Some conceptual chemistry questions

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Thearchitect

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If I have two chemicals mixed in the liquid phase that have the same boiling point, is the beginning boiling point for for both of them now reduced somewhat or does it stay the same?

Let's say I mix hexane with water, and then I bring the heat up to just below the boiling point of hexane(which should be much lower than that of water), it does boil sooner now, right? even though water has a higher bp. Then the hexane can sort of "carry" the water molecules with it? This book doesn't explain this distillation stuff at all.

I'm also having trouble with figuring out how I'm supposed to convert optical rotation into specific rotation. Assuming I used a "length" of 3cm and and solution that was 2g/ml. Is it just (observed rotation)/ ((10/3) cm(1/2) grams per ml). I guess this one wasn't very conceptual. Oh well.

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I guess I sort of get the boiling point thing. But if hexane lowers the BP for water and vice versa when in a mixture, wouldn't that mean they would evaporate at the same rate?? That makes no sense.
 
hexane wouldn't carry the water with it because hexane is very non-polar and water is very polar. It wouldn't mix well to start with more of an emulsion. BP lowering is a colligative property based on the number of moles of something present (this somethign) can be volatile or non volatile. Its behavior is described by raoult's law.

specific rotation (always constant for a given molecule) keep that in mind.
 
DrTacoElf said:
hexane wouldn't carry the water with it because hexane is very non-polar and water is very polar. It wouldn't mix well to start with more of an emulsion. BP lowering is a colligative property based on the number of moles of something present (this somethign) can be volatile or non volatile. Its behavior is described by raoult's law.

specific rotation (always constant for a given molecule) keep that in mind.

I wasn't really asking about that problem in particular. I just used it as an example. Also, I've done several non-polar molecule/water distillations already. They seemed to work just fine. But I'd like to know why , if the BP is reached by the combined vapor pressures, both don't evaporate at the same rate.
 
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Thearchitect said:
I wasn't really asking about that problem in particular. I just used it as an example. Also, I've done several non-polar molecule/water distillations already. They seemed to work just fine. But I'd like to know why , if the BP is reached by the combined vapor pressures, both don't evaporate at the same rate.

Because the evaporation rate is related to the heat of evaporation and the thermal conductivity of the molecule, not the vapor pressure.
 
Medikit said:
Because the evaporation rate is related to the heat of evaporation and the thermal conductivity of the molecule, not the vapor pressure.

When something turns into gas, I'm calling that evaporation. Something evaporates(boils) when its vapor pressure exceeds atmospheric(or other) pressures.
 
I could be wrong here... but wouldn't two different chemicals have to have two different boiling points? I thought each chemical had it's unique set of properties... like a fingerprint.
 
Unfortunately, the mixing of two liquids causes changes in physical properties that are far to complicated to explain in this post (In fact that very topic occupies three-months worth of general thermodynamics). The basic answer is it depends on what you are mixing and how far the mixture deviates from ideality. In addition the new boiling points can only be determined by experiment all equations are empirical only. A good example is an azeotropic mixture of ethanol and benzene: The boiling point of ethanol drops as you add benzene until benzene consists of 58% of the mixture then the BP goes back up again. It is not as simple as you think.
 
MsEvolution said:
I could be wrong here... but wouldn't two different chemicals have to have two different boiling points? I thought each chemical had it's unique set of properties... like a fingerprint.

You are not wrong. But.... two different chemicals have two different boiling points UNTIL you mix them. Then it becomes a mixture with its own set of properties.
 
inshanesworld said:
You are not wrong. But.... two different chemicals have two different boiling points UNTIL you mix them. Then it becomes a mixture with its own set of properties.
What you're describing is called a eutectic mixture, and there are relatively few compounds that form these when mixed. All the others retain their unique boiling point in solution with other, non-reactive compounds, and can be separated fairly easily by distillation.
 
Thearchitect/ said:
If I have two chemicals mixed in the liquid phase that have the same boiling point, is the beginning boiling point for for both of them now reduced somewhat or does it stay the same?

Let's say I mix hexane with water, and then I bring the heat up to just below the boiling point of hexane(which should be much lower than that of water), it does boil sooner now, right? even though water has a higher bp. Then the hexane can sort of "carry" the water molecules with it? This book doesn't explain this distillation stuff at all.


Let me take a stab at it...

If you have two liquids, liq.1 (BP 80deg) and liq2 (BP 130). Then any mixture of liq.1 and 2 will have a BP between 80-130.

So you wanna distill...

Let's say you have a mixture of 50% liq.1 and 50% liq.2. And let's say this mixture first evaporates at 100. The vapor will have more liq.1 in it than liq.2, let's say 60% 1 , 40% 2. This is because liq.1 has a evaporates more easily.

Now, the composition of the remaining mixture also changes. Because more of liq.1 evaporated than 2, the remaining mixture has more liq.2 left in it.

Since it has more liq.2 in it, the BP of the remaining mixture increases a little above 100.

With every evaporation and condensation, the remaining mixture will have more and more liq.2 composition, so the BP will slowly increase. Eventually all of liq.1 will be removed from the mixture.

This is the whole point of distillation of mixture, to separate the more volatile component (liq.1) from the less volatile component (liq.2)
 
I'll let you know as an experienced chemical engineer that it is not that simple. This is due to the fact that when you mix things it completely changes the fugacities escaping tendencies of the mixtures and there is NO WAY to tell, without experimentation, what the new boiling point will be. All empirical equations are based on experimental models.

The only time you could use an estimate is if the two liquids are so similar that they can be treated as ideal mixing. In that case, raoult's law can be used. Otherwise, No.

The other important point to make about distillation is that you can never get a pure species out. That is why distillation towers are so tall. Because the more trays you have the more pure the distillate. The vapor mol fraction of a mix of two substances (even if the temp is at the lower BP of one species) does not ever consist of pure one or the other but a mix of both. (LubDubb hit it right on the head)


Think about it this way.....If I have a pot of water at room temp in a dry climate water is leaving as a vapor. So now lets mix ethanol (a bad example 'cause it forms a 94/6 azeotrope at 1 atm) Don't you think that at the lower boiling point of ethanol there will still be water vapor escaping with the ethanol vapor? of course there would be.

We have huge libraries of book entirely dedicated to the boiling point and vapor pressure properties of different types of mixtures. In fact, an entire branch of chemical engineering is devoted to conducting experiments for companies wanting info on this issue.
 
Tulane, doing MD/MPH. Feel free to PM me if you have any questions about 1st year.

I know it's not that simple, and it's an interesting topic on an intellectual level, but the details are mostly beyond what will ever be useful to me in practice. Those neurons had to be sacrificed to make room for anatomy and embryology.
 
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