Understanding Le Chatelier's Principle

This is the presentation for section 13.3. We're going to be taking a look at shifting equilibrium or Le Chatelier's principle. Remember in the previous section we talked about q and k that if an equal if a reaction is equilibrium you can actually shift it away from the equilibrium or create a new or create a q. So we're going to take a look at Le Chatelier's principle. which basically says that if you apply a stress to a system or to a reaction you can shift the equilibrium or create a q okay so the classic reaction we're going to take a look at is nitrogen plus hydrogen gives ammonia okay if we add ammonia the equilibrium shifts left to offset the stress So what Le Chatelier said was that if a chemical equilibrium is disturbed by adding a reactant or a product, the previous example we saw adding a product, the system will shift in the direction that consumes the added species. Well, this sounds kind of complicated. I prefer to think about this that when we add a reactant or a product, it pushes it to the opposite side. If we remove a chemical, or remove if a chemical at equilibrium is removed then it's going to restore part of the added species or what we're going to find is removing is a pull to the side that it's on. So let's take a look at a couple examples. We're going to kind of concentrate on this N2O4 gives NO2 as a gas. And so adding N2O4, that's an adding. That's going to be a push. So it's going to push it from left to right. If we add the NO2, then it's going to go from right back to left. Notice adding NO2, adding reactant, creates more product. That kind of makes sense. Adding a product creates more reactant. We don't necessarily do this on purpose, and we'll talk about why or how that happens a little bit in a couple slides from now. If we remove the N204, we're going to cause a shift, or we're going to pull it to the left. And if we remove the NO2, we're going to pull it to the right, or it's going to go from left to right. Okay, it's possible to calculate how much it's going to change and we'll do that in Chem 104. We don't do that in Chem 103. We can also change the pressure. So when a system is compressed, the total pressure increases causing the system to shift in the direction that decreases the total number of. moles, which means this is going to compress, is going to squeeze it to the smaller number of moles. If we decrease the pressure, then it's going to shift to the side with the greater number of moles. And if we have the same number of moles of gas on both sides, then we're going to see very little effect. So if you notice, we've got one mole of N2O4 on the left and two moles of N02 on the right. Got to be careful, we're counting up the number of moles of gas. So if we decrease the pressure, it's going to expand to the side with a greater number of moles. So it's going to go from left to right. If we increase the pressure, put pressure on it, we're going to squeeze it from the two to the one or from the higher amount to the lesser amount. which means we're going to go from right to left. When we talk about changing the size of the container, which actually changes the pressure, if we decrease the size of the container, it increases the pressure. But I think this is conceptual a little bit easier. If we make the container smaller, it's going to go to the smaller number of moles. So it's going to go from right to left. It's going to go from the two to the one. And if we increase the size of the container, it's going to expand to the greater number of moles. And so it's going to shift from 1 to 2. The other thing we can do is we can change the temperature. And the trick to this is if we put the energy on the appropriate side, we can treat it as a reactant or a product like we did in our first example. Notice the plus 57.2. So I put that on the left-hand side because that's an endothermic reaction. We're adding energy. So increasing the temperature adds energy. So it's going to push it to the right. So it's going to favor producing more NO2. If I decrease the temperature for an exothermic reaction notice it's got the negative negative means it's leaving its loss so that's why i put the 92.2 kilojoules on the right hand side and so if i decrease the temperature it's going to pull it to the right because i'm going to be removing energy okay again decreasing the temperature favors the nh3 you So, so far we've examined adding or removing a reactant or a product, adding is a push, removing is a pull, compressing or expanding a system, we squeeze to the smaller number of moles of gas or expand to the greater moles of gas, and we can change the temperature. So, and remember, we have to assign the energy, whether it's got a negative or positive sign, and then we can treat it as if adding or removing. a reactant or a product. The one thing that we need to realize is changing the temperature changes the K. And that's why a lot of times you're going to see K and you're going to go, I've seen that reaction before, but why does it have a different K? Well, it's at a different temperature. Catalyst is kind of a trick question. because it increases the rate of both the forward and the reverse reaction the same. Remember catalysis lowers the activation energy so it lowers it the same amount. So in essence what it does is we get the equilibrium faster. We go from Q to K a lot faster but ultimately the equilibrium composition's not changed. Chemists were fooled into thinking that the catalyst was changing the equilibrium because the products were being produced. at a faster rate. But ultimately when it was allowed to come to equilibrium, they ended up with the same amount of reactants and products. So let's take a look at a couple examples. Okay, example one, the volume is decreased. The reason there's no change is there's two moles of gas on the left and two moles of gas. on the right. And remember if the same number of moles of gas change the volume or change in the pressure does not change the equilibrium, okay, or change the position. Again, this is kind of one of those trick type questions you got to be careful with. And so if you'll notice, I've put the kilojoules. on the right hand side added the 92 kilojoules to the right hand side so if i heat this this is adding kilojoules which is going to shift it to the left or it's going to push it to the left and number three we're going to remove the so2 remember removing is a pull so it's going to shift or pull it to the left Again, catalyst is a trick question. There's no change. Just remember, a catalyst allows you to reach equilibrium faster. It doesn't shift the equilibrium. And so in example number five, I'm not going to give you this type of question on your quiz or your exam, but I like this one because it kind of goes through between example number five and number six. I'm going to show you how all of these that what we've been talking about applies to this equation. So if you notice, it's got 30 kilojoules per mole, the delta H. So it's got implied that it's positive. So I'm going to put that on the left-hand side. And then I'm going to take a look at all of these. So if I increase the total pressure. By decreasing the volume, so I'm going to squeeze it to the smaller number of moles, I'm going to go from the three moles, the two NOs plus BR, to the two moles of NOBR. If I add an O, I'm going to shift to the left, okay, by adding product. Would I add product directly? No, but if I don't remove this NO as it's being produced, its concentration is going to go up and then it's going to want to push the reaction in the reverse direction. Removing Br2 pulls to the right. One of the ways I can remove the Br2 is I can chill this reaction down so that gas turns into a liquid. And remember, we don't put solids or liquids into our equilibrium expressions and solids and liquids don't affect the position either. If I lower the temperature... I'm removing the kilojoules. So it's going to pull left. And if I remove NOBr selectively, what that means is, is we don't necessarily do it on purpose, but it may be in the mechanism of the reaction, or it may be in how we're performing the reaction that we accidentally remove the NOBr. And what's going to happen is it's going to pull to the left and we're going to get less products. Well, if you'll notice four of these are left and one is right. And so removing the BR is going to be producing more product. Example number six, again, it's just a continuation of number five. It's that same equation. Notice the 30 kilojoules. So instead of reducing the temperature, I'm going to increase the temperature, which is going to push it to the right. Instead of adding NO, this time I'm going to remove NO. So it's going to pull it to the right. Instead of selectively removing the NOBR, I'm going to push right. And if I compress the gas into a smaller amount, again, I like the idea of squish to the smaller moles. It's going to go from three to two, or it's going to go to, it's going to go left. And if you'll notice, I got right, right, right, left. So the answer would be left for this equation. But like I said, I won't ask you a question. I'd ask you about each of these individually, but not combine them together and have you do all this work for just one question, especially on an exam where you only have a limited amount of time to answer the questions. So let's take a look at a couple of questions that come from the end of the chapter. um in your book it says suggest four ways in which the concentration of hydrazine that's rocket fuel take nitrogen and hydrogen to make hydrazine um and if you'll notice i put the 95 kilojoules on the left hand side because it's implied this is there's a positive in front of it by the delta h so i could add n2 would push it to the right add h2 push it to the right i could decrease the container volume or increase the pressure to squish it from three to one moles and i would need to heat the mixture for the reaction to produce more product number 39 this is asking you actually 39's got two parts okay it's the question at the end of your other chapter in your textbook but This one, we're going to take a look at the temperature first, and then we'll take a look at the increase in temperature, and then we'll take a look at the decrease in volume on the next slide. So again, if you'll notice, I've put the kilojoules in, and if I'm going to increase, I'm going to add kilojoules. So A and B are going to push them to the right because I'm adding energy, and they're endothermic. C and D are both exothermic. Notice the negative sign, which means I add the energy to the right-hand side, which would push them backwards. Now we're going to take a look at the volumes. If you'll notice, I've now added the moles of gas to both sides. And we want to decrease the volume or increase the pressure. A I'm going to squeeze from 4 to 2 so it's going to go right. B, there's going to be no shift because I've got the same number of moles on both sides. C, I'm going to shift left because I'm going to squish or squeeze from the three moles to the two moles. And then be careful of the last one is we don't put solids into our equilibrium expression and they don't influence the position of equilibrium. So we have one and zero. So we're going to squish from 1 to 0 so it's going to shift right. Number 41, we've talked about the reaction of reacting carbon with water. It says what happens to the concentrations of H2CO and H2O in equilibrium if carbon monoxide is removed? Make sure it says H2O on your slot notes. So if we pull to the right, we're going to get more H2. We're going to have a decrease in the CO because we're actually pulling it out. And the H2O is going to be pulled to the right. Notice the carbon is not going to change because it's a solid. If we add a catalyst. We're not going to have a change in equilibrium. All we're going to do is we're just going to get to the equilibrium concentrations faster. We've got ammonia with water to produce ammonium ions and hydroxide ions. If we add sodium hydroxide, if we add a strong base, what we're going to find is we're adding OHs. See, you got an OH and the NaOH and the OH, and so it's going to shift it to the left, so we're going to get more NH3. This is one of those things if you do something by accident. If we add HCl, what we're doing is we're adding hydrogen ion, so we're going to go from NH3 to NH4, or it's also a neutralization reaction. It's going to pull the OH out of solution to create water. If I add NH4Cl, I'm adding NH4 ions and so it'll shift to the left forming more NH3. And this is what I was talking about when you add stuff to a reaction, sometimes it doesn't do what you think it's going to and you can accidentally add products or accidentally remove reactants from the solution. And finally, number 48. Again, we've been talking about chemical reactions, but we can also do this for physical process. This is water as a liquid into a gas. Okay. We're basically boiling water. We add heat or we can reduce the pressure. We reduce the pressure low enough. then that liquid is going to turn into a gas. And there are some planets in our solar system that have water only as a gas because it can't exist as a liquid because there's no atmospheric pressure.

So we're going to take a look at Le Chatelier's principle. which basically says that if you apply a stress to a system or to a reaction you can shift the equilibrium or create a q okay so the classic reaction we're going to take a look at is nitrogen plus hydrogen gives ammonia okay if we add ammonia the equilibrium shifts left to offset the stress So what Le Chatelier said was that if a chemical equilibrium is disturbed by adding a reactant or a product, the previous example we saw adding a product, the system will shift in the direction that consumes the added species. Well, this sounds kind of complicated. I prefer to think about this that when we add a reactant or a product, it pushes it to the opposite side.

If we remove a chemical, or remove if a chemical at equilibrium is removed then it's going to restore part of the added species or what we're going to find is removing is a pull to the side that it's on. So let's take a look at a couple examples. We're going to kind of concentrate on this N2O4 gives NO2 as a gas.

And so adding N2O4, that's an adding. That's going to be a push. So it's going to push it from left to right. If we add the NO2, then it's going to go from right back to left. Notice adding NO2, adding reactant, creates more product.

That kind of makes sense. Adding a product creates more reactant. We don't necessarily do this on purpose, and we'll talk about why or how that happens a little bit in a couple slides from now.

If we remove the N204, we're going to cause a shift, or we're going to pull it to the left. And if we remove the NO2, we're going to pull it to the right, or it's going to go from left to right. Okay, it's possible to calculate how much it's going to change and we'll do that in Chem 104. We don't do that in Chem 103. We can also change the pressure. So when a system is compressed, the total pressure increases causing the system to shift in the direction that decreases the total number of. moles, which means this is going to compress, is going to squeeze it to the smaller number of moles.

If we decrease the pressure, then it's going to shift to the side with the greater number of moles. And if we have the same number of moles of gas on both sides, then we're going to see very little effect. So if you notice, we've got one mole of N2O4 on the left and two moles of N02 on the right.

Got to be careful, we're counting up the number of moles of gas. So if we decrease the pressure, it's going to expand to the side with a greater number of moles. So it's going to go from left to right. If we increase the pressure, put pressure on it, we're going to squeeze it from the two to the one or from the higher amount to the lesser amount.

which means we're going to go from right to left. When we talk about changing the size of the container, which actually changes the pressure, if we decrease the size of the container, it increases the pressure. But I think this is conceptual a little bit easier.

If we make the container smaller, it's going to go to the smaller number of moles. So it's going to go from right to left. It's going to go from the two to the one.

And if we increase the size of the container, it's going to expand to the greater number of moles. And so it's going to shift from 1 to 2. The other thing we can do is we can change the temperature. And the trick to this is if we put the energy on the appropriate side, we can treat it as a reactant or a product like we did in our first example.

Notice the plus 57.2. So I put that on the left-hand side because that's an endothermic reaction. We're adding energy.

So increasing the temperature adds energy. So it's going to push it to the right. So it's going to favor producing more NO2. If I decrease the temperature for an exothermic reaction notice it's got the negative negative means it's leaving its loss so that's why i put the 92.2 kilojoules on the right hand side and so if i decrease the temperature it's going to pull it to the right because i'm going to be removing energy okay again decreasing the temperature favors the nh3 you So, so far we've examined adding or removing a reactant or a product, adding is a push, removing is a pull, compressing or expanding a system, we squeeze to the smaller number of moles of gas or expand to the greater moles of gas, and we can change the temperature. So, and remember, we have to assign the energy, whether it's got a negative or positive sign, and then we can treat it as if adding or removing.

a reactant or a product. The one thing that we need to realize is changing the temperature changes the K. And that's why a lot of times you're going to see K and you're going to go, I've seen that reaction before, but why does it have a different K? Well, it's at a different temperature.

Catalyst is kind of a trick question. because it increases the rate of both the forward and the reverse reaction the same. Remember catalysis lowers the activation energy so it lowers it the same amount. So in essence what it does is we get the equilibrium faster.

We go from Q to K a lot faster but ultimately the equilibrium composition's not changed. Chemists were fooled into thinking that the catalyst was changing the equilibrium because the products were being produced. at a faster rate.

But ultimately when it was allowed to come to equilibrium, they ended up with the same amount of reactants and products. So let's take a look at a couple examples. Okay, example one, the volume is decreased. The reason there's no change is there's two moles of gas on the left and two moles of gas.

on the right. And remember if the same number of moles of gas change the volume or change in the pressure does not change the equilibrium, okay, or change the position. Again, this is kind of one of those trick type questions you got to be careful with.

And so if you'll notice, I've put the kilojoules. on the right hand side added the 92 kilojoules to the right hand side so if i heat this this is adding kilojoules which is going to shift it to the left or it's going to push it to the left and number three we're going to remove the so2 remember removing is a pull so it's going to shift or pull it to the left Again, catalyst is a trick question. There's no change.

Just remember, a catalyst allows you to reach equilibrium faster. It doesn't shift the equilibrium. And so in example number five, I'm not going to give you this type of question on your quiz or your exam, but I like this one because it kind of goes through between example number five and number six.

I'm going to show you how all of these that what we've been talking about applies to this equation. So if you notice, it's got 30 kilojoules per mole, the delta H. So it's got implied that it's positive.

So I'm going to put that on the left-hand side. And then I'm going to take a look at all of these. So if I increase the total pressure. By decreasing the volume, so I'm going to squeeze it to the smaller number of moles, I'm going to go from the three moles, the two NOs plus BR, to the two moles of NOBR. If I add an O, I'm going to shift to the left, okay, by adding product.

Would I add product directly? No, but if I don't remove this NO as it's being produced, its concentration is going to go up and then it's going to want to push the reaction in the reverse direction. Removing Br2 pulls to the right. One of the ways I can remove the Br2 is I can chill this reaction down so that gas turns into a liquid.

And remember, we don't put solids or liquids into our equilibrium expressions and solids and liquids don't affect the position either. If I lower the temperature... I'm removing the kilojoules. So it's going to pull left.

And if I remove NOBr selectively, what that means is, is we don't necessarily do it on purpose, but it may be in the mechanism of the reaction, or it may be in how we're performing the reaction that we accidentally remove the NOBr. And what's going to happen is it's going to pull to the left and we're going to get less products. Well, if you'll notice four of these are left and one is right.

And so removing the BR is going to be producing more product. Example number six, again, it's just a continuation of number five. It's that same equation.

Notice the 30 kilojoules. So instead of reducing the temperature, I'm going to increase the temperature, which is going to push it to the right. Instead of adding NO, this time I'm going to remove NO.

So it's going to pull it to the right. Instead of selectively removing the NOBR, I'm going to push right. And if I compress the gas into a smaller amount, again, I like the idea of squish to the smaller moles.

It's going to go from three to two, or it's going to go to, it's going to go left. And if you'll notice, I got right, right, right, left. So the answer would be left for this equation.

But like I said, I won't ask you a question. I'd ask you about each of these individually, but not combine them together and have you do all this work for just one question, especially on an exam where you only have a limited amount of time to answer the questions. So let's take a look at a couple of questions that come from the end of the chapter. um in your book it says suggest four ways in which the concentration of hydrazine that's rocket fuel take nitrogen and hydrogen to make hydrazine um and if you'll notice i put the 95 kilojoules on the left hand side because it's implied this is there's a positive in front of it by the delta h so i could add n2 would push it to the right add h2 push it to the right i could decrease the container volume or increase the pressure to squish it from three to one moles and i would need to heat the mixture for the reaction to produce more product number 39 this is asking you actually 39's got two parts okay it's the question at the end of your other chapter in your textbook but This one, we're going to take a look at the temperature first, and then we'll take a look at the increase in temperature, and then we'll take a look at the decrease in volume on the next slide.

So again, if you'll notice, I've put the kilojoules in, and if I'm going to increase, I'm going to add kilojoules. So A and B are going to push them to the right because I'm adding energy, and they're endothermic. C and D are both exothermic.

Notice the negative sign, which means I add the energy to the right-hand side, which would push them backwards. Now we're going to take a look at the volumes. If you'll notice, I've now added the moles of gas to both sides. And we want to decrease the volume or increase the pressure.

A I'm going to squeeze from 4 to 2 so it's going to go right. B, there's going to be no shift because I've got the same number of moles on both sides. C, I'm going to shift left because I'm going to squish or squeeze from the three moles to the two moles. And then be careful of the last one is we don't put solids into our equilibrium expression and they don't influence the position of equilibrium. So we have one and zero.

So we're going to squish from 1 to 0 so it's going to shift right. Number 41, we've talked about the reaction of reacting carbon with water. It says what happens to the concentrations of H2CO and H2O in equilibrium if carbon monoxide is removed?

Make sure it says H2O on your slot notes. So if we pull to the right, we're going to get more H2. We're going to have a decrease in the CO because we're actually pulling it out.

And the H2O is going to be pulled to the right. Notice the carbon is not going to change because it's a solid. If we add a catalyst. We're not going to have a change in equilibrium.

All we're going to do is we're just going to get to the equilibrium concentrations faster. We've got ammonia with water to produce ammonium ions and hydroxide ions. If we add sodium hydroxide, if we add a strong base, what we're going to find is we're adding OHs.

See, you got an OH and the NaOH and the OH, and so it's going to shift it to the left, so we're going to get more NH3. This is one of those things if you do something by accident. If we add HCl, what we're doing is we're adding hydrogen ion, so we're going to go from NH3 to NH4, or it's also a neutralization reaction.

It's going to pull the OH out of solution to create water. If I add NH4Cl, I'm adding NH4 ions and so it'll shift to the left forming more NH3. And this is what I was talking about when you add stuff to a reaction, sometimes it doesn't do what you think it's going to and you can accidentally add products or accidentally remove reactants from the solution.

And finally, number 48. Again, we've been talking about chemical reactions, but we can also do this for physical process. This is water as a liquid into a gas. Okay.

We're basically boiling water. We add heat or we can reduce the pressure. We reduce the pressure low enough.

then that liquid is going to turn into a gas. And there are some planets in our solar system that have water only as a gas because it can't exist as a liquid because there's no atmospheric pressure.

Transcript for:Understanding Le Chatelier's Principle

Transcript for:
Understanding Le Chatelier's Principle