 Alright, so let's see if we can get to the bottom of this issue of how to deal with equilibrium constants in molecules versus moles using the same example reaction we've used before to bromine atoms dimerizing to form a bromine molecule. So we know that we can write the equilibrium constant as molecules of product over molecules of reactants all raised to their stoichiometric coefficient. That equation is fine. That's the equilibrium constant that we've just been calling K or maybe KEQ. Now since we're specifically writing in terms of molecules rather than moles, I'm going to call that K sub capital N, the equilibrium constant written in terms of molecules. It would be nice if we could instead think about an equilibrium constant using moles instead of molecules. Perhaps there's some equilibrium constant that is the ratio of moles of Br2 over moles of Br squared denominated in terms of moles rather than molecules. So to keep these two distinct, I'm going to call that one K sub little n, the equilibrium constant denominated in terms of moles. But of course we know the connection between moles and molecules of any species. The moles of Br2 is just molecules of Br2 divided by Avogadro's number. Moles of Br is just molecules of Br divided by Avogadro's number and in the denominator that whole quantity is squared. So if I simplify this expression a little bit, I've got molecules of Br2 in the numerator. I've got molecules of Br in the denominator squared and then I've got Avogadro's number showing up once in the denominator and then in the denominator of the denominator it shows up twice. So that simplifies of course and I can say that this mole based equilibrium constant is molecules of Br2 over molecules of Br squared and I'm left with just one factor of Avogadro's number. Now notice that this molecules of Br2 over molecules of Br squared, that's exactly the original equilibrium constant, the molecule denominated equilibrium constant K sub big n. So I can write the mole based equilibrium constant as the molecule based equilibrium constant, the one we know the value of, the one we know how to calculate the value of just multiplied by Avogadro's number. So for example for the Br2 reaction at 298 Kelvin and in a volume of 1 liter we've seen that the mole based, I'm sorry the molecule based equilibrium constant is 1.6 times 10 to the 7th. If I multiply that by Avogadro's number, 6 times 10 to the 23rd of anything in a mole then that product works out to be, what does that work out to be, 9.6 times 10 to the 30th and now I have units of per mole. So what that means is if we know what the equilibrium constant is, K eq or K sub n, big n, denominated in terms of molecules, we can multiply by in this case one factor of Avogadro's number to figure out what it is denominated in terms of moles. So if we did want to solve the equilibrium problem in terms of moles, if we wanted to use this expression to solve an equilibrium problem, we can now work in terms of moles, we just have to use this quantity as our equilibrium constant rather than this quantity, the K sub big n. That's all fine for this particular reaction, we had to do a little bit of work to figure out our value of K sub little n for this particular reaction. It's going to work out different for every reaction depending on how many factors of NA I have in the numerator and the denominator and how well they do or don't cancel. So let's do this more generally and see what we can say about equilibrium constants even before we know what the reaction is. So we know that the general expression for the equilibrium constant is equilibrium condition is the equilibrium constant in terms of molecules must be equal to products over reactants, molecules of each species raised to their stoichiometric coefficient. That's fine if we want to work in terms of molecules, if we want to write an equilibrium constant in terms of moles, that would be moles of products divided by moles of reactants each raised to their stoichiometric coefficient. So we'll just use the same steps we used over here, everywhere I've written moles, I can write that as molecules of that species divided by Avogadro's number. That's the number of moles, so that's what gets raised to the stoichiometric coefficient. I can pull the Avogadro's number outside of the product, so all I'm going to leave in there is the number of molecules of each species, that depends on this coefficient I, so I can't pull it out of the product. The number of Avogadro's number I'm going to pull 1 over NA out of the product. The question is how many times I pull it out. I pull it out for species I, stoichiometric coefficient of I. So each of the species reactants and products is going to have a different number of 1 over NA's that gets pulled out twice in the denominator here, once in the numerator over here, altogether the total number of times it gets pulled out is the sum of all those stoichiometric coefficients. So I've got 1 over Avogadro's number raised to the sum of all the Avogadro's, of all the stoichiometric coefficients, multiplying this product of molecules to their stoichiometric coefficients. But of course this product of molecules raised to their stoichiometric coefficients, that's exactly the equilibrium constant denominated in terms of molecules. So I can rewrite this as the mole based equilibrium constant is 1 over NA times K sub big N, 1 over NA is raised to the sum of stoichiometric coefficients. I'm going to rewrite the sum of stoichiometric coefficients as delta nu, the total change in the number of stoichiometric coefficients, total change in molecules. So for example, for this reaction, the change in stoichiometric coefficient, which is the same thing as adding up all the stoichiometric coefficients, if I take 1 for the product and minus 2 for the reactant, I've got one product and I lose two reactants every time I gain one product, that sum of 1 and negative 2 is negative 1. So all that means is when I convert two molecules of reactants into one molecule of products I've lost one molecule altogether. So that's the way to think about this delta nu, the change in stoichiometry. How many molecules do I gain or lose when the reaction proceeds? So for the BR2 reaction, that would be negative 1. If I rewrite this equation now, probably a more convenient way to see it would be k sub big n is equal to k sub little n multiplied by Avogadro's number to some delta nu. So either one of these equations is fine, those are equivalent versions of each other. Just to make sure this result matches what we've seen over here, the result we got for the BR2 reaction where the delta nu is equal to negative 1, we found that k sub little n is equal to k sub big n times one factor of Avogadro's number. k sub little n is equal to 1 over Avogadro's number raised to the negative 1. So that ends up with one positive factor of Avogadro's number multiplying k n just like the result we had over here. The work we have on the right just applies to any reaction. So whatever the reaction is, if you know how many molecules have changed going from reactants to products, that tells you how to convert your molecule-based equilibrium constant into mole-based equilibrium constant. That's significant progress. That lets us work in terms of moles rather than in terms of molecules. But there's another step we can make now that we know this technique that will make it even more convenient. So that's coming up next.