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Chemistry: Enthalpy and Entropy Contributions to K

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  • Type: Video Tutorial
  • Length: 14:13
  • Media: Video/mp4
  • Use: Watch Online & Download
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  • Download: MP4 (iPod compatible)
  • Size: 153 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Thermodynamics (8 lessons, $14.85)
Chemistry: Using Free Energy (3 lessons, $4.95)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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We are so close now to making the link between free energy and the equilibrium constant. We've been kind of dancing around this idea for a couple tutorials, now, that these two things are related to each other. So let's review what we actually know. We know that if G° at standard state is zero, then we're at equilibrium at standard state. Now that's a very rare condition. It would be just by coincidence that the reactants and products, starting out with one atmosphere of pressure for all gases and one mole per liter for all solutions, that that was equilibrium. It's very rare that that will ever happen. But if it did, that would tell us that for that condition, we'd have to have the standard free energy equal to zero. But, more realistically, if the standard free energy is greater than zero, then that tells us that starting at standard state conditions, the reaction is going to have to run back towards the reactants in order to get to its equilibrium.
Well, think about that. We started with a value of one for all concentrations and pressures. If we run the reaction backwards, that means the equilibrium constant, wherever it is, is going to have to be, in this case, less than one, indicating again that reactants are favored at equilibrium. In contrast, if G° at standard state is less than zero, the reaction is going to go from left to right, in other words, towards products, or forwards, in order to get to its equilibrium. And therefore, the equilibrium constant is going to have to be greater than 1, once again favoring - remember this guy is just products of reactants - so favoring products, in this case.
So we're starting to understand that connection. What we want to do before we finally formalize this is - in fact, actually, we will go ahead and do this right now. We'll go ahead and connect this idea mathematically, and then I'm going to step back a moment and ask - let's break down G° into its components: H° and S° - how do those two ideas ultimately determine what the equilibrium constant will be?
So first of all, let's just go through the math. I'm just simply going to present for you this as a fact rather than derive it, which you'd fall asleep for anyway. The free energy is equal to minus RT natural log of the equilibrium constant. That's the relationship we've been missing. Now, what this says, if we expand this out to describe G° in terms of H° and T°S°, that H° minus T°S° is minus RT natural log K. Now, I'm going to rearrange this idea to come up with what, to me, is one of the most valuable expressions in chemistry. If we take the natural log of the equilibrium constant - that's equal to . That's just combining these ideas here, rearranging this expression, in other words. But what this says in English is very profound. It says if I want to know about an equilibrium, then the equilibrium constant, and therefore the log of the equilibrium constant - because these are going to move in parallel; the bigger K is the bigger the log of the K is - that that is going to be related to H° and S° the following way. It's going to be related to S° in an absolute sense that whatever S° is, that's kind of set roughly where my equilibrium is going to be. And then, I modify that, depending on what the temperature is, by my H° term. That says at really, really high temperatures - that T is in the denominator - this whole term essentially gets very small compared to this. And my equilibrium is based on wherever there's the best entropy. But at very low temperatures, this becomes a very large number compared to this and my equilibrium depends on where am I at the lowest chemical energy. Very important idea. Really low temperatures, H° is really important; really, really high temperatures, S° begins to dominate, as far as determining where your equilibrium position will be - in other words, whether reactants or products ultimately are favored or not.
So, let's depart from this for a moment and I'm going to remind you of an analogy. Now, I love to talk about analogies. To me they help. Even if they're not absolutely correct, they help kind of cement in place different ideas. If this analogy works for you, fantastic. If it doesn't, just bear with us. But let's go back to the idea of a tomb in some palace or whatever. And in the middle of this room is a pit full of snakes. And these snakes, at real low temperature, we're going to assume, just all hang out in the pit here. But at higher temperatures, they're going to be able to have sufficient energy that some of these snakes can actually get out into the main room area. Now, the analogy is each one of these snakes would represent a molecule, if you will. And the molecule could exist either in the form of the state characterized by the pit or the state characterized by the room. Now, the way this analogy would be set up, this state is lower energy, but this state has the advantage of more area, meaning a greater probability of finding something out here than down here, if these were equal. Just simply saying that the area down here is not nearly as big as the area up here.
At low temperatures, everything is going to be at the lowest energy they can get to. But as we raise our temperature enough, we start to be able to favor this other state, or at least occupy the other state, and the equilibrium constant - the ratio of one state compared to the other - will change to reflect that. In other words, we're going to favor - let's suppose we called this the reactants and this the products, then this would be an endothermic reaction, but it would have a S° that was positive - more disorder up here than down here, going from reactants to products. For that particular condition then, as we raise temperature, we are going to increase the ratio of products over reactants. K should get larger as we raise the temperature, because it's an endothermic reaction. Remember, in Le Châtelier's principle, endothermic reactions raise the temperature; you shift the equilibrium towards the products.
Now, that's going to be true regardless of what entropy is. But the fact that entropy is actually favoring products means that if we got our temperature up high enough, we actually could get the products actually favored. Because there's more area up here, if the temperature gets high enough, the snakes won't care whether they're down here or up here, and we'll end up with the majority of the snakes up in the room rather than in the pit. So in this case, we can actually raise the temperature to the point where the products are favored.
In contrast, imagine we have another room where instead of a pit, we've got a table. So at low temperatures again, the snakes are all going to want to be at the lowest energy place possible and none of them are going to be up on the table. If we raise the temperature up enough, though, we're going to be able to get a few snakes up on the table. We'll never be able to raise the temperature to the point where they'd favor being on the table, because now entropy favors down here. So no matter what we do, we can't ever make products favored - if this is now reactants and products. We can't ever make products favored, but if we get our temperature up enough, at least we can shift the equilibrium in the same way we did before, for an endothermic reaction, to get a few of them, at least, to go to products. You'll never be able to get an equilibrium constant greater than one though - or even equal to one. It's always going to favor reactants.
So now, let's say that mathematically. We'll go back to the equation and this looks a little bit overwhelming at first. But this is the equation that we start with. Now what it says is if we have an endothermic reaction, like our original analogy, H is going to be positive. Now, what we're going to do to make it easier to describe this reaction is instead of plotting K versus T, which is kind of what we're most interested in, mathematically, it will be easier to plot the natural log of K and , because that's going to give us a straight line then. That gives me a form of Y = MX + B. So we have a slope and an intercept then. Our slope will just be and our intercept will be .
So this is the situation where we have an endothermic reaction. And what it tells us is that at very low temperatures - now remember, this is 1 over T, so low temperatures are on this side. At low temperatures, for an endothermic reaction, we're going to heavily favor the reactants. But even if we go to high enough temperatures, as long as we have a S that's negative - in other words, as long as the products are not as ordered as the reactants - then, no matter what we do, although we can get more products, we'll never get to a point where the products are favored, no matter how high we make the temperature. But, if we have an exothermic reaction, then at low temperatures, we actually could reach a point where the products are favored. This dashed line here corresponds to natural log K of zero or K, the equilibrium constant, equal to 1. That's the break-even point, if you will.
So, if it's exothermic and we raise our temperatures, the reaction is going to shift towards the reactants, in accord with Le Châtlier's principle, and eventually, reactants are going to be favored. But at very low temperatures, the products would be favored. So under this circumstance, we can actually change whether products or reactants are favored, whereas again, in the endothermic case, no matter what we do, we'll never be able to favor the products. But the other point I want to come back to is that when we raise temperature, that only talks about how the equilibrium will shift, not what the absolute equilibrium is.
So again, I point out, we can make it shift towards product, but we'll never get to the point where products are favored. It just means that we can move it in that direction.
Now finally, let's look at the reverse situation. This time I have a situation where the entropy is positive, meaning we have more entropy in the products than we do in the reactants. For that situation, if we have an endothermic reaction, at very low temperatures, we actually could favor the reactants, because at very low temperatures, just the fact that it's endothermic - it costs a lot of energy - we're not going to be able to get the products favored. However, if we raise our temperature enough and S is positive - there's more entropy in the products - then the products become favored at very high temperatures. Remember what we said before. High temperatures favors entropy more over S. Low temperatures favor H more relative over S. So as we raise temperature, first of all, notice the slope. We're increasing temperature for an endothermic reaction. The reaction is going to shift towards the products. And in this case, it will shift enough to actually favor the products - in some cases by a great deal. And it's because the entropy for the products, now, is more in disorder in the products than in the reactant. In contrast, finally, if we have the last situation, an exothermic reaction, then no matter what we do, H favors the products, S favors the products. We'll never be able to get an equilibrium constant favoring reactants no matter what we do to temperature. But we can see the same shift in the equilibrium constant, as predicted by Le Châtelier's principle. So if we're here and we go from low temperature to high temperature, it will shift towards reactants, but it will never get to reactants - it will never favor reactants. It will just start to move more in that direction. The equilibrium point will shift but never will reach a point that the equilibrium constant is less than 1, favoring reactants at equilibrium. So this, in my experience, is one of the hardest things for students to differentiate between - where the equilibrium is versus which direction the equilibrium shifts when you change the temperature - very, very different ideas.
Okay, just to illustrate, in this point, with this last example we looked at, the equilibrium will always favor products. However, if we raise the temperature, we don't favor the products by quite as much and the shift is definitely towards reactants, even though we never favor reactants. So that's a tough idea to get a hold of. And so if this helps somewhat - wonderful. If not, you may want to look at this tutorial once more, twice more, five times more. But this is, in my estimation, one of the most important and difficult concepts in what you're learning in general chemistry.
We'll go ahead and stop at this point, but to remind you, what we've done is to link the ideas of entropy and enthalpy to equilibrium and discuss how temperature causes that balance to change.
Thermodynamics
Using Free Energy
Enthalp and Entropy Contributions to K Page [1 of 3]

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