 I'm going to start with something even simpler, single cyclic hexane molecule in vacuum, not interacting with anything. Put it either with lots of his friends in liquid cyclic hexane or in water, and let's compare the three energies along these different paths. Note that the signs are correlated with the direction of the arrow, right? That makes sense. And we would like to know all the components of this process. We already know a little bit about that, so when we moved from the liquid phase to the aqueous phase, we knew that what delta G was. And let's start by putting the delta G, so we have the total earth. I'll write that at the end here. So delta G here was plus 6.7 kcal per mole, but I'm not going to repeat that for all these. Moving from the gas phase to the liquid phase, delta G here is equals to minus 4.4. That's a good process. It loves to do that. And moving from the gas phase to the aqueous phase, then I would have that delta G equals plus 2.3, not favorable. Then we repeat that experiment at multiple temperatures and find out what the entropy is and the enthalpy. So moving from the gas to the liquid phase, you're going to need to trust me here that we got the delta H was equal to minus 7.3. That's very good. And then I'm going to write T delta S just so that you get a number that you can compare to delta G. T delta S here is minus 2.9. Moving from the gas to the aqueous phase, I have delta H equals to minus 7.3. That's interesting. And T delta S equals to minus 9.6. And finally, down here, liquid to the aqueous phase, I have delta H equals to 0. And T delta S equals to minus 6.7. Again, this is what we would get from the experiment. So we know that the experiment tells us the truth, but now we can use the experiment to tell us something what happened on the molecular scale. And in particular what this tells us is that the actual change in enthalpy or energy when moving to either of this medium is, to first approximation, the same. Which is on the one hand very surprising because you would imagine that there should be a difference with hydrogen bonds here, but the numbers tell us that there isn't. On the other hand, it makes a lot of sense because this molecule can't participate in anything but van der Waals-Lenard-Jones interactions, right? And that just depends on the number of neighbors around us. And we have roughly the same amount of neighbors here and here. There is a big difference in entropy though that has to correspond to the difference between these two states. And again, the difference here has to correspond to the difference between those. Delta H here was zero. So with that argument, there should not be any net change in the number of hydrogen bonds formed, but they can be re-divided where they formed. And what I'm predicting here, there should be a very large effect on entropy. So what's happening somehow when I'm taking one hydrocarbon here that is hydrophobic and moving that into water. What somehow happens is that I'm paying with entropy, just as when I formed that single hydrogen bond in a protein. I'm not paying because I'm breaking hydrogen bonds. And that makes a lot of sense if you think about this on a molecular scale. Let's do that.