 Okay. So, quiz five went better than quiz four. Went pretty well in fact. So, that's sort of an A minus right there. And we only have two of these things left. There's one Friday and then a week from Friday, we have the midterm two and then there's one more after midterm two and then we're done with these quizzes. So in principle, if you already have an A on these quizzes, you don't need to take quizzes six and seven because you can drop two quizzes, all right? So, you only need to take them if you want to improve your quiz average. I know some of you will want to, but, okay. So, what we're going to do today is we're going to talk just briefly about some of the things that we talked about on Friday and then we're going to do some examples. We're in the middle of Chapter 17. We're trying to close out this thermodynamic stuff. We should be done with it by the end of the lecture on Wednesday. And then we're going to start talking about kinetics. Okay. So, we've been talking about the chemical potential. All it is is the partial derivative of the Gibbs energy with respect to one of the components, the moles of the number of moles of one of the components that serves as either a reactant or a product in the reaction that we're talking about at constant temperature and pressure where we hold the number of moles of all the other components constant. All right? So, Mu describes how the Gibbs energy is affected by changes in the amount of one component. That's what we've been saying. It is an effect, the Gibbs energy of a single component. And the way I like to think about that because I find this equation to be confusing is if you have a bunch of different components and you've got some total Gibbs energy that we can think of as just being the mass of all of these elements of these components, blue, green, orange. All right? If we were to remove one unit of the blue and weigh again, that difference is tantamount to taking that partial derivative with respect to the blue. While holding green and orange constant, and I think you can see intuitively when you look at it this way that what you're going to end up with is the weight of this unit of blue, consequently, this is telling us, right? This is an effect, the Gibbs energy of a single component if you think about it that way. Right? So, what we know about the chemical potential is that it's extensive, it's got units of energy per mole, just like G, it's pressure dependence and it's temperature dependence mimics that of the Gibbs energy. The concentration dependence is something that we haven't talked about for the Gibbs energy, right? For the chemical potential, here's what it looks like for some component A, all right? There's a standard state for A, all right? And RT log, and that is the activity of A. And the activity is just the concentration multiplied by a factor called the activity coefficient, which we haven't talked about yet, but for the time being, we're just going to assume that this activity coefficient is 1. So, this activity closely approximates the concentration. This is true, this is rigorously true in the limit of dilute A, right? If the concentration of A is very low, then its activity will equal its concentration, okay? And we'll have more to say about the activity later. Okay, so we have this thing called the chemical potential. Now, with this chemical potential, we can begin to understand why G is bowed. In other words, why we can't just connect the Gibbs energy of reactants and the Gibbs energy of products with a straight line, right? Seems logical that you could do that. Isn't the Gibbs energy of these intermediates just a linear combination of the reactant and product concentrations? And if that's true, wouldn't we get a straight line that just connects these two things? And if that's true, there really wouldn't be an equilibrium, would there? Because there'd be no way for there to be an equilibrium. There'd be no dip, no bowing. The bowing is essential in order for there to be equilibrium. And so what we showed is that if we calculate the G when we're half converted, O2 to N2. In other words, we have vessels of O2 and N2. We've allowed them to intermix until the equilibrium is obtained. All right, what we calculate is minus 1,717 joules, right? Even though the free energy of formation, the Gibbs energy of formation of O2 and of N2 are both zero, all right? And so if mixtures of these two things were just a linear combination of zero and zero, we would just have a dash line here. But that's not what we see, right? There's this bowing of the Gibbs energy that occurs as we mix these two components. And the reason that happens is because the entropy of mixing is positive, all right? And so if we take T delta S, that equals delta G, except that it's positive, all right? So that peak would be at plus, well, I think I brought it here. Yes, plus 17, 17 joules. Here it's minus 17, 17 joules because the delta H of mixing is zero for ideal gases, all right? We know delta G is delta H minus T delta S, don't we? Okay, so this positive entropy of mixing is the source of bowing in all types of chemical reactions, right? It's a general feature of chemical reactions that this bowing occurs and consequently because there's a minimum introduced by this bowing, that's where the equilibrium state is. And so this, when I say the positive entry of mixing is the reason for equilibrium, you get what I'm talking about. Okay, the last thing that we talked about on Friday was this thing called the extent of reaction or the progress of the reaction. I call it both things. It's got this funky symbol here which is called psi, I think. Is that right? I'm not very good with Greek letters. I'm just going to refer to it as the E because it looks like an E. All right, if the delta E is 0.1 moles, then as this reaction here progresses from A to B, A is going to get smaller by minus 0.1, B is going to get bigger by that amount. And so we can recast our Gibbs energy versus reaction progress plot instead of having a reaction coordinate here, we can call it the extent of reaction. And what we pointed out is that the slope of this plot at any value of this E is the reaction Gibbs energy. Now you see this delta here, we don't say that. We just call it the reaction Gibbs energy. But when you say the reaction Gibbs energy, you're right the delta sub R. Okay? So the reaction Gibbs energy is just the chemical potential sort of, evaluated with respect to E instead of the number of moles. All right, it looks like the chemical potential except we've substituted E for N. All right, now we call this thing the reaction Gibbs energy. And there's three possibilities. It can be less than 0, 0, or greater than 0. Okay? And so under conditions of constant of the reaction of the, so we understand that a change in the Gibbs energy is given by this expression, the chemical potential of A times the change in its number of moles plus the chemical potential of B times the change in its number of moles. Right? And if we want to exchange E for N, we can recast this expression in terms of the extent of reaction. Okay? And to do that, we have to recognize that DNA is going to be negative because A is getting smaller. All right? And the extent of the reaction is always positive. All right? And so we have to introduce a minus sign here in order to convert from moles to the extent of reaction. Okay? And once we've done that, we can write this expression. And so the reaction Gibbs energy is just the difference in the chemical potential. Right? DGDE, DGDE, that divided by that is just that right there. It's the difference in the chemical potentials. So there are three possibilities as we proceed from A to B, all right? Initially, the reaction is going to be the Gibbs, the reaction Gibbs energy is going to be less than zero. In other words, this slope is going to be less than zero as we proceed from here to here. And then when we get down here, it's going to be exactly zero at equilibrium. And then when we go further, it's going to be greater than zero. We have these funny words that describe that. Okay? Now we haven't said where equilibrium is with respect to the extent of reaction. To get there, let's assume that A and B are ideal gases, then you would call from lecture 16 that we can take the difference between the molar Gibbs energy of the final state and the initial state. That's just equal to this integral right here and for an ideal gas because that's just RT over P. We get RT log P final over P initial. And so we can then define a standard molar Gibbs energy, no free, as one that applies at 1 bar in 298.16 degrees Kelvin. Okay? And so instead of thinking about an initial state and a final state, we're going to convert one of these two guys into a standard state that we reference our Gibbs energy against all the time. Right? And that standard state is characterized by pressure of 1 bar and the temperature of 298.16 degrees Kelvin. Right? It's that guy. We're going to move them over to the right side. Now we're going to call him G superscript zero. Right? The standard molar Gibbs energy. Okay? And so if A and B are gases, we can write their chemical potentials in this form. All right? Parallel to this form, now we're just talking about the Gibbs energy of a single component, A. And so if we look at this difference here, I can substitute this expression in for A and then write an analogous expression to it for B and I get this guy. All right? And now I'm going to collect these two muses together and call that the reaction, standard reaction Gibbs energy. Standard reaction Gibbs energy. Okay? And the P zeroes cancel. We just end up with P B over P A. This quotient here is in general going to be called the reaction quotient. Okay? So now for every value of the extent of reaction across the horizontal axis of our G versus the extent of reaction diagram, we can calculate the reaction Gibbs energy from the Gibbs energy of formation. All right? There's an F there for the product and reaction species. So we can look up these Gibbs energies of formation for the reactants in the products. That allows us to calculate what the reaction Gibbs energy is. At equilibrium, the reaction quotient is given by a special name. So in other words, at equilibrium where this guy is zero, okay, we have this, we're going to call the reaction quotient K instead of Q. Where did we call it Q? Here. All right? That guy is equal to Q, but when this is equal to zero and we're at equilibrium, why? Now we're going to call it K. K only applies when you're at equilibrium. So you have to be talking about equilibrium concentrations or pressures of every reactant in product. Okay? And so that being the case, we can just move this guy over to the left-hand side or move that over to the left-hand side, put a minus sign in front of it, and now we have a relationship between the standard reaction Gibbs energy and the equilibrium constant. This is an important equation. Okay? What does it mean? Well, let's say that the standard reaction Gibbs energy is zero. All right? In other words, the reaction in the product Gibbs energies are the same. All right? If that's the case, if I plug zero in for this guy, all right, I get zero over RT, I get the EXP of zero, that's just one, all right? And what that means is the equilibrium constant's just one. In other words, if we look at a diagram like this one where we're plotting the Gibbs energy as a function of Q, all right, we're going to end up with an equilibrium constant that is here right at the bottom of this well, symmetrically located between A and B, and this curve is also perfectly symmetrical. On the other hand, what if this standard Gibbs reaction energy is negative? We've got a minus sign here, and if that thing is also minus, we're going to have a minus times a minus is a plus, all right? And the EXP of any positive number is going to be greater than one, right? All right? And if it's greater than one, all right, here's that case, all right, standard reaction Gibbs energy is positive, sorry, negative, negative, negative, all right, final minus initial, and so now our curve is going to be skewed, all right? Here's one, all right, we're going to be greater than one. We're on the right-hand side of one. We're over here, all right, so qualitatively the equilibrium constant is going to be greater than one. What does that mean? What does that tell us about the equilibrium state? More of this than that, right? So we can be a little bit more specific about that. If we have some generic reaction, this is just a review from Chem 1, Amols of A, Bmols of B, Cmols of C and Dmols of D, the equilibrium constant is given by this expression where we've written everything in terms of activities, all right, these are the product species, these are the reactant species, and I've taken in every case the activity to the stoichiometric, to the power of the stoichiometric coefficient. Everybody remember how to write these things? All right, and these, because these are activities, each activity, for example, the activity of C is its activity coefficient times its concentration and for D and for A and for B we have the analogous expressions. Now what I told you earlier is for the time being, we're going to assume these gammas are all one, so this whole thing is going to go away, isn't it? It's just going to be one, okay, and that's our equilibrium constant expression written out in terms of concentrations. That's the activity of A, that's the activity coefficient for A, that's the concentration of A, you get the picture, okay? So what this tells us is that we're going to have a product rich state, these are product concentrations, these are reactant concentrations, if K is greater than one, we got more of this than this. That's what this is telling us. We have a product rich state for our system. Conversely, if the standard reaction Gibbs energy is greater than zero, in other words, B is up here and A is down here, that's the positive standard Gibbs energy, now the curve is going to be skewed, the other way K is going to be less than one, okay, and we're going to be reactant rich by virtue of the same thinking, okay? So this picture is self-consistent, we can think about in terms of K, G, standard Gibbs reaction energy, any of those things. Let's do an example, CO, hydrogen, and methanol, all right, are mixed together at 500 degrees K with these concentrations, the partial pressure of CO is 10 bar, H2, 1 bar, methanol, 0.1 bar, and we pass this over a catalyst, this mixture, all right? The question we're going to ask is, can more methanol be formed? Can we adjust the reaction additions to form more methanol? Or are we maxed out, okay? We're given the standard reaction Gibbs energy, 21.21 kilojoules per mole, all right? Now the first thing to say is, catalyst, what does that word mean? How does that affect our calculation? What's the catalyst doing? What's the role of the catalyst here? Anybody know? Speed up the reaction, it's exactly right. It's going to accelerate the reaction, it's going to accelerate the approach to equilibrium, but it does nothing to alter the position of the equilibrium. In other words, inserting the catalyst doesn't change where we're going to end up in terms of the mix of products and reactants, all right? It's not going to do that by definition, okay? So here's what our picture looks like. We're told the reaction, standard reaction Gibbs energy is 21.21 kilojoules per mole, here's what that looks like, all right? We're uphill compared to where we're starting. It's a positive number, all right? So what do we expect? Well, we expect K is going to be less than 1, right? Now I've denominated this axis in terms of the reaction quotient, all right? The reaction quotient is zero when we're starting the reaction because the partial pressure of these guys would be zero at this point, all right? That's not where we are in this problem and this problem we're over here somewhere, all right? But hypothetically, if we were here, the reaction quotient would be zero and if we're here, the reaction quotient would be infinite, wouldn't it? Because we got no, none of, we got only products, no reactants, okay? So we're here, we got none of this so the reaction quotient would be infinite. Okay, so how do you approach this problem? First thing you have to do is write a balanced chemical reaction, boom. CO plus 2 hydrogens gives methanol. Then you write an expression for the reaction quotient and you calculate it, the reaction quotient for where you are right now, the reaction quotient for the conditions that you're given. That's the one I'm talking about. There are an infinite number of reaction quotients, right? All different numbers, these are all different reaction quotients that vary between zero and infinity, all right? So what we want to calculate is the reaction quotient that corresponds to the initial conditions that were given. I should have called it something special. But I didn't, I just called it the reaction quotient. It's the partial pressure of methanol divided by the partial pressure of CO multiplied by the partial pressure of hydrogen squared because there's a 2 in front of the hydrogen, okay? I plug my numbers in, I get 0.01. Reaction quotients and equilibrium constants are always dimensionless, okay? Reaction quotients and equilibrium constants are always dimensionless. I'll show you why in a moment, okay? So now we can calculate K. K is just the XP of minus standard reaction Gibbs energy divided by RT and we were given this in the problem. It's 21.21 kilojoules per mole, so that's 21,210 joules per mole and I can calculate K. Sure enough, it's less than 1, okay? Then I compare Q to K. This is where I am now. This is where K wants me to be. Q is bigger than K. What does that mean? Well, we're not at equilibrium. That's the first thing to say because to be at equilibrium, Q would have to equal K, right? It's bigger. What does it mean that it's bigger? It's on the product side of equilibrium because remember what Q is? Q is products over reactants, right? Okay, so if Q is bigger, that means we've got more products with respect to reactants than we will have at equilibrium. So we are on the right hand side compared to where equilibrium is, right? We're here and here's where equilibrium is, right? Qualitatively. We're at 0.01, equilibrium is at here, okay? And so we can immediately start to make predictions about what's going to happen, can't we? Just looking at this diagram, okay? But let's just do a couple of other things. Let's calculate the reaction Gibbs energy for this progress of reaction, all right? The reaction Gibbs energy is just equal to the standard reaction Gibbs energy plus RT over Q. We calculated Q already. We said it was 0.01, that's the temperature, that's our, that's the standard reaction Gibbs energy. I can calculate now the reaction Gibbs energy at the Q that we care about, 0.01, and that's the number I get, 2.06 kilojoules per mole, all right? The fact that it's positive means that the slope is positive, right? It means that the slope is positive. It means that we're here, all right? That's why it's positive, all right? We already concluded that we were here. We know we're at Q equals 0.01, all right? And so what this means, of course, is that at this mix of reactant and product pressures, the reaction is not, the reaction towards methanol is not spontaneous. In fact, we're going to go back in the other direction, all right? Methanol is going to be consumed and we're going to make CO and hydrogen. That's not what we wanted to do, all right? So that we were asked, all right, can we make more methanol under these conditions? No, you can't. You're going to go backwards. You're going to consume methanol and make these two reactants. Now, what if I increase the concentration of the reactants? All right, I'm going to increase these concentrations and I can just recalculate. Now the Q is 0.001 instead of 0.01. It's gotten smaller. I can recalculate the reaction Gibbs energy and when I do that, now I'm getting a negative number, all right, and the reaction Gibbs energy is the slope, remember, the slope of that green curve, all right? So if that's negative, that tells us that we went from being on the right-hand side to being on the left-hand side. What did I write here? The minus value means that the reaction is now spontaneous and we can make more methanol. We were here under these conditions and now we've moved over to here under these conditions right here, all right? And now the slope is negative. It was positive before and so now we will make a little more methanol. We're going to go from here down to here. It's not going to be a huge amount but we should make a little more methanol under these conditions by adding more, what did I add? More CO or more H2. One, ten, oh, ten, one, I just changed. You get a different Q because the hydrogen partial pressure squared, okay? So we can make predictions and everything is self-consistent. That is self-consistent with that is self-consistent with this thing right here, all right, it all makes sense. It's a little intricate but it all fits together like a puzzle. Consider the following reaction, N2O4 reacts unimolecularly, in other words by itself, to give two NO2s. It's a unimolecular reaction. Calculate delta GR that should be delta RG. R should be over there. At 298.1, calculate equilibrium constant, calculate the equilibrium total pressure of the system consisting of one mole of N2O4 and one liter at 298.15 degrees Kelvin. Got to do all three things. Very similar to a problem that you're going to have on midterm two because it rolls together. A lot of things, doesn't it? Okay, that R should be right there. It's called the reaction. Standard reaction Gibbs energy, all right? Now, we're not given that here for goodness sakes but fortunately you've got your book and there's a table of Gibbs energies of formations in the back of your book and you can just open up that table and calculate this, okay? So, the standard reaction Gibbs energy where that R is actually underneath right there is two times that because that's the product, two times the NO2, the standard Gibbs energy of formation for the NO2 minus the standard Gibbs energy of formation of the N2O4. That's going to give us the standard reaction Gibbs energy. I'm really working hard not to wrap these words around my throat today, all right? I looked that up in the back of your book. I looked that up in the back of your book and I just plug them in here and what I get is that number right there. That's the standard reaction Gibbs energy where that R is actually right there, all right? So, it's thermodynamically uphill. That's what we mean when we say thermodynamically uphill, that's a positive number, okay? This reaction is thermodynamically uphill. We expect Kp to be less than one, right? Because if I think about where the products and the reactants are, the green curve is going to be skewed towards the reactant side and K is going to be less than one. That's my chemical intuition kicking in. I look at that number and I know K is going to be less than one, all right? We want to develop this chemical intuition that maybe we haven't had. Calculate Kp at this temperature. We just calculated that. The standard reaction Gibbs energy. Now I can just rearrange this equation and calculate K. That's pretty easy for me to do. I just plug in that number that I just calculated. There it is right there and I can calculate everything else. Oops, that's my equilibrium constant. Sure enough, it's less than one. That reassures me that I might not have made a mistake, all right? I knew it was going to be less than one before I calculated it. Okay, what else? Calculate the total equilibrium total pressure of a system consisting of one mole of N2O4 and one liter of 298.15 degrees Kelvin. Remember G-CAM, remember doing these problems? You're going to have to be able to do problems like this again. It's the second time you've had to do this probably. Maybe you didn't like it the first time. I hope you like it better this time. I like these problems myself. One mole of this, no moles of this, here's the change. Minus X is the number of moles that are going to be consumed as the reaction proceeds from left to right plus 2X because there's a 2 here, all right? So the total is going to be 1 minus X and plus 2X. Remember this? Okay, here's our equilibrium constant expression. Where did that P0 come from? I know I've got products over reactants so I've got the partial pressure of N2O squared over the partial pressure of N2O4 but then I got an extra P here. Where did that come from? We need to just talk about this one time and then we never need to talk about it again, hopefully. Every one of these pressures is really the partial pressure divided by the pressure that characterizes the standard state. Every one of these pressures. Every time you write a pressure in equilibrium constant expression did anybody ever tell you this in GCAM? No. It's left out and you're just told that equilibrium constants are always going to be dimensionless, right? But the reason they're dimensionless is because this is not the partial pressure of N02, it's the partial pressure of N02 over P0, right? And so if you work out the algebra, if you write PN02 over P0 squared and PN2O4 over P0, all right? You end up with an extra factor of P0 in the denominator that makes this expression dimensionless. You see, you've got P squared in the numerator and you've got P squared in the denominator and all the units are going to cancel, all right? So it's not magical that equilibrium constant expressions have no units. It's because every pressure or every concentration that you input into an equilibrium constant expression is really a ratio between the concentration of the thing that you're talking about and the concentration or pressure of the standard state, okay? Now as you'll see in a second, it becomes important that you not forget about this P0, all right? And you can always insert the right number of P0s just by looking at the expression. If you leave out the P0s and you write the expression the same way that you would in GCAM, you can go back and look at it and go, oh, I can see there's going to have to be extra factors in P0 either in the numerator or the denominator to balance the units. You can stick them in after without thinking about them too hard. That's the pressure corresponding to the standard state, one bar, okay? So now I'm just going to make substitutions from the ideal gas equation. I haven't forgot my P0. I'm going to square this just like that's squared right there and then I'm going to plug for this N. I'm just going to plug in from my expression that I generated right here and so in the numerator I got 2x squared and in the denominator I got 1 minus x right there. Yes, that's from there and that's from there, right? Okay, and then now look at this cluster of variables here. I've got r times T over V multiplied by P0. P0 is one bar, V is one liter, T is 298.15 degrees Kelvin and r in units of liter bars per Kelvin per mole. Which is not a factor that I can remember, all right? That's r in units of liter bars per Kelvin per mole. All of these units cancel but this thing is not equal to 1. It's equal to 24.789. So you've got to divide your equilibrium constant by that factor to get the right answer, all right? The P0 is important. Without that P0 you're not going to get a dimensionless quantity in this pink rectangle here. It's not going to be dimensionless. You see, bar cancels with bar, liters cancel with liters, moles cancel with moles, K cancels with K. Okay? So now I can solve for x in the normal way. I cross multiply, generate a quadratic equation, use the quadratic formula. This C parenthetically will always be negative in these problems. C will always be negative. That's important because that means that this operation underneath this square root sign here is always going to be an addition, okay? If you remember that it's somewhat helpful. Your hands fly across the keys of your calculator because you've done this a billion times, all right? And it's always a plus. Okay? And here's what you get. Of course there's also a negative root that you neglect because it's plus or minus. Okay? So that is the x and now we can calculate finally what these two, what the number of moles are going to be, the final number of moles for both of these two chemical species, 1 minus x, 2x. Here's the final number of moles. But we weren't asked for the final number of moles. We were asked for the final pressure and the final pressure is just the sum of the moles times RT over V, right? Because there's two species. We've got to add them up to get the total number of moles of gas and then RT over V gives us total pressure which is that number right there. This is just taking you back to G chem, all right? But that stuff is important. We do need to remember how to do these problems. Okay, so French guy, you know how this can be important in the future. You remember that name, Le Chatelier. That's a name that you know from G chem. If a chemical system in equilibrium experiences a change in concentration, temperature, volume, or partial pressure than equilibrium is just to counteract the imposed change and new equilibrium is established. That's what he's famous for. He didn't win a Nobel Prize though. He just missed out on that. At the end of his career, he wrote, I let the discovery of the monosynthesis slip through my hands. It was the greatest blunder of my scientific career. Nobody could figure this reaction out in the early 1900s. Everybody was working on it and Le Chatelier even had the right catalyst, all right? Anybody know what the right catalyst is? Silver works, but that's not the catalyst that was actually used. It was iron. The reason Le Chatelier didn't understand this reaction is this is one of the rare exceptions to the Le Chatelier's principle. It doesn't follow his principle. He was thinking about his principle and this reaction doesn't follow it. Why? That's the extra credit question. That's the extra credit question for midterm too. So if you know the answer, in between now and then, you can try and figure it out. Let me say it again. Why does this reaction, which is catalyzed by iron, not follow Le Chatelier's principle? One sentence, right? It's not a long drawn out thing. There are no equations. I want you to just tell me. And who solved this problem? Well, it turns out there's two guys, Fritz Haber and Karl Bosch, right? Fritz Haber figured out the chemistry first and won the Nobel Prize in 1918 for doing that. And then Karl Bosch commercialized the process. And those two guys are the most influential, not scientists, not the most influential scientists. They are the most influential people of the 20th century because they figured this out. According to Nature magazine, a magazine about scientists, right? All right, here's, what am I plotting here? This is the population of the world in units of billions. All right, and this is the year, all right, and here is what the population of the world was doing. And here's when the Haber-Bosch process was invented. Early, about 19, early 1900s. All right, here's what happened to the population of the world. This could never have happened without that. All right, that's why they're the most influential people of the 20th century. What's so important about this reaction? Fertilizer, all right, you need to fix nitrogen. You have to separate nitrogen from its super strong triple bond in order to make proteins because proteins have amino acids in them that don't have triple bonds, right? Half of the protein in your body was prepared synthetically by this reaction, half of it. Half of every protein in your body came from this reaction right here. Isn't that phenomenal? That's called impact, all right? That's a huge impact on your life. Where did that fixed nitrogen come from before the 1920s? You can't see it here, but these are cows and this is a mountain of, exactly what you think it is, all right, on the, most of the world's seagull guano was on the coast of Chile, Peru, and Bolivia, all right? There's a long coastline on the west coast of South America where there are mountains of seabird guano, and that is where all of the fixed nitrogen came from in the world, all right, at this time in the early 1920s. There were wars fought over it, all right? War of the Pacific, read about it, all right? It was fought over bird guano, all right? It was a super important commodity. It's called salt peter, salt peter, right? Potassium nitrate, all right? Critical component in the manufacture of explosives and gunpowder, too, all right? So very important strategically in the early, late 1800s and early 1900s, all right? I think, very interesting. Okay, so Le Chatelier did get the principle right. There are reactions that are, that run counter to what his principle predicts, all right? Almost all cases of reactions that don't obey Le Chatelier's principle have something to do with catalysts. Catalysts mess up the principle, all right? So that's part of a clue, all right? If you want to get the extra credit problem right, you have to think about the catalyst. Why doesn't the catalyst allow Le Chatelier's principle to be obeyed for the Haber-Bosch process? What does the catalyst do to prevent that from happening? It's about the catalyst, okay? Le Chatelier's principle says, position this equilibrium should shift to the left because the shift will result in reduction of the total. So in other words, if you increase the total pressure of a reaction vessel that contains a mixture of these gases, you increase the total pressure, you expect equilibrium to shift to the left because the pressure of the system will be reduced if you do that, right? If I increase the pressure of this reaction vessel, we expect equilibrium to shift to the left because here we've got two moles of NO2 and only one mole of N2O4 and so the pressure of the system will go down if equilibrium shifts from right to left. That's what the Le Chatelier's principle predicts and that's what happens most of the time if there's no catalyst involved. Can we determine the relationship between X and P total for this reaction and verify the prediction of Le Chatelier? Can you work it out, right? In detail and in the one minute that we have left, we don't really have time to do this but let me just click through these slides very quickly and show you. This is what we did a minute ago. We have to calculate the mole fraction of each of these two components. Oh, we're not going to get through this today, so on. All right, so we'll see you on Wednesday.