 Okay, so after spending a little bit of time with the Langmuir model for studying adsorption, we've discovered that it works fine for describing the adsorption of a monolayer onto the surface, but doesn't work well, falls apart completely, in fact, when more than one monolayer can adsorb onto a surface at a time. So in order to study multi-layer adsorption, when more than one layer worth of adsorbate can adsorb onto a material, we need a better adsorption model. One such better adsorption model is this BET adsorption model. Like the Langmuir model, it's named after a person, or in this case, three people. BET stands for Brunauer, Emmett, and Taylor, I'm sorry, Teller, Brunauer, Emmett, and Teller adsorption model, the three scientists that first proposed and developed this model. So this model is, in fact, probably the more commonly used model for studying things like measuring surface area using the technique of adsorption. So it's important to understand what's behind this model as well. So like the Langmuir model, it's a lattice model. So we have some surface. We divide that surface up into lattice sites, specific positions where we can adsorb molecules or not. So I'll continue to label each one of these surface adsorption sites with an X. And then adsorption sites can either adsorb a molecule onto the surface down from the gas phase, or they can have no molecule attached. But now the difference between this model and the Langmuir model is, in addition to adsorbing molecules adsorbing onto the substrate, they can also adsorb on top of each other. So here's a molecule that is adsorbed into the second layer instead of just the first layer. We could, in fact, even have molecules adsorbing into the third layer and so on. So as before, we'll say, let's say we've got M total sites on this surface. And then instead of just being able to talk about surface coverage, what fraction of the surface is covered or not, we'll want to distinguish between, let's say, instead of just theta for surface coverage, theta sub 1 is the fraction of the surface that's covered by only one molecule. So in this case, with six surface sites, only one of those sites is covered by a single molecule, so one-sixth of the surface would be covered by one molecule. There's going to be a theta 2 and a theta 3 for the amount of the surface that's covered, the fraction of the surface that's covered by two molecules or by three molecules and so on. In fact, we can have theta n describing the fraction of the surface that's covered by exactly n molecules. And in addition to these, theta 1 and up, we can have theta 0, which tells us what fraction of the surface has zero molecules adsorbed to it. So this surface site, these three surface sites, have zero molecules adsorbed to them. So in this specific example, half the surface, theta sub 0 would be equal to one-half. Half the surface has no molecules adsorbed. So one thing we can notice immediately about these numbers is if I add those numbers up, how much of the surface is covered by zero molecules plus the fraction covered by one, plus the fraction covered by two, and so on, that has to add up to 100%. Every surface site is either covered by zero or one or two. There's no way to, those are mutually exclusive possibilities. They have to add up to 100%. All right, so that's the basics of our model, what we hope to get out of this model. First, just as a reminder of what the Langmuir model has told us, whether we plot surface coverage or the volume of gas that gets adsorbed to a surface. For the Langmuir model, we found that that behaved like this. It asymptotically approached either full surface coverage, theta equals one, or adsorbed volume equal to the monolayer volume. So that's what was predicted by the Langmuir model. With this BET model, by allowing the model to predict adsorption of more than one molecule onto a specific lattice site, we expect that this ought to maybe behave like Langmuir when there's less than a monolayer absorbed. But it ought to be able to cross above one full monolayer coverage and be able to predict two or three or four monolayers worth of adsorption. So the specific form of that, we won't know until we explore this model a little more closely. All right, so we'll proceed in the same way as we did for the Langmuir model, at least one of the ways we proceeded for the Langmuir model. We can model the adsorption and desorption of the molecules off the surface using kinetics. So let's write down the chemical reactions that would take place when a molecule adsorbs onto the surface. If there's an empty site and one of these gas-phase molecules encounters it, then the system can form an adsorbed molecule. Previously, I would have written A adsorbed, a molecule stuck to the surface or adsorbed onto the surface. But now we need to distinguish between a molecule adsorbed in the first layer or a molecule adsorbed in the second layer, a molecule adsorbed in the third layer, and so on. So I'll write A sub 1 as my notation for a molecule adsorbed onto the first layer. So this would be an A sub 1 molecule, a molecule adsorbed onto the second layer, like this one that would be an A sub 2. So this reaction, there's a forward direction and a backwards direction, each of which has its own rate constant. The only way to get a molecule in the second layer, to adsorb onto the second layer, is if it encounters a molecule that's already adsorbed on the first layer. If this gas-phase molecule binds onto this A1 molecule in the first layer, it can form an A2 molecule in the second layer. So that reaction would be A1 plus a gas-phase molecule could form an A2. And again, that reaction has some forward rate constant that I'll call K2, the rate constant for the adsorption of the second layer molecule. K minus 2 is the rate constant for the desorption of the second layer molecule back into the gas phase. And then we could write down the equations like this for any layer we want. If I have already N minus 1 layers stacked up and I adsorb one more molecule, that will generate a molecule in the n-th layer, and that's the equilibrium, I'm sorry, the rate constant, I'll call KN for adsorption and KN negative N for desorption into or out of that n-th layer of the adsorbed species. Okay, so that sets up our model. That's the reactions we're willing to consider and a picture of what the system looks like and a description of what we're trying to analyze, the fraction of the system that's adsorbed a certain number of layers. So the next step is to ask ourselves what happens at equilibrium. If we're not in the process where molecules are still falling down and accumulating onto the surface, but if we've reached equilibrium, then at equilibrium, it must be the case that the number of molecules in the first layer is not changing. The rate of addition of molecules into the first layer and the rate of disappearance of molecules out of the first layer, that must be zero. They must sum up to zero. So we'll start by saying, and we could do that for molecules in the first layer, molecules in the third layer, molecules in any layer we want, the easiest one to consider first is the rate of appearance or disappearance of molecules out of the empty layer. So an X molecule is a molecule, an X surface site is a site without any molecules attached to it. So if I ask myself how quickly is the rate of unattached surface sites changing, I can create X sites as a product of this reverse reaction, the desorption out of the first layer. So here's an X, the rate at which this negative one reaction is occurring, K minus one times the amount of A1. So that's the rate at which I'm forming X via this desorption reaction out of the first layer. I can get rid of X in the opposite direction. So I'm consuming X, consuming empty surface sites, sorry, that's a positive one. With this reaction, rate constant K1, the reactants in that reaction are the empty surface sites X and the gas phase species. So the amount of gas phase species, I can represent that by the pressure of the gas phase, or it's at least proportional to the pressure of the gas phase. So this is the thing that I want to be equal to zero. So I need for this rate constant times A1 to be equal to this other rate constant times the amount of X and the pressure. So I can, when I rewrite that, I'll say K negative one times A is equal to K1 times X times P. But instead of writing A, what I'll write down, the amount of A1, the amount of molecules that are in the first layer. So remember, we don't count these because these are buried underneath molecules in the second and third layer. So the A1s are molecules, surface molecules that are at the top layer. So this is the only A1 molecule, the fraction of molecules that we call A1 molecules. That's just the total amount of surface M multiplied by the fraction covered by a single molecule. So A1 I can write as M times theta one. Likewise, over on this side, X, the number of unattached surface sites, that's just the total number of sites multiplied by the fraction of them that don't have a molecule attached. And then the P, I'll just leave alone. So now that I've written that equation, I've got an M on both sides that can cancel. The Thetas don't cancel because this is the fraction occupied by single molecules. This is the fraction that's unoccupied. So those are not the same number. So if I rearrange this, I'll rearrange this equation to solve for theta. Let me put that result over here where I'll collect a few of these results. So theta one is going to be equal to K1 divided by K minus one times P times theta zero. So I have a relationship between what fraction of the surface is occupied by a single molecule is some things multiplied by the fraction of the surface that's occupied by no molecules. So that's progress. I have some relationship between two of these variables. To make a little more progress, I can do this same thing not for unattached surface sites, but for singly attached surface sites, sites with a single molecule. So if I ask how quickly are these first layer molecules appearing or disappearing, they appear in this reaction. So overall, the rate of change of these species should be zero at equilibrium. So I'm creating A1 molecules with this first layer adsorption reaction in the forward direction. So that's a K1x, all right, as m theta naught. And then gas phase piece, the amount of that will give me pressure. I can also create A1 molecules not only here, but here. If this second layer reaction proceeds in reverse, if I desorb a second layer molecule into the gas phase, that leaves behind a first layer molecule. So I can form an A1 molecule with this negative two reaction. So let's see, the first one is K1 in the forward direction. I need to include K2 in the negative direction. And then the reactant for that is A2. The amount of A2 is m times theta two. So far so good. So those are the two ways of creating A1 molecules. I also need to account for the fact that I can consume A1 molecules either by doing this reaction in the opposite direction, by desorbing one of these first layer molecules into the gas phase and getting rid of one of these A1 molecules. So that would be a minus K negative one. Reactants in that reverse reaction are A1, which I'll write as m theta one. Or I can consume these A1 molecules by doing reaction two in the forward direction. So that's getting rid of A1s. K2 is in the forward direction. The reactants are A1. The amount of A1 is m theta one. And A, so that's pressure. All right, so there's four terms in this expression. Those have to sum to zero. It's a little more complicated than the previous one. Again, I've got some m's that will cancel. Since I have a zero on this side, I'll just divide through by m and all the m's go away. Notice that two of these four terms, I've got a term after getting rid of the m that looks like K1 theta naught p. That looks just like this K1 theta naught p. I've also got a K negative one theta one. That looks just like this K negative one theta one. The result of this previous equilibrium was that this term has to equal this term. So in my new result, I have one of these terms minus the other one of the terms. So those two terms have to completely cancel. And that's what I've learned by using my first equilibrium result. So after all that's done, all I have left is these other two terms. I'll go ahead and rewrite that one. So I've got K minus two. So zero is going to be equal to K minus two theta two minus K two theta one p. Now I can rewrite, I can solve for this equation in terms of either one of the thetas that I want. If I solve for theta two as a function of theta one, I'm going to get theta two. I'll bring this term over to the other side. I'll get K two and a pressure and a theta one. If I divide by K minus two, that's the result again. So the amount of molecules covered by exactly two, amount of surface sites covered by exactly two adsorbate molecules is this ratio of rate constants times the pressure times the number in the layer below theta one, the fraction in the layer below. Theta one of course in turn was related to theta not. So I could rewrite this one as actually I'll hold off on doing that rewriting for just a minute. So we've done this process for layer one and for layer two. We can imagine doing it again in layer three. You can probably predict what's going to happen now. I'd encourage you to go ahead and write out the kinetic reaction, set it equal to zero and confirm that this is what's going to happen. But if I ask for the n-th layer, again, there's going to be four terms that involve creating or destroying molecule N. I can either adsorb a new molecule under the layer below to make a new A sub N. I can desorb a molecule out of an overfilled layer and expose an A sub N or I can do those reactions in reverse and consume the A sub N. If I write out those four terms, use the results of the previous equilibrium to cancel two of the terms. What I'm left with is that the n-th layer is related to the layer beneath it by this ratio of rate constants, case of N over case of minus N. Now it's time to talk maybe a little bit about the size of these rate constants. I've got essentially an infinite number of rate constants that I've written down. Forward and backward for the first layer, second layer, third layer, n-th layer. But think about what those rate constants mean. Case of one and case of minus one, that's the rate constant for a molecule in the gas phase sticking to an empty surface site. That's related to the binding energy of that surface site. Likewise, desorption, that's probably more closely related to the binding energy of that surface site. Layer two, that's a molecule sticking not to the surface but to a molecule sticking onto a first layer molecule. So that's binding of an adsorbate molecule to an adsorbate molecule. Likewise, layer three is if I bind a gas phase molecule to this surface site, I'll create an A3 molecule, a molecule in the third layer. Again, that's going to be driven by the chemistry of adsorbate binding to adsorbate molecule. So it would seem like a reasonable thing to assume to say that for each of these layers beyond the first one, the rate constant with which a gas phase molecule sticks onto a molecule of its own type doesn't much matter whether it's in the binding to the first layer, second layer, third layer, and so on. So each of those adsorption, forward rate constants should be the same. Likewise, for the reverse reactions, the desorption reaction doesn't matter whether you're desorbing out of the fifth layer or the second layer, those rate constants should all be the same. The only one that certainly will be different is the K1 and the K-1. When I bind to an empty site, I'm binding to this substrate species. I'm not binding to the adsorbate species. So K1 and K-1 are different. All the others should be the same as each other. So what that means is this ratio K2 to K-2, K3 to K-3, Kn to K-n. All those quantities, these two pink boxed quantities, should be the same number as each other. I'll define that as usual, a ratio of rate constants we can describe as an equilibrium constant. So we'll call that the equilibrium constant for this reaction. And the equilibrium constant for binding the second layer, third layer, fourth layer, and so on, those are all going to be the same number. We'll just call those capital K. So I can rewrite theta 2 as K times P theta 1. Theta n is K times P times theta n minus 1. I can't necessarily do the same thing for layer 1 because this ratio of rate constants is different than the others. What I will do, however, that's still going to be some equilibrium constant. It's not going to have the same value as this one. I could call it K prime or K2 or give it a different name. What I'll do instead is say it's going to be some multiple of this rate constant. Whatever the value of this rate constant is, this one might be the same, might be larger, might be smaller. It differs from the other one by some new constant that I'll introduce that's called C. So theta 1 is going to be C times K times P times the occupation of the layer beneath it. So those are what we have for our results of how much the occupation is of each individual layer relative to the one beneath it. That becomes most useful when I work them all the way down. So theta 2 is some constant times pressure times theta 1. But we know what theta 1 is, so I can rewrite this equation as theta 2 being equal to theta 1 is C K P theta naught. If I multiply by an additional K and an additional P, that looks like K squared and P squared. If I do that again for the third layer, I'll get an additional K and additional P. If I do it again for the nth layer, I'll get several more copies of K and P. So if I do the same thing with this equation, theta n is going to be equal to again C times K times P times theta naught. And then every time I've made a jump from theta 1 up to 2 to 3 to 4 to n, I've introduced an extra factor of K and an extra factor of P. So theta n, the fraction of the surface is occupied by exactly n layers of molecules, is this ratio of the two equilibrium constants multiplied by K, the equilibrium constant for the adsorbate, adsorbing onto itself. Race to the n power, pressure race to the n power times theta naught. So if I just knew theta naught, I could bootstrap my way all the way up and learn theta 1 and theta 2 and theta 3 and so on. So that's good. We've solved this Brunauer-Emmett-Teller model in one sense, except we've got two problems. Number one, we don't know what theta naught is yet. And number two, rather than talking about how occupied each layer is, we'd really rather talk about volume, how the volume adsorbed onto the surface depends on the pressure. So we haven't yet written this equation in the form of an isotherm that will help us compare the BET model to the Langmuir model. So that's what we'll do next.