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Author Topic: How will the proportions of a mixture change on boiling?  (Read 2137 times)

Offline Atomic-S

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If there exists Liquid #1 with a boiling point B1 and Liquid #2 with boiling point B2, both fully miscible, and a mixture having a mole fraction F1 of Liquid #1 and 1 - F1 of Liquid #2 is  brought to their joint boiling point and boiled till half of the volume is boiled away, what will the final proportions of the two liquids be in the remaining liquid?


 

Offline chiralSPO

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Re: How will the proportions of a mixture change on boiling?
« Reply #1 on: 14/06/2014 09:30:13 »
This question is pretty complex, and depends on many things.

Knowing their boiling points won't really help--you need to know their vapor pressures over the range of temperatures required. Given two ideal and miscible liquids 1 and 2 with vapor pressure V1 and V2 and mole fractions F1 and F2, and assuming that once a substance is evaporated it is gone, one can set up a differential equation to calculate F1(t) and F2(t) at constant temperature T.

At any given moment, at constant temperature:
F1 + F2 = 1
N1 + N2 = N1+2 (number of moles of each substance, and total)
N1/N1+2 = F1
N2/N1+2 = F2
(d(N1)/dt)/(d(N2)/dt) = F1V1/F2V2

Now, if you are thinking of a distillation such that the temperature is ramped up to keep the mixture at a constant boil, then you also need to know how the respective vapor pressures change with temperature, and whether an azeoptrope is formed etc. etc. etc.
 

Offline Atomic-S

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Re: How will the proportions of a mixture change on boiling?
« Reply #2 on: 21/06/2014 05:26:04 »
It sounds like a non-simple problem, particularly if the liquids are not ideal and V1 + v2 is different than V 1 + 2. Maybe some experiments are in order.
 

Offline Atomic-S

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Re: How will the proportions of a mixture change on boiling?
« Reply #3 on: 21/06/2014 05:40:01 »
Incidentally, this raises another question, namely, in a closed vessel with a mixture of liquids in equilibrium with the mixture of its vapors, will the mole fractions of vapors be the same as the mole fractions in the liquid phase? Of course, that obviously will not be so in general, based on the consideration that if the separate vapor pressures are unequal, then they are unlikely to become equal upon mixing the liquids, but they would (to the first approximation) exist in the ratio F1V1 : F2V2, the Fs referring to the mole fractions in the liquid phase and the Vs referring to the separate vapor pressures of each. But is even this necessarily the case? If there is an additional affinity of the two types of molecules toward one another, resulting in a different overall vapor pressure than F1V1 + F2V2, then what do we use for the vapor pressures for each constituent, in order to calculate their sum? (And I here don't take into consideration any interaction between dissimilar molecules in the vapor phase, which might be significant if the density of the vapor phase starts to approach that of the liquid phase.)
 

Offline chiralSPO

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Re: How will the proportions of a mixture change on boiling?
« Reply #4 on: 22/06/2014 17:01:59 »
This is why chemistry is still very much an empirical science. Simple (and complex) models can only get you so far--at a certain point it is easier to do the experiment, and just check the results against the theory to make sure it makes sense. (most of the equations dealing with non-ideal systems are just variations of the ideal equations with experimentally determined fudge factors built in anyway...)
 

Offline Atomic-S

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Re: How will the proportions of a mixture change on boiling?
« Reply #5 on: 28/06/2014 05:13:37 »
Ah, yes, of course. I should have realized that.
 

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Re: How will the proportions of a mixture change on boiling?
« Reply #5 on: 28/06/2014 05:13:37 »

 

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