 Okay, let's explore this idea of the temperature dependence of the equilibrium constant a little deeper. We've seen the Vanthoff equation, we've obtained this equation that the log of the equilibrium constant changes with temperature proportionally to the enthalpy of the reaction, the enthalpy change of the reaction over RT squared. To look at that from a different point of view, let's go back to the expression we have that relates equilibrium constants to the Gibbs free energy of the reaction. Let me play with that equation a little bit. First of all, again I'll take the log of K so that I don't have an exponential on this side. What I'll have left on this side is a minus delta G over RT. If I move the minus RT over, then on this side I'll just have the delta G. Delta G, of course, has two parts. That's the energetic part and the enthalpy part. There's an enthalpy term, I'm sorry, energy and entropy. There's an enthalpy term and an entropy term. Delta H minus T delta S is the change in the Gibbs free energy. Now what I can do is bring the RT back over to the right side. On the left side I'm just going to have log K. On the right side I've got delta H over negative RT, so I'm going to write that as minus. There's an R in the denominator, delta H in the numerator, and I'm going to write the one over T off to the side, so I've got minus delta H over RT. For the second term I've got minus T delta S divided by minus RT, so the negative signs cancel, the T's cancel, and I have delta S over R. The reason I've written the equation in that form is so that it's a little bit suggestive. I want to know how the equilibrium constant depends on the temperature. Again, this equation tells me not how the equilibrium constant itself, but how the log of the equilibrium constant depends on temperature, and in particular not how it depends on temperature, but it's linearly proportional to one over T. If I think of this equation as being like y equals mx plus b, the equation for a straight line, my dependent quantity, log K is dependent on my independent quantity, one over T, with some slope and some intercept. That's very suggestive. It suggests that if we want to experimentally, for example, understand how the equilibrium constant is depending on temperature, I can do a reaction at one temperature, measure the equilibrium constant, or measure the amounts of reactants and products, and calculate an equilibrium constant, change the temperature, measure another equilibrium constant, and so on. I could make a graph of how the equilibrium constant depends on temperature if I wanted to, but this equation suggests it might be more useful if I ask how the log of the equilibrium constant depends on one over the temperature. If I do that, I'll tend to get data that look something like this. I've got, generally speaking, a straight line. I've built some experimental error into my graph here, but generally speaking, the log of the equilibrium constant will change linearly with not the temperature, but one over the temperature. The slope of that curve will be minus the delta H of the reaction over R, slope given by m in this equation. The intercept, if I extrapolate all the way to low one over T, so I can't get there. So extrapolating to zero for one over T would be like extrapolating to infinite temperatures. So I'll never get there, but in the limit of infinite temperatures, very small one over T, the intercept on this graph is going to be the enthalpy, I'm sorry, the entropy change of the reaction under standard conditions divided by R. So we learned two interesting pieces of thermodynamics about the chemical reaction we're interested in by understanding how the equilibrium constant changes with temperature. We learned the enthalpy and the entropy change of the reaction. So the graph, the equation and the graph will both tell us that the log K is going to change linearly with one over T. It won't necessarily change linearly in a downward direction the way I've drawn it. The slope is negative delta H over R, but delta H does not have to be a positive quantity. We can have reactions that are endothermic for which the reaction's enthalpy change is positive, or we could have reactions that are exothermic where the products are lower in enthalpy than the reactants, so the enthalpy change is a negative number. The graph I've drawn here is for the endothermic case. If the enthalpy of the reaction is positive, this negative sign means the slope will be negative. That means log K will drop as one over T increases. Log K goes up or down when K also goes up or down. So this means that the enthalpy, I'm sorry, the equilibrium constant is increasing as one over T decreases. So when one over T goes down, the logarithm is going up, log of K is going up, so K is going up. One over T goes down whenever the temperature itself goes up. So that means endothermic reactions will have equilibrium constants that get larger as we heat them up as the temperature increases. When the equilibrium constant gets larger, that means the reaction is shifting more towards products as the temperature goes up. So this is the case for endothermic reactions. You might have learned a mnemonic trick to understand how reactions, how equilibrium will shift for endothermic versus exothermic reactions. An endothermic reaction, as the mnemonic goes, is one where heat is a reactant. It's endothermic because I have to put in energy in the form of enthalpy to make the reaction go forwards. La Chateau-Ley's principle says if I increase the temperature, I'm adding more thermal energy to the system so La Chateau-Ley says the reaction should shift a little bit more forward. That's really only just a qualitative and mnemonic trick. The real reason the equilibrium constant goes up when the temperature goes up is because of the thermodynamics we see over on this side of the board. Exactly the opposite is true for exothermic reactions when delta H is a negative number. Negative delta H is a positive number, so the graph won't be, as I've drawn it here, the graph would have a positive slope rather than a negative slope. Exactly the opposite of these conclusions is going to be true that K is going to go down as 1 over T goes down, so K is going to go down as the temperature increases, so the reaction is going to shift backwards as temperature increases. And then La Chateau-Ley's mnemonic trick for an exothermic reaction, one that liberates heat, we can think of heat as being a product of that reaction. Reactions turn into products and generate some thermal energy via the central beam of reaction, so when the temperature goes up we've added more product to the reaction so that's going to drive the equilibrium backwards, says La Chateau-Ley. That makes a connection between something you may have already understood about the difference between endothermic and exothermic reactions, but now we understand at a somewhat deeper level where that comes from. Also, I should point out that this equation, if I take this equation, if I want to make a connection to the Van Hoff equation, another way to get to the Van Hoff equation differently than we obtained it when we first derived the Van Hoff equation, I can take this expression and if I take the temperature derivative of it, then clearly the left side is going to be derivative of log K with respect to T, the right side, if I have an entropy that's temperature-independent, if I'm making a small temperature change so the entropy doesn't change much, there will be no temperature derivative of this term and then the derivative of negative one over T is going to be positive one over T squared and the delta H over R will remain out front. You can see that this is another way of obtaining Van Hoff's equation, and just take the temperature derivative of this equation and we recover the Van Hoff equation. So we've now seen how to get that expression in two different ways.