 Okay, so now we know how to calculate an equilibrium constant from partition functions, and we've seen if we have reactions that only involve diatomic molecules, there's some significant cancellation that occurs, but not every reaction involves only diatomic molecules. So let's take a simple molecule that still involves two gas phase species, but with some monatomic atoms instead of diatomic molecules, so in this case two Br, two bromine atoms combine or dimerize to form bromine gas, Br2, diatomic bromine gas, so we know how to write down partition functions for monatomic species just as well or even easier than we do for diatomic species, so we can also write down the equilibrium constant for this reaction, that's going to be partition function of products raised to the stoichiometric coefficients over partition functions of reactants raised to their stoichiometric coefficients, so as long as we can write down partition function of Br2, which we've already done, partition function of Br atoms, which is even easier, then we can determine what this equilibrium constant is going to be, so for Br2 I've got a translational component, I've got a rotational component that looks like temperature divided by, I'll just say the symmetry number for bromine is 2, I've got a vibrational component, sorry this should be rotational constant for Br2, so still in the numerator I've got a vibrational component e to the minus vibrational temperature of Br2 over twice kT over 1 minus an exponential of the vibrational temperature of Br2 over kT, that's the vibrational term, I have a bond association energy term and I have an electronic partition function which is just 1 for Br2, that's the numerator, the denominator luckily is not quite as ugly, partition function for Br, I can write the translational piece 2 pi m kT over h squared raised to the 3 halves, but I'm just going to square it in the denominator so that 3 halves becomes a 3, also I've got a volume that shows up in the translational partition function and that gets squared because the whole partition function is being squared, for the rotational part and vibrational part and for the bond association part I don't have to worry about those at all, these is a monatomic atom so it doesn't have any rotational states, it doesn't have any bond to vibrate, it doesn't have any bond to dissociate and worry about a bond association energy, so I don't have to worry about these terms at all, I do have to worry about the electronic degeneracy, in this case let me say the degeneracy for Br2 is 1 and the g that I'm going to square for bromine atoms because bromine atoms have an unpaired electron in them they're doublets, the degeneracy is 2 for a bromine atom, all right so now we have a little simplification we can do, it's not nearly as much cancellation as we saw for the reaction that involved only the diatomic molecules, for example I've got a volume in the numerator and a volume squared in the denominator, I can cancel one of them but that means I'm left over with one volume in the denominator, likewise the 2 pi kT and h squared those are only going to partially cancel, I've got all those constants to the three-halves up top and to the third in the denominator, so after that cancellation, a minimal amount of cancellation I see that I've got, if I ignore the masses, an extra factor of 2 pi kT over h squared to the three-halves in the denominator after these cancel so I'll put that back up in the numerator and flip it upside down, I've got a 1 over v that didn't end up cancelling, I've got the rotational terms that are purely from the Br2 molecule, vibrational terms also purely from the diatomic molecule, bond association only matters for the diatomic molecule and then this ratio of degeneracies is a 1 in the numerator and a 4 in the denominator, so as before we can plug in some numbers and see what value this quantity has, let's say this partition function let's calculate at a temperature of 298 kelvin so every time I need to plug in a temperature I'll use 298, we know many of these constants I should, oh I've forgotten the masses so this was everything, not including the masses I've got to insert one more term here which is mass of Br2 to the three-halves in the numerator and a mass of Br cubed in the denominator, those masses, the mass of Br79.9 I believe, 79.918, sorry 79.918 grams per mole for the mass of bromine, double that for the mass of bromine molecules. Rotational constant, vibrational constant, dissociation energy we've already seen in the previous video where Br2 was one of the reactants so I'll use the same values there, if I do this arithmetic keeping the pieces separate at first, this contribution from these constants and the masses works out to be a very small number but notice that we have to be careful with units so the units of these h-squareds and k's and t's and masses don't completely cancel like they did in the previous case and the net quantity that we end up with has units of volume which is going to have to cancel the units of volume in this 1 over v term so if I do that in units of liters it turns out to be this ridiculously small number of liters that hasn't yet included this 1 over v term I haven't told you what volume we're going to use the rotational term is 1300 the vibrational term 0.58 the bond association energy contributes a very large contribution 2.1 times 10 to the 34th that number is unitless and then we have a 1 over 4 from the electronic degeneracy so notice we're in a different situation now than we were for the HBR formation reaction because this of this lack of cancellation we've got this 1 over v term we can't finish the arithmetic until we know what volume we're interested in so if I throw some bromine atoms into a box with volume 1 liter and I use 1 liter here then I can evaluate the constant in fact if we do that if we say volume equals 1 liter then the value of the equilibrium constant that we get is now a unit less 1.6 times 10 to 7th strongly towards product bromine not surprising perhaps that bromine atoms will strongly prefer to dimerize and form BR2 molecules or let's say I did this not in a volume of 1 liter let's say I make a box box as big as a large room let's say that's 50 meters on a side so fairly large auditorium scale room 50 meters on a side cube that number and I get a volume that's in meters cubed if I use that volume here doing my unit calculation appropriately and I calculate with the equilibrium constant is then the value I get is a small number so this K is less than 1 this K tells me the reaction is not any longer strongly towards product there's the volume is large enough that the bromine atoms have a more difficult time finding each other and they have a more difficult time reacting to dimerize to form BR2 so the equilibrium constant depends not only on the temperature in ways that we have seen before but it also depends on the volume being able to calculate the right value of the equilibrium constant will depend on what our reaction conditions are not just the temperature but also the volume so the other thing this tells us about these equilibrium constant calculations is when we have a reaction where the number of molecules on the product side is not the same as the number of molecules on the reactant side we have two molecules turning into one molecule power of two in the denominator power of one in the numerator that meant these cancellations didn't happen as significantly as they did in the previous example and not only did we not have all these terms canceling but that meant that we had to worry about units a little more than we did in the case where the number of molecules is the same on the reactant product side so what this means is when we have a reaction where the molecularity changes to molecules becoming one molecule it complicates our calculation of the equilibrium constant it also complicates unfortunately the math that we have to do when we solve equilibrium problems and so we'll see how that works coming up next