 So we've seen that the Boltzmann probability distribution tells us how likely a system is to occupy a particular state, the jth state of a system. If we know the energy of that state, we can calculate p sub j, as long as we also know beta. And we've seen that beta is inversely proportional to the temperature, 1 over kT, where k is a proportionality constant. And if we want a numerical value, for now we're using Boltzmann's constant k. And this quantity q is just the sum of all these terms we've called Boltzmann factors, the sum of all these e to the minus energies, either e to the minus beta energy, or if we'd rather use temperature, which we usually would, it's e to the minus energy divided by kT. So we're now in a position where we can do some numerical examples and see what this Boltzmann probability distribution means for some real chemical problems. So let's work an example. And let's go back to the example of a butane molecule, butane C4H10. So we've worked a few examples with butane before, and I've just given you experimental values for the energies, or actually for the probabilities, not for the energies. But we're now in a position to see where those probabilities come from. So let's make sure we understand a little bit about the confirmations of butane. So butane is C4H10 and butane is a linear hydrocarbon molecule. I can write the structure of butane like this, and I have not drawn the hydrogens on these two of these carbons, because I want to draw them in three dimensions using bond and wedge notation. So I've intentionally drawn the three-dimensional structure here so you can see that I've put the methyl group on one side of the molecule with a dihedral angle on 180 degrees away from the methyl group on the other side of the molecule. So this would be the conformation that we would call anti or trans, because this methyl group and this methyl group are on opposite sides up and down from each other. If you prefer a Newman projection, this would be the Newman projection for that butane molecule, where if I'm looking at the molecule from this side of the molecule on the near side, the methyl group looks like it's pointing upwards and on the far side, the backside of the molecule, the methyl is pointing downwards. So either way, you think about the structure. The three-dimensional structure has this form we call anti. The other way we can draw that structure, if I continue to show the methyl group on the left side pointing upwards, instead of putting the methyl group down on the other side of the molecule, I can put it in one of these two positions coming forwards out of the plane of the glass board here. So the other two substituents are hydrogen. The Newman projection in this case in the front methyl group points upwards and in the back. Again, if I look at it from this direction, it's rotated clockwise by 60 degrees. So that would be the conformation we call gauche and in particular, gauche plus. And the third conformation we need to think about is the gauche minus. And the only difference between the gauche plus and gauche minus is instead of rotating forwards by 60 degrees, it's rotated backwards by 60 degrees. So we can start with the Newman projection and the three-dimensional conformation would look like this. So again, looking from this side, one of the near methyl group is pointing straight up. On the backside of the molecule, it's rotated counterclockwise by 60 degrees. So those are the three different conformations of the butane molecule. I can tell you the relative energies of those molecules. So the energy, let's say we'll label those EA for the anti-EG plus and EG minus, the energy of the gauche plus and the gauche minus states. If the energy of the anti-state is 0, if that's what I'm calling the 0 of energy, then the energy of the gauche plus and gauche minus states are both equal to each other. And those are both 3.6 kilojoules per mole higher than the anti-state. So these two states are higher in energy than the anti-state. So I can draw an energy diagram that shows what that looks like if I draw those energies on a ladder. The anti-state is the lowest energy. The two gauche states are higher in energy by some amount of energy, 3.6 kilojoules per mole elevated by 3.6 kilojoules per mole, but they're the same energy as each other. So I have three states in total. My question then, after all this setup, my question is, what is the probability that I'm in the anti-state? And what are the probabilities that I'm in the gauche plus and minus states at room temperature? So if I know the temperature and the energies, I can calculate the probabilities using the Boltzmann distribution. So the first step in doing that would be to calculate this quantity q. So let's calculate q. That's the sum of these Boltzmann factors. So I need to calculate the Boltzmann factors for each of my three states, for the anti, for the gauche plus and for the gauche minus. And then q will be the sum of those Boltzmann factors. It'll be e to the minus e anti over kt, e to the minus e gauche plus over kt, and e to the minus e gauche minus over kt. We can stick numbers in and see how that works. Some of the numbers are relatively easy. The anti is 0. So my 0 of energy is here. So when there's a 0 up in the exponent, that first exponential, that first Boltzmann factor is 1. So e to the 0 is 1. The only one that's going to take any work is the gauche terms. So each of these terms looks like e to the minus energy of the gauche state, 3.6 kilojoules per mole. I need to divide that by kt, but first let's get kilojoules per mole into the same units as k is going to have, Boltzmann's constant. So there's 1,000 joules in a kilojoule. And to get rid of the moles here, I can use Avogadro's number. There's 6.022 times 10 to the 23rd of something in a mole. So that gets rid of the moles in this energy in kilojoules per mole. And after cancellation, moles have canceled, kilojoules have canceled, and I'm left with just joules. So that's still up in the exponent. I need to divide by kt. So I'm dividing by Boltzmann's constant since I want a number. I'll use 1.38 times 10 to the minus 23rd joules per Kelvin in the denominator of the exponent. And I also need the temperature, 298k in the denominator of the exponent as well. So that whole exponential represents this Boltzmann factor. We need to plug those numbers into a calculator. And let me go ahead and write what that equals before I think about the third Boltzmann factor. If I plug in e to the minus 3.6 times 1,000 divided by Avogadro's number divided by Boltzmann's constant divided by the temperature, 298. What I get is 0.23 with no units. And let's just double check the units in the denominator, 1 over Kelvin cancel to Kelvin. So I have a joules in the denominator that cancels a joules in the numerator. There's no units left in the exponent as it should be because it's a unitless exponent. So e to that quantity in the exponent with a negative sign comes out to 0.23. So I haven't yet considered the third exponential, the third Boltzmann factor. But that's just going to be just like the second one, e to the minus g plus. The energy of the Gauch plus state, 3.6 kilojoules per mole, is exactly the same as the energy of the Gauch minus state, also 3.6 kilojoules per mole. So the third Boltzmann factor and the second one are identical. So I really just need to include that one twice. So I can include it 0.23 for Gauch plus, the same number 0.23 for Gauch minus. And that gives me, adding the numbers in my head, would give me 1.46. I know from having worked out on the calculator with rounding, these are more like 0.23 and some change. So when I add them together, it rounds up to 1.47. So this quantity, q, turns out to be 1.47. That's our intermediate result. That just gives me the value of q, which I need to calculate the probabilities. But we're almost done with the calculations. If what I want to know is the probability that I'm in the anti-configuration, then p of any state is 1 over q times the Boltzmann factor for that state. So p anti is 1 over q e to the minus e anti over kt. But again, the anti-energy makes things easy. That energy of 0, e to the 0 is just 1. So this is going to be 1 for the Boltzmann factor divided by 1.47 for q. The number we've just obtained for q. And when I do that division, that turns out to be 0.68. So there's a 68% chance that a butane molecule at room temperature will be in the anti-conformation. That should sound familiar when we've worked problems before. I've told you that there's a 68% chance that a butane molecule is in the anti-conformation. Now we see why that's true. It's the Boltzmann distribution and the energies of the states that tell us why that's true. If we want to know the probability that the molecule is in the Gauch plus conformation, really we know that these two have to add up to everything that's not 68%. So we could take a shortcut to the answer, but we could also say it's 1 over q e to the minus e Gauch over kt. We've already calculated that Boltzmann factor. That's 0.23. So it's 0.23 for this Boltzmann factor divided by 1.47 for q. That works out to be 16%. So 68% of the molecules are anti. 16% of them are Gauch plus. An equal number, 16% of them will be Gauch minus. And that adds up to 100% of our molecules. So that's the numerical answer to the question we've been asking. What's the probability of being in each state? Given that we know the energies, given that we know the temperature, they work out to these particular values by using Boltzmann's equation. Let me point out one more thing, which is what Boltzmann's distribution tells us about probabilities as a function of energy. So we started out several video lectures ago knowing first just that the probability decreased when the energy increased. We knew that if a molecule has high energy state and a low energy state, it's less likely to be found in the high energy state than the low energy state, mainly from everyday intuition about how we know molecules behave. That certainly is true. What we've now seen is that actually probability looks like e to the minus energy. It's a decaying exponential function of the energy. So probability drops as an exponential function of the energy. So the higher the energy is, the Boltzmann factors tell us that the ghost state is less likely to be found than the anti-state because of this exponential. And now that we also know that beta is equal to 1 over kt, the latest evolution in our understanding of this probability not only is it a decaying exponential function of the energy, but that function depends on temperature in this way. It's e to the minus energy over temperature with Boltzmann's constant in there to do a unit conversion. So if this is what it looks like at some temperature t, if I do it at a hotter temperature, so still the probability is decaying as the energy increases, but it doesn't decay as quickly because the temperature is large. If I do something at a hot temperature, relatively similar probabilities for two states with different energies. On the other hand, if I do this at a cold temperature, where we know at cold temperatures, it's quite likely that we'll find molecules in the ground state as opposed to the excited states. And the reason for that is the steepness of this exponential for cold temperatures since temperatures in the denominator, that exponential factor is large, and it's a steep decay in the exponential. So now we know not just that high energies are less probable, not just that the energy decays exponentially, but the slope of that exponential, the decay constant, and that exponential depends on temperature in a particular way. So what we're ready to do next is continue with examples talking about how to calculate probabilities if we know the energies. But turns out we can do something a little more powerful than what we've done here and calculate probabilities for some states even when we don't know all of the energies that the molecules can have.