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LeChatelier's Principle


Let's talk about the Le Chatelier's principle and how we can affect the position of equilibrium and the balance between reactants and products. So Le Chatelier's principle basically says that a system a chemical system is gonna try to fix whatever you do to it. So what kinds of systems stresses can we create? Well, the ways that we can stress the system, there's a couple, and the easiest way to visualize them is with a little seesaw and the first way we'll talk about is if you were to add reactants.

This is gonna make the fundamental issue is that Q the reaction quotient becomes too small, right? Because we said that the law of mass action equilibrium constant is basically equal to products over reactants. So if you change the value of the reactants, then you will no longer be at equilibrium and the overall fraction will get smaller.

So that's the fundamental issue. But visually, it's much faster just look at it and say, we started off balanced with, we'll call these reactant boxes and product boxes. And we're gonna say that if you add reactants, it's like putting another box onto this side. So is this seesaw balanced?

You can probably guess and see it for yourself that visually, nope it's not, right? So what would happen is it would shift like this, right? To balance the system, Le Chatelier's principle says to fix this, what we're going to do is we're going to move some of these reactants to products. We're gonna perform the Ford reaction, we're gonna shift to the right.

So we say the the fix here is we shift to the right. Aka, we're going to make products and yeah, okay, so we have two and a half boxes and two and a half boxes. Boom seesaw is balanced. So the next one to tackle is essentially the opposite. If you take out reactants Q gets too large but what does the seesaw say?

Well, the seesaw says that if we remove reactants, let's go ahead and take one way. That's clearly not balanced. So how do we fix it? Well, some product has to break down and go backwards to reactants. So now, take half product box put over here and now I have two and a half on both sides, right?

So to balance this out when you remove reactants, we have to shift to the reactants inside, we say we shift left. All right what are the other changes? Well we could add products or remove products which again create very similar issues with Q and the fix is, let's take a look at the seesaw. Well the seesaw says that for the first one if we had more products, the way to balance it out is to shift left, right?

We have to take some of this and move it over. So this would be a shift left to reactants. If we remove products, take them away and we're gonna say, well, there's not enough now we have to take some react and move it over. So now we're gonna shift to the right. So these changes all kinda makes sense when you think about them from a seesaw's point of view.

The next change is gonna be a little bit different. So these changes are different because we're talking now about pressure and we can't treat pressure just with the seesaw. It's always gonna depend on the exact equation you're given, so let's work from an equation. Use this one.

When you decrease pressure, the issue is, there is not enough pressure, you need more. So how do we make that work? We make that work by shifting to the side, that we'll have more pressure and what is pressure? Well, it's collisions between molecules.

So which side has more collisions? Aka, which side has more moles of gas, so it shifts the side with more moles of gas. The reactor inside has a total of 1, 2, 3, 4 moles of gas, the product side has two mils of gas. So in this particular situation, when we decrease the pressure we will shift left because that is the side that has more moles of gas, okay?

Don't get stuck in thinking left or right when it comes to pressure, think where are the moles of gas that I need to fix the problem. So here I decreased the pressure, I got more pressure by going to the side with more moles. What happens if you increase pressure? Well, it's gonna be the same but different, right?

If we increase pressure now the primary issue is, theirs too much. So how do we decrease the pressure? We shift the sides with less moles of gas. So we shift to the side with less moles of gas and that will solve our problem. There's tricky language that can be used.

So instead of saying decrease pressure or increased pressure, you might see increase volume or decrease volume. So if you increase the volume of a container that will decrease the pressure because the molecule spread out. If you decrease the volume of a container that will increase the pressure because the molecules are pushed closer.

Another way you might see it is as a piston that is either expanding or compressing. These would also be ways to change the pressure. So let's try a question. The human body maintains homeostasis despite many chemical disturbances. PH is regulated via several chemicals but in an important one is bicarbonate ion.

What will happen to the concentration of H+ in the following system, if sodium bicarbonate is added? A, H+ will increase. B, H+ will decrease. C, H+ will remain constant. D, the H+ change will depend on the amount of sodium bicarbonate removed.

Go ahead and pause the video, give it a shot and then come on back. So we just need to treat this like a seesaw. What's gonna happen to H+? So we're looking to see what happens here, if sodium bicarbonate is added. So we're gonna increase the left hand side right? So we have some system that's balanced.

If I add more reactants, well now we have too much so it's gonna shift to the right to balance things back out. In order to shift to the right, we have to use up some hydrogen ion and some of the sodium bicarbonate that we added. So overall The H+ ion will decrease, because we're shifting away from it, we're going to the right hand side.

Let's try a similar question, but it is a little different, so be careful. The human body maintains homeostasis despite many chemical disturbances. PH is regulated via several chemicals, but an important one is bicarbonate ion. In the following system at equilibrium, what will happen with the concentration of carbon dioxide if the PH is lowered? A, additional CO2 will be produced.

B, less CO2 will be produced. C, carbon dioxide is a different phase than the acid and will be unaffected. D, changes in heterogeneous systems cannot be assessed. Go ahead and pause the video, try it and come on back for the answers. So in this one, we're looking at carbon dioxide which is a gas so this is true and we're talking about if the pH is lowered.

So when pH is lowered you need to remember what is pH it's the measurement of H+ ion. That means that when the pH goes down, the hydrogen ion concentration is actually increasing. So we're adding over here, take a quick look at the seesaw. Well, if we have a balanced situation and I add more reactants, we still shift to the right just like the previous problem but the wording now is a little bit different.

So carbon dioxide will have more of it produced. So we'll have additional CO2 being produced. When it comes to temperature, we again have a situation that's kind of similar to pressure, where you don't always know exactly what's going to happen. But the good news is we can use the seesaw here as a mnemonic. So let's take a look and see what happens when we increase temperature.

If I increase temperature, that's gonna be like quote unquote, it's not really adding heat, okay? Now heat is not normally a part of a reaction unless we're talking about endothermic reactions or exothermic reactions. As a quick refresher endothermic reactions have a positive delta H, and exothermic reactions are negative.

For endothermic reaction, essentially heat is going in. So is that heat acts like a reactant. So heat is a reactant for endothermic and exothermic reaction heat is coming out. So heat is like a product. All right, so it I add heat to an endothermic reaction that would like this, let's say we had say A + B yields or it's in equilibrium with C, if it's endothermal, delta H = +10.

Heat + A + B yield C is our reaction. So for this reaction, if I increase the temperature, I add heat, it's like adding reactants, which on the seesaw, we said looks like this. Well now this is unbalanced, to fix it, I have to shift to the right. So the fix when you have an increased temperature on an endothermic reaction, is to shift to the product side.

If I decrease temperature again, let's decrease temperature let's do it on a endothermic reaction. It's going to look like this. So I'm decreasing temperature, I'm taking away heat, so I'm gonna have less reactant. So how do I fix this?

I'm gonna shift over here to the left. So we can predict easily if we know that heat can be treated as a reactant or a product from Le Chatelier's principle, exothermic reactions are the opposite, right? For an exothermic reaction, heat would be a product and then you would add or subtract from the seesaw based on what's happening with temperature to figure out those switches.

All right, so at this point, we've talked about how to change the concentrations of reactants and products by adding or removing several things, changing pressure, and changing the temperature. However, there's two things you should be aware of that can change that do not do anything. And you might think that they do, right?

They're kind of deceptive. These two things are adding a catalyst which we know makes reactions go faster and adding a noble gas element to a mixture. So if we add a catalyst, think what what a catalyst does. It increases the rate of reaction. However it increases the rate of both the forward and the reverse reactions, both directions get faster.

So catalysts will help you get two equilibrium faster, but they don't change the position of the equilibrium and they don't change the concentration of any system that's at equilibrium. The noble gas change is a little bit more deceptive. So let's take a slightly closer look at that, and then see how it works out. To investigate it, we have to have an equation, boom.

If we add a noble gas element, well usually, the question will go something like this. They'll say, you have maybe 2 atmospheres of nitrogen, 3 atmospheres of oxygen and 2 atmospheres again NO2. And will say that the system is that equilibrium and that the equilibrium constant is equal to whatever this works out to be.

Then they'll say the change on the system the stress on the system is going to add 0.5 atmospheres of neon gas to the system. So quite often people think, we've increased pressure, we move to the side with less pressure but this is neon, it's noble gas. Think about what happens because of pressure? You get collisions between molecules.

So when we add neon we'll get collisions with neon, right? Nitrogen hit neon, oxygen hits neon, NO2 hits neon but neon is a noble gas. Neon doesn't react, so you increase the pressure but you can't produce any more molecules of product and you can't break down the product using neon. It's just not in the equation. So when you add a noble gas to assist on the equilibrium It doesn't do anything.

It just changes the total pressure in the container, but it doesn't change the pressure of any of the individual parts. So the position of equilibrium will not change. Let's try another question. Fluoride in toothpaste bonds strongly to enamel, however,if ingested it may bond strongly to other chemicals within the body.

In the following equilibrium system containing fluoride ion, what will be the effect of adding Calcium fluoride? A, the concentration of sodium fluoride will increase. B, the concentration of sodium chloride will increase. C, no change in concentrations will be observed. D, the concentration of calcium chloride will decrease.

Go ahead work on work on it and I'll talk to you about the answer in just a moment. So this one is a little bit trickier. We haven't actually addressed a change in something that's solid. Remember what goes into an equilibrium equation we said it's the products over reactants and we don't include solids. The reason is because solids and liquids, they can't change concentration, they don't affect the position of equilibrium cuz they're always constant.

So even though you add more calcium fluoride, the actual concentration doesn't increase, which is weird, right? Cuz you've added it, but keep in mind the density of calcium fluoride is set, so you don't change the concentration within the system. So because we're adding a solid, we're adding something that is not part of the law of mass action.

No change is gonna be observed. So here we have the summary for this lesson. As long as you've absorbed all the major points here, you understand how to use the seesaw, and you can understand the predictions that it produces, you're all set.

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