 Okay, so we've obtained this condition for chemical equilibrium. We know a chemical reaction is going to be in equilibrium if this condition holds. Stoichiometric coefficients multiplied by chemical potentials adds up to zero for all the components of the reaction. To see what that feels like for a real example system, let's go back to our example of H2 and Br2 reacting to form HBr, and see if we can learn something interesting about that particular equilibrium reaction. So this equilibrium condition tells us, if I just use the stoichiometric coefficients from this reaction, stoichiometric coefficient for H2 is negative one, so I've got negative chemical potential of H2, negative chemical potential of Br2, and positive twice the chemical potential of HBr have to add up to zero. That's my equilibrium condition. Another way of writing that same reaction, let's move reactants and products to opposite sides of this equal sign, so I can say chemical potential of H2 and Br2, if they're on the opposite side from the product, chemical potential of H2 and Br2 have to add up to twice the chemical potential of HBr. That looks essentially just like this chemical reaction written in terms of chemical potentials rather than chemical species. Now what can we do with this? We know a fair amount about chemical potentials, in particular for this reaction at room temperature anyway, H2, Br2, HBr, those are all gaseous species. The chemical potential of a gas, we know how to write down the chemical potential of a gas in particular, we know that chemical potential is going to depend on the pressure of the gas. As this reaction goes forward, I destroy some H2 and Br2, I generate some HBr, the amounts of each one of those reactants and products are going to be changing, their pressures might be changing. If I recall that the chemical potential of a gas, in particular an ideal gas, at a particular pressure is equal to its standard state chemical potential plus RT natural log of its partial pressure relative to standard pressure, that's an expression I can use to write down the chemical potential of each of these gaseous species. So just substituting that expression in here, I can say chemical potential of H2, which is Mu0 plus RT log of hydrogen's partial pressure over P0, add that to the chemical potential of Br2, which is Mu0 for Br2 and RT natural log Br2 pressure relative to standard pressure and then on the right hand side of the equation I've got twice standard chemical potential for HBr and twice RT natural log pressure of HBr relative to standard pressure. So there's a nice long expression that tells me what this chemical equilibrium condition looks like for HBr, for this HBr reaction. We can clean that up a little bit by collecting, let's say, all of the chemical potential terms on one side. Put those on the right hand side, so I've got twice MuHBr0, if I pull these two over to the right hand side of the equation I've got subtracting chemical potential of H2 and chemical potential of Br2 under standard conditions. Then on the left side of the equation I'm going to keep all the RT log pressure terms. So I've got RT log pH2 over P0, RT log pressure of Br2 over P0, and I get a negative sign when I bring this minus 2 RT natural log pressure of HBr over P0. Now that I've got all these RT natural log terms together I can collect those into one. I've got natural log plus a natural log minus a natural log. So that's going to be a product, natural log of the product, pH2 over P0, PBr2 over P0. The negative sign here means this one's going to be upside down, pHBr underneath P0, and the 2 out front means I need to square it. So I've combined all those terms on the left side into one natural log term. On the right side I've got a difference in chemical potentials. I've got chemical potential of products, two chemical potentials of HBr minus chemical potential of reactants. So products minus reactants with stoichiometric coefficients, that sounds like the change in chemical potential when the reaction proceeds, so we could call it that. I'm actually going to call it change in molar free energy, just recalling the fact that the chemical potential is partial molar Gibbs free energy. So the free energy change of this reaction is the same as the difference of chemical potentials for products minus reactants with the right stoichiometric ratio. So let's see what we can do next. If I'm going to move the RT over to the right side. So I've just got a natural log. These P0s, P0s twice in the denominator, twice in the numerator, those are going to cancel. So all I'm left with is pressure of H2, pressure of Br2 over pressure of HBr squared. A word of caution here, that cancellation of the P0s, the two in the denominator and the two in the numerator happen to cancel in this case exactly only because of the coincidence that I've got two molecules on the reactant side and exactly two molecules on the product side. If this chemical reaction had ended up generating a net number of molecules of product or consuming some number of net molecules, then that cancellation wouldn't happen, that cancellation doesn't always happen. In fact, that's one source of complication that we'll run into when we start to deal with more complicated problems. And for the moment, those P0s all canceled. This ratio, pressure of H2, pressure of Br2 over pressure of HBr squared, that looks nice except I'd rather have it the other way around. I'd rather have products divided by reactants, right now I've got it upside down. So I'm going to write, if I rewrite it one more time, I can say if I flip it upside down, pressure of products HBr over pressures of reactants H2 and Br2, I've turned this side upside down. When I invert what's inside the log, I change the sign, so I'm going to move that sign over to the right side. Now I can undo this natural log by exponentiating both sides. So if I take e to the ln of this quantity, I'll just get this ratio of pressures, pressure of HBr squared over pressure of H2 and pressure of Br2. When I exponentiate the right hand side, I get e to the minus delta G over RT. So that should begin to look a little bit familiar if you've dealt with equilibrium constants before at the general chemistry level. This quantity e to the minus delta G over RT, that the delta G is the delta G for this particular reaction, this H2 and Br2 forming HBr, that reaction. That particular reaction has a Gibbs free energy associated with it. If I know that value and I calculate this quantity e to the minus delta G over RT at a particular temperature, I'm going to get a number. That number we call the equilibrium constant. So this expression that I have here, this is still the equilibrium condition. Remember we started with this expression saying this is the condition that must be satisfied if this reaction is at equilibrium. I've just manipulated that equation in a number of different ways to arrive at this result, which looks very different, but is still the equilibrium condition. If it happens to be true that the ratio of the pressures of these different components, HBr, H2 and Br2, that particular ratio of pressures happens to equal this particular constant, then we know the reactions at equilibrium. H2 and Br2 are in equilibrium with HBr. So this is just a different way of describing this same equilibrium condition. We've done it here with a fair amount of work for the specific case of H2 and Br2 forming HBr. We could do that every time we run into a new reaction, but of course it's going to be more useful to do it once and for all for the general case to see what is the equilibrium condition written in terms of some equilibrium constant for an arbitrary reaction with arbitrary stoichiometric coefficients. So now that we've seen this example for the HBr reaction, we'll do it for the more general case coming up next.