Vapor Pressure Contradicting Colligative Props?

[justadream]

This is why I got from TBR:

Adding impurities:

1) MP decreases (freezing point depression)

2) BP increases from boiling point elevation (TBR page 83),


But for BP, TBR says that if the thing you add in is volatile, BP decreases. (TBR page 85).



Also, TBR then discusses on page 86 how the types of forces between the molecules mediates vapor pressure in mixtures. Thus, if the molecules in a mixture repel, you have increased VP. If the molecules in a mixture attract, you have decreased VP. As such, how can you make blanket statements like statements 1 and 2 above without knowing the attractive/repulsive nature of molecules that are mixed together?

---

Members don't see this ad.

 

[Czarcasm]

This is why I got from TBR:

Adding impurities:

1) MP decreases (freezing point depression)

2) BP increases from boiling point elevation (TBR page 83),


But for BP, TBR says that if the thing you add in is volatile, BP decreases. (TBR page 85).



Also, TBR then discusses on page 86 how the types of forces between the molecules mediates vapor pressure in mixtures. Thus, if the molecules in a mixture repel, you have increased VP. If the molecules in a mixture attract, you have decreased VP. As such, how can you make blanket statements like statements 1 and 2 above without knowing the attractive/repulsive nature of molecules that are mixed together?

I'm terrible explaining this stuff, but hear me out. It might help to explain this with numbers:

At 1 atmospheric pressure (760 mmHg), two pure liquid solutions in a flask have the following vapor pressures:

Pure A: 70 mmHg
Pure B: 380 mmHg
(just assume 1 mole each)

If you recall Raoult's Law, mixing both of these solutions together will produce a total pressure equal to mole fraction of the vapor pressures for each liquid in their pure state. So in other words, if we were to mix 1 mole of each, the total pressure would be:

1 mole (A or B)/ (2 total moles) = 0.5 mole

0.5 mole fraction A x 70 mmHg = 35 mmHg
0.5 mole fraction B x 380 mmHg = 190 mmHg

For the mixture, the total pressure = 35 mmHg + 190 mmHg = 225 mmHg. Notice this is considerably higher than the vapor pressure of pure A alone. We went from 70 mmHg (pure A) to 225 mmHg (mixture). If the atmospheric pressure is still 760 mmHg and we already have a vapor pressure of 225 mmHg, then the amount of heat needed to get more molecules to vaporize would be less (resulting in a lower boiling point temperature).

This is entirely different when the colligative property are ionic salts added to a liquid solution (since vapor pressure of salts are negligible or non-existent). In that scenario, it might help to consider that each individual salt is increasing the intermolecular attraction of the liquified molecules, preventing their escape into the gas phase, making it more difficult to vaporize and therefore more difficult to boil (higher BP).

 

[NextStepTutor_1]

This is why I got from TBR:

Adding impurities:

1) MP decreases (freezing point depression)

2) BP increases from boiling point elevation (TBR page 83),


But for BP, TBR says that if the thing you add in is volatile, BP decreases. (TBR page 85).



Also, TBR then discusses on page 86 how the types of forces between the molecules mediates vapor pressure in mixtures. Thus, if the molecules in a mixture repel, you have increased VP. If the molecules in a mixture attract, you have decreased VP. As such, how can you make blanket statements like statements 1 and 2 above without knowing the attractive/repulsive nature of molecules that are mixed together?

It is presumed, when making these "blanket terms" labeled as 1 & 2, that the solute or impurity is completely dissolving into solution. And generally this is the case. Take for instance salt and water, adding a large amount of salt to a pot of water increases the boiling point and decreases the melting point.
In a question, it should either be made apparent that the solute is attracted to the atoms/molecules of the solution (dissolves) or not (creates a layer) by the substances the experiment is using, ex. salt and water, or through the language used in the question, ex. completely dissolves, forms a layer, etc. Deciphering this info from the passage or question will allow you to use these statements to help you choose the correct answer.

 

[justadream]

@Czarcasm @NextStepTutor_1

So should I think of "boiling point elevation", "freezing point depression", "vapor pressure depression" as applying only to when you dissolve some type of solute in a solvent? In these cases, since the solute dissolves completely, the colligative properties should hold without exceptions (at least for MCAT)?

And with regard to Czarcasm's example, I think I mostly understand it. So then the deviations occur when, for example, when you add Substance A to Substance B, substance A reacts with substance B (to make the resulting substance it easier or harder to boil)?

 

[NextStepTutor_1]

So should I think of "boiling point elevation", "freezing point depression", "vapor pressure depression" as applying only to when you dissolve some type of solute in a solvent? In these cases, since the solute dissolves completely, the colligative properties should hold without exceptions (at least for MCAT)?

Yes, this is a great mental schema of A-->B-->C (bp increase, mp decrease, vp decrease) to have when thinking about adding solute to solution.

 

[sillyjoe]

This is why I got from TBR:

Adding impurities:

1) MP decreases (freezing point depression)

2) BP increases from boiling point elevation (TBR page 83),


But for BP, TBR says that if the thing you add in is volatile, BP decreases. (TBR page 85).



Also, TBR then discusses on page 86 how the types of forces between the molecules mediates vapor pressure in mixtures. Thus, if the molecules in a mixture repel, you have increased VP. If the molecules in a mixture attract, you have decreased VP. As such, how can you make blanket statements like statements 1 and 2 above without knowing the attractive/repulsive nature of molecules that are mixed together?

I was actually speaking with my physics professor recently and he said a great way to think about this is in terms of entropy. When you dissolve salt into water it increases it's entropy because it is no longer in the organized crystal form. If you freeze it, it becomes a crystal again and thus more organized. It loses that entropic favorability so relative to pure water, it wants to freeze less and the freezing point depresses.

You can apply the same thinking to boiling point elevation. If you vaporize the solvent, the solute forms a crystalline solid and loses the entropic favorability so the boiling point relative to pure water elevates.

I just thought this was a interesting perspective.

 

[justadream]

@sillyjoe

I understand your freezing example but I was thrown off by the "If you vaporize the solvent, the solute recrystallizes and loses the entropic favorability so the boiling point relative to pure water elevates."

How would something recrystallize when it's being vaporized (heated)?

 

[sillyjoe]

@sillyjoe

I understand your freezing example but I was thrown off by the "If you vaporize the solvent, the solute recrystallizes and loses the entropic favorability so the boiling point relative to pure water elevates."

How would something recrystallize when it's being vaporized (heated)?

Sorry didn't mean recrystallize. Just mean crystallized.

When the water is boiled away the solute/salt forms a crystalline structure again which is entropically less favorable than the disordered state of the aqueous solution. As a result, the boiling point of the water is elevated to account for the extra energy input needed to overcome the -entropy of the salt forming a crystalline structure.