Understanding Le Chatelier's Principle

What we're going to look at next is how reactions respond to change. And this all falls under everyone's favorite Frenchman Le Chatelier's principle. So something that we know, we know that a system not at equilibrium will move toward equilibrium. So what happens if we have a reaction at equilibrium and we disturb it or perturb it or... put some pressure on it in some way. So this is what Le Chatelier's principle covers. We will do this first qualitatively, but we will also do it quantitatively. And Le Chatelier's principle says when a chemical system at equilibrium is disturbed, the system shifts to minimize the disturbance and return to equilibrium. So the important parts we have to start at equilibrium. Then we do something to it. Well, when we do something to it, it's now no longer at equilibrium. And we're back to what we came into this video knowing, which was that a system not at equilibrium will move toward equilibrium. So the reaction after we have done something to disturb it will return back to equilibrium from whatever this new place is that we have ended up in based on the type of disturbance. So I have a reminder here first that is going to be give you some space to practice with calculations. And then we will talk about the different types of disturbances to reactions and how that affects things. So the reminder is that for any reaction, the equilibrium concentrations can change or be different. But the equilibrium constant is a constant. So the thing that matters is the K value, the Kc, the Kp for the reaction as written at a given temperature. Now the exact amounts could be different. So I have an enormous table here for the reaction 2SO2 gas plus O2 in equilibrium with 2SO3 gas at 1000 Kelvin. The Kp is 0.0415. So in this table, what I have for you is three different trials. So each row is a trial. I have a column for each of the reactants and products with some of the pressures filled in. And the KP is completely filled in, right? The KP is 0.0415 no matter what. So what I would like you to practice doing is... write the equilibrium constant expression for this reaction and show that all of these values are correct. So show that the different pressures listed all go to a KP of 0.0415. So let's talk about the different disturbances. You could add or remove a reactant or a product. So this is essentially it's changing the concentration or changing the pressure. So I'm going to do N2O4 in equilibrium with 2NO2 as my reaction that I'm looking at. You'll notice it's just the reverse version of the reaction that we looked at in an earlier video, but now N2O4 is on the left, so it's my reactant. Okay, so if this reaction is at equilibrium and I one, my first thing, I add NO2 to the flask. So I increase the amount of product. So in my balance of products over reactants, I have too much product to be at equilibrium. So in order to get to equilibrium, it's going to use up some of the NO2. So the reaction would shift In this case, to the left, it would use up NO2 to make N2O4, and it would go until it's at equilibrium. Now, numerically or mathematically, if you like to think about this with the numbers instead of, I mean, it's still qualitatively with the numbers, but if you have a better feel for the numbers. Products over reactants, if I'm adding NO2, I'm increasing the numerator. And K is a constant, so the only way to get the ratio of products over reactants, the numerator over the denominator, to be back to this one constant value, well, if I've increased the numerator, the only way for it to do it is if I increase the denominator. But since it's a chemical reaction, the only way to increase the denominator is by taking from the numerator. So the amount that I take from the top and add to the bottom will vary depending on how far we've pushed it away from the equilibrium, but mathematically that's what we're doing. Okay, so let's imagine if we started from equilibrium and then removed NO2. So we're decreasing the amount of product. In order to get back to equilibrium, the reaction makes more product by shifting to the right. So my reaction shifts to the right, which uses up some product. Reactant makes product and would put me also mathematically back into balance. What if I added N2O4? So if I increase the amount of a reactant, I now have too much reactant, too high a value in my denominator. the reaction would shift right and create more product and decrease the amount of reactant in order to get back to equilibrium. What if I removed N2O4? So if I decreased the amount of reactant, my reaction would shift to the left. So it would use up some of the product and make some reactant in order to get back to equilibrium. Okay, what if we changed the pressure? and the volume. So I have pressure and volume here because within the context of the ideal gas law these are inversely related. So if I decrease the volume the result is an increase in pressure and if I increase the volume the result is a decrease in pressure. So when a change is a change in volume We have to use the resulting change in pressure to determine what would happen to our reaction, because the pressure is how we are communicating within the equilibrium constant and the equilibrium constant expression. And even if we're not doing it directly through pressure, right, the moles and volume concentration was directly related to the pressure. All right, so Let's look at this with a sample reaction. If I have N2 plus 3H2 in equilibrium with 2NH3, I've also kind of broken this down. Like let's picture the individual molecules. So my molecule of N2 with the two black atoms, my molecules of H2 with the two green atoms, and then my products, two molecules of NH3, which right? Molecules scale it up. We're really talking about moles of gas. On the left, I have a total of four moles of gas. One is N2, three are H2. On the right, I have two moles of gas. Both are NH3. So what I'm imagining is if I have this cylinder with a piston with a reaction inside of it, I'm at equilibrium. On the left side, if I pull the piston up and increase the volume, so my reaction has a bigger volume to move around in, an increase in the volume is a decrease in the pressure. When the pressure decreases, the thing I'm trying to do in correcting it is I need to increase the pressure. So we did one thing, whatever we're doing. according to Le Chatelier's principle, is to undo what was done. So I decreased the pressure, my reaction is going to increase the pressure. The way that the pressure increases, because p is nRT over v, my only option is to increase the number of moles. So the way the reaction shifts would be to the left, because it has a higher number of moles. On the right side here if I do the opposite, so I push this piston down I decrease the volume. A decrease in volume will result as an increase in pressure. So I've increased the pressure. My reaction will shift in the direction that decreases the pressure, which is the least number of moles, which for this reaction is to the right. Now, here's another example. Carbon monoxide plus water in equilibrium with carbon dioxide and H2. This reaction does not respond. to changes in volume or changes in pressure. I mean, we mean the overall pressure of the entire reaction. And the reason it doesn't is because there's nowhere for it to shift. If I increase the pressure and I want to shift to a side with fewer moles, both sides have two moles. If I decrease the volume, decrease the, or, and increase the pressure, I can't. I don't know which one I said first. So the opposite of whatever. It can't decrease or increase the number of moles by shifting either way because it has the same number of moles of gas on both sides. So now the question is, what if I don't change the volume of the container? What if I change the pressure in the container by adding something like argon? So I'm going to add argon gas to my reaction. within this container, and argon is considered an inert gas, right? It's a noble gas. These are things that aren't going to react with anything that we have. What would happen? The pressure inside the container has increased, but we see no change. So the total pressure inside the container increased, but the partial pressures of the gases remained the same. So since the volume remained the same, the components of the reaction are present in the same concentration and at the same pressure that they were at before. So there's no change to the reaction because we may have changed some things about the container that we're looking at, but for this reaction, we have not changed any of its conditions. What about temperature? So we've been giving you temperatures with all of your equilibrium constants, and equilibrium constants are temperature dependent. We'll learn how to calculate that a little bit later. But on a reaction, sometimes heating it up helps, sometimes heating it up doesn't help. So what we need to know about a reaction in order to determine how changing the temperature would affect it. is we need to know if it's endothermic or exothermic. And because of the order in which we teach these things, thermodynamics is after all of this. So while you might know these terms, we haven't explicitly defined them. So I will give you a definition here and what you're going to look for. And then we'll learn more about what endo and exothermic mean and how to get these values from. data about reactions. So if a reaction is exothermic, this means the reaction gives off heat. And it is, this is not perfect. This is just conceptual to help you with figuring out directions for Le Chatelier's principle. But if a reaction is giving off heat, you can kind of think about heat as a product of the reaction. So A plus B in equilibrium with C plus D plus heat. So if this is exothermic, heat is a product. The way that you will see that indicated, if it doesn't say the word exothermic, it will say that delta H is less than zero. So delta H is the change in enthalpy, which again, something that's going to come up in a later module. But if the change in enthalpy is negative. That means it's giving off heat. So for this reaction, if you increase the temperature, which would be the equivalent of adding heat, that's like adding a product. So my reaction would shift left. Now, the result really is that you get a decrease in the equilibrium constant value. But what we would just observe is... that the reaction shifts to the left. For something that is endothermic, it means the reaction absorbs heat as it goes. So this would be like saying A plus B plus heat. in equilibrium with C plus D. So if you increase the temperature, which also, so this would be indicated by saying the change in enthalpy is positive, so anything greater than zero. Increasing the temperature of an endothermic reaction, that's like increasing a reactant, my reaction would shift to the right. Now it's really an increase in the equilibrium constant value, but what we would see is a shift to the right of this reaction. And then I just have a note, final thought reminder here, a change in temperature changes the actual value of the equilibrium constant. And we will learn how to calculate by how much that happens a little bit later.

So this is what Le Chatelier's principle covers. We will do this first qualitatively, but we will also do it quantitatively. And Le Chatelier's principle says when a chemical system at equilibrium is disturbed, the system shifts to minimize the disturbance and return to equilibrium. So the important parts we have to start at equilibrium. Then we do something to it.

Well, when we do something to it, it's now no longer at equilibrium. And we're back to what we came into this video knowing, which was that a system not at equilibrium will move toward equilibrium. So the reaction after we have done something to disturb it will return back to equilibrium from whatever this new place is that we have ended up in based on the type of disturbance. So I have a reminder here first that is going to be give you some space to practice with calculations.

And then we will talk about the different types of disturbances to reactions and how that affects things. So the reminder is that for any reaction, the equilibrium concentrations can change or be different. But the equilibrium constant is a constant.

So the thing that matters is the K value, the Kc, the Kp for the reaction as written at a given temperature. Now the exact amounts could be different. So I have an enormous table here for the reaction 2SO2 gas plus O2 in equilibrium with 2SO3 gas at 1000 Kelvin.

The Kp is 0.0415. So in this table, what I have for you is three different trials. So each row is a trial. I have a column for each of the reactants and products with some of the pressures filled in. And the KP is completely filled in, right?

The KP is 0.0415 no matter what. So what I would like you to practice doing is... write the equilibrium constant expression for this reaction and show that all of these values are correct.

So show that the different pressures listed all go to a KP of 0.0415. So let's talk about the different disturbances. You could add or remove a reactant or a product.

So this is essentially it's changing the concentration or changing the pressure. So I'm going to do N2O4 in equilibrium with 2NO2 as my reaction that I'm looking at. You'll notice it's just the reverse version of the reaction that we looked at in an earlier video, but now N2O4 is on the left, so it's my reactant.

Okay, so if this reaction is at equilibrium and I one, my first thing, I add NO2 to the flask. So I increase the amount of product. So in my balance of products over reactants, I have too much product to be at equilibrium.

So in order to get to equilibrium, it's going to use up some of the NO2. So the reaction would shift In this case, to the left, it would use up NO2 to make N2O4, and it would go until it's at equilibrium. Now, numerically or mathematically, if you like to think about this with the numbers instead of, I mean, it's still qualitatively with the numbers, but if you have a better feel for the numbers.

Products over reactants, if I'm adding NO2, I'm increasing the numerator. And K is a constant, so the only way to get the ratio of products over reactants, the numerator over the denominator, to be back to this one constant value, well, if I've increased the numerator, the only way for it to do it is if I increase the denominator. But since it's a chemical reaction, the only way to increase the denominator is by taking from the numerator. So the amount that I take from the top and add to the bottom will vary depending on how far we've pushed it away from the equilibrium, but mathematically that's what we're doing.

Okay, so let's imagine if we started from equilibrium and then removed NO2. So we're decreasing the amount of product. In order to get back to equilibrium, the reaction makes more product by shifting to the right.

So my reaction shifts to the right, which uses up some product. Reactant makes product and would put me also mathematically back into balance. What if I added N2O4? So if I increase the amount of a reactant, I now have too much reactant, too high a value in my denominator.

the reaction would shift right and create more product and decrease the amount of reactant in order to get back to equilibrium. What if I removed N2O4? So if I decreased the amount of reactant, my reaction would shift to the left.

So it would use up some of the product and make some reactant in order to get back to equilibrium. Okay, what if we changed the pressure? and the volume. So I have pressure and volume here because within the context of the ideal gas law these are inversely related.

So if I decrease the volume the result is an increase in pressure and if I increase the volume the result is a decrease in pressure. So when a change is a change in volume We have to use the resulting change in pressure to determine what would happen to our reaction, because the pressure is how we are communicating within the equilibrium constant and the equilibrium constant expression. And even if we're not doing it directly through pressure, right, the moles and volume concentration was directly related to the pressure.

All right, so Let's look at this with a sample reaction. If I have N2 plus 3H2 in equilibrium with 2NH3, I've also kind of broken this down. Like let's picture the individual molecules. So my molecule of N2 with the two black atoms, my molecules of H2 with the two green atoms, and then my products, two molecules of NH3, which right?

Molecules scale it up. We're really talking about moles of gas. On the left, I have a total of four moles of gas. One is N2, three are H2. On the right, I have two moles of gas.

Both are NH3. So what I'm imagining is if I have this cylinder with a piston with a reaction inside of it, I'm at equilibrium. On the left side, if I pull the piston up and increase the volume, so my reaction has a bigger volume to move around in, an increase in the volume is a decrease in the pressure.

When the pressure decreases, the thing I'm trying to do in correcting it is I need to increase the pressure. So we did one thing, whatever we're doing. according to Le Chatelier's principle, is to undo what was done. So I decreased the pressure, my reaction is going to increase the pressure. The way that the pressure increases, because p is nRT over v, my only option is to increase the number of moles.

So the way the reaction shifts would be to the left, because it has a higher number of moles. On the right side here if I do the opposite, so I push this piston down I decrease the volume. A decrease in volume will result as an increase in pressure. So I've increased the pressure.

My reaction will shift in the direction that decreases the pressure, which is the least number of moles, which for this reaction is to the right. Now, here's another example. Carbon monoxide plus water in equilibrium with carbon dioxide and H2.

This reaction does not respond. to changes in volume or changes in pressure. I mean, we mean the overall pressure of the entire reaction. And the reason it doesn't is because there's nowhere for it to shift. If I increase the pressure and I want to shift to a side with fewer moles, both sides have two moles.

If I decrease the volume, decrease the, or, and increase the pressure, I can't. I don't know which one I said first. So the opposite of whatever. It can't decrease or increase the number of moles by shifting either way because it has the same number of moles of gas on both sides. So now the question is, what if I don't change the volume of the container?

What if I change the pressure in the container by adding something like argon? So I'm going to add argon gas to my reaction. within this container, and argon is considered an inert gas, right? It's a noble gas.

These are things that aren't going to react with anything that we have. What would happen? The pressure inside the container has increased, but we see no change. So the total pressure inside the container increased, but the partial pressures of the gases remained the same.

So since the volume remained the same, the components of the reaction are present in the same concentration and at the same pressure that they were at before. So there's no change to the reaction because we may have changed some things about the container that we're looking at, but for this reaction, we have not changed any of its conditions. What about temperature?

So we've been giving you temperatures with all of your equilibrium constants, and equilibrium constants are temperature dependent. We'll learn how to calculate that a little bit later. But on a reaction, sometimes heating it up helps, sometimes heating it up doesn't help. So what we need to know about a reaction in order to determine how changing the temperature would affect it.

is we need to know if it's endothermic or exothermic. And because of the order in which we teach these things, thermodynamics is after all of this. So while you might know these terms, we haven't explicitly defined them.

So I will give you a definition here and what you're going to look for. And then we'll learn more about what endo and exothermic mean and how to get these values from. data about reactions. So if a reaction is exothermic, this means the reaction gives off heat.

And it is, this is not perfect. This is just conceptual to help you with figuring out directions for Le Chatelier's principle. But if a reaction is giving off heat, you can kind of think about heat as a product of the reaction.

So A plus B in equilibrium with C plus D plus heat. So if this is exothermic, heat is a product. The way that you will see that indicated, if it doesn't say the word exothermic, it will say that delta H is less than zero.

So delta H is the change in enthalpy, which again, something that's going to come up in a later module. But if the change in enthalpy is negative. That means it's giving off heat. So for this reaction, if you increase the temperature, which would be the equivalent of adding heat, that's like adding a product.

So my reaction would shift left. Now, the result really is that you get a decrease in the equilibrium constant value. But what we would just observe is...

that the reaction shifts to the left. For something that is endothermic, it means the reaction absorbs heat as it goes. So this would be like saying A plus B plus heat.

in equilibrium with C plus D. So if you increase the temperature, which also, so this would be indicated by saying the change in enthalpy is positive, so anything greater than zero. Increasing the temperature of an endothermic reaction, that's like increasing a reactant, my reaction would shift to the right.

Now it's really an increase in the equilibrium constant value, but what we would see is a shift to the right of this reaction. And then I just have a note, final thought reminder here, a change in temperature changes the actual value of the equilibrium constant. And we will learn how to calculate by how much that happens a little bit later.

Transcript for:Understanding Le Chatelier's Principle

Transcript for:
Understanding Le Chatelier's Principle