this  will be the most useful and practical solution justify the use of anhydrous alcohol instead of standard ethanol for the conservation treatments we need to perform).

To all of you: Can you please start mentioning some laws, formulas, principles……….??

For example…. Do I have to use the Clausius–Clapeyron relation to calculate the evaporation speed? Henry’s law?
What law do I have to use to calculate the absorption of water in anhydrous alcohol from air ? Can I use the Raoult’s law or do I have to use something more specific for a Non-ideal solution?

Thanks again for your *delete me*
Roberto