Deviations from Raoult's Law/ Azeotropes

[Medgen]

Everywhere I look it seems to say that a positive deviation from Raoult's law will result in a mixture that has a lower boiling point than either of the pure compounds from which it was made... but the graph indicates otherwise. The graph indicates that you could have a vapor pressure that is below that of one of the pure compounds and thus a higher boiling point. Can anyone explain why sources from wikipedia to EK seem to miss this? Or am I missing something?

To see what type of graph I am referring to: http://www.google.com/imgres?imgurl...1&ndsp=29&ved=1t:429,r:5,s:0&biw=1367&bih=694

Not sure if that link works but any graph showing a deviation from Raoult's law shows what I refer to.

Thanks!!!!

---

Members don't see this ad.

 

[Philosophy]

Everywhere I look it seems to say that a positive deviation from Raoult's law will result in a mixture that has a lower boiling point than either of the pure compounds from which it was made... but the graph indicates otherwise. The graph indicates that you could have a vapor pressure that is below that of one of the pure compounds and thus a higher boiling point. Can anyone explain why sources from wikipedia to EK seem to miss this? Or am I missing something?


Not sure if that link works but any graph showing a deviation from Raoult's law shows what I refer to.

Thanks!!!!

Here's my common sense account for azeotropes deviations. In a mixture of A-B. If A-B intermolecular attractions are stronger than A-A attractions and/or B-B intermolecular attractions, the vapor pressure will be depressed and the mixture will be relatively more stable than an ideal solution and have a higher BP than predicted. This is the case for Ethanol and Water, and incidentally accounts for why only 95% purity can be achieved in distillations of H20/EtOH mixtures. (Right graph that you refer to)


If A-B IM attractions are weaker than A-A and/or B-B, the liquid form will be less stable than in an ideal mixture, the VP higher at a given temp, and thus the mixture will not need as high a temp to boil. (Left graph that you refer to)

Hope this is useful to you!

 

[SN2ed]

Thread moved. These types of questions belong in the Study Q&A forum.

 

[SaintJude]

Here's my common sense account for azeotropes deviations. In a mixture of A-B. If A-B intermolecular attractions are stronger than A-A attractions and/or B-B intermolecular attractions, the vapor pressure will be depressed and the mixture will be relatively more stable than an ideal solution and have a higher BP than predicted. This is the case for Ethanol and Water, and incidentally accounts for why only 95% purity can be achieved in distillations of H20/EtOH mixtures. (Right graph that you refer to)


If A-B IM attractions are weaker than A-A and/or B-B, the liquid form will be less stable than in an ideal mixture, the VP higher at a given temp, and thus the mixture will not need as high a temp to boil. (Left graph that you refer to)

Hope this is useful to you!

Can anyone explain further how Philosophy was relating azeotrope deviations to intermolecular forces? The AAMC is moving more towards testing analytical skills, so I wouldn't be surprised if they included the OP's above graph link to ask us to interpret a deviation from the ideal situation.

 

[syoung]

Here's my common sense account for azeotropes deviations. In a mixture of A-B. If A-B intermolecular attractions are stronger than A-A attractions and/or B-B intermolecular attractions, the vapor pressure will be depressed and the mixture will be relatively more stable than an ideal solution and have a higher BP than predicted. This is the case for Ethanol and Water, and incidentally accounts for why only 95% purity can be achieved in distillations of H20/EtOH mixtures. (Right graph that you refer to)


If A-B IM attractions are weaker than A-A and/or B-B, the liquid form will be less stable than in an ideal mixture, the VP higher at a given temp, and thus the mixture will not need as high a temp to boil. (Left graph that you refer to)

Hope this is useful to you!

Basically BP is when the vapor pressure of the solution is higher than the atmosphere pressure, so molecules can enter gas phase right? So lets say with water and ethanol. Water boils at 100 C and ethanol is at 74 C. Well the reason why H2O boils so high is because of its high H-bonding (a type of IM force). Similar to Ethanol, but ethanol is a bit bigger than H2O, so it can't form as strong of H-bonding between its molecules, plus it has some C-H's too. Now when you mix them, you change the interactions between the molecules, from their normal. Water is mixed with Ethanol, so it can't have a BP of 100 anymore, as EtOH is interfering with its normal way of bonding water molecules. Ethanol is mixed with H2O now, so it can't have a BP of 74, because the smaller H2O molecule is bonding better with EtOH than EtOH is with itself. So in this case, because the IM bonding in the mixture is stronger than EtOH's, it goes below the R's law of an ideal mixture.

My 2 cents, if i'm not mistaken.