 So as we've used the Boltzmann distribution, we've discovered that it's useful to be able to compare the difference in energy between two systems to kT, this ratio delta E over kT. And the reason that that's useful is because this ratio delta E over kT shows up in the exponential of this relative probability distribution. So for example, suppose that that ratio delta E over kT is one or two or three, some relatively small number. So that means that difference in energy is a few times the value of kT. So we say that's equal to a few kT. Then what that means is e to the minus one or minus two, either minus three, those are small numbers. They're less than one, but they're not ridiculously small. They're not vanishingly small. So they might be 30%, 10%, 5%. So that means the population in this upper state, p sub i, relative to the population of the lower state, is a small number but not vanishingly small. So we get the probability of occupying state A. That's less than the probability of occupying state B because its energy is higher if delta E is larger than zero. But not so small that we can ignore state I. On the other hand, let's go to the extreme of saying that difference in energy is large and all I need for it to be significantly larger than kT is have it be more than just a few kT. Maybe even 10 kT doesn't have to be 100 kT or 1,000 kT, maybe just 10 kT, more than just a few kT. Then that ratio, delta E divided by kT, if that ratio is 10, e to the minus 10, that's all of a sudden a pretty small number. So that means that the probability of being in state I is quite a bit smaller than probably being in state J, often small enough that we can completely ignore it. If e to the minus 10 is a very small number, much less than one, the probability that molecules are in state I is much smaller than the probability that they're in state J, so we can often ignore that probability. On the other extreme, however, let's say that state I is still the higher energy state, but the difference in energy between state I and state J is small compared to kT, so that ratio delta E divided by kT is quite small, much less than one, so that means the exponent that I'm evaluating here is e to the minus some small number, e to the minus 0.01, e to the minus 0.01, something like that, e to the minus negative but small number is a value close to one, so that means the probability of being in state I is about the same as the probability of being in state J. It'll be a little bit less because the energy is a little bit higher, but for most practical purposes, the probability of occupying those two states are very similar to each other. That tells us immediately if we just know how many kT's this difference in energy is, whether it's a few or many or a tiny fraction of kT, that immediately tells us something qualitative about how populated these two states are. In order to use that, however, we need to know what is the value of kT, and of course the answer to that question depends on what the temperature is, but if we pay attention in particular to room temperature, 298 Kelvin or 300 Kelvin, temperature is near where we will do most of our work at room temperature, then the product of Boltzmann's constant times the temperature, 1.38 times 10 to the minus 23 joules per Kelvin, if I multiply that by something in the vicinity of 300 Kelvin, the units cancel, joules cancel, I'm sorry, Kelvin cancel. What I'm left with is just joules, kT is a measurement of energy as we'd expect because it's divided by energy in this unitless exponent, so if we multiply this out, that quantity turns out to be in units of joules, 4.1 times 10 to the minus 21st joules, so that's a small number, that's a significant number because that is essentially the yardstick, the benchmark against which we compare other energies to find out whether states are populated or relatively unpopulated at room temperature. If I convert that to different units, if I convert it to units of kilojoules per mole, turns out if I multiply by Avogadro's number and divide by 1000 to get kilojoules, that'll work out to be about 2.5 kilojoules per mole, or if I prefer to work in kilocalories per mole, a non-SI unit, it's still relatively common for describing energies of chemical reactions, it works out to be about .6 kcal's per mole. Those values are useful, worth memorizing in fact, because if you're given an energy, let's say for example, I tell you that a carbon-hydrogen bond in a typical alkane, the bond energy for that bond is about 400 kilojoules per mole. That's a value, it's got a numeric value, it's got a unit. Any time you see a value, you can ask yourself, is that significant, is it a large number, is it a small number, what does it mean? Until this discussion, that number might not have meant too much to you, it's just a value, an energy with a particular unit. But now that we know how to compare energies to the value of kT, 400 kilojoules per mole, how does that compare to kT? Well kT is in units of kilojoules per mole is only two and a half. So 400 is quite a few multiples of kT, it's more than 100 kT. So that's much, much larger than kT, certainly more than just a few kT. So we're in this circumstance, the probability, so what probabilities am I talking about now? What two things differ in energy by 400 kilojoules per mole? That's the energy of an associated CH bond relative to an associated or a covalently bonded CH bond that takes 400 kilojoules per mole to break that bond. So the dissociated atoms have more energy by 400 kilojoules per mole than the covalently bonded atoms. So what this tells us, knowing that that energy is many, many, many kT, that tells us that at temperatures like 300 Kelvin, at room temperature, there's almost no probability that the molecule is going to be found in the higher energy state. CH bonds don't spontaneously dissociate and have their hydrogens dissociate off of the carbon because that energy is so much higher than kT. That's a good thing. I'm thankful that most of the CH bonds and the organic molecules that make up my body are not at much risk of dissociating at room temperature. I'll give you another example, a different type of bond. So what we've learned from this is CH bonds in general, covalent bonds, CH bonds in particular, covalent bonds more generally have enough energy, so much larger than kT that they don't spontaneously break at room temperature. A different type of bond, let's consider the hydrogen bond between two water molecules, for example. So this is not a covalent bond, this is the hydrogen bond. That hydrogen bond energy is about six kilojoules per mole. So is that a large number? Is that a small number? It depends what we compare it to. The most useful thing to compare it to again is kT. So in this case, how big is a six kilojoule per mole hydrogen bond relative to kT? Well that's just a couple of kT. Two kT would be five, three kT would be seven and a half. So it's between two and three kT. That falls in this ballpark of being a couple of kT if you kT. So it's true that the more energetic state, so again the more energetic state is two water molecules with that hydrogen bond broken. The lower energy state is the state with the hydrogen pointing towards the oxygen and with the hydrogen bond formed. The higher energy state, the dissociated molecules are less likely than the associated state, but not dramatically so. So in this case e to the minus two, two and a half, three, that's a number that might be down around five or ten percent. So what that means is in a sample of water, liquid water for example, maybe five or ten percent of the molecules have their hydrogen bonds broken at any particular point, but that's a significant fraction, a fraction that we can't afford to ignore. And in fact that's what causes liquid water to be liquid water. If all the hydrogen bonds were formed and permanent the same way covalent bonds were breaking relatively quite infrequently, then we wouldn't have liquid water that diffuses. But if five or ten percent of those water molecules can break their hydrogen bonds at any given point and then move on to form a hydrogen bond with a different water molecule, that's what leads to the properties of water being a liquid and diffusing at room temperature. So this tells us quite broadly if we compare energies, the size of an energy to KT because of the way that the energy and KT show up in this exponential in the Boltzmann distribution, we can immediately get at least a qualitative sense even without doing much math about whether the two states that we're interested in are roughly the same probability. One is somewhat more probable than the other or one is so improbable that we can ignore it.