 How are you guys? Okay. So, quiz five is Friday. It's mainly going to focus on the stuff that we've done in the last two lectures, including today. And that turns out to be stuff that's in these two sections mainly in chapter 16. All right? We're going to jump around a little bit because we're trying to condense this material into the smallest number of lectures possible without throwing out too much stuff that's important. And we're really only being partially successful. We are throwing out stuff that's important constantly. But we've got to talk about kinetics and chemical dynamics, and so we've got to leave some stuff out. And so if it feels like I'm jumping around and skipping stuff as I go forward, I am. A lot of stuff is getting left behind. Okay? I can only apologize for that. If we had 15 weeks, we would still probably not have enough time to do everything. Okay? So, if you look back at the last two lectures, you'll do a pretty good job, I think, of answering the problems on quiz five. I'm going to pull the questions right out of the last couple of lectures. So what we're going to do today is review some of the key concepts from Monday. We're going to talk about this thing called the chemical potential for the first time. And I am really not happy with how I explain this. I've been working on this for the last two days. I still don't like the way this is getting explained. You're the victims. We're going to talk about the Gibbs-Duhem equation. We're going to explain the bow. I'll show you what I mean by that in a minute. Okay? And that'll sort of be the end of chapter 16. We're leaving out a whole bunch of stuff in chapter 16. That's really important. Okay? But we're just going to leave it out. So it's mainly sections 16-3 and 16-4. Okay. So we're going to call this lecture the chemical potential, but we're really not doing such a good job of explaining this concept, I'm afraid. So the first time I put this table up, I think, was last Friday. We were trying to show that these state functions, the internal energy, the enthalpy, the Helmholtz energy, the Gibbs energy can be used to tell us whether a chemical process is spontaneous or not. When that chemical process is governed by these constraints, constant temperature and pressure, constant temperature and volume, constant pressure and entropy, all right? And what we concluded is that, you know, we can derive all of these things for these constraints, but this guy right here and this guy right here aren't super useful to us because in the laboratory, we can't control the entropy. These two guys are more useful to us because we can carry out an experiment under conditions of constant temperature and volume. We can constrain the volume using one of these fancy par bombs, okay? And we can do an experiment in here and that's not a difficult thing to do. So as chemists, the Helmholtz energy tends to be a useful thermodynamic marker for us. But even more useful is the Gibbs energy because the Gibbs energy assumes constant temperature and pressure and constant pressure is enormously convenient for us because we live in constant pressure. And so if we do an experiment that's open to the atmosphere, the pressure will stay constant automatically, okay, and we can watch chemistry occur at a particular temperature and we can predict whether it's spontaneous or not based on the Gibbs energy. Do calculations that pertain to chemistry under these conditions, okay? So what does DG of P and T, what was P and T held constant less than zero mean? It means typically we're going to have a profile for the Gibbs energy that looks like this qualitatively, of course, all right? The position of these two end points is going to be higher or lower. The depth of this well is going to be higher or lower but qualitatively it's going to look like this. So we sort of want to try and understand the qualitative features of this plot. That's sort of what this lecture is supposed to be about today, okay? So we've got reactants on this side, pure reactants, pure products, and in the middle we have a mixture of reactants and products, and there's a preferred mixture, right, a preferred number of products compared to reactants, right, and if you have any other mixture of reactants and products along this green curve, they will spontaneously evolve in the direction of this, of these blue arrows, because the Gibbs energy gets smaller in that direction, and the Gibbs energy tells us what direction things are spontaneous in. So things, you roll downhill in this direction until you get here, you roll downhill in this direction until you get here, and at that point there's no further change in the system. You're at equilibrium, and this derivative, right, the change in the Gibbs energy with this reaction coordinate here is equal to zero at that point. Okay, so the Gibbs energy is special. We're going to pay more attention to it. We want to think about how it varies with temperature, how it varies with pressure. We want to be able to calculate how the Gibbs energy varies with temperature and pressure. We've got this equality that we defined weeks ago. The Gibbs energy is equal to the enthalpy minus temperature times the entropy from this equation. It is apparent if I just take the derivative of G with respect to T in this equation, I get minus S immediately, and that's the temperature dependence of the Gibbs energy at constant pressure, full stop, right? It's minus the entropy. So immediately we can conclude since we know that the entropy is always positive, G decreases with increasing temperature. The Gibbs energy goes down with increasing temperature, and that's odd. We don't expect energies to go down with increasing temperature. Energy's always seem to go up with temperature, but this is an exception to that rule, one of the things that makes this a little counterintuitive. And the rate of change of G with T, the size of this differential here is going to be greatest for systems having high S, right? And that's gases. Now, this first statement, you know, I say it like it's a foregone conclusion that we all understand that the entropy is positive, but we haven't really explicitly said what the third law of thermodynamics is. The third law of thermodynamics, the first law of thermodynamics is conservation of energy. The second law of thermodynamics is the direction of spontaneity increases entropy in the system and the surroundings combined, right? So the first law is about energy. The second law is about entropy and the spontaneity of a process. And the third law simply says that the entropy of a perfect crystal at absolute zero is zero. In other words, if you make the system cold enough so that it can only occupy one state, the atoms are locked into the positions defined by a crystalline lattice, right? The entropy of that system is zero by definition. That's what the third law of thermodynamics says. So now we know all three. Gibbs postulated this, but we can also derive it using statistical mechanics. Remember S equals K log W. Well, if W is one, in other words, if there is only one thermally accessible state for the system, right? And at a temperature that's low enough, that's always going to be a crystalline state, right? All the atoms are occupying positions in a crystal. Y log of one is zero, right? So the entropy is zero once you get down to a temperature where there are no more excited states of the system that are accessible thermally. You've only got the ground state, that's it. And so the entropy of that ground state is zero. We never said this, but now we've said it. Okay? So we said that the temperature dependence of the Gibbs energy is just minus the entropy. So for gases, the entropy is large. And so this is minus S, right? Okay? And so we've got a large negative slope here. For liquids, we've got a negative slope that is not quite as large as for gases and for solids. We have a negative slope that is even less large than it is for liquids. Okay? And so qualitatively, this is what we expect the Gibbs energy to do as a function of temperature. Completely counterintuitive because it's going down. Now, we can also think about the temperature dependence of this quotient, the Gibbs energy divided by the temperature. And the reason people do that is because there's a compact mathematical form for what that temperature dependence is, all right? And so if we take the derivative of this quotient now, we have to use the quotient rule to do that. And we walk through these, this algebra on Monday, all right? But we get down to here, so what are we doing? We've got 1 over T times the derivative of G with respect to T and then G times the derivative of T with respect to T. And after we do all the algebra, we make some substitutions, we get this equation right here. The derivative of this quotient with respect to the temperature is given by minus H over T squared. And that's a very simple mathematical form that ends up being useful to us, all right? It allows us to make a measurement of H and derive information about G from that, all right? And in the laboratory, that turns out to be a convenient way to make this measurement. So, this equation is an important one. It's called the Gibbs-Helmholtz equation. So, there's really two equations that pertain to the temperature dependence of the Gibbs energy. There's DGDT, which is just minus S. And there's the Gibbs-Helmholtz equation, two equations that talk about the temperature dependence of G. This can also be Delta G and Delta H, Gibbs-Helmholtz equation also works in that form. We don't need to know the absolute Gibbs energy, the absolute enthalpy. Okay. So, what about pressure? Well, if we start with this expression for DG. And we said DT equal to zero, all right? We want to know what happens, how G depends on the pressure at constant temperature. So, we get rid of this term. We just have DG equals VDP. And we can integrate that from some initial pressure to some final pressure. And this is the whole story in terms of the temperature dependence, a pressure dependence, all right? This is the only thing that we need to know, all right? And the question is, how do we evaluate this integral? It depends on whether the molar volume or the volume is constant as a function of pressure or not. All right? It's quasi-constant as a function of pressure for liquids and solids because they're virtually incompressible. So, it takes enormous pressures to change the volume of a liquid or solid. And so, under those conditions, we can get a really good idea of how the free energy is going to change, the Gibbs energy. By just moving this V out front, we know what the molar volume of the liquid or the solid is. And we can just treat this as a delta temperature, all right? We move the molar volume out front. These are now molar quantities. We just attached an M to everything, okay? And if it's a liquid or a solid, this is going to give us an excellent approximation to what the Gibbs energy change is that we're looking at as a function of pressure. Really simple. If we're talking about a gas, we can just make a substitution for this molar volume from the ideal gas equation, move the RT out front, and we've got log Pf over Pi. And that's the equation that we're going to use, all right? So, we can calculate the change in the Gibbs function just by substituting the initial pressure and the final pressure into this equation. Pretty easy. Okay? So, there's hardly any change in the Gibbs energy for liquids and solids as a function of pressure. That's what this plot is supposed to show. You see how flat these lines are? The Gibbs energy is not changing. But for gases, there's a pronounced increase in the Gibbs energy as you increase the pressure. And that, the equation for that purple line there is right there, all right? We can calculate what the Gibbs energy is as a function of that final pressure if we know what the initial pressure is. Okay. Yes. So, this is what I just said. The temperature dependence of the Gibbs function, there's really two equations that we want to know about. One is just the simple derivative of the Gibbs function with temperature. That's minus the entropy if we evaluate that derivative of constant pressure. And then there's the Gibbs-Hemholz equation. All right? The temperature dependence of the Gibbs, the pressure dependence of the Gibbs function is always given by this integral right here. And we just evaluate the integral differently if we're talking about liquids and solids. Now, we might know in detail what the volume dependence of a liquid or a solid is. Right? There are equations that describe that. And if we know that, we can just substitute those equations for this VM, leave it inside this derivative, leave it inside this integral rather, and evaluate the integral and get the exact change in the Gibbs energy. But the rest of the time, if we don't know what the change in the volume is for a liquid or solid, we can use this simple equation to tell us how the Gibbs energy changes and for gas is it's this guy. All right? Pretty simple equations. Okay. Now, as we transition how we're briefly into Chapter 16, we seek answers to two nagging issues. The first is, we haven't really said anything about how individual chemical species contribute to G. Right? We said the Gibbs energy is very important. It allows us to predict the spontaneity of a reaction under conditions of constant temperature and pressure. But we haven't said how individual chemical species factor into the Gibbs function. We haven't explained how to do that. So how do individual reactant and product species contribute to the Gibbs function? Can we calculate the Gibbs function and DG from their concentrations? And the second thing is, doesn't matter, matter. In other words, we haven't said much about open systems and an open system, not only energy can be exchanged but matter can be exchanged. All right? So when there are matter fluxes in and out of the system, how does that affect the Gibbs function? We haven't said anything about that. Okay. So how do individual reactant and product species contribute to G? We know we have pure reactants here, pure products here. How does this work? And how is G affected by transfers of matter in and out of the system? We're talking specifically about open systems now. All right? This problem only exists with open systems. There are no exchanges of matter possible here or certainly here. All right? So we haven't said anything about this stuff. How does that, how is the Gibbs energy affected by matter fluxes? Okay. So let's do an experiment. And before we do this experiment, I'm just going to admit I think pedagogically what I'm about to do is completely useless. What we're trying to do here is derive the chemical potential. All right? And if you don't like this derivation, I don't blame you. I don't like it either. All right? And you don't need to know it. But I'm going to show it to you anyway for purposes of completeness. It makes me feel better. What's that? I don't know if tradition is the right word. I mean there's got to be a better explanation for the chemical potential than what I'm about to show you but this is what, this is what you're going to get. Two bulbs. One contains hydrogen, one contains deuterium. These are just two isotopes of the same element for goodness sakes. All right? We'll call this one container one and this one container two. This process that we're about to do, this valve is now closed. All right? The process that we're about to do is going to occur at conditions of constant temperature and pressure. All right? For that to be true, the pressure of these two things and these two bulbs has to be identical. Since G is an extensive variable, what does that mean? What's an extensive variable? What's an example of an intensive variable? Yes, temperature. All right? Is temperature additive? I mean if I take a solution that's room temperature and another solution that's room temperature and I mix them together, do I get a solution that's 600 degrees Kelvin? No. Temperature's not an extensive variable. Mass is an extensive variable. All right? And the Gibbs energy is an extensive variable. In other words, I can think about the Gibbs energy in container one, the Gibbs energy in container two. The total Gibbs energy between container one and container two is the sum. That's not true of the temperature, is it? Okay. So it's an extensive variable. That means I can write the total Gibbs energy in terms of the Gibbs energy of container one, Gibbs energy of container two. Yes, this G1 represents the Gibbs energy of all the D2 and H2 in container one and this G2, you get the idea. Okay, we understand the temperature and pressure dependencies of G already but we should also expect a concentration dependence. So the Gibbs energy in container one, once I open this valve, see how it's open now? And these two gases can mix. I have to consider moles of H2 and moles of D2. This notation is confusing. N1 refers to H2, N2 refers to D2. I need to fix this and likewise, the free energy in the second container also contains contributions from these two gases, all right? We can expect it to be a function of the two gases. We know after we open the valve, the two isotopes will spontaneously mix. That means that the total Gibbs energy must be going down because if it's a spontaneous process, we know that the Gibbs energy is a marker for that process, Gibbs energy will be minimized for any spontaneous process. It will go down. Okay, so we know after we open the valve, the two isotopes will spontaneously mix. DGs are the sum of H2 and D2, all right? So these are expressions for the left side and the right side, the left bulb and the right bulb, container one and container two. In the left bulb, how much is the free energy going to change? Well, it's going to, as N1 flows out and N2 flows in, all right? There's going to be the rate of change per unit of change in the number of moles of N1 multiplied by the number of moles of N1 that flows in. And then there's going to be the change, this is the rate of change of the Gibbs energy with the number of moles of N2 multiplied by the total number of moles of N2 that flows in or out. Okay, we've got an expression that's down here for the right-hand side that's analogous, all right? It considers the change in the number of moles of 1 and the change in the number of moles of 2, hydrogen and deuterium, hydrogen and deuterium. Now, the other equation that we can write is simply that there's a relationship between Dn1 and minus, in other words, if hydrogen flows from bulb 1 to bulb 2, all right, we're going to have minus dehydrogen from bulb 1 and plus dehydrogen in bulb 2, aren't we? All right, and so there's a mathematical relationship between Dn1 in bulb 1 and Dn1 in bulb 2, right? They have to be related to this equation right here. All right, so for component 1, Dn1 equals minus Dn1. For component 2, Dn2 equals minus Dn2, I mean it's intuitively obvious that that would have to be true, but if I now combine these equations with these equations, I get these equations right here. What are these? All right, this is everything pertaining to component 1, everything pertaining to hydrogen and this is everything pertaining to deuterium, all right? Here I've taken, here's a hydrogen term and here's a deuterium term, all right, and here's a term for Dg1 and Dg2, all right, I've added these two things together and I've reorganized these two terms so that I've got only hydrogen terms in that first thing. Let me show it to you again. These are only hydrogen terms and these are only deuterium terms. I've just done a little algebra. If you sit down with a piece of paper, you'll see this is actually pretty straightforward to do this, all right? This equation I submit to you as counterintuitive as it is, is important. Why? The mixing of H2 and D2 is kind of, I wanted to, okay. So we know that if we open this valve, these two gases are going to mix and when they stop mixing, Dg equals 0, right? We're at equilibrium, right? How can Dg equals 0? Look at this thing, all right? When is Dg going to be equal to 0? Well, there's only two ways that can happen, right? One way is if Dn1 and Dn2 are both 0. In other words, nobody ever opened the valve. The valve stayed closed. There's no change in the number of moles flowing between these two bulbs, all right? All the hydrogen stays on one side, all the deuterium stays on the other side and under those conditions, you're at equilibrium with the valve closed. Nobody ever opened the valve. Well, that's not the case that we care about. We want to open the valve, all right? The other possibility is that this difference and this difference are both 0, all right? If that's 0 and that's 0, I think you'll agree that Dg will equal 0, so that has to equal that and that has to equal that. Now, that's profound, although it's not obvious that it would be. So, what we just said, if that has to equal that and that has to equal that, these two things have to be equal to one another and these two things have to be equal to one another. What is this, all right? This is the rate of change of the Gibbs energy with respect to the moles of hydrogen in the left bulb and this is the analogous thing in the right bulb. The derivative of the Gibbs function with respect to hydrogen has to be the same on the two sides, all right? The derivative of the Gibbs energy with respect to this component hydrogen and the same thing has to be true for deuterium, right, before we're at equilibrium, all right? So, this thing is somehow important. It's a partial molar quantity, that's what we call it, all right, that's what it's, and so if we take the derivative of any thermodynamic function with respect to the number of moles is called a partial molar quantity. It's a partial derivative of this thermodynamic function, the Gibbs function with respect to the number of moles of some component, that's what a partial molar quantity is, all right? So, this is obviously a partial molar quantity. I'm not telling you anything if I tell you that, all right? But it's important enough that we give it a special name. We call it the chemical potential. And so, instead of saying the partial derivative of the Gibbs function with respect to hydrogen at constant temperature pressure and all the other components in the system, all right, we just write mu sub 1, all right? And what has to be true then in order to be at equilibrium is that the chemical potential of hydrogen on the left side and the right side of that bulb have to be equal to one another. Otherwise, this equality will not be satisfied and we will not be at equilibrium. Do you feel enlightened now about the chemical potential? No, that's why I think this is probably not terribly useful to do, but so we have just demonstrated that the spontaneous mixing of hydrogen and deuterium will continue until the chemical potential of either isotope and container one is equal to the chemical potential that isotope and container two. It's this thing, this chemical potential that has to be the same on both sides before we're at equilibrium, right? The uniform chemical potential of each component of a multi-component system is a requirement of equilibrium. In other words, you could think about this room as being in equilibrium. And you could ask yourself, you know, in this room, where is the water? Well, there's water everywhere. There are waters in these polyester fibers on this chair and there's water inside this plastic that makes up the back of your chair. Believe it or not, there's water in the plastic. It's between the polymer fibers. And there's water in the air that we're breathing and there's water in those acoustic tiles, all right? And if we're in equilibrium in this room, the chemical potential of the water in those tiles and in this polyester and in that plastic and in the air has to all be equal, right? The chemical potential of the water everywhere in this heterogeneous system has to be equal to one another. Otherwise, we're not in equilibrium with the water. The water, if the chemical potential of the water is higher in these polyester fibers, the polyester fibers will be outgassing water because DG will go down if that's the case, all right? So if you've got a heterogeneous chemical system that has different phases, different chemical compounds in it, all right, if it's at equilibrium, the chemical potential of each component has to be the same everywhere in that heterogeneous system. That's a very important thing to understand, all right? The chemical potential is the thing that has to be the same everywhere in this heterogeneous chemical system. This room is just an example of that. So your book shows this diagram. This is meant to represent two phases, all right? Let's say ice and water or liquid water and gaseous water, all right? It turns out that if you're at equilibrium, right, the chemical potential of water in ice is exactly the same as the chemical potential of the water in liquid water, all right, at equilibrium, that has to be true, not only that, but the chemical potential over here in the liquid water has to be the same as the chemical potential over here and the chemical potential here in the ice has to be the same as the chemical potential here. The chemical potential has to be the same everywhere, even though there's two different phases here, liquid and solid, yes. Well, in that example, we're maintaining temperature and pressure constant, okay? So is it independent of the kinetic energy? No, because the chemical, the kinetic energy, the temperature is a marker of the kinetic energy and we know that for a gas, the Gibbs function is not independent of the temperature, right? Eight and no, the chemical potential also depends on the temperature in exactly the same way that the Gibbs function does, yeah, right. Yes, okay, so we use the Gibbs function here, we express the chemical potential in terms of the Gibbs function, it turns out that there's, you could equally well express this chemical potential in terms of the enthalpy, the Helmholtz energy or the internal energy, right, it can be defined with respect to any of those parameters if you satisfy these conditions of constant volume and temperature, entropy and pressure and so on and so forth. The other, the number of moles of the other components in the system other than I are also held constant when you evaluate this partial derivative, okay? Now, our expression for these, okay, so we've encountered this expression for the Gibbs energy before but it turns out to generalize this for open systems we have to include the possibility of fluxes of material in and out of the system and that's what this term right here does. What is this? This is the sum of the chemical potential times the change in the number of moles of whatever component it is that's leaving or being added to the system. If it's added, that's a summation, if it's subtracted, it's a subtraction. All right, we add rather the chemical potential times the number of moles for each species that's being added or subtracted from the system, that is an additional term in terms of understanding what the Gibbs energy is, right? So we simply can add, we just use the chemical potential to add to the Gibbs energy for each chemical component that we're talking about. It's part of why the chemical potential is so convenient. So under these conditions of constant temperature and pressure this equation becomes, if temperature and pressure are constant then the change in the Gibbs energy is just given by these fluxes of material in and out of the system, right? Okay, now, so we said, can we calculate G and DG from the concentrations of individual chemical species? That's what we're doing here, okay? And can we compensate for the addition to subtraction of material from the system in terms of understanding what happens to G? Yes, we just have to include this term right here and we can figure out if G is getting larger or smaller based on whether we're adding or subtracting material from the system if we know what the chemical potential is. Okay, yes, yes, yes, okay. Now, if this equation is true, we know at equilibrium DG is equal to zero, right? If DG is equal to zero then this thing has to be equal to zero, all right, at equilibrium. And so it seems intuitively obvious that that would have to be the case, but that's given its own name. That's called the Gibbs-Duhem equation, all right? So this equality has to be satisfied at equilibrium. All right, in other words, things can move around in a system at equilibrium, but this equality has to be satisfied. In other words, this summation cannot change. It's got to be equal to zero. This equation just says that the chemical potential summed over all the moles of products and reactants is a conserved quantity at equilibrium. Yes, that's all that says. So we're going to do an example of what that means. For example, if you have A getting converted into B, let's say this is a nice summarization reaction. The Gibbs-Duhem equation says that at equilibrium that plus that has to equal zero because the summation only contains two terms. There's only two chemical species, A and B, all right? And so this equality has to be satisfied, right? So a mixture of ethanol water is prepared with a mole fraction of water of 0.4. Ethanol water mixture. If a small change in the mixture composition results in an increase in the chemical potential of ethanol by 0.35 joules per mole, I don't know how that would have been accomplished, but hypothetically, if you could increase the chemical potential of ethanol by 0.35 joules per mole by how much will the chemical potential of the water change? Or if we change the chemical potential of one component of this two component systems, how will the chemical potential of the other component respond? All right? We know this is a Gibbs-Duhem problem because we're at equilibrium. Does it say that anywhere? A mixture of ethanol water at equilibrium, okay? So we're going to use the Gibbs-Duhem equation. It's a two component system, so we've got two components. We've got an ethanol component and a water component. We just solve for the change in the chemical potential of the water, all right? And if these derivatives are small, the first assumption to make is that they are small, all right? Then we can just treat this as the change in the chemical potential of the water. That's the change in the chemical potential of the ethanol. Oh, we're told what that is, all right? Now we need to know this mole fraction. These two moles, all right? What I've written here is the mole fraction. Everybody understand what the mole fraction is? It's the moles of component one divided by the total moles of all of the components in the system. And so in this case, that would be the mole fraction of ethanol, all right? That's not the mole fraction. That's the number of moles of ethanol. But the mole fraction is just the number of moles of ethanol divided by the total moles in the system. So we usually call that X sub-ethanol, all right? And I think you can see that if I take the quotient of the two mole fractions, the total number of moles will cancel and I just get moles ethanol divided by moles water, okay? And so this 0.6 over 0.4, that's just these two mole fractions. That's the change in the chemical potential of the ethanol. And so I can calculate what the chemical potential of the water is going to do. It's going to go down by minus 0.53 joules per mole. That's just the Gibbs-Duhem equation. Very simple. Okay. So once we understand about the chemical potential, that the equilibrium is determined by the system looking for the, oh, sorry. You're right. This should be mu dn. This is the, yeah, you're right, sorry. Yeah, this should be mu dn. This should be mu dn. You're absolutely right. Oh, sorry. Yes. All right? So you can actually recast this equation in either of two forms or it can be mu dn or nd mu, right? And so how you want to use it depends on the information that you're given. Yes. So one of the consequences of the chemical potential is in understanding this diagram right here, all right? So once again, this is the Gibbs energy on this axis and this is the temperature on this axis and we're talking about three phases of a particular component, let's say water. Okay? And this, the slope of these lines is going to be dictated by these derivatives. In other words, the slope of this line is going to be equal to minus the entropy for vapor phase water. This is minus the entropy of liquid water and this is minus the entropy of solid ice. Now, in principle, the system can choose three different Gibbs energies at any temperature that you choose on this axis, all right, except where there's a phase transition or the system can choose any one of three energies. It always chooses the lowest of these three energies. It minimizes the chemical potential. Okay? The chemical potential of liquid water is actually equal to its Gibbs energy, all right? And its lowest for the solid at a temperature that is below the melting point of ice. So at 253 degrees Kelvin, we're 20 degrees below the melting point of ice, water chooses to exist in a solid state and the reason is that the chemical potential of the solid is less than the chemical potential, the liquid or the gas. That's why from a thermodynamic standpoint, water as ice at temperature is below 273 degrees Kelvin. All right? If we go up to 273, the chemical potential of liquid water and ice are equal to one another. And at that point, if we stay in the solid state, the chemical potential of ice will be higher than that in water, all right? And that's thermodynamically unfavorable. The system can evolve towards a new equilibrium by melting the ice to form water, okay? And so the system now follows this new equilibrium line, which is the liquid phase line. And so if we heat the system up to 330 degrees Kelvin, that's above the melting point of ice. We're tracking now the liquid phase. The system stays in the liquid phase because everywhere along this line, it is the minimum Gibbs energy that's possible for the system, all right? When we arrive at the boiling point of water, 373.16 degrees Kelvin, the chemical potentials of liquid and gas, phase, water are equal to one another. And only at that point can you have the coexistence of these two phases, all right? Steam and liquid water. In this case, liquid and solid ice, all right? This phase coexistence can only happen at equilibrium at this phase transition temperature. And then if you go to still higher temperatures above the boiling point, only gas phase water is possible at, for example, 390 degrees Kelvin. And so the way the system evolves is shown by this red line because the red line indicates the minimal chemical potential of the water at each one of these temperatures. At these phase transitions, the chemical potential of two phases are equal to one another. And in principle, they can coexist in the chemical potential across the phase boundary for steam at the boiling point in the liquid and in the gas phase is exactly the same. By definition, it has to be. And the same thing is true for liquid water, the chemical potential of the water in the liquid and in the solid. At that point, at 273.16 degrees Kelvin, the chemical potential is the same in the ice and in the water. That's why you can have coexistence of those two phases. Yes, we're just tracking the lowest chemical potential for the system as we track along this red line. Okay. Finally, I want to talk about this. We're going to call this the bulb. So I think this plot is actually very counterintuitive. All right, why do I say that? Look, if you've got reactants here and you've got products here, right, the shortest distance between two points is a straight line, if this is the Gibbs energy of the products and this is the Gibbs energy of the reactants, why would the Gibbs energy of the mixture of the reactants of the products ever be less than this? In other words, if this is the Gibbs energy of the products and this is the Gibbs energy of the reactants, why wouldn't the Gibbs energy of some mixture of the reactants of the products just lie somewhere along this dashed line, right? Isn't it intuitively obvious that that should be the case, of course if it was, there would be no equilibrium in chemistry because the equilibrium is defined as dG equals zero, all right, and there's no minimum in this plot. It's a straight line. And so the reaction would just go boom, wouldn't it? Or if products had a higher G function than reactants, there would be no reaction, all right? There would be no equilibrium possible. You would either be fully in the product state or fully in the reactant state, and if they had exactly the same Gibbs energy, you'd have sort of a pseudo equilibrium where you could have any composition. But no composition would be preferred. So let's go back to this equation here, this experiment rather. We've got oxygen and nitrogen. We've got half a mole of each. The valve is closed. Now we open the valve, all right? Here's, are we out of time? Welcome back to this on Friday. Thank you.