 So the reaction rate k was equal to k0 multiplied by e raised to, I'm going to say delta f divided by rt, to save some writing. With the exponent here and 1 over t up in the exponent, it's going to be quite useful to start, try to plot the logarithm of k versus 1 over t. And that might go from say 1 over 350 kelvin to 1 over 300 kelvin. So this is called, that is 1. If I now do that for folding and unfolding, I'm going to get two curves. Do you see that they're going to be quite beautiful and fall almost exactly on straight lines? So first I have one curve process that goes faster. This is going to be folding. Folding will go faster when the temperature is lower, which is a bit strange. Most processes will do the opposite. While unfolding, that will go faster when the temperature is higher, as you normally expect from a chemical process. But given how beautiful and near these are, we would like to know what the derivative here is. Well, we can calculate that. It's not that hard. So let's see what we can do. We have l and k versus 1 over t. So I'm going to start down here in time. I need a bit of space. So I want to calculate the derivative of l and k with respect to 1 over t. That will lead to two parts. Leave some room in the nominator first. The derivative of 1 over t, that's minus 1 over t squared. So I'm going to get minus 1 over t squared falling out and then dt. The derivative of l and k, that's going to be ln derivative of l and k0 plus ln and exponent cancel. And then it's going to be a minus sign. So it's going to be minus delta f divided by rt. But here's the beautiful part. k0 is a constant that describes the fundamental process. So that is roughly going to be 0, the derivative of it, that is. And minus 1 over t2. Those two minus signs cancel and then I get t2 in front of the expression. So here I get t2 multiplied by the derivative of delta f divided by rt with respect to t. This is still a non-trivial expression but it's not that bad anymore. That's the derivative of a quotient. We have delta f and then divided by rt. But to make that slightly easier we can think of that as delta f multiplied by 1 over rt. So that is going to be t2 multiplied by first the derivative of delta f with respect to t. Multiplied by 1 over rt plus, that is going to be minus sign in a second. So don't write out the sign yet. Delta f multiplied by the derivative of 1 over rt. That's going to be minus 1 over rt squared, right? Minus 1 over rt squared. I could have put the r in front of everything. And now things are actually starting to look pretty nice. I know that you might not think that yet. It says a t2 there in front. I will multiply both these terms by t. And then I can cancel that t2 against both those t2s, right? And then I'll move the 1 over r in front of everything. So we have 1 over r multiplied by... What is the derivative of delta f with respect to t? Well, you know that. We went through that a couple of lectures ago. That's minus s. If you don't remember, we watched an old lecture on what we derived it. So here it says minus s multiplied by t. I'm going to say minus ts. Minus, what was delta f? Well, sorry, not just delta s, f, delta s. Because this is delta f there, right? Well, delta f was delta e minus t delta s. Minus minus is plus, so it's t delta s. Do you see how those two terms cancel? So this is going to be minus delta e divided by r. With delta e, now if I'm looking at folding, this delta e is when I go from the unfolded state up to the transition barrier. This is... I bet you don't see how beautiful this is. The derivative here is positive when I'm folding, and this was unfolded. So when I'm folding, if the derivative is positive, that means that the energy difference is negative. So what I have just shown, that the energy of the unfolded state, based on the experiment, is larger than the energy of the transition state. Cool, right? We can't determine the structure of it, but I can say something about what the energy of it must be based on this curve. If I go in the other direction, the derivative has the other sign. So that means that the energy of the folded state must be lower than the transition state. So I will just continue. So that the energy of the native state must be lower than the energy of the transition state. So as I'm folding, this was what I hand-waved about before, and I'm proving it based on experiments. The energy of the unfolded state is highest, then we drop a bit to the transition state, and then we drop even more to the native state. But it's not that the transition state is going up in energy. If only I had a way to do this for entropy, too. It turns out we do. The book does not go through that, so normally I would just write this down, but since the book does not go through it, I figured I should show you.