 All right, so let's pay a little more attention to this difference that takes place when we have a reaction that has the same number of molecules on each side compared to a reaction like this, the second one that has a different number of molecules on the reactant side than the product side. So again, on the reactant side if I add up the stoichiometric coefficients, remembering that the ones on the product side are positive, on the reactant side they're negative, and the top reaction those add to zero in the bottom they don't. So what that means is when I go to write down an equilibrium constant, if we calculate the value of the equilibrium constant we can do that with partition functions. We're done doing that for now, thankfully. If we go to solve an equilibrium constant and say that equilibrium constant that we've calculated must be equal to the number of moles of, sorry, number of molecules of product raised to stoichiometric coefficients divided by the number of molecules of reactants raised to their own stoichiometric coefficients, we have been fairly cavalier about whether we write that down in terms of molecules or in terms of moles. I can write that equation either as molecules of HBr squared over molecules of H2 and molecules of Br2, or I can write down moles of HBr squared over moles of H2 and moles of Br2. I can do that for the top reaction because I have the same number of quantities N in the numerator as I do in the denominator. I can't do that for the second reaction. The equilibrium constant is molecules of Br2 over molecules of Br squared, products over reactants using number of molecules. It's not going to be the same thing if I write down moles of Br2 over moles of Br. That might or may not be a confusing statement at this point. Sometimes I can, sometimes I can't. If I have the same number of molecules on each side, it doesn't matter whether you use molecules or moles. If you have a different number of molecules on each side, then it does matter. In case that statement is not clear, in case it's not clear why, I can't use moles in the second reaction. We can work an example that will make it clear that that is the case. Let's suppose we want to focus on that second reaction, bromine atoms becoming Br2 molecules. Let's say at 298 Kelvin in a volume of one liter because we've already seen under those conditions the equilibrium constant for that Br2 formation reaction is 1.6 times 10 to the 7th. We've calculated the value of the equilibrium constant under those specific conditions. Let's say initially we've got some Br atoms. I place a mole of, am I going to use a mole? Yes, let's use a mole. So I can say the initial amount of bromine atoms, actually no. Let's do Br2 molecules. The initial number of Br2 molecules, bromine gas, is one mole or 6 times 10 to the 23 molecules. Either one of those is the same number. Initially the amount of bromine atoms I have is zero. Likewise, the number of moles of bromine atoms I have is zero. Now that's enough information to say once we reach equilibrium, how much bromine, how much Br2 will we have? So we can do that problem either in terms of molecules or we can do it in terms of moles. Start by doing it in molecules because that's where we know the conditions are true. The first equation that we derived, the equilibrium condition relating the equilibrium constant to the amount of products and reactants that was derived in terms of molecules. So that equation is definitely correct. We can say the equilibrium constant must be equal to molecules of HBr2 over molecules of H2 molecules of Br2. We know these initial amounts, H2 and Br2, sorry we're using this reaction. Molecules of Br2 divided by molecules of Br2. We know the initial amounts but we're not interested in the initial value of this ratio. We're interested in the value when we reach equilibrium. So when we reach equilibrium, we're not going to have N0 molecules of Br2 anymore. We're going to have N0 plus the extent of reaction. So when the reaction proceeds forwards, I gain one molecule of Br2. As many times as it proceeds forward, I've gained that number of molecules of Br2. Don't be confused by the fact that I'm starting out with no Br atoms. So of course the reaction can't go in the forward direction. The value of the extent of reaction, that can be a positive number or a negative number. So even though if we look ahead, we know that the reaction is going to go backwards. There's no Br atoms, there's a lot of Br2 molecules, so the reaction must go backwards to reach equilibrium. All that means is this extent of reaction is going to be a negative number. The reaction will have proceeded backwards by some amount. So we'll end up with a negative extent of reaction, which is going to reduce the total amount of Br2 molecules. In the denominator, I've got the number of molecules and the number of atoms of Br squared. I started with zero. When the reaction goes forward, I lose twice the extent of reaction of Br, and then I square that quantity. So all together, N0 plus Xc over 4 Xc squared. That's going to be the value of this ratio of molecules over molecules squared when I'm at equilibrium. Or at any point, when I'm at equilibrium, this ratio will exactly equal the equilibrium constant. So now I just have an algebra problem to solve. I need to solve for the value of Xc that makes this equation true. So if I rearrange this to get rid of the fraction and say that I've got 4 Xc squared times K on the left-hand side. If I bring this Xc and this N0 over to the other side, I've got a minus Xc and a minus N0. That all is going to equal zero. So now I've got a quadratic equation to solve. I'm solving this for the value of the extent of reaction. I've got a quadratic term and linear term. So quadratic equation tells me that value of Xc is going to be minus B plus or minus root B squared minus 4ac over 2A. So B is negative one. So negative negative one is one. And then inside the square root one squared is one. I need to subtract 4 times Ac. So 4 times negative 4k N0. And that's all over twice A. A is 4k, so the denominator is twice that, or 8k. All right, so that's as far as I can go with algebra before plugging numbers in. We've got all the numbers we need to plug in here. N0 is 6 times 10 to the 23rd. K in the numerator and denominator is this equilibrium constant. So if I do that, I'm of course going to get two solutions, the positive solution and the negative solution. If I work that out, I'm going to get either 9.7 times 10 to the 7th or negative 9.7 times 10 to the 7th. That's the unitless quantity for the number of times the reaction is preceded to get to equilibrium. I've got two solutions. Only one of those makes sense. Like we talked about before, I've got no BR, so the reaction cannot proceed forward. There's no way for the reaction to proceed forward, so the positive extent of reaction, that one is physically nonsensical. So the only value for the extent of reaction that makes sense is this value negative 9.7 times 10 to the 7th. Let me point out also if I want to convert that to moles, if I divide that by Avogadro's number, that'll be negative 1.6 times 10 to the minus 16th in units of moles. So the reaction is going to proceed backwards with a negative sign 97 million times, or I can think of it as proceeding backwards 1.6 times 10 to the minus 16 moles of times in the backwards direction. That number, the extent of reaction isn't terribly interesting. What we really wanted to know is how much of the BR2 and the BR are there after we reach equilibrium. So the amount of BR2 at equilibrium is the initial amount plus the extent of reaction. So 6 times 10 to the 23rd plus this negative number. So 6 times 10 to the 23rd plus 10 to the 7th. 10 to the 7th looks large, but it's tiny compared to 10 to the 23rd. So this is only going to change the amount of BR2 way out at the 16th digit of this decimal number or so. So that's still going to be 6 times 10 to the 23rd, or almost all of the initial 1 mole of BR2 is still going to be there because the reaction proceeded backward just a tiny little bit. More interesting is the amount of BR that I've got. Negative twice the extent of reaction is the amount of BR. So if I take this number and double it, I'll get 1.9 times 10 to the 8th, or in moles that would be 3.2 times 10 to the minus 16th moles. So the total amount of BR that gets formed as these BR2s decompose and reach equilibrium is I've got 190 million atoms of BR or this tiny number of moles. 3.2 times 10 to the minus 16 moles of BR and that tells me that the reaction has reached equilibrium under these conditions. So we've worked an example. The purpose of doing that example was to illustrate why we can do the problem in terms of molecules, but we can't do the problem in terms of moles. So let's do the problem again once through a little quicker, now using moles instead. So if we try to work that same problem, so I'll put a question mark over my equal sign because we're not sure that this is going to be true. If I say K is equal to moles of BR2 over moles of BR squared, set that equal to K, the algebra is going to work out exactly the same. I've got initial amount plus extent of reaction divided by negative 2 extent of reaction squared. I get the same quadratic equation, just written in terms of moles instead of molecules. The extent of reaction is now going to be 1 plus or minus 1 plus 4 K little n, not over 2 K. And when I solve that expression, you can see already that we're going to get a different answer because the only thing we've changed is this big n has become a little n. When I plug in one mole instead of 6 times 10 to the 23rd, then I'm going to get a different answer. If I go ahead and do that in terms of moles, what I'll find is that the solutions I get are positive 0.013 moles or negative 0.0012 moles. And again, the positive solution is nonsensical because the reaction can't proceed forward if we don't have any reactants to begin with. So this is the solution I get, negative 0.0012 moles. That's not the same number either for the extent of reaction nor if I calculate the number of atoms of BR2 that I've got, negative 2 squiggle. That's going to be the negative signs cancel, 0.00224 moles is not the same value as 3.2 times 7 to the minus 16 moles. So this example is just there to illustrate that we can't in many cases write down the equilibrium constant directly in terms of moles of products and moles of reactants. To be correct, we need to do it in terms of molecules of products and molecules of reactants, which makes things a little more cumbersome in terms of the magnitudes and the numbers that we end up using. But it turns out there is a way that we can rescue things. We can still work in moles the way we want to the way we're used to doing if we change our understanding of what this equilibrium constant is. So that's the next step we'll be to talk about what change we would have to make to the equilibrium constant to allow us to proceed using moles rather than molecules.