In the previous concept development study, we developed the second law of thermodynamics, an absolutely beautiful description that predicts what kinds of processes will occur spontaneously. In this concept development study, we're gonna develop or apply this to make predictions regarding equilibrium. And in the process, we're gonna produce one of the most amazing results actually available in all of the field of Physical Chemistry. And it will involve the introduction of something that chemists call the free energy. Quickly, we're gonna review what we learned in the previous concept development study. The second law of thermodynamics tells us that the entropy of the universe increases for any spontaneous process where the entropy is k times the logarithm of W. Where W is the number microstates describing whatever the macrostate is that we are interested in. We broke down the entropy changes in the universe to be equal to the entropy changes of our system. Whatever the reaction process is that we're interested in looking at plus the entropy changes that result in the surroundings as a consequence of whatever change it is that occurs in the surrounding, I'm sorry, in the system. And what we have observed then from the second law of thermodynamics is that the sum of the entropy change in the system plus the entropy change in the surroundings must be greater than 0. We applied this in a specific case by considering the entropy change in the surroundings to a rise from enthalpy changes or energy changes in the system. And we replaced the entropy of the surroundings with minus the enthalpy change of the system divided by the temperature. A simple rearrangement of this particular equation in which we multiply both sides of the equation by minus the temperature, by multiplying by minus one we reverse the sign. We produce this inequality which is a slightly different form of the second law of thermodynamics that doesn't require us to calculate the entropy of the universe. But allows us instead to calculate both enthalpy changes and entropy changes just relating to the system. In chemistry, we frequently redefine these latter quantities by introducing a new term that we call the free energy, and we denote it by G. G is typically called the Gibbs free energy, that's why it's G. Gibbs is Josiah Willard Gibbs, a physical chemist from more than 100 years ago. Free energy, we're not really gonna describe too much why we call it free energy other than to note that one, it does have energy units. Notice that G = H- another term, and H has units of energy. And secondly to know that it's kind of a funny sort of an energy because it's an energy that includes an entropy term. In fact, as we'll see in a little while, the free energy is really a measurement of entropy changes rather than energy changes. And we should always keep that in mind as we're studying this. Why did we define G in this particular way? The answer is if I take delta G, that is the change in G for any process going from say, reactants to products. Provided that I keep the temperature constant, I can easily just say that delta G = G at the end- G at the beginning. Which = H at the end- T times S at the end- H at the beginning- T times S at the beginning. And it's easy enough to rewrite this as H2- H1- T(S2- S1), which means delta G = delta H minus T delta S. But if we remember from the previous slide which I'll back, back up to here, we'll see that remember, delta H- T delta S for a spontaneous process must be < 0. So in fact, we can defined the Gibbs free energy in this way because this quantity, delta H- T delta S is a particularly interesting quantity for us. In particular, we can now say that our criterion for a spontaneous process is that we have to see a negative change in the Gibbs free energy. Delta G < 0 for any spontaneous process. One conclusion that that immediately leads us to is that if we had delta G > 0, the reverse process would have delta G < 0. And therefore the reverse process would be spontaneous. The only way to have an equilibrium process is therefore for delta G to be neither greater than zero nor less than zero. Therefore, delta G would have to be = 0, and we have created a simple criterion for the determination of equilibrium. And we'd like to now see how we can apply that to specific equilibrium cases. Over the course of this next couple of lectures, we're gonna apply this criteria to understand phase equilibrium, and we're gonna apply it to understand reaction equilibrium. Before going any further though, since we're gonna use this delta G term quite a bit, Want to make sure we understand what the relationship is between the free energy and the entropy change of the universe. Here are these same equations that I wrote a little while ago, delta S of the universe = delta S of the system + delta S of the surroundings. We can replace delta S of the surroundings, assuming that the only change in the surrounding results from energy flow out of the system into the surroundings. Or energy flow out of the surroundings into the system, dependent upon whether it's an endothermic or an exothermic process. Notice I can multiply both sides of this equation by minus the temperature to produce the equation- T * the delta S to the universe =- T * the delta S of the system + the delta H of the system. And again, looking back at our equation over on the pad here, we notice that this term on the right, delta T- delta H of the system, is in fact what we now call delta G. What that means is, that delta G, when we observe it, is really just delta S of the universe, except scaled by the temperature and multiplied by a minus one. So when we observe delta G, what we're really doing is making a measurement of what we think the change in the entropy of the universe is gonna be. That minus sign is important. Because, remember, if delta S of the universe always increases, then that means delta G will always decrease for any spontaneous process. So it's worth keeping in mind that our criterion for a spontaneous process is that delta G is gonna be less than zero. Let's apply this to a specific case. It's a case that we discussed in the last concept development study, the condensation of water at one atmosphere pressure of water. So, here's the process. Water gas goes to water liquid. We did a calculation in the previous concept study at delta H0- T delta S0. Remember that zero refers to standard conditions, meaning one atmosphere pressure. Then delta H0- T delta S0 = delta G0. And we did that calculation in the previous study, and we found that the value is negative and that is < 0. And correspondingly then, the condensation of water, when the pressure of the water vapor is one atmosphere, at 298K is in fact a spontaneous process. And we know this to be true. Thus the air is very much super saturated with water if the pressure of the water is that high. Under what conditions would we expect to see equilibrium? Well one of the things we also observed was that 373 degrees Kelvin, when we calculated this term here, we discovered it's = 0. So again, at equilibrium, delta G = 0. In this case what we have looked at then is that for one atmosphere pressure of water vapor, the equilibrium must occur at 373 Kelvin. Of course that's 100 degrees centigrade, the boiling point, so we already knew that that was gonna be the equilibrium. But, what about equilibrium at 298K? That should be possible as well, because we know that there's an equilibrium vapor pressure at every temperature. And in the case of water vapor we know that that equilibrium vapor pressure is 23.8 torr. So why is it that we can get delta G = 0 at 298K so that we have equilibrium? It must be the case, since we know that the equilibrium vapor pressure varies with the temperature. It must be true that the entropy of the gas phase must vary with the pressure, so that it is possible for us to achieve equilibrium at pressures other than one atmosphere and temperatures other than the boiling point. But why would this be true? Why would the entropy of the gas vary with the pressure? Well, we're gonna dig through a little bit of math here, so stick with me closely. Let's imagine that once again what we have done is to create our old friend the cylinder fitted with a piston, such that we can easily change the volume of a trapped gas inside that cylinder. We're gonna trap a fixed number of moles of gas, and we're gonna hold them at constant temperature, so these two numbers will be fixed. And the question we're gonna pose is imagine that we move this cylinder up and down so that we vary the volume. How would that vary the entropy of the gas? To answer that question, we need to write down an expression that tells us what the entropy of the gas is, and remember that the entropy of the gas needs to be the logarithm of W. Let's imagine we have one molecule. If we have one molecule, then the number of ways in which that molecule can be arranged inside this volume can be said to be proportional to, at least, if not equal to, the volume of the gas itself. Because the volume is a measure of how many locations there are in the gas where that one molecule could be. So, the entropy for one molecule in a volume V is k times the logarithm of V because the number of micro-states, the number of places that I could put that one molecule, is the volume. What if I have Avogadro's number of molecules? Well, there are V places to put the first molecule, and V places to put the second molecule, and V places to put the third molecule, and so forth. So, the entropy then becomes k times the logarithm of the volume raised to the Avogadro's power. The reason for this is that if I have two particles, particle one can be in any number of V places, and particle two can be in any number of V places. So the number of places that I can arrange two particles is V times V. For three particles, it's V times V times V. For four particles, it's V times V times V times V, and so forth, until we've got Avogadro's number. By using the properties of logarithms, I can now rewrite this as Na times k times the logarithm of V. And it turns out that this Boltzmann's constant k times Avogadro's number is our old friend, the gas constant. I won't show that that's true here, but we're gonna take that as a given. We now have an expression for the entropy of a gas at volume V as a function of the volume V. Notice that the volume V increases, then the entropy increases. And it increases because the larger the volume I've captured over here, the more places are that I could put the molecules. Now what we would like to do is write an expression in which we can relate the entropy for a larger volume to the entropy at a smaller volume, so S(V2)- S(V1). Well that must be = R logarithm V2- R logarithm V1, which can be re-written as R logarithm of V2 / V1. We're almost there now. What we would really like to know, if we go back and look at our slide back over here though, is we wanna know how this varies with the pressure, not with the volume. But of course, we know something about Boyle's Law, and remember N and T are fixed, so we could use Boyle's Law. So if I write S(P2), the pressure when the volume is V2,- S(P1), the pressure when the volume is V1, must be = R times the logarithm. Remember, pressure and volume are inversely proportional to each other. So V2 / V1 = P1 / P2. And that is in turn =- R times the logarithm of P2 / P1. So, on the next slide actually, I think I have written that equation down, that the variation of the entropy as a function of the pressure is given by the expression that I've just arrived back over here. Let's think about this equation for a minute. What this actually tells me is that if P2 < P1, the pressure is smaller, then P2 / P1 is a number less than one. The logarithm of P2 / P1 is a number < 1, negative, and therefore this is gonna be a number > 0. That tells me S(P2) > S(P1) when the pressure is lower. Why does the entropy go up when the pressure goes down? The answer has to do with this diagram back over here. For a fixed number of moles and a fixed temperature when the volume increases, the pressure goes down. And when the volume increases, the entropy increases, just as we've reasoned here. So, it makes sense, a decrease in the pressure will increase the entropy. Let's actually use this expression to calculate the entropy at a different pressure. We're gonna apply this to the water case again. One of the measurements we've already looked at is that when the pressure P1 is equal to one atmosphere, the entropy of that is 188.8 joules per mole Kelvin. That's just from tabulated data that we looked at in the previous concept study. I can now take this number and plug it in here, assuming that P1 = 1 atmosphere. And then I can try to imagine for example, if I we're to take, P2 = the equilibrium vapor pressure of water, which is 23.8 torr. And plug that number in for P2, we can do that calculation, that's a straightforward calculation. If I do that calculation, what I find is that the entropy at 23.8 torr is a larger number than it was at one atmosphere of pressure. Again, why did it go up? Because in order to achieve a pressure of 23.8 torr, I must have evaporated one mole of gas into a much, much larger volume. We can now take this number and actually calculate the value of delta S for this process. So again, the delta S of the process is gonna be the entropy we've just calculated, and here's the entropy of the liquid. So the process liquid going to gas is gonna be 217.6- 69.9, multiply that by the temperature. And add it to the enthalpy change for this process. And when we do that, what we will actually discover, put those numbers in the calculator yourself, delta G is actually = 0. Let that sink in for a minute. What that means is, we've actually proven that when the pressure of the gas in the equilibrium here is 28.3 torr, then the overall entropy change for gas condensing, or gas evaporating, is, for the universe, is 0. In other words, that delta G is equal to 0, and remember delta G is proportional by the temperature to the entropy change of the universe. That means that the entropy change of the universe is 0 for equilibrium between gas water and liquid water, provided that the gas is at 23.8 torr. We've exactly predicted the experimental conditions. In fact, we could have started off with delta G = 0 and derived from that that the appropriate pressure would be 23.8 torr. We're gonna do that kind of calculation in the next lecture in which we're gonna be able to show what the relationship is between the vapor pressure and the temperature. Just by using this last expression we've developed here, as well as this expression we have developed for the relationship between entropy and pressure.
