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I am working with a company to help improve their reaction process where they use SDA 2B Alcohol (Karl Fischer titration has it at 99.66% EtOH, 0.3375% toluene). Specifically, I am attempting to determine what theoretical %wt or %vol of water is pulled from atmospheric air into pure ethanol given an infinite amount of time? Does it ever reach an equilibrium state and, if so, what is it?

My gut guess is ~95.5%wt EtOH/4.5%wt H2O, the azeotropic concentration.

Assume constant relative humidity and room temperature. The ethanol is stored in a closed tote (~220 gallons). Initially, the tote is full of as-shipped ethanol (negligible air volume). However, half the ethanol is removed as part of a batch run. So, really, what I'm trying to get at is what happens to the leftover ethanol that is now sitting in a closed tote with ambient conditions.

Please let me know if additional details will help answer the question. Thanks!

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    $\begingroup$ Following gut feelings it is much more like a "equilibrium concentration as a function of temperature, humidity level", and most probably around room temperature it doesn't do anything with the azeotropic concentration. Pleas,e also note that the "equilibrium situation" also assumes plenty of alcohol in the air, which happens most probably faster than the humidity going into alcohol. $\endgroup$
    – Greg
    Commented Jul 17, 2014 at 0:03
  • $\begingroup$ @Greg I believe this would suffice as an answer. Humidity won't be drawn in, but alcohol will certainly vaporize. $\endgroup$
    – Martin - マーチン ♦
    Commented Jul 17, 2014 at 3:17

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You have a closed system with (say) 100 gallons of (say) pure ethanol and 100 gallons of moist air. There is not enough water in the air to go into the ethanol and reach the azeotropic ratio -- not by 2 or 3 orders of magnitude (see below).

If you know the relative humidity of the air at the time the tote was opened (and the temperature), you can find the absolute humidity (the mass % of water vapor in the air) and so estimate the maximum amount of water that went into the tote. Since 100 gallons is about 20 times the volume of a mole of gas, assuming an absolute humidity of 10% (it's very likely way less) the answer is about 2 moles, or about 40 grams. In equilibrium (in the closed system) almost all of the water vapor will very soon go into the ethanol, so you have 100 gallons of ethanol (with a bit of toluene), plus 40 g of water. That's 0.01%. Whether this much water ruins your alcohol or not, you will know better than anyone.

Every time you open the tote, you're potentially letting in 40g of water. I say "potentially" because, since the gas content is not instanty exchanged, much less water than that might go in. But still, if you open it often, the percentage will keep going up -- asymptotically to the azeotropic ratio, but since the alcohol will also evaporate, as has been pointed out, you probably won't ever get near that ratio.

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  • $\begingroup$ Hmm, as the alcohol evaporates the approach to a higher (azeotropic ratio) will increase I think. Granted the amount of air (with water) that enters when opened is small but some alcohol is leaving and diffusion will occur. The amount of air entering when pouring/pumping out alcohol however may be the half or even more if batch sizes are bigger and less than half remains sometimes. About 50g/1000l @ 40degC is the maximum you can get in with foggy tropical air. $\endgroup$
    – KalleMP
    Commented Sep 5, 2016 at 21:47

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