 So, we're going to be talking about vibrational partition functions, but let me just say that you did well on quiz 2 again, very happy with that, 69 of you got an A out of 107, it's outstanding. So, there'll be another quiz on Friday and then a week from Friday, we have midterm one that's coming right up. So, we'll have more to say about that. Okay, so we were talking about this symmetry number on Friday. This is a confusing topic, it isn't discussed very much in your book, all it is, is it's the number of indistinguishable orientations of a molecule, period. The number of indistinguishable orientations of a molecule. So, for simple molecules, you don't need any fancy thought process to discover what the symmetry number is. You can just take the molecule, label the atoms, turn it around, and figure out for yourself how many indistinguishable orientations there are. For example, SO3 is a molecule we talked about on Friday, now I've taken it and I've labeled the oxygens so we can keep track of them. Let's figure out what the symmetry number is for SO3. I think everyone can see that if I take this guy right here and I rotate him by 120 degrees, the 120 degrees rotation rotates this guy down here, this guy goes over here, and this guy is going to end up over here, and so that's 120 degree rotation right there, I just took the molecule and I went er, right, and if I do it again, this three is going to rotate down here, er, okay, so there's three orientations shown on the screen. Is that it for this molecule? What if I take this guy and I flip him over like that? What did I do? I moved, I flipped him over so that the two is down here now, the one is up here but the three didn't move. All right, I took the guy and I flipped him over like that, all right, with these guys I took this guy, I rotated, I rotated, now I flip. All right, does this look exactly like this? It does. All right, the oxygens are all in the same place but this is not the same as this, this, or this, all right, I kept the labels on these oxygens. Does everyone see that? The reason I could do this is because this molecule is flat, I think you can see if I do this flipping thing with ammonia, I'm not going to get something that's indistinguishable because these three guys, in the case of ammonia, are hydrogens, they're either going to be sticking out of the screen or sticking in the screen and when I flip the molecule over, it's going to look totally different, all right, so the fact that I can flip this guy over and get an indistinguishable orientation is only because he's flat and then once I've got this, I can do the 120 degree rotation business again and so I end up with one, two, three, four, five, six indistinguishable orientations for this molecule that are possible. I don't think there are more than that, all right, so you can just physically count them. The symmetry number is just the number of indistinguishable orientations, right, that's what it is. Now I have a little recipe for finding this that I tried to tell you about on Friday and the reason I like this scheme is because it works when the molecule gets bigger and more complicated and you don't want to label every atom and turn it around and flip it over and count and count and count, all right, this gives you a faster way to find the symmetry number. What we do is we identify an n-fold symmetric axis. What does that mean? Well, here's an axis for SL3 that's two-fold symmetric, right, I say it's two-fold symmetric because I can do 180 degree rotation of this guy and I get the same orientation. So it's a two-fold symmetry axis. Then you decide whether this axis contains a mirror plane right, does this axis contain a mirror plane? If I put a mirror right here, does this side of the molecule look just like this side of the molecule? No, right, so this axis does not contain a mirror plane and then you count the number of these axes and you multiply by that number and then you multiply by two if the answer to two was yes. How many axes are there? One, two, three, count the number of these axes, in this case there's three, right, multiply by the n from the n-fold and then you multiply by two if there was a mirror plane. Okay, so in this case we've got three axes times two-fold symmetric times one because we're not going to multiply by two, two times three is six. Symmetry number is six. Everyone see that? Three steps, identify a symmetry axis, decide whether it contains a mirror plane. In this case it doesn't. I can't put this bar anywhere along this axis and get a mirror plane, okay? If I could do that then I would be multiplying by two and the symmetry number would be 12. Okay, let's do this guy, aluminum chloride. Once again I can, because the molecule is small, I can put a label on every single atom and I can turn it around and rotate it and I can figure out how many indistinguishable orientations there are. I can just do it myself without using any formulas, right? Here's the aluminum chloride, I labeled all the atoms. Now if I rotate this guy like this, okay, the one is going to rotate down here and the two is going to be up here, the four is going to rotate around to where the three is, the six is going to rotate up like that. So I think you can see if I rotate the molecule like this, I'm going to get this guy right here. Everyone see that? And now if I take that guy and I rotate him like that, what happens? The six is going to rotate down to where the one is, right? The five is going to rotate up to where the two is, all right, but the three and the four are not going to rotate, right? The three and the four are going to stay right, the three is going to stay in the front because I'm rotating around like this, I'm sort of rotating around that axis that goes right through the three and the four, okay? So can everyone see that if I do that rotation, I'm going to end up with this guy and he's different from this guy and different from this guy, but you can see they're all indistinguishable, if I didn't have these chlorines labeled, you couldn't tell the difference, could you? Okay, and then I can, with this guy right here, rotate him like that again, all right? And if I do that, five rotates down, six rotates up, four rotates back, sorry, four rotates the front, three rotates the back, you get the idea, yes, yeah. Now if I take this guy, I'm going to prove it to you in a second, what if I take this guy and I rotate him like I did that guy and two goes down here, one goes up there, right? If I rotate like that, the one is going to go up, the two is going to go down, I get this guy right here if I do that, there are no other orientations, okay? And if you don't believe me, mess around with it, all right? Four possible orientations, the symmetry number for this molecule must be four, right? How do we find it with the shorthand method? Well, choose an axis, there's two possible symmetry of the axis that you can choose. In fact, there's three, we could use yours also, all right? Your axis goes up and down, right through the center of the molecule, right? Okay, the one I'm using goes right through these guys, these two guys right here, these two aluminums, okay? But we could also use one that goes right through these chlorines right here, all right? There are in fact three orthogonal axes that we could use for this guy, all right? But let's just use this guy for the sake of argument, okay? Is there a symmetry plane along this axis? Now, of course, if I had some artistic talent, I could draw it properly, but is there a symmetry plane there or not? How many people think there's a symmetry plane? How many people don't see it? Okay, well, that's pretty good. See, yet the answer is yes, there is, all right? And if I could draw this properly, you'd see that this side of the molecule looks exactly like this side of the molecule if I could just draw this properly, okay? So, there's one axis, there's two-fold symmetry, and the mirror plane says that that has to be a yes, the side where there contains a mirror plane, yes, and so we multiply by two. Two times two is four times one is four. Everyone see that? And if I had chosen this axis, it would still be yes to the mirror plane question, all right? It would still be two, right? Because I'd be rotating, I'd have to rotate this only, it's only got two-fold symmetry there, all right? So it would still be one axis times two-fold symmetry times two, and if I chose this axis like this, it's still going to be one axis times two-fold symmetry times two. You should get the same answer every which way you look at it, otherwise, if these answers are not self-consistent, there's a mistake somewhere, yes. No, there's only one axis like this, there's two of the two aluminums, there's only one way to draw an axis through those two aluminums, right? If I wanted to draw the axis through these two chlorines right here, there's only one way to do that, okay? If I want to draw the axis through the center of the molecule up and down like this, there's really only one way to do that, all right? So there's really only one of each one of these, yes, there couldn't be more than one mirror plane. That's impossible. That's like some kind of fun house, two mirror planes, no, you can have the most one, but you might have none. Let's do benzene, okay? Once again, we can do the labeling experiment, call that one A, B, C, D, and so on, all right? I can rotate them down here, that's two orientations, you can't tell the difference, all right? If I go all the way around and I flip them over like a pancake and I do it again, I'm going to have 12 orientations, does everyone see that? All right, if I just rotate them by 60 degrees, boom, boom, boom, boom, I'm going to get six, and then I flip them over like a pancake and I do it again, I'm going to get 12, right? All right, how does the Penner method give this? All right, there's one, two, three equivalent axes that I can draw through the carbons. Everyone see that? One, two, three, does this axis contain a mirror plane? Right there, yes, okay? So the calculation is three axes times two-fold symmetric times two, two times two is four times three is 12, boom. Everyone see that? I could also draw the axis right through the center of the molecule coming out of the screen, I could choose that as an axis too, all right? How many of those axes are there? One, is there a mirror plane? Yes, it's right in the plane of the screen, it's right there. It goes right through, cuts the molecule right in half, okay, there is a mirror plane and so if I do that calculation, one axis times six-fold times two is 12, you get the same answer, right? It's six-fold symmetric now, one, two, three, four, five, six, for that axis it's six-fold symmetry, not two-fold. Everyone see that? Okay, we're not going to beat this to death anymore, but you should be able to look at a molecule and figure out what its symmetry number is. Any which way you want, label the atoms, rotate it around in your head, cut out the model, okay? On the exam it'll be easier if you can do it my way. Okay, now estimate the rotational partition function for HCL at 25 degrees and 250 degrees, see? It's B value is 10.59 wave numbers, rotational constant, okay? Actually pretty big, rotational constant, pretty big energy for a rotational constant. Okay, here's our choices, here's the equations that we flashed on the screen on Friday, we're going to use this guy, I hope you've noticed that this guy and this guy are basically the same equation because if the linear molecule lacks a center of symmetry, sigma equals one, okay? So these are just the same equation for goodness sakes, okay? But in this case, the molecule lacks a center of symmetry. So we're going to use this guy. That's the rotational temperature, theta sub R. The rotational temperature is just B over K, B in units of joules, of course, we're going to use this equation, okay? And so the rotational temperature here, I can calculate what it is, 10.59, convert to joules, divide by K, I get 15.23 degrees Kelvin. That's the rotational temperature, pretty cold, huh? Actually, it's usually colder than this, all right? But that's pretty cold, 15 degrees Kelvin. Okay, so this temperature is much lower than either one of these target temperatures. That's 25 degrees C, not K, all right? That's 250 degrees C, okay? So we expect to have lots of thermally accessible rotational states, don't we? All right, we're at temperatures that are way higher than the rotational temperature of this molecule, right, 15 degrees. Okay, so let's calculate what the partition function is. This is our equation, it's just KT over B, so there's KT. This is just unit conversion for B, all right? 10.59, convert it to joules, 10.59, convert it to joules. Only thing that's different is the temperature. That's the partition function I get. Do those numbers look right? What do you think? They look about right? I mean, we expect something significantly bigger than one, don't we? We're at temperatures at 25 degrees C, we're at 298 K, the rotational temperature is 15, all right? So lots of rotational states are going to be occupied. 15 degrees is when we start to have multiple rotational states occupied, all right? We're way above that. We're going to have lots of rotational states occupied qualitatively. That looks about right. Notice that 250 is not 10 times 25 in Kelvin units, all right? It's about a factor of 2, all right? And that's also what we see here. So we probably didn't make a math mistake here, okay? That was easy. What about methane? Let's calculate its partition function at 298 degrees K. It's rotational constants 5.2412, wave numbers. These rotational constants are often known to 4, 5, 6 sig figs, all right? Because in spectroscopy you can measure the frequency very, very, very accurately, right? It's not too often in physical chemistry. We can measure something with this kind of precision, all right? But with spectroscopy you can measure rotational constants, all right, with very high precision. Five sig figs, all right? We know it to within one part in 10,000. We need the big boy for this, all right? He's a nonlinear molecule, methane, okay? The moments of inertia, though, are all the same because of the symmetry of the molecule. So let's just be cubed, okay? We need the symmetry number. What is it going to be for methane? Hmm, think back. Four vertices. What's the symmetry around each vertex? Three, did someone say three? Four vertices, each of them is three-fold symmetric. Is there a mirror plane along those axes? No. So what's the symmetry number? Right, 12. Okay, so we plug everything in. This is 1 over 12, blah, blah, blah, blah, blah, unit conversion, blah, blah, blah. 36.711 is the partition function that I calculated. Does that seem about right? Yeah, it's higher than what we saw for HCL, all right? But B is a lot smaller, all right? B is half as big as it was for HCL. So at this temperature, we expect Q to be larger. It's about twice as large, all right? But, hey, B is half as big. So it makes perfect sense, okay? These rotational partition functions are pretty easy. We just, we've only got these three equations. And in fact, it's really only two equations. Okay. Now, vibration. Yes? That is the rotational constant about each of the three principal axes of the molecule, X, Y, and Z. And so in the case of methane, because it's four hydrogens and it's tetrahedrally symmetric, those three rotational constants are all going to be the same. Now, if it was chloromethane, if you had one chlorine and three hydrogens, then there would be one unique rotational axis that had a different B value. So it would be A, B squared, okay? And you'd have to be told what they are, right? You'd have to be told. That's why we need that big equation. Okay. So, harmonic oscillator. Remember, way back when we were talking about it. Here's the harmonic potential. One-half K, X squared, or R minus RE, R is the equilibrium bond distance, everyone remember Hooke's law? Evenly spaced energy levels. This is what they are. All right. New plus one-half times H new, or V, rather, plus one-half times H new. That's the vibrational quantum number. Okay. We have our normal expression for the partition function. We just plug this energy in and we've got the partition function, right? It always works this way. And then all we do is we simplify this, right? But we'll always start with this. This is always the expression for the partition function no matter whether we're talking about translation, rotation, vibration. You can derive the partition function by just plugging the energy into this expression and simplifying it, always. Now, to make this a little bit simpler, we often neglect the zero point energy. Later on, if we calculate the energy using the equations that I'm going to show you, we can always add in that zero point energy at the end. We know what it is. We can put it back in if we want to make sure that we don't neglect it. Okay? So if we neglect the zero point energy now, the one-half goes away, we get this guy right here. So if I write that series out, all right, it looks like this. V equals zero, one, two, three, and so on. Okay? Now, there's a nice closed form expression for what this series sums to, right? It's going to sum to this if that's an X. Of course, it's not an X, it's an H nu over kT. Okay? So we're going to use the same form. We're just going to substitute an H nu over kT and that's our vibrational partition function right there. Really simple expression. How big is it? Well, let's just say a typical vibration is 2,000 wave numbers. Now, that's actually a pretty energetic vibration. All right? But 2,000 wave numbers. I think you'll agree, all right, that's not as energetic as a proton vibrating. It's got a lower energy than that. Okay? And so that's what the frequency turns out to be. And so if I plug in that number and I calculate what Q is, I get almost exactly one for the partition function. All right? At room temperature now, the partition function is 1,00064. Okay? So this is totally different than what we saw with translation and vibration, translation and rotation. With translation, we had millions of accessible states, didn't we? We had a partition function that in one dimension, if we had a one micron box, was 67,000. Remember that? Make it a three-dimensional box. You've got millions of accessible states. With rotation, we get 20, 30, 40 rotational states that are accessible at room temperature. With vibration, we get one. All right? What does that number mean? It means that molecules are almost always in their ground vibrational level at room temperature. All right? It is unusual for the vibrationally excited states of a molecule to be excited at room temperature. All right? That would only happen if you had really heavy atoms like I2, right? Iodine has a vibrational energy of 214 wave numbers. How much thermal energy is there at room temperature? 200. Okay? So Iodine is excited. All right? But my goodness, I mean that's 226 grams per mole atoms, bowling balls, all right? They're barely moving, all right? Very low energy. Doesn't take much energy to excite that vibration at room temperature by golly. It is excited, all right? But it's the exception. Okay? So at room temperature, just one vibrational state is thermally accessible in general. There's exceptions, but in general, all right? Very different from translation or rotation. Now, let's do a calculation. The triatomic molecule chlorine dioxide has three vibrational modes at frequencies of 450, 945, and 1100 wave numbers. What's the partition function of 298 degrees Kelvin? Okay? How much thermal energy is there at room temperature? 200 wave numbers. What is the lowest energy vibration here? 450, okay? So we want to keep that in mind because what do we expect the partition function to be here? Just looking at this problem before we calculate anything. We're expecting partition number, partition function of 10, 100, how about 1? It should be close to 1, 1 and change, right? 1 and something, right? 1.0 something maybe, right? Is that what our intuition is telling us? Yeah. Because the thermal energy is way lower than that. Okay. Let's do it. There's our molecule. How many vibrational modes does it have? Three. All right? But it's a good question to ask. Did I say anything about the degeneracy of these three states here? I didn't say anything about it, but we know that it's a nonlinear molecule, so that's 3n minus 6. And it's 3, 3 times 3 is 9 minus 3. So there's only three vibrational states in this molecule. If I give you three energies, you know they're all non-degenerate. Right? Okay. So it's 3n minus 6, it's 3. And so the partition function is just going to be that times that times that. What's the partition function? What's that guy? All right? I just take my expression for the partition function. I plug in my beta, my H, my C, all my constants. That's the temperature that we're talking about to get 1.12. Actually a little higher than we were expecting. We expected 1 and change, but 1.0 something. All right? It's a little higher than we expected. Okay? And if I do the same calculation for 945 and 1100, I should get smaller numbers, yes. Smaller than that. Okay? And so the overall partition function is that times that times that or that. Right? Now, do these numbers all make sense? Well, you know, first of all, they're all 1 and change. Secondly, that one's bigger than that, which is bigger than that. All right? If these aren't in the right order, that's a dead giveaway that we've made some kind of mistake. Okay? And so qualitatively, yes, we can live with that number. It seems about right. It's a little higher than what we were expecting maybe, but just barely. Okay? Now, if we make it really hot, this is 200,000 degrees. This is 100,000 degrees. All right? Then you can, in principle, have lots of vibrationally accessible states and they increase quasi-linearly with temperature. Of course, the molecule falls apart way before you can do this. All right? There's no such thing as a bond association energy that is large enough to allow you to see this. But if you could, if the molecule didn't fall apart at these temperatures, this is what you would see, all right? This is for a 1,000-wave number mode. This is for a 100-wave number mode. The vibrational temperatures are 144K and 1,400 degrees K. Okay? We can always calculate the vibrational temperatures just h nu over K. Yes, it would fall apart. That's only about one electron volt for most molecules. So, 298 degrees Kelvin is a high temperature for translation, a high temperature for rotation, but a low temperature for vibration. We want to keep that in mind. Right? Lots of translational states, lots of rotational states, but in general, very few vibrational states are thermally accessible at temperatures near room temperature. What about the energy? Well, we've got this nice expression that we derive chemistry's least intuitive equation. All right? All we have to do is plug in our partition function here and here. We can calculate the energy directly from that, okay? And so, you can do that, simplify it. There's your equation, 13.39. That's the energy for N molecules, right? That's the vibrational quantum. This equation emits the zero point energy, so we can add it in, all right? We can calculate the energy using this equation and then add in h nu over 2 to make that zero point correction if we're worried about it. If there are multiple states, we have a term like this for every state. In other words, more than one h nu, energy is additive. Makes sense. At temperatures that are so high, blah, blah, blah, blah, we can, if the temperature is much higher than the characteristic vibrational temperature, we can approximate the denominator as the first two terms that find infinite series. The infinite series looks like this. Take the first two terms, 1 minus 1 is zero. H nu cancels with h nu, so we just end up with big N over beta. N is the number of molecules, so the total energy is just NKT and the reason that's interesting is because that's what we were calculating when we did the equi-partition theorem. We wrote the classical Hamiltonian for vibration. It had two terms, and so it's 2 times KT over 2 is just KT per mode, remember this? Okay, so the equi- the equi-partition theorem tells us that we've got RT in terms of energy per mode or R versus in terms of the heat capacity, right? So we've got 3R over 2 or 5R over 2, 7R over 2, remember this? Yeah, well it turns out this was right, right? It's the high temperature limit. If you take the partition function and you look at the high temperature limit, you get the equi-partition prediction, right? For the contribution to the heat capacity and the contribution to the internal energy. Okay, here's a midterm exam question from a couple of years ago. You're going to have one like this on yours. On this exam, almost everyone got A wrong, but A is easy. What is the answer to A? A dichlorine oxide molecules cooled to 4 degrees Kelvin. How much vibrational energy does it retain at that frigid temperature? That's what they said. How much energy does that molecule retain at 4 degrees Kelvin? How much vibrational energy does it retain? We're physical chemists. Yes, 1 1⁄2 H nu. How many H nu's are there? 1, 2, 3, right? This molecule has 3 vibrational modes. You can't suck the zero point energy out of any one of them, even at 4 degrees Kelvin, right? The molecule is going to keep bending a little bit. It's going to keep asymmetrically vibrating and it's going to keep symmetrically vibrating even at 4 degrees Kelvin, right? Okay, so the answer is zero point energy at 680, zero point energy at 330, and the zero point energy at 973, calculate that, you get 991 wave numbers. It contains 991 wave numbers of energy even at 1 milli Kelvin, right? You can't remove that energy from the molecule. Quantum mechanics says you can't do it. And you guys have had 20 weeks of quantum mechanics, right? Does everyone see what I'm talking about here? You can't remove the zero point energy from any mode. Okay, I convert that to joules. Da, da, da, da, da, da, da, da, da, da, da, da. Okay, now one mole of Cl2O is warmed in a one liter container to 2,000 degrees Kelvin. What fraction of these molecules is the 680 wave number of vibrational mode excited? You're going to have something very similar to this to answer. Here's what the equations page of that exam looks like, okay? Equations are intentionally disorganized on this page. The purpose being that you have to fish out the right one and it will not be obvious which one to use. So if you don't know what you're doing, you could be in trouble. You need this guy and this guy. That's the partition function we just derived, right? For vibration, all right? And that is the normal equation. That's the Boltzmann distribution law, all right? In the denominator, the Boltzmann distribution law is the partition function we just derived this form of the partition function for that equation. So we're just going to plug that in for that right there. Okay, and what do we want? We want n little n over big n, don't we? We want little n over big n. We want to know what fraction of the molecules. Okay, now, here's the confusing part. Am I going to use in this partition function the overall partition function for the molecule? In other words, q, q, q, multiplied together or am I going to just use q680? Because I'm only asked about the 680 wave number mode. I'm asked is the 680 wave number mode excited? All right, you see that conundrum? Can everyone see that the difficulty in the calculation here is what are you going to use for this q down here? Are you going to use the overall partition function for the molecule which has three terms in it? Or are you just going to use the q680? You've got to decide. Here's how you make the decision. If you are asked to calculate the fraction of the molecules for which the 680 mode is excited, it's in its first excited vibrational level and the other two modes are anything, then you only use q680 down here. In other words, if the identity of these other two modes isn't specified, they can be anything. All we want to know is the 680 excited or not. Then you just put q680 down here. But if on the other hand you're asked, let's say that you're asked, I want to see if the 680 mode is excited and these other two are not. I want to know what fraction of the molecules have 680 excited and the other two modes, unexcited. So you've specified what the other two modes are doing. Hey, that's a different story. You've got to include all three. You with me? So if you specify all three, you need all three in the denominator. If you specify one, I only want to know what one is doing. I don't care about the other two. You only use the one that you care about. I don't think your textbook does a good job of explaining this fine point. So for one moment I just want to know what's right, it's my most excited. Okay, so this is our equation right here. All right, plug in the numbers. This is what the denominator is equal to right here. I just calculated that's what this is right here. I'm just putting the exponent rather. It's minus 0.4890086 and so when I plug that in here and here I get 23.7% of the molecules will have the first excited state excited. Okay, at 2,000 degrees Kelvin, does anyone have any intuition about whether this number makes sense? Because I don't. All right, 0.24%, 2,000 degrees Kelvin. All right, in order to figure it out we need to know what the characteristic vibrational temperature is at 680, all right? So we can calculate that 978, now we compare that with 2,000, does 2,000 higher than that? Yes. Okay, so we expect there to be an appreciable population in this vibrational state. All right, we're half of the way, all right? At 2,000 degrees, all right, we've put more energy into the molecule than necessary, right, than necessary to populate this state. So we expect this state to have appreciable population. Everyone see what I'm talking about? 2,000 degrees, 2,000, that is how much thermal energy it would take to start to populate this vibrational level, all right? At that temperature we would expect the vibrational level to just start getting populated, all right, and we're at 2,000 degrees, all right? So we expect there to be an appreciable population, all right, 24% could be, all right? It doesn't sound unrealistic. Everyone see what I'm talking about? Okay, now what if we did, all right, everybody remember by the way what these numbers up here, these superscripts in front of the element? Everybody remember what those are? I didn't use them here, but now I'm using them. What are those? What is that 35 right there? Yeah, it's the atomic mass. So I'm talking about an isotopically pure sample of chlorine oxide, all right? The 16 is the mass of the oxygen in atomic mass units, all right? The 35 is the mass of the chlorine. How do I convert that 35 into the actual mass per chlorine atom? How do I do that? What if I want to know the mass of a single isotopically pure chlorine 35? What do I do? What are the units of that 35? Grams per mole, all right? So if I want to know how much one chlorine 35, I just take 35, divide by Avogadro's number, right? Yes, that's what you do. Okay, it's got three vibrational modes, boom, boom, boom, all non-digital. So what fraction of these molecules have the 680 excited, the 330 excited, but not the 973? Now we're going to specify all three. I want one quantum of energy in that guy, one quantum of energy in that guy, and I want that guy to be in his ground state, all right? I could ask that question, right? If I ask the question that way, I'm asking for all three, so you've got to use the overall vibrational partition function down here. See that? And then that energy right there is the total energy, right, two tricky things. All right, you're going to use the overall vibrational partition function because I'm telling you I want all three of these things to be a certain way, any which way, any way I ask you for them to be. If that was a 2 and that was a 5, it would still be the same. I'm still telling you how I want those three states to be configured, all right? And then that energy is the total vibrational excitation energy in the molecule, the total, all right? So I said 1, 330 and 1, 680 add them together. If I said 2, it would be 2, 680s and 2, 330s, all right? The energy here is the total vibrational energy for all of the different excitations that you want to consider. You see that? I think this is rather confusing. I'm sorry? What if there was degeneracy? Yeah, so if 680 was doubly degenerate, there'd be two of these terms, right, two of the singly degenerate terms multiplied together for 680, okay? And we did the calculation, you get a smaller number, blah, blah, blah, blah, yes, it makes sense because it's smaller than it was before. It should be a lot smaller because we're now talking about a tiny subset. First of all, there's two things excited. The other thing, we're insisting that that be unexcited so there's a tiny fraction of molecules that are going to meet these specifications. Okay? All right, we'll see you on Wednesday.