

Chemistry: Vapor Pressure Lowering



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  • Type: Video Tutorial
  • Length: 12:16
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 131 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Physical Properties of Solutions (14 lessons, $22.77)
Chemistry: Colligative Properties (5 lessons, $9.90)





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We've been looking at a variety of different physical equilibria. We started with considering the equilibria involved with phase changes, going from a liquid to a vapor, going from a solid to a liquid. Then we started to look at what happens in mixtures. What happens when we have a solute dissolved in that material? How does that affect what happens with the equilibrium of solute and solution and that equilibrium? Now we're going to look at the two ideas together. What happens when we take a solute and put it in a solvent, another material? What happens to the properties of that solvent? What happens to its vapor pressure? What happens to its boiling and freezing points? These are examples of colligative properties and we'll define that in just a moment. But again, our focus now is going to turn to how dissolving a solute in a different material affects the properties of that host material, of that solvent.
Let me start by telling you something that may not be very intuitive to you. It turns out that if you take a solvent that has a vapor pressure that we can measure, and remember that's an equilibrium between the liquid and the vapor, and you put a solute in. You dissolve a solute into that liquid phase. You actually change the vapor pressure of that material. So, again, looking at a cartoon here, what I'm saying is, if we focus on the equilibrium between a liquid and its vapor, and remember what vapor pressure was. We could measure that equilibrium, vapor pressure. It turns out that if dissolve, and I illustrate by these little red dots here, if we dissolve a solute into the liquid, we actually lower the vapor pressure of the solvent. Now, that's a tricky idea, how that works. Let's make sure we understand the experiment first and then we'll try to get into what on earth is going on here.
If we look at a plot of vapor pressure, so again, we're measuring the vapor pressure, as a function of how much solute we put in. And for the sake of simplicity, I'm going to assume our solute is ideal. What I mean by that is, I'm going to insist that the solute be absolutely non-volatile, for the moment, so we don't have to worry about that pressure, and also that it has no significant attractions to the solvent. I'm going to assume that those no strong bonding taking place at all between the solute and the solvent, just that we're dissolving it in there. And remember that's going to primarily be an entropy concept then, that this is just dissolved in there simply because there's more room for the solute to move around in, more disorder. Yet we see again a decrease in the vapor pressure.
So, let's go on to this. This again is referred to as Raoult's Law, the notion that that vapor pressure decreases in a linear fashion as we increase the amount of solute. And as I mentioned, this is an example of what we call a colligative property. Now here's the interesting thing about a colligative property. Colligative property depends only upon the concentration of the solute, but not on the properties of the solute itself. In other words, in principle, it makes no difference what the solute is, what the chemical properties of the solute are, that they will all obey the exact same laws. And it depends again only on concentration. So that says with respect to Raoult's Law, that it makes no difference what you dissolve in the solution, nonetheless, you will see the same exact behavior, that the vapor pressure of the solvent will drop as you dissolve more and more of the solute, whatever it is. So, Raoult's Law mathematically is the following: that the vapor pressure of the solvent is equal to the vapor pressure of the pure solvent (that's what this little symbol is here) times 1 minus the mole fraction of the solute. That's going to give us the same thing. Now, just rewriting this a little differently, using our definition of what a mole fraction is. Remember that's the amount of, in this case, mole fraction of solute is the amount of solute divided by total moles. So, another way to describe that then is the vapor pressure of the solvent is pure vapor pressure times the mole fraction of the solvent. Everything else, in other words. So, again, notice that these two things are exactly the same. Mole fraction of solvent is indeed 1 minus the mole fraction of the solute. Putting it simply, that's saying that 2/3 is 1-1/3. That's just the same way of doing this. I'm just talking about fractions here. Now, using either of these expressions because they are saying the exact same thing, again, that's describing a linear dependence on vapor pressure. The more stuff we put in, the more this drops down. And the big question again is, why?
So, let's return to this picture very quickly. Again, in this ideal solution, the main thing going on here is solute is getting dispersed into solvent. It's free to wander throughout solution. That increases the entropy of the system by having that solute free to move around. Think what happens as we start to remove some of that liquid and put it into the gas phase. By doing that, we're reducing the volume available for the solvent to get lost in. That's actually going to decrease our entropy relative to if the solute wasn't there at all. Because of that entropy change, that actually makes the solution phase a little bit more stable than it would have been if it was pure solvent. And that's what is responsible for this vapor pressure lowering. In other words, we're shifting this equilibrium between liquid and gas, the more solute we put in the liquid, the more we stabilize the liquid, through entropy now, the more it wants to stay in the liquid phase. And so our vapor pressure decreases a little bit as a result.
Let's actually work a problem where we illustrate that. Let's suppose, then, that we have in this problem listed the vapor pressure of pure benzine is equal to 1.25 atmospheres, and suppose now that we dissolve naphthaline, just another hydrocarbon, in the benzine, and the molecular weight is listed here and the mass is specified. We dissolve it in 78 grams of benzine, calculate the vapor pressure of the benzine in the above solution. Our expectation, again, is that the vapor pressure for benzine will decrease. Now, here's an aside. Naphthaline actually has its own vapor pressure. That won't affect the vapor pressure of benzine whatsoever, so we're not going to worry about that. All we're worried about is how much naphthaline is dissolved. And that's going to be what dictates what the vapor pressure of the benzine is.
So, what do we need to do first? We need to figure out the moles of naphthaline we're talking about. So, 6.40 grams divided by its molar mass gives us 0.0499 moles of naphthaline that we've dissolved in the solution. The mole fraction of benzine then is the moles of benzine divided by the total number of moles. So that's going to be, as it turns out, the numbers happen to work out so that that's exactly 1 mole of benzine, at least to three significant figures, divided by the moles of benzine plus moles of naphthaline. And that's our mole fraction again. That gives us, then, a mole fraction of benzine of 0.95, about 95%, 0.952 is our mole fraction of benzine. And then, finally, plugging it into Raoult's Law, the vapor pressure of the benzine then is going to be 0.95 times the vapor pressure of the pure benzine. So, notice in a way all Raoult's Law is doing is giving you a percentage of the total. It's saying, if there's 95% benzine in the solution, then the vapor pressure is only going to be 95% of its total value if it were pure. That's essentially all Raoult's Law is saying. That is what's causing this vapor pressure lowering. So, multiplying 0.95 times the vapor pressure again of pure benzine gives a value of 0.119. Notice again, it's less than the vapor pressure of the pure liquid. And that's exactly what has to happen if we're dissolving a solute, according to Raoult's Law.
Now, I've said all along that we're counting on the fact here that benzine and naphthaline have no significant interactions with each other. In other words, I'm assuming that Delta H is zero for the process of dissolving the naphthaline and the benzine. Well, what if that is not true. What if there is in fact a significant attraction between solute and solvent? Or, what if the solute disrupts significant attractions between solvent and solvent? Then we have something called a non-ideal solution, much in the same way that we talked about a non-ideal gas, in that we are now forced to not ignore anymore attractions between molecules. That's going to cause deviations from Raoult's Law. Raoult's Law will no longer be completely independent of what you're dissolving. In fact, there is no simple law that is going to describe it, but we can qualitatively predict how it's going to deviate from the ideal situation. Again, much like we did with ideal gasses.
So, let's see if we can predict what's going to happen here. Let's first of all consider a situation where when we dissolve the solute, we have very significant interactions between solute and solvent. So we have a lot of attraction. Delta H is going to be negative in that there is an overall release of energy when we dissolve it. That's going to stabilize the solution phase even more than just due to entropy. By stabilizing the solution phase, we drop from the ideal situation, the green line here, down to a little bit lower energy. That means that we have to pay an even higher price to go from the liquid phase to the vapor phase. If we have to pay more cost, our vapor pressure arguably is going to be a little bit lower than what we expect for the ideal solution. And, so, indeed, Raoult's Law starts to deviate from ideal behavior, that's the green line, in that the more solute we put in, the vapor pressure drops but it drops a little bit faster than Raoult's Law predicts. Conversely, if we have an endothermic situation, where a way to think of that is, as we put in the solute, we're causing more damage breaking bonds between solvent and solvent than we are by making up by solute/solvent. Then we have a situation where, overall, it costs us some energy to make the solute. That causes the energy to be higher. That means the difference in energy between let's say, instability between that solution and the vapor phase is less than the ideal case. That means the vapor pressure is probably a little higher than we would expect for the ideal situation. As a result, the deviation in Raoult's Law is positive, meaning there is a vapor pressure lowering, but it's not quite as significant as it would be for the ideal solution. So, again, we get this deviation in the positive direction rather than the negative interaction. By looking at that, it actually tells us about what's going on in solution. In other words, if we look at Raoult's Law and see how it deviates, or rather we do an experiment, see how the experiment deviates from Raoult's Law, that gives us some insight as to whether the overall process of dissolving is endothermic or exothermic. And so, again, we make that tie between the molecular level and talking about bonds and whether they are being broken or made and the entropy involved in mixing and so forth, and the macroscopic level where we're talking about changes in something like vapor pressure, in this case.
Physical Properties of Solutions
Colligative Properties
Vapor-Pressure Lowering Page [1 of 2]

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