Vapor pressure

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BestDoctorEver

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I am having trouble understanding that higher boiling point decreases vapor pressure... Can someone explain?

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Something with a high boiling point is made up of molecules that are relatively happy to be in liquid phase. It takes more energy to get up to the point where you're knocking them loose into gas phase.

Vapor pressure is a measure of how much liquid will evaporate to gas phase, in a closed system. Higher vapor pressure means more molecules are jumping into gas phase, creating a higher partial pressure of that compound.
 
Something with a high boiling point is made up of molecules that are relatively happy to be in liquid phase. It takes more energy to get up to the point where you're knocking them loose into gas phase.

Vapor pressure is a measure of how much liquid will evaporate to gas phase, in a closed system. Higher vapor pressure means more molecules are jumping into gas phase, creating a higher partial pressure of that compound.


I had a question about a passage on vapor pressure and figured I'd just add to this one.

Here's the question:

If the vapor pressure of a solution is higher than predicted by Raoult's law, then the temperature of the solvents when put into the solution will:

A. Increase due to the energy absorbed by the breaking of bonds
B. Increase due to the energy released by the formation of bonds
C. Decrease due to the energy absorbed by the breaking of bonds
D. Decrease due to the energy absorbed by the formation of bonds

According to the graphs in the passage, an increased vapor pressure should mean a deltaH>0. deltaH>0 should mean that we're putting in energy to form bonds, right? So wouldn't that mean that temperature would decrease because energy is absorbed by the formation of those bonds, D? Where am I messing up in my thought process?
 
I had a question about a passage on vapor pressure and figured I'd just add to this one.

Here's the question:

If the vapor pressure of a solution is higher than predicted by Raoult's law, then the temperature of the solvents when put into the solution will:

A. Increase due to the energy absorbed by the breaking of bonds
B. Increase due to the energy released by the formation of bonds
C. Decrease due to the energy absorbed by the breaking of bonds
D. Decrease due to the energy absorbed by the formation of bonds

According to the graphs in the passage, an increased vapor pressure should mean a deltaH>0. deltaH>0 should mean that we're putting in energy to form bonds, right? So wouldn't that mean that temperature would decrease because energy is absorbed by the formation of those bonds, D? Where am I messing up in my thought process?

Right off the bat, remember that you have to spend energy to break a bond (endothermic), and energy is released as a bond is formed (exothermic). So I'd eliminate D.

Still a tough question though, but I'll try to reason it out for funsies, then someone can come correct me maybe. Also for funsies.

Raoult's law describes an ideal solution where vapor pressure of each component depends on its own vapor pressure when it's pure, and it's mole fraction in the ideal solution.

So if the total vapor pressure of a given solution is HIGHER than Raoult's law would suggest, that means the solvent and solute aren't binding that favorably. Still some binding to have solubility, but it's binding that isn't as favorable as the binding that existed in the pure compounds. They leave the solution and become vapors more easily than Raoult predicts. If they were happier together and bound more tightly to each other than they did to themselves, you'd see a vapor pressure LOWER than Raoult's prediction as fewer molecules would vaporize.

I'm a little thrown by the odd wording of, "the temperature of the solvents when put into the solution will" but I guess they're asking what happens to the diluent when you further dilute an established solution? In which case I'd say, since the solvent and solute interact unfavorably, dilution must be favorable. More solute-solute bonds can form, and energy is released, making the solution feel warm. Is it B? :p
 
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I am having trouble understanding that higher boiling point decreases vapor pressure... Can someone explain?

I was having trouble with this recently. The way I'm trying to process it in my brain is (I think this is correct, someone else comment please)

1. A liquid boils when its vapor pressure == local atmospheric pressure. So for water @ 100C its vapor pressure = 760 torr / 1 atm.

2. Now imagine you have a liquid with a really low vapor pressure. Say you want to get it to boil in the lab at constant pressure of 1 atm. You need to make the vapor pressure of that liquid equal 1 atm. The further away that is (lower), the more you're going to need to boil / increase temperature / increase kinetic energy of the molecules (3/2 RT?) to get vapor pressure of the liquid to equal atmospheric.

The boiling point of water falls as you increase altitude (atmospheric pressure drops). You're decreasing the gap between vapor pressure of the liquid and vapor pressure of the surroundings, lowering the boiling point. At 8000 meters the boiling point of water is only 75C.

Of course Raoult's Law deviations can be fed into this. A non-volatile addition will result in a partial pressure of mole fraction volatile * pressure of volatile (aka the non-volatile contributes nothing except reducing the mole fraction of the volatile component from 1.0 to something less than 1). A volatile addition will be the sum of the partial pressures of each. Standard negative and positive enthalpy deviations apply, with a negative delta H causing a negative deviation (you've created a more stable solution, with a lower vapor pressure and thus higher boiling point) or positive delta H causing a positive deviation (you've created a solution that is not as stable, with a higher vapor pressure and a lower boiling point).

If the vapor pressure of a solution is higher than predicted by Raoult's law, then the temperature of the solvents when put into the solution will:

A. Increase due to the energy absorbed by the breaking of bonds
B. Increase due to the energy released by the formation of bonds
C. Decrease due to the energy absorbed by the breaking of bonds
D. Decrease due to the energy absorbed by the formation of bonds
Ok, so we have a positive deviation from Raoult's law. As mentioned above, the higher vapor pressure is going to cause a drop in boiling point, and the positive deviation indicates a positive delta H, endothermic reaction. So C and D should be out, because the solvents aren't heating the reaction system. Remember that breaking bonds does not release energy, it takes energy (biological reactions are coupled). I would think it would be B as well, though now my brain is a little confused, as an exothermic solution formation also releases heat when it forms new bonds, losing energy to the surroundings and dropping the solution system to lower, more stable energy state.

tl;dr - I think it's A or B, what is the correct answer / reasoning?

e: what's frustrating is the amount of little stuff to remember for a topic like this that may not even make it onto your particular MCAT based on your luck that day. Spent your time reviewing inorganic chem, solutions / colligative props / acids bases, etc? Have fun with some optics and Lensmakers' Equation!
 
I was having trouble with this recently. The way I'm trying to process it in my brain is (I think this is correct, someone else comment please)

1. A liquid boils when its vapor pressure == local atmospheric pressure. So for water @ 100C its vapor pressure = 760 torr / 1 atm.

2. Now imagine you have a liquid with a really low vapor pressure. Say you want to get it to boil in the lab at constant pressure of 1 atm. You need to make the vapor pressure of that liquid equal 1 atm. The further away that is (lower), the more you're going to need to boil / increase temperature / increase kinetic energy of the molecules (3/2 RT?) to get vapor pressure of the liquid to equal atmospheric.

The boiling point of water falls as you increase altitude (atmospheric pressure drops). You're decreasing the gap between vapor pressure of the liquid and vapor pressure of the surroundings, lowering the boiling point. At 8000 meters the boiling point of water is only 75C.

Of course Raoult's Law deviations can be fed into this. A non-volatile addition will result in a partial pressure of mole fraction volatile * pressure of volatile (aka the non-volatile contributes nothing except reducing the mole fraction of the volatile component from 1.0 to something less than 1). A volatile addition will be the sum of the partial pressures of each. Standard negative and positive enthalpy deviations apply, with a negative delta H causing a negative deviation (you've created a more stable solution, with a lower vapor pressure and thus higher boiling point) or positive delta H causing a positive deviation (you've created a solution that is not as stable, with a higher vapor pressure and a lower boiling point).


Ok, so we have a positive deviation from Raoult's law. As mentioned above, the higher vapor pressure is going to cause a drop in boiling point, and the positive deviation indicates a positive delta H, endothermic reaction. So C and D should be out, because the solvents aren't heating the reaction system. Remember that breaking bonds does not release energy, it takes energy (biological reactions are coupled). I would think it would be B as well, though now my brain is a little confused, as an exothermic solution formation also releases heat when it forms new bonds, losing energy to the surroundings and dropping the solution system to lower, more stable energy state.

tl;dr - I think it's A or B, what is the correct answer / reasoning?

e: what's frustrating is the amount of little stuff to remember for a topic like this that may not even make it onto your particular MCAT based on your luck that day. Spent your time reviewing inorganic chem, solutions / colligative props / acids bases, etc? Have fun with some optics and Lensmakers' Equation!


My reasoning was the same as yours and catburr but I came out with C as the answer. I don't think its A because absorption to break bonds wouldn't raise temperature. I don't understand why you would pick B if you know that the mixing of these two substances was endothermic. If the solute-solute bonds are more stable than the solute-solvent bonds than the energy used to break the solute-solute bonds would be greater than what you would get back from the formation of solvent-solute bonds. Overall the reaction would need more energy and would take it from the system.
 
My reasoning was the same as yours and catburr but I came out with C as the answer. I don't think its A because absorption to break bonds wouldn't raise temperature. I don't understand why you would pick B if you know that the mixing of these two substances was endothermic. If the solute-solute bonds are more stable than the solute-solvent bonds than the energy used to break the solute-solute bonds would be greater than what you would get back from the formation of solvent-solute bonds. Overall the reaction would need more energy and would take it from the system.

Hmm, the reason I'm leery of C and D is that if there was a positive deviation, the reaction was endothermic, shouldn't the temperature of the solvents now increase because it's taking in energy from its surroundings?

Or is it that the temperature of the solvents are decreasing because that energy is being 'transfered' into the new solution?
 
Hmm, the reason I'm leery of C and D is that if there was a positive deviation, the reaction was endothermic, shouldn't the temperature of the solvents now increase because it's taking in energy from its surroundings?

Or is it that the temperature of the solvents are decreasing because that energy is being 'transfered' into the new solution?

Why do you say temperature of solvents and energy of its surroundings? There is only one temperature which is from the surroundings. The reaction is going to take energy from the surroundings and thus lower the surroundings temperature. Atleast that is how I think of it. Enthothermic reactions are always going to lower the temperature of the surroundings, why should this be any different.
 
Why do you say temperature of solvents and energy of its surroundings? There is only one temperature which is from the surroundings. The reaction is going to take energy from the surroundings and thus lower the surroundings temperature. Atleast that is how I think of it. Enthothermic reactions are always going to lower the temperature of the surroundings, why should this be any different.

The question refers to temperature of the solvents, so maybe I'm being confused by that phrasing or something. I agree that the reaction is going to suck in energy and lower the temperature of the surroundings. So would the temperature of the solvents when put into the solution decrease because their energy is being put into breaking bonds (choice C)?

And I think I understand your last post better now as to why B is wrong; the temperature of each solvent can't increase, because that would indicate energy being released from solution formation, which we know isn't the case because we're dealing with an endothermic reaction here.
 
I am having trouble understanding that higher boiling point decreases vapor pressure... Can someone explain?

Think about all of the vapor coming off of water when you boil it. If the boiling point was higher, do you think there would be as much vapor?
 
i picked /B/

my rationale:
vapour pressure higher than Raoult's prediction --> higher mole fraction in solvent (from raoult's eq) && more solvent in vapour form --> molality ("m") is greater than expected

when more of solvent is added (thats my interpretation of the question), gms of solvent increases, so molality ("m") decreases --> boiling pt elevation is less(er) --> more of solvent can exist in vapour state now

since in original state, there was more of vapour in the soln, it is safe to assume that energy was absorbed by solvent molecules to allow conversion from liquid to vapour state i.e. either solute-solvent interaction or solute-solute interaction or solvent-solvent interaction is/are exothermic... since the question asks about 'temp of solventS", latter would make sense.. so, more solvent being added, more energy is released due to bond formation

combining these two, addition of more solvents would cause increase in temp due to bond formation :)
 
i picked /B/

my rationale:
vapour pressure higher than Raoult's prediction --> higher mole fraction in solvent (from raoult's eq) && more solvent in vapour form --> molality ("m") is greater than expected

when more of solvent is added (thats my interpretation of the question), gms of solvent increases, so molality ("m") decreases --> boiling pt elevation is less(er) --> more of solvent can exist in vapour state now

since in original state, there was more of vapour in the soln, it is safe to assume that energy was absorbed by solvent molecules to allow conversion from liquid to vapour state i.e. either solute-solvent interaction or solute-solute interaction or solvent-solvent interaction is/are exothermic... since the question asks about 'temp of solventS", latter would make sense.. so, more solvent being added, more energy is released due to bond formation

combining these two, addition of more solvents would cause increase in temp due to bond formation :)

A mole fraction of anything will always decrease the vapor pressure of a pure solvent. Let's say the vapor pressure of pure hexane(l) was 80 torr. Mixing it into a solution with other solvents will decrease it's vapor pressure since it's now a mole fraction of what it use to be (1/1). For example, mixing 2 moles hexane(l) with 8 moles of octane(l) results in a mole fraction of 2/10 or 1/5 for hexane. A mole traction of 1/5 times 80 torr = 16 torr. In other words, the vapor pressure for a solvent in a mixture will always be less than it would be in a pure solution. This will always be the case for each solvent in a mixture. This is what Raoult's law tells us.

I'm not sure what this has to do with molality since it's irrelevant to the question being asked. The question isn't asking about a colligative property (ie. Boiling Point Elevation or Freezing Point Depression), so there's no need to even consider molality.
 
I am having trouble understanding that higher boiling point decreases vapor pressure... Can someone explain?

You have this statement wrong. A higher boiling point does NOT decrease the vapor pressure. Vapor pressure is independent of boiling point. If you're not sure why, just think about what the boiling point would be at the surface of the earth versus Mount Everest.

Boiling Point: atmospheric pressure = vapor pressure

As you increase altitude, atmospheric pressure decreases, while vapor pressure for a particular solution doesn't change. As a result, less heat is needed to reach the point where vapor pressure = atmospheric pressure (boiling point) at higher altitudes.

Another thing to note is that vapor pressure is entirely dependent on intermolecular forces. As intermolecular forces increase in strength, there is a less tendency for molecules to escape into the vapor phase. This explains why the vapor pressure of octane(l) has a much higher vapor pressure than does water(l) at room temperature (london dispersion forces vs. all intermolecular forces in water). It also explains why the boiling point is much lower for octane (at sea level) than it is for water.

Higher Vapor Pressure = Lower Boiling Point
Lower Vapor Pressure = Higher Boiling Point
 
I'm not sure what this has to do with molality since it's irrelevant to the question being asked. The question isn't asking about a colligative property (ie. Boiling Point Elevation or Freezing Point Depression), so there's no need to even consider molality.

true.. mention of molality was simply an extension of my thought process (more so a reiteration to solidify understanding) :)
 
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Whoa, everyone got excited about this one. Now I really want to know the answer...

I don't understand why you would pick B if you know that the mixing of these two substances was endothermic. If the solute-solute bonds are more stable than the solute-solvent bonds than the energy used to break the solute-solute bonds would be greater than what you would get back from the formation of solvent-solute bonds. Overall the reaction would need more energy and would take it from the system.

I picked B just based on trying to interpret (probably incorrectly) the wording of the question, "adding solute to the solution." If you have a highly concentrated solution, there's a lot of unfavorable bonds there. Add in more solute, you dilute it. More new favorable solute-solute bonds. Not saying it's the best of logics, but was just my train of thought. :)
 
Whoa, everyone got excited about this one. Now I really want to know the answer...

It is a very good question and I would absolutely expect it as a discrete or part of a vapor pressure / solution passage.

Here is the thought flow I think is absolutely critical to the question (requoting the question here because there are a lot of posts):

Question said:
If the vapor pressure of a solution is higher than predicted by Raoult's law, then the temperature of the solvents when put into the solution will:

A. Increase due to the energy absorbed by the breaking of bonds
B. Increase due to the energy released by the formation of bonds
C. Decrease due to the energy absorbed by the breaking of bonds
D. Decrease due to the energy absorbed by the formation of bonds

1. vapor pressure higher than predicted by Raoult's Law == endothermic reaction, positive delta H.
2. An endothermic solution formation results in weaker bonds and a higher vapor pressure, and thusly a lower boiling point. Heat is not removed from the solution as in an exothermic formation, so the energy level of the bonds in the solution remain high (and more unstable, weaker).

So now I guess, you see what the 'temperature of the solvents' means. This would be the average kinetic energy of the molecules of each of the solution components.

Now my brain is saying, "ok, you tossed in these two solvents, and they still sucked in energy from their surroundings to form a solution. So their kinetic energy has gone up. Discard C and D."

Now I get confused between A and B though. A says that the solvents gained energy from breaking bonds; breaking bonds would consume energy. Maybe the energy taken in from surroundings broke these bonds and increased kinetic energy of solvent molecules?

B says that the temperature increased because of energy released from forming bonds. Yeah, forming bonds can release energy, this is the idea behind exothermic solution formation: new solute-solvent bonds are formed, excess energy released as heat to the system because we're assumedly at constant pressure, making delta H = q. But the reaction was endothermic, not exothermic... :confused:
 
It is a very good question and I would absolutely expect it as a discrete or part of a vapor pressure / solution passage.

Here is the thought flow I think is absolutely critical to the question (requoting the question here because there are a lot of posts):



1. vapor pressure higher than predicted by Raoult's Law == endothermic reaction, positive delta H.
2. An endothermic solution formation results in weaker bonds and a higher vapor pressure, and thusly a lower boiling point. Heat is not removed from the solution as in an exothermic formation, so the energy level of the bonds in the solution remain high (and more unstable, weaker).

So now I guess, you see what the 'temperature of the solvents' means. This would be the average kinetic energy of the molecules of each of the solution components.

Now my brain is saying, "ok, you tossed in these two solvents, and they still sucked in energy from their surroundings to form a solution. So their kinetic energy has gone up. Discard C and D."

Now I get confused between A and B though. A says that the solvents gained energy from breaking bonds; breaking bonds would consume energy. Maybe the energy taken in from surroundings broke these bonds and increased kinetic energy of solvent molecules?

B says that the temperature increased because of energy released from forming bonds. Yeah, forming bonds can release energy, this is the idea behind exothermic solution formation: new solute-solvent bonds are formed, excess energy released as heat to the system because we're assumedly at constant pressure, making delta H = q. But the reaction was endothermic, not exothermic... :confused:

Still think the answer is C.

I think that the temperature of the solvent and the temperature of the solution are the same thing. If temperature of the solvent is the kinetic energy of the solvent molecules then the solvents would use that kinetic energy to form unfavorable solute-solvent bonds. This would reduce the kinetic energy of the solvents and therefore the temperature. Note that the solute-solvent bonds would produce energy since its a bond formation but that energy would be smaller than that needed to break the solvent-solvent bonds. Where does the energy for the breaking of solvent-solvent bonds come from? I believe its the kinetic energy of solvent and solute molecules which at the end would reduce the temp. 5
 
You don't need more information, all you need to know was that the reaction was endothermic. Several people arrived at that conclusion without more information.

Hmm yeah, guess I over thought the question. The only thing you had to do was realize that the positive deviation indicates an endothermic reaction, and then figure out that the energy absorbed (since its endothermic) goes into breaking the old bonds.

Weird wording with the temperature of solvent thing, but I think I get it now. All this for one little concept that may never even appear on my MCAT, awesome.

Appreciate the help!
 
Right off the bat, remember that you have to spend energy to break a bond (endothermic), and energy is released as a bond is formed (exothermic). So I'd eliminate D.

Still a tough question though, but I'll try to reason it out for funsies, then someone can come correct me maybe. Also for funsies.

Raoult's law describes an ideal solution where vapor pressure of each component depends on its own vapor pressure when it's pure, and it's mole fraction in the ideal solution.

So if the total vapor pressure of a given solution is HIGHER than Raoult's law would suggest, that means the solvent and solute aren't binding that favorably. Still some binding to have solubility, but it's binding that isn't as favorable as the binding that existed in the pure compounds. They leave the solution and become vapors more easily than Raoult predicts. If they were happier together and bound more tightly to each other than they did to themselves, you'd see a vapor pressure LOWER than Raoult's prediction as fewer molecules would vaporize.

I'm a little thrown by the odd wording of, "the temperature of the solvents when put into the solution will" but I guess they're asking what happens to the diluent when you further dilute an established solution? In which case I'd say, since the solvent and solute interact unfavorably, dilution must be favorable. More solute-solute bonds can form, and energy is released, making the solution feel warm. Is it B? :p

Lol, sorry, it's C. The explanation is that because it's endothermic, there will be a decrease in temp. I'm still confused about this one.
 
Lol, sorry, it's C. The explanation is that because it's endothermic, there will be a decrease in temp. I'm still confused about this one.

I was overthinking it then, with all my "dilute the unfavorable solution" business. :) The question is basically asking, when you mix these things together, how does it feel. You just have to figure out, in a roundabout way, that the process of mixing is endothermic. So how does endothermic feel to the touch? It feels cold, because energy is being sucked out of the surroundings into the molecules, so that those unfavorable reactions can occur.

It's easy to get tripped up into thinking, well, if endothermic means the molecules are sucking in energy, why is it I don't feel warmth when I touch them, if they're so full of energy now? I like to think that it's because they aren't "expressing" their high energy state with heat, they're expressing it by creating an unfavorable bond. Kind of an unscientific description maybe, but it helped me to remember endothermic = cold surroundings, exothermic = warm surroundings.
 
I keep getting mixed up with bond breaking being endothermic!!! It's such a simple concept to lose points over.
 
This one is really simple. Since you are getting vapor pressure than you will have to input heat to break the bonds. Therefore, the rxn is ENDOTHERMIC. Endothermic rxns tend to absorb heat from the surrounding (in this case solvent); thus, there is a decrease in temperature of the solvent.

Answer is C...
 
This question just tore me apart on TPR #1, glad I finally understand it now though!

H is positive so it’s endothermic, pulls in heat, to break bonds, freeing up molecules which then escape into gas phase (vapor pressure). So any solvent will experience a decrease in temp (solvent is the surrounding)
 
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