Understanding Equilibrium and Constants

All right, what is the definition of equilibrium? What's the definition of equilibrium? Yeah. The rate of the reaction in the forward direction equals the rate of the reaction in the reverse direction. That is the definition. Now notice you've got to be careful here, because that's what the definition of equilibrium means. And you should know what it is, but you should also know what the definition of equilibrium is not. The forward rate equals the reverse rate. However, it does not mean that the forward rate constant equals the reverse rate constant. That's not true. So because you're going to get... get a question on the test in all likelihood that says, which of the following is true about a system at equilibrium? Forward rate equals reverse rate. That's true. Forward rate constant equals reverse rate constant. Not necessarily true. So that might be one of the false answers they try to get you on. You might get another one. Concentration of reactants equals concentration of products. Is that true for any system at equilibrium? No. For some systems, you have way more products. For other systems, you have way more products. systems you have way more reactants and for some reactions you got a fair amount of both it totally depends the key is is it's the rates that are equal the reactants are turning into the products just as fast as the products are going backwards and turning into reactants and if those rates are happening at the same time then what must be true What's happening to the overall concentrations? They're not changing anymore. So when you reach equilibrium, concentrations are no longer changing. And that's tricky. There are two types of equilibrium. We talk about only one type in chemistry. We talk about them both in physics, though. In the physics class, we might talk about a static equilibrium. And that looks like this. In a static equilibrium, everything is stopped. That's not what we mean at all in chemistry. One of the other false definitions of equilibrium you might get on a multiple choice test is everything is stopped. That's not true in chemistry. In chemistry, we don't talk about... static equilibrium at all. We talk about a dynamic equilibrium where the forward reaction is still going on, the reverse reaction is still going on, they're just going on in both directions at the same rate. They have equal rates. That way for every every reactant that turns into product, you have a product that turns back into reactant. Everybody cool with that? All right. So there's your definition of equilibrium. So, and we often define equilibrium constants to go with this. We said that some equilibrium, some equilibria will have more products than reactants at equilibrium. Some will have more reactants than products, but some are going to have both. And the way we figure that out, by looking at a single number, is by looking at what we call the equilibrium constant. What would be the expression, I'm going to count on the fact that you guys know this, what would be the expression for the equilibrium constant for this reaction? Good. It's products over reactants raised to the power of their coefficients. So notice, this is capital K now. We're not talking about lowercase k right now. This is capital K. It is an equilibrium constant. It is the ratio of products over reactants. In the last chapter we dealt with lower case k, which was the rate constant. Notice we also use capital K for what? Elf. Kelvin. And if you took a physics class, we'd use K for a whole bunch of other stuff. It's a conspiracy. If we use K for enough things, we know you're going to get confused. And if you get confused, then we can weed you out and keep you from going to med school. Good times, right? So everything's a conspiracy, right? So here's the deal. What do these brackets mean? It means molar concentration of. So these are molarities. Well, gases, we can measure their molar concentrations, but for gases, it's often easier to measure their concentrations in what units or in what way? As pressures. And so for gases, we can also talk about what we. we call Kp, and it uses pressures instead. Still the same ratio, products over reactants raise the power of their coefficients, but we're using pressures. And so for gases you can have Kc, which is molar concentrations, or Kp, which uses partial pressures. Now here's the deal, what did I leave out of both of these expressions either way? I left the solid out. Solids and liquids do not show up in these expressions. Solids and liquids are known to have what we call as an activity of one. And so if I put carbon down here but it's equal to one, well what's anything you know multiply? or divided by one. It's self and it doesn't affect anything. And so we just leave solids and liquids out of these expressions entirely. Gases will show up, aqueous species show up, but no solids, no liquids. And that's tricky. Notice back in the last chapter with rate laws, in a rate law, a solid or a liquid might show up in a rate law sometimes. Nothing says they don't, nothing says they can't. But in equilibrium constants, they never show up. No solids, no liquids. Cool. So what's nice about this is by looking at one number, we can get an idea of whether this particular equilibrium happens to favor the reactants or the products. So it turns out, you don't know this, but I do, that this reaction heavily favors the products. At equilibrium, I'm going to have a boatload of carbon monoxide, but I'm not going to have a whole lot of O2 in the system at all. So what would I predict about the Kc value in that case? It's going to be a big number. We usually say not just greater than 1, but greater than greater than 1. It's a lot bigger than 1. And that means, again, that this reaction favors the products. There will be more products than reactants at equilibrium. So what if I had a reaction that favored the reactants? What would I expect out of my k value in that case? Yeah, if you've got a lot of reactants but not very many products, so then your denominator is going to dominate, and that's going to be a lot smaller than 1. Somebody give me a number that is a lot, lot bigger than 1. 1000. Somebody give me another one. So, how about I, let's try this. You tell me if a number is a lot, lot bigger than 1. A billion. A trillion. A quadrillion. Negative 4. Cool. By the way, you can't have a negative equilibrium constant. How about 0.3? How about 2? 5? 50? 157? Cool, where's the cutoff? Well, there is no cutoff. Here's the problem. When does a number just get from being a little bigger than 1 to being a lot bigger than 1? There's no set cutoff, turns out. And the cutoff would actually be different for different reactions. And so it's really annoying. So what they usually do to be on the safe side is most textbooks, including yours, says a lot bigger than one means bigger than a thousand. That's what it translates into. And a lot smaller than one means less than one. 1,000th, 10 to the minus 3. The truth is, though, the truth is that you can't really define a set cutoff for every reaction. It'd be different for different reactions. So they just kind of overestimate and say, well, if you're beyond these, you're for sure okay. If a K is bigger than 1,000, that's definitely favoring the products. If it's smaller than 1,000, definitely favors the reactants. So they're just playing it safe. So just FYI. So if I told you an equilibrium constant was equal to 4, what would be true? Well, if your equilibrium constant's not too much bigger or smaller than one, then you're going to have a fair amount of both reactants and products present at equilibrium. Cool? All right. And so that's nice. By looking at one single number, I can kind of tell that when this reaction's done, what am I going to have? All products? All reactants? Or some of both? So one thing to note about these, these equilibrium constants are constants. So when we say that, what we mean is that they are constants for a given temperature. You change the temperature, that's the one thing that'll change it. But nothing else changes it. You can change concentrations, you can change pressures, all that lovely stuff, and the K value doesn't change. The only thing that changes is temperature. Now notice this is different than the rate constant. What were the two things we could change that would change the rate constant? Activation energy and temperature. But notice, how do you change the activation energy? By adding a catalyst. But if you recall, we said that a catalyst does not shift the equilibrium at all. Which means that a catalyst does not change your k value at all. It does not shift the equilibrium. And so it's only temperature that can change this thing. That's it. Most of you guys will forget that before the next half hour is over. One thing to note, KP and KC here are not necessarily equal. So there's one case where they will be equal, but they're not necessarily equal. And the way they're related... It's through this equation. Kp is equal to Kc times RT to the delta n power, where delta n is the change in the number of moles of gas. And when we look at the change in anything, it's always final minus initial. What is the change in the number of moles of gas for this reaction? It's 1. We've got 2 moles of gas at the end. as products, we only have one mole of gas here. Notice the solids don't count for this, it's just change in moles of gas, and 2 minus 1 is positive 1. So if we were trying to look at the relationship between Kp and Kc here, for this reaction, delta N would be positive 1. Now where that comes from, so it turns out, we can derive that from PV equals nRT. We'll rearrange it just a little bit to look like this. In PV equals nRT, the ideal gas flow, what's n? Moles of gas, what's V? Volume, usually measured in units of? Liters, what's moles over liters? Malarity. And so pressure is equal to malarity times RT. So pressure and malarity are not equal. They're proportional. You double the malarity, you double the pressure. But they're not equal. There's that extra factor of RT. That's where that extra factor of RT comes from right here. If you notice, if we take these partial pressures, and instead of pressures, we plug in molarity times RT. Well, here we'd have molarity squared times RT. RT squared. But on the bottom, we'd only have one RT. And so we'd be left with an extra RT. That's where they come from. When would KP be equal to KC? What would have to be true of that reaction? No change in moles of gas. If you had equal number of moles of gas on the reactants and product side, that is the one case where Kp would equal Kc. It's not true in this case, but it would be if you had equal moles of gas on both sides of the reaction. Sometimes you'll get a question involving a calculation with this. Sometimes you'll get a question that just says, for which of the following reactions does Kp equal Kc? And you're just looking for the ones that have equal moles of gas on the reactants and product side. So, well, notice where does this, I'm glad you brought that up actually, where does this r come from? It comes from pv equals nrt. And if it comes from pv equals nrt, guess what r value you've got to use? This is the one place this semester where you're going to use that same R value you used last semester, 0.08206 liter atmospheres per mole Kelvin, because it comes from Pivnert. Pivnert is where we use this R value, but every other time this semester when R comes into context, it's going to be... talking about an equation involving energy, and you're going to use that other R value, 8.314 instead, joules per mole Kelvin. But this one place this semester, we use this value of R. Again, both values are going to be on the front of your exam. You need to know which one to use and when. Thanks for bringing that up, by the way. All right. So you're given a reaction, next step. And after the reaction, you're given the value of Kc. Okay. And then below it there are three more reactions that are somewhat similar to this one, and I want to know the Kc value for each one of those. So the first that I want to know the value for is this one. We've got to figure out the Kc value. Well first off, how does the second reaction compare to the first one? Everything's flipped around backwards. Everything's flipped around backwards. And every once in a while, students are like, oh, so instead of 278, does it equal negative 278? But can you ever have a negative equilibrium constant? Why not? Let's look at this for a sec. What's Kc? What's the expression for this first one here? Somebody help me out here. SO3 squared, awesome. SO2 squared times, awesome. Notice, can a concentration ever be negative? Can I like say, hey, I got a beaker and, you know, I pour a little Kool-Aid in there, and if you asked me how much Kool-Aid was in there, I'd be like, negative 5 molar. There's less than none in there. so it makes no sense you can't have negative concentrations there's no such thing lowest concentration could ever have a 0 right when you have none and so if these can never be negative in this ratio can never be negative either so that's why you can have negative equilibrium constants so there's no way that this reaction's equilibrium constant is coming out to negative 278 that's impossible but notice what's the expression for this one It's SO2 squared times O2 all over SO3 squared. How does this expression relate to the first one? It's inversed. So notice, if this first one... is equal to 278, which, awesome, is the same as 278 over 1. Then this second one is going to equal 1 over 278. Cool? So, next reaction you're given is this one. How does this reaction compare to the original? Everything's cut in half. Everything's cut in half. And so a lot of students see that and they're like, oh, so this guy's Kc should equal half of 278. But that again is wrong. Where do the coefficients show up in the expression? as the exponents. And so it's not that we need to multiply this by one half, we need to take it to the one half power. And so as we'll see in a minute, this Kc is going to equal 278 to the one half power. Or what's that the same as? Square root, same as square root. So if you look here, the the Kc expression is going to equal SO3 all over. SO2 times O2 to the one-half power. And if you look here, going back to the original expression, if I were to take this entire expression to the one-half power, what does the one-half power do to the exponents inside? It multiplies them all by one half. So instead of being to the second power, after factoring that in, it would be to the first power. That one, instead of being squared, would also, factoring that in, be to the first power. and where O2 was, you know, had an exponent of 1, well, now it would be 1 half. And so if everything in here equals 278, well, square rooting it gets me this expression, and so it's equal to the square root of 278, or 278 to the 1 half power. Cool. So the last one here. There's this guy. How does this reaction relate to the original? It's reversed and cut in half. And so, when we reversed a reaction, it was one over. When we cut a reaction in half, it was to the one half power or square root. Now, since it's reversed and cut in half, we've got to do both. And it doesn't matter which order you do it. You can do one over one. over 278 and then take it all to the 1 half power or you can take your 278 to the 1 half power and then do 1 over it really doesn't matter take your pick either way this means the same thing mathematically it's 1 over the square root of 278 Cool. If you ever get hosed on one of these problems, what should you probably do to figure it out? The long way, write out the expressions for both reactions and compare them, and go from there. If it's doubled, you'd square it. If you triple all the coefficients, it'd be cubed. If you tripled and reversed it, it'd be at a negative 3 power, and so on and so forth.

That is the definition. Now notice you've got to be careful here, because that's what the definition of equilibrium means. And you should know what it is, but you should also know what the definition of equilibrium is not. The forward rate equals the reverse rate. However, it does not mean that the forward rate constant equals the reverse rate constant.

That's not true. So because you're going to get... get a question on the test in all likelihood that says, which of the following is true about a system at equilibrium? Forward rate equals reverse rate.

That's true. Forward rate constant equals reverse rate constant. Not necessarily true.

So that might be one of the false answers they try to get you on. You might get another one. Concentration of reactants equals concentration of products. Is that true for any system at equilibrium?

No. For some systems, you have way more products. For other systems, you have way more products.

systems you have way more reactants and for some reactions you got a fair amount of both it totally depends the key is is it's the rates that are equal the reactants are turning into the products just as fast as the products are going backwards and turning into reactants and if those rates are happening at the same time then what must be true What's happening to the overall concentrations? They're not changing anymore. So when you reach equilibrium, concentrations are no longer changing. And that's tricky.

There are two types of equilibrium. We talk about only one type in chemistry. We talk about them both in physics, though. In the physics class, we might talk about a static equilibrium. And that looks like this.

In a static equilibrium, everything is stopped. That's not what we mean at all in chemistry. One of the other false definitions of equilibrium you might get on a multiple choice test is everything is stopped.

That's not true in chemistry. In chemistry, we don't talk about... static equilibrium at all. We talk about a dynamic equilibrium where the forward reaction is still going on, the reverse reaction is still going on, they're just going on in both directions at the same rate.

They have equal rates. That way for every every reactant that turns into product, you have a product that turns back into reactant. Everybody cool with that? All right.

So there's your definition of equilibrium. So, and we often define equilibrium constants to go with this. We said that some equilibrium, some equilibria will have more products than reactants at equilibrium. Some will have more reactants than products, but some are going to have both. And the way we figure that out, by looking at a single number, is by looking at what we call the equilibrium constant.

What would be the expression, I'm going to count on the fact that you guys know this, what would be the expression for the equilibrium constant for this reaction? Good. It's products over reactants raised to the power of their coefficients.

So notice, this is capital K now. We're not talking about lowercase k right now. This is capital K.

It is an equilibrium constant. It is the ratio of products over reactants. In the last chapter we dealt with lower case k, which was the rate constant.

Notice we also use capital K for what? Elf. Kelvin. And if you took a physics class, we'd use K for a whole bunch of other stuff. It's a conspiracy.

If we use K for enough things, we know you're going to get confused. And if you get confused, then we can weed you out and keep you from going to med school. Good times, right? So everything's a conspiracy, right? So here's the deal.

What do these brackets mean? It means molar concentration of. So these are molarities. Well, gases, we can measure their molar concentrations, but for gases, it's often easier to measure their concentrations in what units or in what way? As pressures.

And so for gases, we can also talk about what we. we call Kp, and it uses pressures instead. Still the same ratio, products over reactants raise the power of their coefficients, but we're using pressures.

And so for gases you can have Kc, which is molar concentrations, or Kp, which uses partial pressures. Now here's the deal, what did I leave out of both of these expressions either way? I left the solid out.

Solids and liquids do not show up in these expressions. Solids and liquids are known to have what we call as an activity of one. And so if I put carbon down here but it's equal to one, well what's anything you know multiply?

or divided by one. It's self and it doesn't affect anything. And so we just leave solids and liquids out of these expressions entirely.

Gases will show up, aqueous species show up, but no solids, no liquids. And that's tricky. Notice back in the last chapter with rate laws, in a rate law, a solid or a liquid might show up in a rate law sometimes.

Nothing says they don't, nothing says they can't. But in equilibrium constants, they never show up. No solids, no liquids.

Cool. So what's nice about this is by looking at one number, we can get an idea of whether this particular equilibrium happens to favor the reactants or the products. So it turns out, you don't know this, but I do, that this reaction heavily favors the products.

At equilibrium, I'm going to have a boatload of carbon monoxide, but I'm not going to have a whole lot of O2 in the system at all. So what would I predict about the Kc value in that case? It's going to be a big number. We usually say not just greater than 1, but greater than greater than 1. It's a lot bigger than 1. And that means, again, that this reaction favors the products. There will be more products than reactants at equilibrium.

So what if I had a reaction that favored the reactants? What would I expect out of my k value in that case? Yeah, if you've got a lot of reactants but not very many products, so then your denominator is going to dominate, and that's going to be a lot smaller than 1. Somebody give me a number that is a lot, lot bigger than 1. 1000. Somebody give me another one.

So, how about I, let's try this. You tell me if a number is a lot, lot bigger than 1. A billion. A trillion. A quadrillion. Negative 4. Cool.

By the way, you can't have a negative equilibrium constant. How about 0.3? How about 2?

5? 50? 157? Cool, where's the cutoff?

Well, there is no cutoff. Here's the problem. When does a number just get from being a little bigger than 1 to being a lot bigger than 1? There's no set cutoff, turns out. And the cutoff would actually be different for different reactions.

And so it's really annoying. So what they usually do to be on the safe side is most textbooks, including yours, says a lot bigger than one means bigger than a thousand. That's what it translates into.

And a lot smaller than one means less than one. 1,000th, 10 to the minus 3. The truth is, though, the truth is that you can't really define a set cutoff for every reaction. It'd be different for different reactions. So they just kind of overestimate and say, well, if you're beyond these, you're for sure okay. If a K is bigger than 1,000, that's definitely favoring the products.

If it's smaller than 1,000, definitely favors the reactants. So they're just playing it safe. So just FYI.

So if I told you an equilibrium constant was equal to 4, what would be true? Well, if your equilibrium constant's not too much bigger or smaller than one, then you're going to have a fair amount of both reactants and products present at equilibrium. Cool?

All right. And so that's nice. By looking at one single number, I can kind of tell that when this reaction's done, what am I going to have?

All products? All reactants? Or some of both? So one thing to note about these, these equilibrium constants are constants.

So when we say that, what we mean is that they are constants for a given temperature. You change the temperature, that's the one thing that'll change it. But nothing else changes it.

You can change concentrations, you can change pressures, all that lovely stuff, and the K value doesn't change. The only thing that changes is temperature. Now notice this is different than the rate constant.

What were the two things we could change that would change the rate constant? Activation energy and temperature. But notice, how do you change the activation energy? By adding a catalyst.

But if you recall, we said that a catalyst does not shift the equilibrium at all. Which means that a catalyst does not change your k value at all. It does not shift the equilibrium.

And so it's only temperature that can change this thing. That's it. Most of you guys will forget that before the next half hour is over. One thing to note, KP and KC here are not necessarily equal. So there's one case where they will be equal, but they're not necessarily equal.

And the way they're related... It's through this equation. Kp is equal to Kc times RT to the delta n power, where delta n is the change in the number of moles of gas. And when we look at the change in anything, it's always final minus initial.

What is the change in the number of moles of gas for this reaction? It's 1. We've got 2 moles of gas at the end. as products, we only have one mole of gas here. Notice the solids don't count for this, it's just change in moles of gas, and 2 minus 1 is positive 1. So if we were trying to look at the relationship between Kp and Kc here, for this reaction, delta N would be positive 1. Now where that comes from, so it turns out, we can derive that from PV equals nRT. We'll rearrange it just a little bit to look like this.

In PV equals nRT, the ideal gas flow, what's n? Moles of gas, what's V? Volume, usually measured in units of? Liters, what's moles over liters?

Malarity. And so pressure is equal to malarity times RT. So pressure and malarity are not equal.

They're proportional. You double the malarity, you double the pressure. But they're not equal.

There's that extra factor of RT. That's where that extra factor of RT comes from right here. If you notice, if we take these partial pressures, and instead of pressures, we plug in molarity times RT.

Well, here we'd have molarity squared times RT. RT squared. But on the bottom, we'd only have one RT.

And so we'd be left with an extra RT. That's where they come from. When would KP be equal to KC? What would have to be true of that reaction?

No change in moles of gas. If you had equal number of moles of gas on the reactants and product side, that is the one case where Kp would equal Kc. It's not true in this case, but it would be if you had equal moles of gas on both sides of the reaction.

Sometimes you'll get a question involving a calculation with this. Sometimes you'll get a question that just says, for which of the following reactions does Kp equal Kc? And you're just looking for the ones that have equal moles of gas on the reactants and product side.

So, well, notice where does this, I'm glad you brought that up actually, where does this r come from? It comes from pv equals nrt. And if it comes from pv equals nrt, guess what r value you've got to use? This is the one place this semester where you're going to use that same R value you used last semester, 0.08206 liter atmospheres per mole Kelvin, because it comes from Pivnert. Pivnert is where we use this R value, but every other time this semester when R comes into context, it's going to be...

talking about an equation involving energy, and you're going to use that other R value, 8.314 instead, joules per mole Kelvin. But this one place this semester, we use this value of R. Again, both values are going to be on the front of your exam.

You need to know which one to use and when. Thanks for bringing that up, by the way. All right. So you're given a reaction, next step. And after the reaction, you're given the value of Kc.

Okay. And then below it there are three more reactions that are somewhat similar to this one, and I want to know the Kc value for each one of those. So the first that I want to know the value for is this one.

We've got to figure out the Kc value. Well first off, how does the second reaction compare to the first one? Everything's flipped around backwards. Everything's flipped around backwards.

And every once in a while, students are like, oh, so instead of 278, does it equal negative 278? But can you ever have a negative equilibrium constant? Why not? Let's look at this for a sec.

What's Kc? What's the expression for this first one here? Somebody help me out here. SO3 squared, awesome.

SO2 squared times, awesome. Notice, can a concentration ever be negative? Can I like say, hey, I got a beaker and, you know, I pour a little Kool-Aid in there, and if you asked me how much Kool-Aid was in there, I'd be like, negative 5 molar.

There's less than none in there. so it makes no sense you can't have negative concentrations there's no such thing lowest concentration could ever have a 0 right when you have none and so if these can never be negative in this ratio can never be negative either so that's why you can have negative equilibrium constants so there's no way that this reaction's equilibrium constant is coming out to negative 278 that's impossible but notice what's the expression for this one It's SO2 squared times O2 all over SO3 squared. How does this expression relate to the first one?

It's inversed. So notice, if this first one... is equal to 278, which, awesome, is the same as 278 over 1. Then this second one is going to equal 1 over 278. Cool?

So, next reaction you're given is this one. How does this reaction compare to the original? Everything's cut in half.

Everything's cut in half. And so a lot of students see that and they're like, oh, so this guy's Kc should equal half of 278. But that again is wrong. Where do the coefficients show up in the expression?

as the exponents. And so it's not that we need to multiply this by one half, we need to take it to the one half power. And so as we'll see in a minute, this Kc is going to equal 278 to the one half power.

Or what's that the same as? Square root, same as square root. So if you look here, the the Kc expression is going to equal SO3 all over.

SO2 times O2 to the one-half power. And if you look here, going back to the original expression, if I were to take this entire expression to the one-half power, what does the one-half power do to the exponents inside? It multiplies them all by one half. So instead of being to the second power, after factoring that in, it would be to the first power.

That one, instead of being squared, would also, factoring that in, be to the first power. and where O2 was, you know, had an exponent of 1, well, now it would be 1 half. And so if everything in here equals 278, well, square rooting it gets me this expression, and so it's equal to the square root of 278, or 278 to the 1 half power. Cool. So the last one here.

There's this guy. How does this reaction relate to the original? It's reversed and cut in half.

And so, when we reversed a reaction, it was one over. When we cut a reaction in half, it was to the one half power or square root. Now, since it's reversed and cut in half, we've got to do both.

And it doesn't matter which order you do it. You can do one over one. over 278 and then take it all to the 1 half power or you can take your 278 to the 1 half power and then do 1 over it really doesn't matter take your pick either way this means the same thing mathematically it's 1 over the square root of 278 Cool.

If you ever get hosed on one of these problems, what should you probably do to figure it out? The long way, write out the expressions for both reactions and compare them, and go from there. If it's doubled, you'd square it. If you triple all the coefficients, it'd be cubed.

If you tripled and reversed it, it'd be at a negative 3 power, and so on and so forth.

Transcript for:Understanding Equilibrium and Constants

Transcript for:
Understanding Equilibrium and Constants