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# CHEM 125a: Freshman Organic Chemistry I

## Lecture 36

## - Bond Energies, the Boltzmann Factor and Entropy

### Overview

After discussing the classic determination of the heat of atomization of graphite by Chupka and Inghram, the values of bond dissociation energies, and the utility of average bond energies, the lecture focuses on understanding equilibrium and rate processes through statistical mechanics. The Boltzmann factor favors minimal energy in order to provide the largest number of different arrangements of “bits” of energy. The slippery concept of disorder is illustrated using Couette flow. Entropy favors “disordered arrangements” because there are more of them than there are of recognizable ordered arrangements.

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html## Freshman Organic Chemistry I## CHEM 125a - Lecture 36 - Bond Energies, the Boltzmann Factor and Entropy## Chapter 1. Chupka and Inghram’s Determination of Graphite’s Heat of Atomization [00:00:00]
Okay, so how do you get the energy from graphite to an atom? One way is spectroscopy, and we talked about that last time, that if you see what the minimum amount of energy you can put into CO Now Professor Sharpless said — you know, he talked about increasing dimension of carbon; start with an atom, go to a line, polyacetylene and then to a bunch of double bonds, and finally to a saturated hydrocarbon — he said that carbon atoms are very hard to come by. You need a really, really low vacuum to get it. Right? That is, the equilibrium constant for carbon atoms coming together to form bonds is very, very favorable. Right? So it’s hard to get atoms. So how are you ever going to get atoms? Well let’s think about that problem here. So we know that the equilibrium constant at room temperature is k. Then you know ΔE. So, but if the equilibrium you’re trying to measure is between graphite and carbon atoms, and the energy to take a carbon atom out of graphite is 170 kilocalories/mol, then the equilibrium constant is 10^{-127}. Right? And that means, since there are only something of the order of 10^{80} atoms in the universe, there would not be, at room temperature, a single carbon atom at equilibrium with graphite if everything in the universe was graphite. Right? So it’s not a very favorable equilibrium constant. What could you do about it, if you wanted to measure the equilibrium constant, and in that way get the energy difference? Any knobs you can twist? Lucas?
Now, of course, exactly what one means by the concentration of graphite needs to be — you scratch your head a little bit about that. Right? But you don’t actually need to know it, because if you could measure the pressure of the C atoms at equilibrium with graphite, at very high temperature — call that the pressure of carbon, right? — that is some constant, and that constant will include whatever this concentration of graphite should be. So B. So multiply both sides by that concentration of graphite. Right? So we get some constant on the right, and then that heat of formation, in the exponent, that we want to find. So that means we could write — if we took the log of both sides. You have the log of the pressure of carbon atoms, is the log of whatever B is — that’s some constant — minus this other term. And what that means is that if you — that that minus the heat of formation of the carbon atom, divided by So that’s not easy to do. But it was done in 1955 by Chupka and Inghram. And this is the sketch of their instrument, and I’ll show you how they did it. First there’s a graphite cylinder. Okay? And around it is a can made out of tantalum. Now why tantalum? Because that’s about the highest melting thing you can get; it’s the highest melting metal. So 3293 Kelvin is its melting point. So you could really heat the heck out of this thing. Okay? And now you surround that with wires, and those wires are made of tungsten, because that’s very high melting too. And so you connect wire — you connect electricity to the tungsten, and also to the tantalum. So electrons boil off the tungsten and bombard the tantalum, and when the electrons hit it, they heat it up. So you can heat the heck out of this tantalum can by bombarding it with electrons. And that, of course, heats the graphite that’s inside. Now, so inside there’s going to be a gas of carbon, in equilibrium with the graphite. Now it won’t just be C atoms. It’ll also be C Now what they did, Chupka and Inghram, was to drill a tiny hole, in the top of this thing. And that will allow a little bit of the gas to escape; not so much that you destroy the equilibrium inside, just a really slow leak. Right? So now, if you could measure the amount of these different carbon species leaking out, you could know what the pressure of them was inside the can. Okay? Now, so there’s a beam of carbon species coming out, gaseous carbon species. This is all at very, very high vacuum. Right? And then up there, an electron beam comes across and hits these carbon things, and knocks electrons out of them. And that converts them into C+; C
Now you know which was the right one, measured by spectroscopy. It was the one that said 171. So this experiment, actually measuring the equilibrium between carbon atoms and graphite, by a really gargantuan kind of experiment, is what settled the question finally. So that when you look at the appendix of this book, Streitwieser, Heathcock and Kosower, you find that there are heats of formation for atoms and radicals, measured by spectroscopy and things like this. And there you see carbon, 170.9. And that was done by Professor Chupka, who was in this department and who used to come and tell people how he did this experiment. But as you can see, he passed away in 2007. So thanks to Professor Chupka. And the nice thing is, once that’s done, it’s done. Now you know it and you just plug it in when you burn your stuff and want to know what its energy is. You can get it relative to carbon atoms in the gas phase. ## Chapter 2. Calculating Equilibrium Constants from Bond Dissociation Energies [00:14:20]Okay, now how good are these spectroscopic experiments? Well this is a neat thing. The heat of atomization of methane, measured in the ways we’ve just been talking about, is 397.5 kilocalories/mol. Now that comes from mating a carbon atom with four hydrogen atoms; that is 397.5. Right? So we know what the average bond energy. There are four such bonds. So each one’s worth 99.4 kilocalories/mol. So about 100 kilocalories/mol for a C-H bond; that’s convenient to remember. Okay? But that’s not how much it costs to take a single hydrogen atom away from methane. Taking a single hydrogen atom away from methane, the so-called “bond dissociation energy”, which is the actual experimental energy it takes to do some particular process — average bond energy is just an average, but the individual ones are not the same — taking a hydrogen away from methane is 104.99, plus or minus 0.03 kilocalories/mol. Close, but not the same thing. And then you have CH So these are very good experiments. So we know, through heroic spectroscopy and this work of Chupka and Inghram, we know what these energies are, bond energies, and bond dissociation energies. So here are average bond energies in a table that you have at the end of this organic chemistry text. And it says a carbon-hydrogen bond is 99; and now you see where we get that. And you see that a carbon-carbon bond is 83. But the second carbon-carbon bond, in a double bond, is only 63. Right? Why is it weaker? Why is the second bond of a carbon-carbon double bond weaker than the first? Pardon me? Devin, what do you say?
sp. So the second one is probably more than 20 kilocalories weaker than the single one. But at any rate it’s 146, that you add up to get a double bond, and 200 for a triple bond; which means the third bond is worth only 54 kilocalories/mol. Okay? And in C=O, it’s about the same as C-C. So C-O is eighty-six. But the double bond, notice, is different in this case. Now the second bond is stronger than the first. So the carbonyl group is an especially stable group. Okay? So you have the question, can you sum up these average bond energies and get useful heats of atomization? So can you look at a structure and say how stable it’s going to be?^{2}Okay, so let’s try it. So here’s heats of atomization by additivity of average bond energies. So we have these average bond energies from the table. A C-C single-bond is 83. A C-H is 99. C double bond C is 146, 86, 111 and so on. And we’re going to sum them all up to get the heat of atomization — compare it with the actual heat of atomization. Okay, so for ethene, there are four C-H bonds, there’s one C-C double bond. Add them up, and you get 542. The actual heat of atomization is 537.7. So there’s an error of 4.3 kilocalories/mol, which is less than 1%. That’s pretty good. But on the other hand, ethene probably entered in to determining these average bond energies. Right? So it’s not 100% fair. Okay? How about cyclohexane? Now we have 6 carbon-carbon single bonds and 12 carbon-hydrogen single bonds: 1686 versus 1680.1. An error of only 5.9, less than half a percent error. So pretty good, by adding up bonds. Cyclohexanol. Remember we had quite a bit of trouble with these partly oxidized things before, when we were trying to just do it on the basis of the elements. But if you add bonds together, you get within 0.3% of the right value. Or if you do glucose, which has lots of oxygens in it, then you get within again less then 1%; 0.7% of the right value. So this is pretty darn good; very impressive, very small errors, to predict these. But the question is, is it useful? How accurate does it have to be, to be useful? So why do you need to know the values? Because you want to know equilibrium constants. You want to know which direction a reaction will go, for example. Okay, so we know that the — if we want to calculate an equilibrium constant, we can do it at room temperature with this 3/4ths trick. So the calculated equilibrium constant is whatever we’re calculating here, for energy, between two things. We have two things: calculate the energy of these, the energy of these, compare the energies, and that’ll give us the equilibrium constant, according to this formula. But notice I’m doing it on the basis of calculation. Now so that’s — whatever — the calculated energy is whatever the true energy would be. But there’s also some error in there. Okay? But if you add two exponents, that’s the same as multiplying two things together. So the calculated equilibrium constant is the true equilibrium constant — that’s the first part, the part that it has with So if you want the error to be small, that factor to be small, then So let’s try it with the equilibrium between a ketone and the so-called enol, which is an isomer of a ketone in which a hydrogen has been taken from the methyl group on the right and put on the oxygen, and the double bond moved. Okay? So that’s a very important equilibrium that we’ll encounter when we talk about the chemistry of ketones. Let’s see what the equilibrium constant — should there be more ketone or more enol? Do you have a guess right at the outset of which one would be more stable? I would guess the ketone, because of what I just told you, that the C-O double bond is remarkably stable. Right? In the other case you have a C-C double bond. Okay, let’s see. Now we could add together all the bonds. But most of them, most bonds are the same between the starting material and product. We only need to compare the ones that change. Okay? So we’ve highlighted in red the bonds that change between the two forms, the two isomers, because we’re interested in the difference in energy between these two, to get the equilibrium constant. Okay, so these are the numbers I took from the table, that you see on the top left there: 179 for C-O double bond and so on. And I sum them up and that’s 361, for those bonds. And the new set of bonds, in the enol, sum to 343. So the ketone indeed is more stable, it appears, by 18 kilocalories/mol. So 18 kilocalories/mol means that you have a factor of 10 Now why? Well one thing is that we — that those bonds that we cancelled, the C-H bonds that didn’t change, in fact did change between the starting material and the product. Why could I say that they did change? In both cases, on both sides, there are single carbon-carbon, carbon-hydrogen bonds. How can I say they change? Angela?
sp, on the carbon, as you go across. And the ^{2}sp’s on the right should be more stable. Okay? So the ^{2}sp-H should be stronger. So these things that I was saying cancelled do not actually cancel, if we take hybridization into account. So that’s one factor. And there’s another as well, which is you have that unshared pair, on the top right here, on the oxygen, is adjacent to a double bond. That means that this high HOMO can be stabilized by the ^{2}π* low LUMO; it’ll overlap. That isn’t a possibility here, where the unshared pairs on the oxygen did not overlap with the π* orbital. So you get intramolecular HOMO/LUMO mixing in the enol that you don’t get in the ketone; which will help stabilize the enol, with that — we could draw that resonance structure. So those two things together make up that 9 kilocalorie error; or at least we can think — they contribute to it at least. So constitutional energy, what we would get by adding bonds together, has to be corrected for various “effects”, we’ll call them, such as resonance, that’s what we just looked at, like this HOMO/LUMO thing, such as hybridization changes, or such as strain, as in the case of axial methylcyclohexane, that we looked at. So there are lots and lots of these corrections that you have to apply to this model, where you add together bond energies in order to predict the energies of a particular molecule.But for many cases now, you can do a pretty good job of predicting these things, and actually not do so bad at predicting relative energies of isomers, and therefore equilibrium constants. And these effects, of course, are a polite name for error. Right? They’re correcting — various ways of correcting errors that you think there should be in this scheme of just adding bonds together. Now, energy determines what can happen. Things always move toward equilibrium. Right? So if the ratio of two things is something, but the equilibrium ratio is different, the ratio will always move toward that, toward the equilibrium, if it’s in isolation. But there’s another equally important thing is how fast will it go there? And that, as we’ve seen before, can be approximated as 10 ## Chapter 3. The Boltzmann Factor: How Is Temperature Related to Energy? [00:27:55]Now both of these things suggest that being low in energy is good. Right? You favor things that are low in energy. But you might ask why? That’s not what people say about money. They don’t say the less money you have the better. Right? Why the less energy the better? This is a really interesting case, and it has to do with statistics. And especially at Yale we should talk about this, because in 1902, when Yale celebrated it’s bicentennial, they published a number of books showing off the scholarship of Yale; as you can see here. And the most important of those books, by about 500 miles, was this one: Now when you do this, you get exponents. And the organization of our presentation here is going to have to do with three different ways in which statistics enters into exponents, for purposes of doing equilibrium. So there’s the Boltzmann Factor; that’s what we’ve been talking about, the S = k lnW . So log relates to an exponential, and we’ll see why that is. And here’s his tombstone in the cemetery in Vienna. And you’ll notice, up at the top there, S = k ln W. Okay?So what Boltzmann considered was the implication of random distribution of energy. Suppose you have a certain amount of energy, in a system, but it’s distributed at random. Right? So purely statistically. Then how should it be distributed? How much energy should any particular molecule have, is the question. And we can visualize this in a simple case, which is very like what he did, except he did it analytically and in a much bigger system. But just using four containers, which are like molecules, and each one can have a certain amount of energy in it. And we’ll consider the energy to be — to come in bits. He used that idea, that there were bits of energy to be distributed among molecules, or degrees of freedom within molecules. He didn’t think that energy came in bits, but it made it possible to do the statistics, and then he just took the limit when these bits get very, very, very, very small, so that it becomes like a continuum of stuff, like a whole sand of energy bits. But anyhow, let’s just count up how many different ways there are of putting three different bits of energy — or actually not different, they’re all the same — but three bits of energy into the red container. Okay? So if you — how many complexions — that’s what, any particular arrangement he called a complexion — how many different complexions have a certain number of bits in the first container? Well suppose you put all three of those energy bits into the first container. How many different ways are there of doing that? Just one. Okay? But suppose you put only two into the first container? Now how many different ways are there of arranging it so that there are two in the first container? How many different ways? There are three: 1, 2, 3. So there’ll be three ways of putting two bits in the first container. How many of putting one in the first container? Well we put one there; there’s 1, 2, 3, 4, 5, 6 ways of doing it. Okay? So there’s six ways of putting one in there. And how many of putting none at all in there? 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Okay? So there are 10 ways of doing that. So let’s make a graph and see how many ways — what’s the probability that you’ll have a certain number of bits of energy in the first container? Okay, it looks like that: 10, 6, 3, 1. And what does that curve look like, if you made a plot of that? What type of curve does it look like? Is it a straight line? Anybody got a name for it, or something that looks a little bit like that?
So it turns out that if you do this, the average energy is ¾ ΔH comes just because that’s what you expect. If you randomly distribute things, it’ll come out that way.## Chapter 4. Entropy and the Tendency toward “Disordered Arrangements” [00:36:24]Now how about the entropy factor? And this one is fun. Feynman, in his wonderful
[Movie plays]
So how can you have a disordered arrangement, if the shepherd sees a dragon? Okay? The situation favored at equilibrium, by entropy, is one where particles have diffused every which-away, not into a coiled up piece of paper like the yellow thing, or not into a dinosaur. Every which-away; the key word is ‘every’. That’s what’s statistical about it. A disordered arrangement is a code word for a collection of random distributions, whose individual structures are not obvious. So if a thing looks like, you know, a regular lattice like that, I say, “Ah ha, that’s a regular lattice.” But if it looks like this, I don’t say it’s exactly that; I say it’s disordered, by which I mean I can’t tell the difference between that one and this one, or this one, or this one. Right? So there are a whole bunch of those arrangements that I count together when I say ‘disordered’. It’s a collective word. So if all of them are equally likely, it’s much more likely to have disordered — many, many, many arrangements — than the particular ordered ones that we’re thinking about, even if they’re all equally likely. So that’s the idea. It is favored at equilibrium because it includes so many individual distributions. So entropy is actually counting, in disguise. You count all these different arrangements, or all the different quantum levels, and the more you have, under a certain name, the higher the entropy associated with that name is. So, for example, a very common value of the entropy difference between two things is 1.377 entropy units. That seems a weird number, right? Now 1.377 happens to be G is? That’s the Gibbs Free Energy, which includes both heat, both the kind of things — bond energy that we’ve been talking about — and entropy is included in there too. So we can split it apart, into the part that has to do with heat — or enthalpy, the ΔH between the two things, gauche and anti — and TΔS, the part that has to do with entropy. I suspect you’ve seen this G=H+TS before; H-TS before. But let’s just split it apart. Since they’re in the exponent we can multiply two things together. So we have the first part, the one we’ve been dealing with, 3/4thsΔH. Right? And then we have the red part, that has to do with entropy. But you can simplify that. How can you simplify the part that has to do with entropy, right off the bat?
Okay, we’re going to stop here. And just so everybody’s on the same page, we’ll have the final lecture on Wednesday. But then I’ll be here at class time on Friday too, and we can have a discussion then, to review for the exam. And I’m willing to have another one. I forget. When did I say? On Monday night. Now do people have — there are not exams at night, are there; or are there? Does anybody have a — is Monday night okay to have the review? It’s probably the best time to have it, so you have a full day after that before the exam. So I’ll get a room for next Monday night, a week from tonight, for a review session. But also I’ll be here on Friday at lecture time. So we’ll see you. If anybody wants to see this experiment, we’ll do it. [end of transcript] Back to Top |
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