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Le Chatelier
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0:02
Le Chatelier's principle is what this lesson is all about. He gets credit for the concepts that we're going to look at. This is a very descriptive discussion. We're not going to be in to the world of calculations, we're just going to be examining the equilibria and Le Chatelier's principle is all about taking a system that's in equilibrium, disturbing that equilibrium, and seeing how the equilibrium or the reaction responds to get back to equilibrium. Let me give
0:32
you an a common example. To go to a restaurant where it's a really busy night, so Friday night, it's a very popular restaurant, they don't take reservations, and you're sitting in the lobby waiting to go in. We've talked about this equilibrium every time somebody leaves somebody will enter and you have this exchange. If you're looking at it from above down at the reaction, which is the people in there, it looks like the lobby is full, it
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looks like the restaurant is full all the time. Let's say that this equilibrium has been established and it was established when one of the server's wasn't at work yet, showing up late, so they had several tables that were closed down and nobody could be seated at those tables. Finally this person comes running in is ready to start serving. Well what's going to happen for a little bit? That was a disruption of the equilibrium. Now a good disruption if you're out in the
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lobby wanting to get in, but for a while you could have more people entering then exiting. You're going to have a more rapid entrance and slower exiting. I mean exiting is really not changing its speed, but the entering into the restaurant, having a seat is happening at a greater rate until those spots are filled and then we're back to an equilibrium. If the opposite happens in your equation system is such that everybody is entering and leaving and
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somebody over in the back corner ends up having a big food fight, I don't know and it closes down four or five tables. For a while you're going to have a much more big exiting happening nobody's going to be coming in and sitting in those seats. It disrupts that equilibrium and there's going to be more rapid exit than there's going to be entrance until the equilibrium gets reestablished. We're going to be looking at chemistry examples of this,
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chemical equilibria. This balance is very delicate and can get knocked out quite easily, but it will work to re-establish that equilibrium. The conditions that we're going to see, whatever it happens, we're going to be able to predict will this reaction proceed towards products towards the right in order to get to equilibrium or will it go towards reactants, towards the left, in order to get to equilibrium and that's what you're going to be saying.
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Which way will it proceed to get to equilibrium? Here is the statement of Le Chatelier's principles that you can get down into your notes. If an external stress is applied to a system at equilibrium, so you're stressing it, knocking it out of equilibrium, the system will adjust in such a way to try to relieve that stress. So that's it, you're trying to relieve the stress that got created in order to get back into equilibrium. We're going to look
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at three possible scenarios I think here. One is a changing concentration. If you go and dump more in and bring the concentration up to something that was in equilibrium, it's not going to like it. Or you remove something from the reaction chamber. It's not going to like that. It's changing the concentration. Changing in pressure and volume. This would be if the gases are involved. When you squeeze in and compress it, the pressure is going to go up. It's not gonna
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like that stress. If you're going to expand it, the pressure is going to drop. It's not going to like that stress so it's going to adjust to get back to where it was. The last one is changes in temperature. When you change the temperature, it is going to knock it out of equilibrium and it has to reestablish. We're going to look at those three scenarios one at a time. Let's start with changes in concentration. We see here back to my equilibrium between N2O4 and
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NO2 and I chose one of those pictures we had earlier as we started this unit. The dark blue, the first dark blue patch we have there is it's in equilibrium, but what happened here? Let me go over to the screen and write on it. What's happening here is we're dumping in and increasing the amount of NO2, so we're putting in a bunch of I mean sorry in N2O4. Well, that's bringing this way up. Bringing more of that in, we've dumped some in. Well it's not going to
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like that, so it's going to start using it up. It's going to use it up and produce some NO2. So what's going to happen? I'm going to be dropping this down as I use it up. I'm going to be producing NO2 as it's moving in that direction. Now, does it go back to the exact same values that it had before? No, it is at very different values, but it's going to go back to its same K. We know that the Kc value is the concentration of NO2
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squared over the concentration of N2O4. It's not going back to the same concentrations, but it's going back to the same equilibrium constant. It's trying to get back to its target. This is its target, that K value and it'll shift to do that. That's how adding something will affect it. When a substance (this is a key statement) when a substance in a reaction is added, what's it going to do, it's going to shift so it can use up some of that
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added substance. It's not always right or always left. Let's go back to this reaction here. In this case, we put in the N2O4. It wanted to use it up, so it shifted it to the right. But the statement doesn't say...oops it's going the wrong direction. The statement doesn't say it always shifts to the right, it says it's going to shift to try to use up what was added. The reverse obviously would be true. If you take something out, it's not going
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to like that anymore. What is it going to do? It's going to shift to form some more of that substance. It's always just trying to alleviate the stress and get back to the same equilibrium constant value. Let's look at an example. Here is an equilibrium. That equilibrium has been established. It's got some K value that it likes and then the forward and the reverse reaction are happening at the same exact rate. Let's think what's going to happen. Which way is this going to shift if you go and you
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add some NO2. You're adding NO2. Well, if you add NO2, it's going to shift to the left because it's going to try to use up what it added. Let's try another one. What if you remove something? Let's remove some of the Br2. So we're taking out Br2 out of the system. Maybe we're dropping the temperature so it's liquefying it so it's no longer bromine gas in there and we're being able to pull that bromine off out of the
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system. What is it going to do? Well, it's going to try to make more so it's going to shift to the right to try to produce some more. Here is a situation where I've got this reaction that we were looking at earlier and I have removed some N2O4. When I remove it and take it out, it's going to shift to try to make more. So, I see it's coming up as it makes more. As it makes more of the N2O4, it is using up some of the NO2 until it gets back into equilibrium. Does
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it get back to the exact same values? No, it just gets back to the same K. The products, NO2 squared, over reactants raised to the power of their coefficients. I's got to get back to the same K value, not back to the same exact concentrations, but you can figure out which way it has to shift to try to achieve that value by knowing that when you remove something, it tries to make more. Alright, changes in volume and pressure,
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that's the next thing to look at. If you're dealing with gases, as the volume goes down, what happens to pressure? This was Boyle's law. We learned about that in the gas chapter. As the volume goes down, the pressure goes up. Alright, P1 V1 equals P2 V2. This is what's happening here. When the pressure goes up or the volume goes down consequently and there must be gases involved for this to be a thing. So your pressure goes up, it's going to
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shift in the direction where there's less gas moles. Less gas molecules. If you have less gas, then you have a lower pressure. That's going to relieve the extra pressure formed. So, opposite statement. When the pressure goes down or the volume goes up, this is a vice-versa, the equilibrium will shift in the direction of more moles of gas to relieve the stress. You might think well the pressure goes down, the stress is relieved. No, it's not going
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to like that anymore. It liked the pressure it had before and it will shift to try to get back to the same Kp value. It can't get back to the exact same pressures, but it'll try to get back to the same Kp value. Let's see if we can apply that one. Let's have a couple exceptions. Let's have a situation where maybe we've got A gas plus B gas going to 2C gas. The delta N is zero. I have two moles of gas on this side. I have two moles of
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gas on this side. Two minus two is zero. If there's no change in the moles of gas, then shifting is not going to help it at all. Increasing the pressure, decreasing the pressure by changing the volume is not going to have any effect on the equilibrium because shifting is not going to be able to cause any relief of that stress. The other one that we see is sometimes you might see a scenario where they go and dump in an inert gas. Let's say we have this system in a
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reaction chamber at equilibrium and I go dump in some argon gas. Argon is a noble gas. It's going to increase the pressure of the whole chamber, but it's not going to change the partial pressure of these individual gases in the chamber. It will have no effect on it. While pressure generally shifts towards the side with less moles, here are two scenarios where it's not going to change the equilibrium at all. Here is our example. Consider this
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reaction. I want to go into this system and I'm going to compress it. What does that mean? The volume is going down. What is going to happen to this system? If the volume is going down, the pressure is going to go up and it's going to shift towards the side with fewer moles of gas. Which side has fewer moles of gas? Well this side has three moles of gas and this...oops it disappeared on me... this side has two moles of gas, so it is going to shift to
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produce more of the SO3. The last possible scenario is changes in temperature. To determine the effect of change in temperature, you must know if the reaction is exothermic or endothermic. It's as it's written. We know these are reversible reactions, so they can be written either way and change whether it's endothermic or exothermic, but you're looking at a reaction and you've got to know whether it's endothermic or exothermic. Here's
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the statement and it's not what I use, but it's a true statement. I'll show you how I actually think about it. A temperature increase favors an endothermic reaction. You pump in heat, it's going to go in favor an endothermic reaction. A decrease in temperature is going to favor the exothermic reaction. Now once you have one learned, you don't you know the other one is just the opposite. This is true and this is the way it works, but this is how I think
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about it instead. To help me remember this, what I do is I write heat as a reactant if it's endothermic because I'm putting in heat or I write it as a product if it's exothermic. Let's give you an example up here at the top. Let's say I have this reaction, A gas going to B gas and it has a delta H of 50 kilojoules per mole. For that reaction, 50 kilojoules. Alright, is that endothermic or is that exothermic? A positive delta H is endothermic, so I'm going to write heat
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as going in. It's a reactant, it's going in. That's the first statement and then I think of raising the temperature as adding heat and then I'm back to my concentration argument. If I add this, it's going to want to use it up so it's going to shift to use it up. It would shift it to the right. Raising the temperature favors an endothermic reaction. If I lower the temperature, it says they are thinking lowering the temperature as removing heat. If I take heat out by
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lowering the temperature, it's going to try to make more. It's going to go in the opposite direction. Going in the opposite direction is an exothermic reaction. So, that fits the statement that I had earlier, but I think of it this way. Here is your example to try. First of all, I have a reaction written up there. Is this reaction endothermic or is it exothermic? It is exothermic. If it's exothermic, it is giving off heat and that's a negative
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value. I can write plus heat on the product side. Heat is being produced, given off. Now I want to think about this which way will equilibrium shift in this case if the temperature is increased. Will it shift to the right or to the left? Hopefully you said to the left. If you are increasing the temperature, you're adding heat. It will shift to try to use it up. It'll shift towards the product I mean the reactants. Let's ask this question -
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does that match the statement and this may come up later I thought I had it up here. Does that match the statement that we had earlier which said that adding heat favors the endothermic reaction? Well this reaction is exothermic as written in this direction. If we add heat, we're pushing it in the opposite direction which is in the direction of being endothermic right. Because as written it's exothermic. When temperature has changed, really what's
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happening is it changes the value of the equilibrium constant. In our other two scenarios of changing pressure and changing concentration, the K value didn't change. You knocked it out of equilibrium unit try to get back to that target, but in this scenario the K actually changes and it has a very different K and it has to get and find that K value so it shifts to get there. Let's think of this reaction. We knew that raising of the the temperature
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shifted this reaction to the left because heat was a product. We added heat When we added the heat, when we raise the temperature, it shifted it to the left. Since K is products over reactants, what is going to happen to the magnitude of K as you shift it to the left? We know that K is going to be N2 times O2 over NO squared. Now we can do it in terms of concentrations or pressures. If it's shifting it to the left, what is it going to do to it? You're raising this
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number. You're lowering the numerator, raising the denominator so K is going to decrease. As you increase the temperature, the K will decrease for an exothermic reaction.
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Remember that statement that a temperature increase favors an endothermic reaction and a temperature decrease favors an exothermic reaction. This is exothermic as written. When we raise the temperature, it is pushing it in the endothermic direction, so it fits that that bill. You can't know raising the temperature what it's going to do to its K value until you know whether it's endothermic or exothermic. If heat we'e on the other side, if heat
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were here, we'll put a big if because it's not, raising the temperature would shift it to the right and it would make the K get bigger instead. Now that takes us through Le Chatelier's principle. We have three scenarios to look at. Changing in concentration, we know it's going to shift away from what we add or towards what we use up or what we take out. We have changing pressure and how it'll shift to relieve that stress. We have changing temperature and
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seeing which way it'll shift to get back to its new K value. That's Le Chatlier's principle. Now what's up next? It's a very important lesson where we're going to start just working a lot of different calculations. I'm going to model for you how to deal with problems. We have done several already with our unit where we're trying to calculate our K value. We're trying to determine the equilibrium constant and there's a lot of varieties of how that can work, but we
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know we just need to know products over reactants raised to power their coefficients and we'll be able to get those values. Then there is the opposite. You know the K, but you want to obtain the equilibrium concentrations. It's a little bit more involved. This is a very important unit and you need to watch it when you're ready to really focus hard on what I deal with and give yourself time to start and stop and think about my word. I just want to warn you when
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you go to our next lesson that you need to give yourself plenty of time, a fresh mind, and not be rushing through it because we're learning it here but we're going to using it over the next couple of chapters as well and we want to get it earlier so that we're feeling good about it as we proceed.
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