 Okay. So we're going to be talking about equilibrium. We're going to have a little bit more to say about mu because mu is important and then I'm going to talk about the bow and then I'm going to talk about the equilibrium constant. This is chapter 17 material. We've already tunneled through chapter 16, leaving out lots of stuff along the way. Non-ideal solutions, we don't have time to talk about them. They're very important and there's lots of other stuff there as well. We're already in chapter 17. How do individual reactants and products contribute to G? There could be a whole slew of reactants here. There could be a whole slew of products. How do we attribute the G for a reaction to individual reactants and products? As chemists this is obviously an important thing for us. We talked about this thought experiment right here. We have two isotopes of hydrogen in two different bulbs separated by a valve that's initially closed. Then we open it and when we open it we can describe the change in the free energy on the left side in terms of what's happening with component 1 and what's happening with component 2. In this case component 1 is H2 and component 2 is D2. So we can write two equations for the left-hand side, two equations for the right-hand side, two equations. Two terms. Here's the left-hand side, here's the right-hand side, here's component 1, here's component 2. Same thing over here. On the right-hand side component 1 and in component 2. And then if we consider that anything that leaves the right-hand side has to enter the left-hand side and anything that leaves the left-hand side has to enter the right-hand side we can write a second pair of equations, these two guys. All right and what this says is we're only talking about component 1 here, let's say hydrogen, the partial derivative of the free energy on the left side and the partial derivative of the free energy on the right side, those two things have to be equal to one another at equilibrium. And the same thing has to be true in terms of component 2. All right, so this could be for hydrogen, this could be for deuterium, these partial derivatives have to be satisfied, these equalities that involve these partial derivatives. All right, so these are so-called partial molar quantities. Whenever you take the partial derivative of a thermodynamic state function with respect to the number of moles of something it's called a partial molar quantity. Molar because you're taking the partial derivative with respect to the number of moles, okay, and the chemical potential is a special case of a partial molar quantity that involves the Gibbs energy, right, the Gibbs energy, the partial of the Gibbs energy with respect to component 1 or component 2 or component 3 would be the chemical potential of component 1 or component 2 or component 3. Mu describes how G is affected by changes in the amount of one component, right, it is in effect the G of a particular component. It is the Gibbs energy of a single component, all right. Why didn't they just call it the Gibbs energy of component 1? I don't know, it would have been a lot easier because that's what it is, right? It's just the Gibbs energy of component 1, hydrogen or deuterium. Now, it's not obvious to me when I look at this partial derivative it looks a lot more complicated than just the Gibbs energy of component 1. Why did they write it this way? This really is the Gibbs energy of component 1, all right. Do other people have a problem seeing that? Because I know I always have. So the way I like to think about it is shown here, all right. Here's a system that contains three different components and the total G for this system, think about it as just being a balance and there's a scale here, we call this the G meter. We know the free energy, excuse me, the Gibbs energy is an extensive quantity, don't we? All right, and so this is a beautiful analogy for that. We're weighing these components, it's like a G meter, this is an extensive, this is going to be an extensive quantity, isn't it? It's just like mass, okay, and so what we said is the partial derivative of G with respect to N and here we're going to pay attention to N blue, all right. We're going to try and think about the chemical potential of blue here, all right, at constant green, orange, P and T, all right. And the way to think about that derivative is that if you remove a little bit of blue or add to it, a unit of blue, here we've subtracted a unit of blue, this partial derivative is just this difference and what we're left with is the block that used to be here, that's the chemical potential of the blue. For me it's a lot easier to think about it this way, all right. Now it's clear, yes, this is just the chemical potential of the blue, all right. Now it's not the chemical potential of the blue always, it's the chemical potential of blue exactly when the composition of the system is as I'm showing it here. There's green here, there's orange here, and there's a bunch of other blue here, right. The chemical potential I'm talking about is specific to the rest of the stuff being there too. The chemical potential could be different if there was different amounts of stuff here, right. It would be as though this balance could be different, this arrow could point in different directions even for a single blue block if the rest of these guys, maybe if there's only three green guys or six orange guys, the chemical potential could change a little bit, but we're always talking about the chemical potential of this pure component that we're talking, whatever it is that we're talking about, in this case it's the blue blocks. Does this make any more sense? I don't know, it does for me. Okay, so what do we need to know about the chemical potential? It's extensive just like G. It's got units of energy per mole just like G. It's pressure dependence, it's the same as G, and so is the temperature dependence. Okay, so I can say the partial derivative of the chemical potential of any component with respect to its partial pressure is just equal to the molar volume of that component, and if I integrate that I'm going to get an equation that looks just like the equation that we write for G, except now it pertains to a single component. And the same thing with the entropy. But how does it depend on concentration? How does the chemical potential depend on concentration? We haven't talked about that even with respect to G. Well, here's the equation. The chemical potential of some species A is given by the standard chemical potential that applies under defined standard state conditions, which in the case of a gas would be a pressure of one bar and a temperature of 298.16 degrees Kelvin. In the case of a solute in a solution, the standard state is an activity of one molar. We haven't said what activity is yet. So it's equal to this mu zero, which is the chemical potential for that pure compound in its standard state plus RT log of the activity of A. What's that? Well, to beg the question and not answer it, the activity of A is just equal to the activity coefficient of A times its concentration. Now I haven't told you what the activity coefficient is. Right? That's the activity coefficient. That's the concentration. So now we have a relationship between the chemical potential and the concentration of A. We all know what the concentration is and the concentration could be a pressure. Right? There'd still be an activity coefficient. All right? So if it's in the gas phase, it's a pressure. It's the partial pressure, of course. And if it's in the liquid phase, it'd be a concentration. And this activity coefficient, it turns out, can vary between zero and one. And we're going to talk more about that later on. But it's sort of a distraction for us now. So we're not going to talk about it right now. For our purposes, we're just going to equate for the time being the activity with the concentration. We're just going to pretend for the time being that this activity coefficient is always one. Okay? That's an assumption. It's an incorrect assumption most of the time. We'll talk about activity coefficients later. Okay? So we can relate the chemical potential to the concentration through an equation that looks just like this. Okay. Now, we already understand that G is minimized upon approach to equilibrium. We already talked about this. Equilibrium is at the bottom of this curve. But we didn't talk about why this is bow shaped. Because it's not intuitive that it would be. I mean, if this is the Gibbs energy of the reactants, this is the Gibbs energy of the products, why doesn't the Gibbs energy of every intermediate mixture of reactants and products fall in this dash line up here? It's the simplest solution to this problem, isn't it? And if that were the case, there would never be any equilibrium, would there? Because there would never be any curvature to this thing. No matter where you put this dot and where you put this dot, you're always going to have a straight line and it's never going to have a minimum where you would expect to find equilibrium, is there? All right. So the fact that there's a bow to this curve is sort of essential to how chemical systems work. Chemical systems have equilibrium. We need to understand that. This is not what happens. Okay. So let's do a simple calculation. I've got O2 on the left side, N2 on the right side. Half a mole of each. Now I'm going to open the valve. This reaction coordinate is the mole fraction of N2. When I open the valve, I'm going to have 0.25 moles of O2 and 0.25 moles of N2 after everything comes to equilibrium. Okay? And so I'm, the mole fraction is simple. The moles of N2 is 0.25 over the total number of moles is 0.5 and so 0.25 over 0.5 is 0.5 and so we're right here on this diagram now. All right? We're right here after all the mixing has happened. Okay? So let's calculate G. Well, G is just equal to the number of moles of N2 times the chemical potential of N2 and the number of moles of H2, except H2 is not in there. That should be O2 times the chemical potential of O2. It's all O2 and N2 here. Okay? And so what's the chemical potential of N2? Well, it's just given by the, because we're talking about a gas, that's the partial pressure of N2. This is the, this is one bar. We can just sort of ignore it. Your book usually does. Just call it one. Okay? And this is it, analogous term for O2, all right? We multiply by the number of moles. Yes, this is the chemical potential of N2 at one bar and 298.16 degrees Kelvin and that's one bar. Yes, your book just writes, it gets rid of the P0, just calls it one. Don't be confused by that. Okay? So since P is the same on both sides, we'll just call it P prime. At equilibrium, the total pressure on the two sides is the same. We'll just call it P prime. Okay? So after opening the valve, we're going to have P prime over 2 for nitrogen and P prime over 2 for oxygen because half the oxygen is going to go into the left valve, left bulb, half the oxygen is going to go into the, you know what I mean? Half the nitrogen is going to go, all right? So the pressure is going to go down by a factor of 2, right? Yes, this is all for oxygen, this is all for nitrogen. Okay? So the difference, the delta G final minus initial is just going to be P prime over 2 divided by P prime for nitrogen, same thing for oxygen and if I put the numbers in, half of them all, total of nitrogen, 8.31451, 298.16 Kelvin, all right? If there's a term for nitrogen, a term for oxygen, that's why there's two blocks here, all right? When I plug it into my calculator, I get minus 1,717 joules. Okay? So I can look at my plot then, all right? This delta is 1,717 joules when I'm at 0.5, that's not on the line. It's nowhere close to the line, the line's at 0, why is that? The line's at 0 because the mu 0 for nitrogen and the mu 0 for oxygen are both 0, turns out, look in the back of your book, there's a big table of standard free energies of formation, standard Gibbs energies of formation rather, you can look up nitrogen and oxygen, any other diatomic gas like that gives energy of formation is 0, all right? So we're measuring versus 0, okay? In other words, the curve has got to go through this point right here, it's got to go through that point and that point, so it's got to be a curve, all right? There's no way it can be a straight line between these two data points, we're just proving that. But the question is why? In other words, we know delta G is delta H minus T delta S, which of these two things is responsible for this curvature? Well, there's no change in the enthalpy during this reaction if these are idos, what is there? There's no change in the enthalpy if these are idogases, it's got to be in that term right there, it's got to be the entropy and of course when you think about it, as you mix these two gases together, right, and that you allow each of these two components to have more volume to occupy, the entropy of the system is going to go up, that's the driving force for this curvature that we're showing here, all right? And in fact if I plot the entropy, it looks like that, all right? Now what I've plotted here is actually T times delta S because I wanted to show you that it's exactly the exact positive reflection of this guy. If I put a mirror here, you get this, that's because I multiply by T, so that is going to be numerically equal to the depth of the delta G well, isn't it? It's going to be numerically equal. So entropy, I think this is moderately profound, all right? If you didn't have entropy acting in this process, there wouldn't be any equilibrium at all because if there's no curvature of this G versus progress of reaction curve, there is no equilibrium, all right? The universe would be fundamentally different than it is, all right? We have this curvature in delta G because of the entropy of mixing, that's what this is, the entropy of mixing. Okay, now this is one of my favorite molecules. It's called resveratrol. Why would it be one of my favorite molecules? Where is resveratrol found in our environment? One of the nice things about resveratrol is it makes things live longer and if you're my age, you're very interested in that even though it's only been tested in yeast, fruit flies, and worms. I'd like to think that human beings may also benefit from the same biochemistry that's going on with these guys, all right? Mice do well on resveratrol, all right? They don't get cancer as often when they eat a lot of it. Unfortunately, scientists feed them enormous fun, all right? And then they don't get cancer for God's sakes, all right? Here's the mouse without eating huge amounts of resveratrol, in acetone, mind you, happy day. The reason we humans like it is because it's found in red wine, California Pinot Noir's, oh my. All right, so this could mean that red wine is good for you, but it really cares. It's absolutely good. All right, why do we even mention it? Well, because there are two isomers, a resveratrol, and if you hit resveratrol with 350 nanometer, likely you get isomerization around that double bond right there. So there's a system transversion. And so this is like the simplest chemical reaction, all right? An isomerization reaction is like the world's simplest chemical reaction. Unimolecule, usually like or the stimulant. Isomerization reaction, see if we can figure out whether they can deliberately take the generic version of A goes to B, okay? Now, let's say that we're starting off with N sub A moles, that's why I'm calling it zero, because this is the starting point, and N sub B, this is zero, because that's the starting point. And if there's a change now, some reaction occurs, some of this A is going to get converted to B and some of this B. So if the reaction is going from left to right here, we're going to lose A and make B. And so this thing called the progress of the reaction or the extent of the reaction is going to tell us how far the reaction has proceeded from left to right, okay? So if that's minus point one, that's plus point one, and the units here are moles, the extent of the reaction has units of moles. It's not dimensionless, okay? And so if we start off with one mole of A and zero moles, B, we can convert this axis here into an axis in terms of the progress of the reaction, right? And if that's the case, this would be one and this would be zero, the progress of the reaction is zero when we got all A and it's one when we got all B, makes sense? One mole, right? The progress of the reaction always has units of moles. Okay, so we define this all for this plot of any value of the progress of the reaction as delta R of G. This is the notation you're both using. So this is the reaction gives free energy. So if I look at this point right here, all right, the reaction gives free energy is the partial derivative of the G function with respect to the extent of the reaction, right? It turns out you have to denominate everything in terms of the extent of the reaction in order for this to work properly, okay? So here, delta R of G is less than zero, yes, the slope is negative, here delta G of R is equal to zero, yes, the slope is zero, here delta G, delta R of G is greater than zero, yes, the slope is positive, okay? So we've got this thing called the reaction gives free energy. Except we're not supposed to use the word free. Okay, so delta R, that's meant to imply an incremental change. The conditions of constant pressure and temperature if the reaction advances by some small amount, all right? A will change by d moles of A, B will change by d moles of B, okay? Notice that the extent of the reaction is positive by definition, okay? So here, if we have a reaction from A to B, d and A would be a negative quantity, wouldn't it? Because A would be getting smaller. B would be getting bigger, so that would be positive, that would be negative, and so in order for this to be equal to this, I have to have a minus sign there because that's going to be positive by definition. Everyone see that? Okay, so these two things are equal to one another, and so if that's the case, I can factor out this, what is that, chi? Factor that out, and I've got this expression right here. And so in other words, this Gibbs reaction energy, right, is just going to be proportional to the difference in the chemical potentials. So delta G of R should be delta R of G, of course, is just the difference between the chemical potentials. It's just that, right? When I take DG, D chi, I get that guy. Okay, so it's obvious from this plot, but let's say it anyway, the Gibbs reaction energy changes continuously monotonically with the progress of the reaction or the extent of the reaction, therefore the composition of the reaction mixture. Yes, this is a smooth and continuous function. Okay, so that means that there's three types of reactions, reactions that are exergonic, exergonic, where delta R of G is less than zero, endergonic, where delta R of G is greater than zero, and equilibrium where it equals zero. And if you don't know what this word means, as always, Wikipedia can help you, means the releasing energy in the form of work, if it's an exergonic reaction, it can release energy in the form of work. You can do work, all right, if delta R of G is less than zero. All right, if you want delta R of G to be greater than zero, you've got to put work into the system to make that happen. Okay, but we still have not learned anything more about where equilibrium is located in terms of the extent of the reactions. To get there, let's assume A and B are ideal gases, then you will recall from lecture 17, isn't this lecture 17? Let's assume A and B are ideal gases, then you'll recall from lecture 16, yes, that's right, right? So the difference in the free Gibbs energy between some initial state and some final state is just given by this integral, where that's the molar volume, and we can convert that, we know if we're talking about an ideal gas, plug in chug, here's the expression as a function of pressure, okay, and we can define a standard molar Gibbs free energy as one that applies at a pressure of one bar and 298.16 degrees Kelvin, that's what that is right there. Okay, and so if A and B are gases, we can write their chemical potentials in this form. For example, because this is just the chemical potential of a single species, and it's going to follow exactly this form, I can write this equation right here by analogy to that one, and this is the standard chemical potential which once again applies at one bar and 298.16 degrees Kelvin, okay? So the Gibbs reaction energy, we said, is the difference between these two chemical potentials, and so if I write an equation like this for A and another one for B, I can take the difference, there's the B one, there's the A one, and when I take the difference, yes, I already said that. I've got mu B zero minus mu A zero, that's that, all right? And then the P zero's are going to cancel, I'm going to end up with P B over P A, and this is the standard now, Gibbs reaction energy. It applies at one bar and 298.16 degrees Kelvin, okay? And we have a name for this, right? This quotient between the partial pressure, B and the partial pressure of A is called the reaction quotient, something that you heard about in general chemistry, low of those many years ago, all right? We call it Q, the reaction quotient. Okay, so for every value of the extent of the reaction across the horizontal axis of our G versus the extent of the reaction diagram, we can calculate the Gibbs reaction energy from the Gibbs reaction energy to the product reaction species present at that extent of the reaction, yes. And so finally, at equilibrium, the reaction quotient is given by, so when this is zero at the bottom of that well, right, where you've got a horizontal line tangent to the curve, we can call this K, right? The reaction quotient isn't Q anymore, here it's Q, all right? Under the special case where that's zero, we can call the reaction quotient K, K or K equilibrium, all right, and these are then equilibrium values for these two partial pressures. And from this, I can just move this guy to the left-hand side and I've got minus RT log K. So there's a relationship between the free energy of the reaction, the standard free energy of the reaction, standard because this is a zero, and the equilibrium constant. For example, let's say that the delta G, the reaction is zero, right? If delta G is zero, Y can solve for K, delta G is zero, exponential of zero is one, yes, all right? In my extent of the E, in my equilibrium constant for the reaction is one, all right, which is going to be correspond to equal partial pressures of B and A, right, because the equilibrium constant is the partial pressure of B divided by the partial pressure of A, okay? Here's the reaction, the standard reaction free energy difference between A and B. It's zero, all right, the equilibrium constant's right in the middle. This is going to be a nice symmetrical curve, isn't it? Yes. And on the other hand, if this standard Gibbs reaction energy is less than zero, in other words, the final Gibbs energy minus the initial Gibbs energy is as shown here, all right, this guy is less than zero, now the curve is going to look like that. Now equilibrium's got to be over here, all right, and that means that it's going to be greater than one. It has to be greater than one. What does that mean for the equilibrium constant? Well, if we look at the equilibrium constant, here's the equilibrium constant refresher for a generic reaction, A moles of A plus B moles of B goes to C moles to C and D moles of D. We can write the equilibrium constant terms of the equilibrium activities of each reactant and product species, yes, it's just the activity of C taken to the C power times the activity of D taken to the D power and so on, products over reactants, right? We can write this in terms of activity coefficients and concentrations, yes, these are activities and these are concentrations and these are activity coefficients. And so what this means is that you've got a product rich mixture. You've got more product than you do reactant because you're on the right-hand side of one. And if the opposite is true, if you're going uphill from A to B, you're going to end up with an equilibrium constant that has the opposite. It's reactant rich. All right, we'll come back to this on Monday. Have a good weekend.