teodor
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Which formula do you use to correct boiling point depending on the atmospheric pressure
I am doing several experiments with simple (Fischer) esterification and I noticed that today my water-ethyl acetate azeotrope passes something around
69C. I checked my notes which I made a week ago and it was almost 71C at the same stage. Now the barometer shows 738mm Hg. The last time I believe it
was 770+.
I found different formulas/calculators on the Internet which use different coefficients.
And first I've checked Vogel's book but I feel his formula is completely wrong (or I don't understand it). In the first edition it was:
Dt = 0.001 * (760 - p) (t + 273)
So the temperature is in K, unlike other similar formulas I see on the Internet. And it gives very huge values. It was not corrected in the 3rd
edition of the book, so I suspect I don't understand something.
So, practically speaking, what formula do you use to get the precision like 0.1C in your calculation (for the case of normal atmospheric (not vacuum)
distillation)?
[Edited on 1-12-2021 by teodor]
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Triflic Acid
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Well, I live in a high-altitude desert, so I've always had the problem that water boils here at 97C. Drives me insane. If you want to correct for
pressure, you'll need to run this a few times on days with different pressures, and then calculate the energy of vaporization. Then use the
Clausis-Clapeyron Equation. Or, if you want to just wing it,
"A general rule of thumb is that for pressures within 10% of one atmosphere, a 10mmHg drop in pressure will account for a 0.3C - 0.5C drop in
boiling point."
There wasn't a fire, we just had an uncontrolled rapid oxidation event at the power plant.
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vmelkon
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Quote: Originally posted by teodor  |
Dt = 0.001 * (760 - p) (t + 273)
So the temperature is in K, unlike other similar formulas I see on the Internet.
[Edited on 1-12-2021 by teodor] |
I think the temperature is in °C. Notice the +273.
So if you insert -273 °C, you get Dt = 0.
I guess the pressure is in mm Hg.
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CharlieA
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Dt = 0.001 * (760 - p) (t + 273)
So at room temperature (25 deg C) and 750 mm Hg pressure:
Dt =0.001/mm Hg * (760 mm Hg - 750 mm Hg) (25 deg C = 273 deg C)
Dt = 0.001 (10 mm) (298 deg C) = 2.98 deg C
This seems too large by a factor of 10 or so.
I assume Dt meant delta-T, the difference in T. For a lower p, you would expect a lower T and therefore a negative Dt.
Where did you get the original equation? Could it be something like: 0.0001(p-760)(t+273)?
...or am I barking up a wrong tree here?
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Plunkett
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The Antoine equation which is a semi-empirical derivation of the Clausius-Clapeyron Equation lets you calculate the vapor pressure of pure substances
over a range of temperatures. It has the form log10(P)=A-B/(C+T) where P is pressure, T is temperature and A, B, and C are empirical constants. You
can easily rearrange the equation to calculate boiling point as a function of pressure. Antoine constants are available for most common organic
compounds. Just search "NIST *compound* antoine constants" and you should be able to find the constants in the NIST WebBook. The Antoine equation
only works for pure substances, but if you have a mixture of similar compounds (e.g., heptane, and hexane) you can combine the Antoine equation and
Raoult's law to calculate the bubble and dew points of the mixture. Raoult's law only works for ideal mixtures. I do not know how you would
calculate the bubble and dew point of a non-ideal mixture like water-ethyl acetate within 0.1 °C without thermodynamics software.
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unionised
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No it isn't.
That's why you have to add 273
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macckone
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I live at high altitude and have moved several times.
At 5000ft (1500M) even azeotrope ratios change, at 8000ft it is even worse.
From a practical standpoint, if the boiling point is stable you are collecting a single fraction, when it is changes rapidly you are at a different
fraction.
A change in pressure can throw off the boiling point by as much as a degree.
Things like DCM have to be cooled to be used, might as well go with butane that is easier to get here.
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