Lecture 15. Getting to know the Gibbs Energy.
Penner, Reginald
 In Chemistry 131C, students will study how to calculate macroscopic chemical properties of systems. This course will build on the microscopic understanding (Chemical Physics) to reinforce and expand your understanding of the basic thermo-chemistry concepts from General Chemistry (Physical Chemistry.) We then go on to study how chemical reaction rates are measured and calculated from molecular properties. Topics covered include: Energy, entropy, and the thermodynamic potentials; Chemical equilibrium; and Chemical kinetics. 
 Okay, how are you guys doing? You got those mid-quarter daldrums. I know how you're feeling. Okay, this is a chapter 16 topic. We're really cranking away on chapter 16 here. Edging our way towards the end of thermodynamics, which I hope will come probably in the middle of next week. Now we posted the quiz scores this morning. This was the hardest quiz. It might have been too hard. First question was pretty easy. Didn't you think? So these guys got A's. These guys got B's. There's a few C's here. We usually don't have hardly any C's. Okay, quiz five is Friday. The first two are supposed to be easy. The first two problems. In this case, the second one was actually not so easy. It was the body temperature one. Okay, I also posted the key in case you want to look at that. So we won. And what's even sweeter is we beat SC. Very, very sweet. I'm always amused by the fighting anteater. The anteater is the most docile animal that we know of in the animal kingdom. But to dress him up for the athletic department, here he is fierce. There's no such thing as a fierce anteater. In nature, they're not known to be fierce. Okay, so we're going to review a little from Friday. Why does that say Wednesday? We're going to talk about how the Gibbs free energy varies with temperature and pressure. We'll do a couple of examples. Just sort of easy ones to ease into this subject. Okay, we'll do a bunch more examples on Wednesday. All right, so on Friday, not Wednesday, we said, look, there's three types of systems. Okay, and uniquely for this guy right here, it's an isolated system. There's no energy or matter exchanged with the environment, the surroundings. We don't have to consider anything except the system when we think about the spontaneity of processes that occur within it. Okay, it's blocked off to the surroundings. It doesn't even know about them. And so we can say any process that has an entropy, positive entropy change, is going to be a spontaneous process for an isolated system. We don't even have to think about the surroundings. They're not part of our thought process. But we don't have isolated systems in chemistry too often. They're almost always in communication with the environment. And so we have to consider open systems and closed systems as well. And in those cases, because there is communication with the surroundings, it's the total entropy, surroundings plus system that matters. Okay, in terms of figuring out if this process is spontaneous. Notice that the focus is totally on entropy. We're not saying anything about the energy. The energy can do anything it wants. We're only focusing attention on the entropy to understand whether these processes are spontaneous or not. Okay, so we've got to have this term in here for the surroundings. We didn't need it for the isolated system. Okay, so now we're just going to do a little algebra. We're going to move the surroundings over to the right hand side, put a negative sign in front of it. And then we're going to remember that dS is minus dQ over T. And so we can make that substitution for the surroundings right here. And then we're just going to remember that Q is a conserved quantity. In other words, if plus Q is heat entering the system, that heat had to come from the surroundings. And so the surroundings has to be minus dQ. There's conservation of Q. And then we have to think back and remember that dU is dW plus dQ. And so we can just solve for dQ in this expression. dU minus dW. And if we plug that guy in for dQ, we get this guy right here. And if we consider only pressure volume work, this dW is PDV. Okay, and so then we're going to multiply by T surroundings and move it over to the left hand side. So we get rid of this T surroundings now. Now it's over here. And we just have dU plus PDV. We're going to drop that cis subscript. If you don't see a subscript, just assume we're talking about the system. Okay, so this is the pink equation. It only took us about four steps to get there from conservation of entropy. Or not conservation of entropy from entropy dictating which process is spontaneous or not. We kept coming back to this pink equation on Friday in deriving different thermodynamic state functions from it. In fact, we showed that if the process occurs under conditions of constant volume and constant entropy, then it's the internal energy that tells us whether the process is spontaneous or not. And if instead the process occurs under conditions of constant pressure and entropy, why it's the enthalpy that's going to tell us whether the process is spontaneous or not. But unfortunately, we don't encounter these two sets of conditions very often. It's virtually never the case that the entropy is constant. If you're a chemist, you can ask, you know, how do you do that? How do you do an experiment with constant entropy? I don't know the answer. So when you're doing an experiment and you want to understand whether it's spontaneous or not, the chemistry that you're looking at, it's unlikely that you're going to be paying attention to these two variables to figure that out. They're not going to help guide your decision-making process in figuring out whether your chemistry is spontaneous or not. That's what we care about here. So we need some other state functions. And we talked about one, all right? The Helmholtz energy, all right? In chemistry, temperature is frequently constant but not the entropy, all right? Constant temperature, lots of constant temperature chemical processes that we can think about. So let's consider the case where Dt is zero and the volume is zero. We're going to need for two things to be zero. Otherwise, we're not going to end up with a state function. And so we'll also define a new state function A, which is going to be called the Helmholtz energy. It's going to be defined as the internal energy minus Ts. And so if we take the derivative now to get Da on the left-hand side, we're going to get Du and we're going to have Dts. And so we can split that into two terms, Dts minus Sdt. And then we can just solve for Du. And this expression, Du is going to be equal to Da plus Tds plus Sdt. And of course, the next thing that we do is we plug this thing into the pink equation, put this expression for Du into the pink equation, and then look for the terms that are going to cancel for us, all right? So we've got Da, Tds, Sdt, and our expression for pressure volume work from the pink equation. Okay, and the first thing that we notice is that we got Tds here. We got Tds here. Now these two Ts are different in principle. This is a T for the surroundings. This is a T for the system. But as we converge on equilibrium, these two temperatures will become very, very close. And under those conditions, we can expect these two terms to cancel for us. And then under conditions where we said Dt is 0 and Dv is 0, there's Dv. We can cancel out that term. And Dt, we can cancel out that term. All right, and we're just left with Da. There's nothing else left. And so it's going to be Da is less than 0. And so this Hemmholtz energy is going to be a state function that we can use to tell whether the chemistry that we care about is spontaneous or not when temperature and volume are held constant. And in the laboratory, we can enforce those limits. We can do an experiment at constant temperature, maintaining the volume constant, let the pressure do whatever it wants. Okay, we need one of these things. All right, it's got a defined volume and it's built like a tank and so even if the pressure changes a lot, we're going to enforce constant pressure. And in principle, it's the Hemmholtz energy that will tell us whether a reaction in this parbomb is going to be spontaneous or not. Right, we would want to, if we do an experiment in here, we would want to use the Hemmholtz energy to figure out whether it's spontaneous or not. Now, if you do undergraduate research, how many people have done undergraduate research? How many people have seen a parbomb? All right, a few of you, who do you guys work for? You all work for the same person? Oh. He's got a parbomb in his lab. Okay, so it's not completely impossible that you would use one of these things, right? They are in rather common use but I would say probably 99.9% of all the chemistry that we're likely to do is not going to be in a parbomb. All right, 99.9%. So, we need a different thermodynamic function. The Hemmholtz energy is fine but constant volume is inconvenient for us to use because we needed parbomb to do it in many cases. In chemistry, it's even more useful to make predictions about processes occurring at constant pressure and temperature because that's dead easy. Right, we live in an environment of quasi-constant pressure. Okay, and so we can do chemistry that's open to the environment and make predictions about whether it's spontaneous or not. To that, we're going to use something called the Gibbs energy. All right, we're going to define it as H minus Ts. Okay, we're going to do the same kind of algebra we did before, Dg is DH minus Dts and so we've got two terms here now. And then we're going to think back to Friday when we wrote an expression for DH. We said DH is Du plus PDV plus VDP. Okay, and so we can just plug that in for DH here. Now, we've got this long thing here that's equal to Dg. Okay, and so once again, we're just going to solve for Du. All right, put all of this other stuff on the right-hand side. And then once we've got Du, we're going to just plug it into the pink equation again. There it is. Put all of this stuff in for Du. We get this long thing here. Okay, and some of these terms are going to start canceling for us. As usual, PDV, PDV, right? Remember these two pressures are not in principle identical. That's the system pressure. That's the surroundings pressure. All right, but in the limit of equilibrium, they will be the same. All right, Tds, Tds, same idea there. And then because we're talking about G, we're going to make P constant. And so we're going to lose that guy. And Dt, so we're going to lose that guy. So everything cancels out here except Dg, which is going to be less than or equal to zero. And so that's going to be the state function that we're going to want a key on most of the time as chemists. All right, now if you're a physicist, if you're some other kind of scientist, these other state functions might be more important to you under other sets of conditions. But for chemists, it's all about the Gibbs free energy. The Gibbs energy, we're not supposed to call it the Gibbs free energy anymore. It's just the Gibbs energy. Okay, now I know that's tedious. But this is important, right? This is actually one of the more important things in thermodynamics that we need to be able to understand. All right, here's where we're doing chemistry. And this guy right here, we're open to the atmosphere and the Gibbs function is going to tell us whether this blue stuff here is going to react spontaneously. All right, we don't need the par bomb. Okay, so today and last Friday, we've taken the consideration, we've taken the condition for spontaneous change for non-isolated systems. We consider the total entropy change, system plus surroundings. That's got to be greater than or equal to zero. And from that, we derived all of these different conditions that apply for these different constraints. Volume and entropy, temperature and volume, pressure and entropy, temperature and pressure. We've got these four different. And what I've told you today is look, these two are not super useful to us. As chemists, these two are more useful. And this one is way more useful than that. All right, we derived these all. We didn't have to assume anything. Very proud of that. It's hard. Okay, now these conditions here also serve to tell us whether this system is proceeding towards equilibrium. It not only tells us whether the chemistry is spontaneous, it'll tell us whether the system is proceeding towards equilibrium or not. For example, what I'm plotting here is the Gibbs energy for some chemical process. And on this axis, I have the reaction coordinate. So this represents reactants right here and this represents products. This is 100% products. This is 100% reactants. But as you move along the axis in this direction, we're converting reactants into products. That's what I mean by a reaction coordinate. Sometimes we'll call this reaction coordinate X or chi. Okay, reactants getting converted to products, very generic. What does DG that should be P and T? How did that happen? Less than zero. There it is, P and T. All right, so first of all, this difference here between the reaction and product Gibbs functions, that's the delta G reaction. Makes sense. Now, let's consider a process that starts right here and ends right here. Or we can ask, is such a process going to be spontaneous or not? Well, we have a criterion here if we just change that V to a T. All right, we know DG should be less than zero. Okay, and so we can say G final minus G initial is that DG final minus initial. Is that going to be less than zero or greater than zero? What do you think? Yeah, it's a small number minus a bigger number, and so that difference is going to be negative, isn't it? All right, and so we would predict that's a spontaneous process. Yes, DG at constant T and P is less than zero. What about this guy? Same conclusion. What about this guy? No, final minus initial is going to be a positive number now. All right, so that's not going to be a spontaneous process going from here to here. No. All right, and what about that guy? Yes, final minus initial is going to be negative again, and so that should be spontaneous. All right, so basically what we're concluding is that if you're over here, we're going to go spontaneously downhill in this direction, and if you're over here, you're going to go spontaneously downhill in this direction, and that this minimum here in the Gibbs energy is going to indicate the equilibrium position of this reaction. It's the point where DG over DX, where X is now my reaction coordinate, is equal to zero. At that point, there's no more driving force for spontaneous change. We're at equilibrium. Okay. Now, amongst these four thermodynamic potentials, U, H, A, and G, G will be by far the most important to us. Yes, yes, yes. How does G depend on temperature? All right, how does the Gibbs energy depend on temperature? Well, that's a rather important thing for us to understand, because as chemists, if we want to accelerate a reaction, G is going to tell us whether the reaction is spontaneous or not. All right, we want to understand how temperature will influence that spontaneity. G is equal to H minus TS. We know that, and so we can take the derivative immediately, DG, DT. Even I can take this derivative. I get minus S. Okay, and so what this tells us is two things. First of all, since we know that S is always a positive quantity, there's no such thing as negative entropy. S is always a positive quantity. All right, that tells us that G has to decrease with increasing temperature, because that derivative is always going to be negative. All right, that's kind of surprising. The Gibbs energy is going to go down as the temperature goes up. That's counterintuitive. Don't all energies go up when you increase the temperature? Not this one. All right, the Gibbs energy goes down as you increase the temperature. Not only that, but the rate of change of G with temperature is greatest for systems having high entropy. The higher the entropy, the greater the change in G is going to be with temperature. Well, what kind of systems have high entropy? Well, gases, my laser pointer's dying, gases have the highest entropy. And so the rate of change of the free energy with temperature is going to be the highest, then liquids, then solids. Solids have the lowest entropy. Okay, so this plot is right out of your chapter 16. Gases, biggest slope, right? Here's the Gibbs energy on this axis. Here's temperature. It's going down. For every single one of these guys, it's going down, right? But it's going down at a rate that depends on the state. Gases show the largest decrease in Gibbs energy with temperature. Liquids next, solids show the least. Okay, so a couple things are, I mean, one thing that's surprising for sure is that the Gibbs energy goes down with temperature. All right, it's an unusual energy, isn't it, that goes down with temperature. Okay, so we can evaluate this derivative. And then we can go back to this equation right here, and we can just say, we can solve for minus s. All right, if I solve for minus s here, I'm going to get g minus h over t, just solving for minus s in that equation right there. Okay, so I've got dg over t at constant p is g minus h over t. And then we can rearrange that, just split this into two terms, move g over t to the left-hand side. I don't know why we actually did that, because I don't think we need this result here. All right, maybe we'll come back to this in a second. But let's just look at this for a second. This is the derivative of g over t. I don't know, that has anything to do with you. This is just the derivative of g over t. If I use the quotient rule to evaluate this derivative, I've got 1 over t times the derivative of g with respect to t, and I've got g times the derivative of t with respect to t, right? Two terms in my quotient rule expansion. Okay, now, the derivative of 1 over t is just minus 1 over t squared, right? Okay, and so that's that derivative right there. And this guy, if we factor out 1 over t, so I'm going to pull the 1 over t out of both of these two terms and put it right there, all right? Now I've got this expression here, and that is just the entropy, right? The derivative of the Gibbs energy with respect to t at constant p, that's the entropy. Okay, and so I can plug that in to this expression right here. I still got g over t, and then I can move, maybe that's why I did it. No, all right, forget that. This, all right, is just plugging in for g from that equation two slides ago. Okay, and so this is s over t. This is s over t, so we're going to get rid of the s over t. We're just going to be left with h over t squared, all right? And this is your equation 15.62b. This is the Gibbs-Hammholtz equation, which is important because it allows us to measure h by looking at the temperature dependence of g. And the temperature dependence of g is something we're going to be able to measure experimentally, all right? And so we can get h directly from that using this Gibbs-Hammholtz equation. Now, if this is a delta h and that's a delta g, this equation still is fine. Okay, so now let's ask some questions about the Gibbs function. We already asked a question about the temperature dependence. We said the Gibbs function goes down with increasing temperature, surprising. The rate at which it goes down depends on the entropy, all right? Higher the entropy, the faster the temperature rate of change of the Gibbs function. What about pressure? All right, we've got this expression here for dg. And if we want to look at this at constant temperature, we can say dt is zero, all right? And so that term is just going to go away. We've got dg is Vdp. And so to find out what the free energy, free energy, the Gibbs energy change is, there's the Gibbs energy at p final minus the Gibbs energy at p initial. I can just integrate this Vdp, all right, from initial to final pressure. And if I know what that is, if these are molar quantities, of course, there's going to be an m there, an m there, and an m there, all right? If these are all molar quantities. And so this equation is not super useful to us unless we know how this volume changes with pressure, all right? But one thing that's obvious is for phases like solids and liquids that are essentially incompressible, Vm is virtually constant, independent of pressure, all right? There's not much compressibility of a solid phase or a liquid phase, right? And so there isn't much pressure dependence of Vm. And so we can write a simpler expression. If we can pull Vm into the front of this integral sign, right, because it's constant, then the integral just turns into this. And we've got an extremely simple expression that allows us to evaluate the pressure dependence on the Gibbs energy, right? It's just the molar volume times the change in pressure. So if you look, once again, this is the Gibbs energy on this vertical axis here. And this is the pressure on this horizontal axis, right? And for liquids and solids, you get virtually a horizontal line. Because they're incompressible, right? Their volume doesn't depend on pressure very strongly. Interestingly, as the pressure gets higher, the Gibbs function goes up a little bit. All right? With gases, there's a much stronger effect. Gibbs energies of gas depend strongly on the pressure. And you might expect them to because gases are far from incompressible. They're highly compressible. All right? So their molar volume is highly dependent on pressure. Consequently, their Gibbs energy is highly dependent on pressure. In fact, their Gibbs energy goes up with increasing pressure. Now, we can actually figure out what this is for ideal gases very readily. All right? We can just substitute for Vm from the ideal gas equation. Move the RT out front. All right? So that's going to move out front. We've got 1 over P. So we're just going to log P final over P initial. That is the equation that describes the change in the Gibbs energy for an ideal gas as a function of pressure. All right? So change the pressure. We have very simple equation. Probably should go on your equations page for quiz five. What is with this projector? What is with my laser pointer? Too cheap to buy new batteries for it. Please, please. OK. So this is what the volume is doing as a function of pressure for an ideal gas. It's following this purple line here. OK? And so if we want to evaluate this integral, we're going to be integrating from some initial pressure to some final pressure. This is the area underneath this curve. So this is the Gibbs energy. All right? And it's obvious that as we make PF higher and higher and higher, this integral is going to get bigger, right? And so it's obvious that the Gibbs energy is going to go up. Just based on that. Now we can define a standard molar Gibbs free energy. All right? We're constantly, there shouldn't be the word free in here. We're trying to get rid of the word free. Should be the standard molar Gibbs energy. All right? That's defined at a defined pressure, which is one bar. All right? That's how we define the standard Gibbs energy. Or the standard entropy. Or the standard anything. If it says standard, it's one bar. OK? And so in this case, we can write this expression. It just follows directly from this guy right here, except that we've now defined a particular Gibbs energy that applies when the pressure is one bar. OK? So this initial guy is now that guy. OK? So all this plot shows. Here's the molar Gibbs energy, and here's the pressure. And what we said earlier is that as I increase pf, I'm going to make this integral larger. And so that tells us the Gibbs energy's got to go up with increasing pressure. All right? What I just showed on this slide right here is that we're going to define a special Gibbs energy at one bar. And so that's what this becomes. This becomes one bar. Imagine this is one bar, and now we're integrating to higher pressure from that. So the same intuitive picture applies. If we move this to higher pressure, the integral is going to go up. And that's why this plot is going up, up, up, up. But you can see that it's got downward curvature. It's got downward curvature because here there's a big change in the Gibbs energy. Smaller, smaller, smaller. This has got upward curvature. So if we can integrate this guy, increases in pressure are going to have a progressively smaller and smaller effect on the Gibbs energy. And that's what we're seeing here. That's why there's downward curvature of this guy. All right? So the Gibbs energy goes down as we increase the pressure and up as we increase, sorry, goes down as we increase the temperature and up as we increase the pressure. Is all of this confusing? Yes. Absolutely. I mean, if you don't think so, you're just not paying attention. OK. Should we do some examples? The change in the Gibbs energy of 25 grams of methanol, mass density 0.791 grams per cubic centimeter, when the pressure is increased isothermally from 100 kilopascals to 100 megapascals, oh, I should say, calculate the change in the Gibbs energy. No verb. Calculate the Gibbs energy. When we subject this methanol to a change in pressure, that's a change by a factor of 1,000. It's a big change. But so your first thought process should be, it's a liquid, it's incompressible, let's use the simplest equation we can. And the simplest equation for any liquid or solid is just V. M times delta P. V. M times delta P. I immediately get the change in the Gibbs energy. This is what I did. And so if I'm willing to give up these molar, forget molar, I'm just going to calculate the difference. It doesn't ask me for the change in the molar Gibbs energy, it just asks me for the change in the Gibbs energy. So here I need, I'm going to calculate the change in the Gibbs energy. The volume of my 25 grams of methanol, there's the density and so I can calculate the volume. That's 10 to the 5, that's 10 to the 8, so there's my factor of 1,000 and so when I plug these numbers in I get 3.157 times 10 to the 9 cubic centimeters per cubic centimeter pascals. That's the delta G that I get. I don't like these units but I can convert them later. Now, oh, so I converted them right here. 3.157 times 10 to the 9 centimeters cubed pascals, no I don't like that and so I can use the definition for, here's the conversion factor for pascals to ATMs, there's the conversion factor from cubic centimeters to later, now I've got later atmospheres and so then I can use the ratio between the two different Rs to do the unit conversion to get joules. Tidious, but when I do that I get 3.157 times 10 to the 3 joules, or roughly 3 kilojoules. That's what the change in the Gibbs energy is going to be. Now, when I looked up the answer in the key it had something far fancier. It used the isothermal compressibility of methanol, which you can look up in the back of your book. That's the isothermal compressibility, it's 1.26 times 10 to the minus 9 per pascal, well for goodness sakes. We can go through and do it with the isothermal compressibility and see how different our answers will be. The isothermal compressibility is defined by equation 14.59, it's 1 over V times the derivative of volume with pressure constant temperature, and we can just linearize this differential, final minus initial, because we know the change in volume is going to be small. Pressure difference is not small actually, but we're going to linearize that also. So we're going to make two approximations, we're going to say the initial volume is just the final volume also, it's not going to change much. So that's going to become VI, so I know what that is. And the final pressure is much, much larger than the initial pressure, and so that difference right there I can just approximate as the final pressure. The initial pressure is just a rounding error on the final pressure, literally. So that's the equation I'm going to use. And so if I just do a little algebra, split this into two terms, now solve for VF, I get this expression right here, which can be further simplified to give me this expression right here, and then what I've got to do, sadly, is integrate that, so I'm going to plug that in for VM and do the integral. And so that's what I did here. I can calculate what this VI is, that's the mass, that's the density, and here are the two derivatives I'm going to do. I'm going to actually pull this out front, and then I'm going to take the derivative of dp, well that's pretty easy to do, and then kappa t times the derivative of p, and so I can run these two integrals. Now this is just going to be 10 to the 8 minus 10 to the 5. This guy is 1 half kappa tp squared, and when I plug the integration limits into these two antiderivatives, this guy ends up being almost 10 to the 8. That's the rounding error that gets subtracted from him, and these are the results for these other two terms. And so what I calculate is 2.96 times 10 to the 3 joules, which is also about 3 kilojoules, but if I do this carefully considering the isothermal compressibility of methanol, I get a slightly different answer. It's different by one part in 30 roughly. That's wrong, that's right, but pretty darn close to being right. So it depends on the precision that you need on the quiz. That might be good enough, unless this is also an answer. That would be cruel, wouldn't it? One part in 30 difference in the answer? No, I would never do that. Everyone see how to do that? Isothermal compressibility makes it a little bit more complicated. So the last thing to say today is that, and this may be completely obvious to everybody, is that if you want to know the standard Gibbs free energy for a reaction, like this reaction right here, what you do is you look up on a table, the Gibbs standard, that free shouldn't be there again, the Gibbs energy of this stuff, the Gibbs energy of this stuff, and the Gibbs energy of this stuff, and the standard energy difference is that minus that plus that. I can look this up in tables. Now, if I don't have a table of Gibbs energies, well I've only got a table of entropies and enthalpies, which is not terribly uncommon, you've got to use this equation right here. You can look up these enthalpies, you can look up these entropies, and if you know the temperature at which the reaction is happening, you can figure out whether the reaction is spontaneous at that temperature or not. You can look up the delta G. This is the delta G reaction. So in the case of H, we've got this delta sub R H, what that notation means is that R is the reaction, right? This is the enthalpy change for the reaction. This is the delta H formation. That's what we're going to look up in the table. What we want is we want to look up the delta H formation for all the reactants and all the products. Add up the reactant delta H formation, subtract, sorry, add up the product delta H formations, subtract the reactant delta H formations to get the delta H for the reaction. This zero means standard. What does that mean? Pressure equals one bar. Temperature equals 298.16 K. What is that new? It's just a stoichiometric coefficient in front of these. So that would be one, one, and one in this case. We need to do the same thing for the entropy. Here's the enthalpy, here's the entropy. We still get stoichiometric coefficients. We're going to look up these standard entropies, products minus reactants. So we can do that. Here's, I've done it by Golly. That's for this guy right here, which is what? Propionic acid. And these two guys are for those two in no particular order. Okay, so the delta H is just that, an evaluation of that, all of those numbers. Okay, and you can do the same thing for the entropy, from a table of standard entropies. You do exactly the same thing. And the delta G then is just the delta H minus T times the delta S, where we put plug in the T. All right, we get 72.6 kilojoules per mole at the delta G. We conclude that this reaction should be spontaneous at this temperature. Okay, I think that's all I got. 105. Are there questions about this stuff? I know it's not riveting. All right, we'll do, hopefully, it's going to get more interesting, I think.
 Okay, how are you guys doing?  You got those mid-quarter daldrums.  I know how you're feeling.  Okay, this is a chapter 16 topic.  We're really cranking away on chapter 16 here.  Edging our way towards the end of thermodynamics,  which I hope will come probably in the middle of next week.  Now we posted the quiz scores this morning.  This was the hardest quiz.  It might have been too hard.  First question was pretty easy.  Didn't you think?  So these guys got A's.  These guys got B's.  There's a few C's here.  We usually don't have hardly any C's.  Okay, quiz five is Friday.  The first two are supposed to be easy.  The first two problems.  In this case, the second one was actually not so easy.  It was the body temperature one.  Okay, I also posted the key in case you want to look at that.  So we won.  And what's even sweeter is we beat SC.  Very, very sweet.  I'm always amused by the fighting anteater.  The anteater is the most docile animal that we know  of in the animal kingdom.  But to dress him up for the athletic department,  here he is fierce.  There's no such thing as a fierce anteater.  In nature, they're not known to be fierce.  Okay, so we're going to review a little from Friday.  Why does that say Wednesday?  We're going to talk about how the Gibbs free energy varies  with temperature and pressure.  We'll do a couple of examples.  Just sort of easy ones to ease into this subject.  Okay, we'll do a bunch more examples on Wednesday.  All right, so on Friday, not Wednesday, we said, look,  there's three types of systems.  Okay, and uniquely for this guy right here,  it's an isolated system.  There's no energy or matter exchanged  with the environment, the surroundings.  We don't have to consider anything except the system  when we think about the spontaneity of processes  that occur within it.  Okay, it's blocked off to the surroundings.  It doesn't even know about them.  And so we can say any process  that has an entropy, positive entropy change,  is going to be a spontaneous process  for an isolated system.  We don't even have to think about the surroundings.  They're not part of our thought process.  But we don't have isolated systems in chemistry too often.  They're almost always in communication  with the environment.  And so we have to consider open systems  and closed systems as well.  And in those cases, because there is communication  with the surroundings, it's the total entropy,  surroundings plus system that matters.  Okay, in terms of figuring out if this process is spontaneous.  Notice that the focus is totally on entropy.  We're not saying anything about the energy.  The energy can do anything it wants.  We're only focusing attention on the entropy  to understand whether these processes are spontaneous or not.  Okay, so we've got to have this term in here  for the surroundings.  We didn't need it for the isolated system.  Okay, so now we're just going to do a little algebra.  We're going to move the surroundings over to the right  hand side, put a negative sign in front of it.  And then we're going to remember  that dS is minus dQ over T. And so we can make  that substitution for the surroundings right here.  And then we're just going to remember  that Q is a conserved quantity.  In other words, if plus Q is heat entering the system,  that heat had to come from the surroundings.  And so the surroundings has to be minus dQ.  There's conservation of Q. And then we have to think back  and remember that dU is dW plus dQ.  And so we can just solve for dQ in this expression.  dU minus dW. And if we plug that guy in for dQ,  we get this guy right here.  And if we consider only pressure volume work,  this dW is PDV.  Okay, and so then we're going to multiply by T surroundings  and move it over to the left hand side.  So we get rid of this T surroundings now.  Now it's over here.  And we just have dU plus PDV.  We're going to drop that cis subscript.  If you don't see a subscript,  just assume we're talking about the system.  Okay, so this is the pink equation.  It only took us about four steps to get there  from conservation of entropy.  Or not conservation of entropy from entropy dictating  which process is spontaneous or not.  We kept coming back to this pink equation on Friday  in deriving different thermodynamic state functions  from it.  In fact, we showed that if the process occurs under conditions  of constant volume and constant entropy,  then it's the internal energy that tells us  whether the process is spontaneous or not.  And if instead the process occurs under conditions  of constant pressure and entropy,  why it's the enthalpy that's going to tell us whether  the process is spontaneous or not.  But unfortunately, we don't encounter these two sets  of conditions very often.  It's virtually never the case that the entropy is constant.  If you're a chemist, you can ask, you know,  how do you do that?  How do you do an experiment with constant entropy?  I don't know the answer.  So when you're doing an experiment and you want  to understand whether it's spontaneous or not,  the chemistry that you're looking at, it's unlikely  that you're going to be paying attention  to these two variables to figure that out.  They're not going to help guide your decision-making process  in figuring out whether your chemistry is spontaneous or not.  That's what we care about here.  So we need some other state functions.  And we talked about one, all right?  The Helmholtz energy, all right?  In chemistry, temperature is frequently constant  but not the entropy, all right?  Constant temperature, lots  of constant temperature chemical processes  that we can think about.  So let's consider the case where Dt is zero  and the volume is zero.  We're going to need for two things to be zero.  Otherwise, we're not going to end up with a state function.  And so we'll also define a new state function A,  which is going to be called the Helmholtz energy.  It's going to be defined as the internal energy minus Ts.  And so if we take the derivative now to get Da  on the left-hand side, we're going to get Du  and we're going to have Dts.  And so we can split that into two terms, Dts minus Sdt.  And then we can just solve for Du.  And this expression, Du is going to be equal  to Da plus Tds plus Sdt.  And of course, the next thing that we do is we plug this thing  into the pink equation, put this expression for Du  into the pink equation, and then look for the terms  that are going to cancel for us, all right?  So we've got Da, Tds, Sdt, and our expression  for pressure volume work from the pink equation.  Okay, and the first thing that we notice is  that we got Tds here.  We got Tds here.  Now these two Ts are different in principle.  This is a T for the surroundings.  This is a T for the system.  But as we converge on equilibrium,  these two temperatures will become very, very close.  And under those conditions, we can expect these two terms  to cancel for us.  And then under conditions where we said Dt is 0  and Dv is 0, there's Dv.  We can cancel out that term.  And Dt, we can cancel out that term.  All right, and we're just left with Da.  There's nothing else left.  And so it's going to be Da is less than 0.  And so this Hemmholtz energy is going to be a state function  that we can use to tell whether the chemistry  that we care about is spontaneous or not  when temperature and volume are held constant.  And in the laboratory, we can enforce those limits.  We can do an experiment at constant temperature,  maintaining the volume constant,  let the pressure do whatever it wants.  Okay, we need one of these things.  All right, it's got a defined volume and it's built  like a tank and so even if the pressure changes a lot,  we're going to enforce constant pressure.  And in principle, it's the Hemmholtz energy  that will tell us whether a reaction  in this parbomb is going to be spontaneous or not.  Right, we would want to, if we do an experiment in here,  we would want to use the Hemmholtz energy  to figure out whether it's spontaneous or not.  Now, if you do undergraduate research,  how many people have done undergraduate research?  How many people have seen a parbomb?  All right, a few of you, who do you guys work for?  You all work for the same person?  Oh.  He's got a parbomb in his lab.  Okay, so it's not completely impossible that you would use one  of these things, right?  They are in rather common use but I would say probably 99.9%  of all the chemistry that we're likely to do is not going  to be in a parbomb.  All right, 99.9%.  So, we need a different thermodynamic function.  The Hemmholtz energy is fine but constant volume is inconvenient  for us to use because we needed parbomb to do it in many cases.  In chemistry, it's even more useful to make predictions  about processes occurring at constant pressure and temperature  because that's dead easy.  Right, we live in an environment of quasi-constant pressure.  Okay, and so we can do chemistry that's open to the environment  and make predictions about whether it's spontaneous or not.  To that, we're going to use something called the Gibbs energy.  All right, we're going to define it as H minus Ts.  Okay, we're going to do the same kind of algebra we did before,  Dg is DH minus Dts and so we've got two terms here now.  And then we're going to think back to Friday when we wrote an  expression for DH.  We said DH is Du plus PDV plus VDP.  Okay, and so we can just plug that in for DH here.  Now, we've got this long thing here that's equal to Dg.  Okay, and so once again, we're just going to solve for Du.  All right, put all of this other stuff on the right-hand side.  And then once we've got Du, we're going to just plug it  into the pink equation again.  There it is.  Put all of this stuff in for Du.  We get this long thing here.  Okay, and some of these terms are going to start canceling for us.  As usual, PDV, PDV, right?  Remember these two pressures are not in principle identical.  That's the system pressure.  That's the surroundings pressure.  All right, but in the limit of equilibrium,  they will be the same.  All right, Tds, Tds, same idea there.  And then because we're talking about G, we're going to make P constant.  And so we're going to lose that guy.  And Dt, so we're going to lose that guy.  So everything cancels out here except Dg, which is going to be less  than or equal to zero.  And so that's going to be the state function that we're going  to want a key on most of the time as chemists.  All right, now if you're a physicist, if you're some other kind of scientist,  these other state functions might be more important to you  under other sets of conditions.  But for chemists, it's all about the Gibbs free energy.  The Gibbs energy, we're not supposed to call it the Gibbs free energy anymore.  It's just the Gibbs energy.  Okay, now I know that's tedious.  But this is important, right?  This is actually one of the more important things in thermodynamics  that we need to be able to understand.  All right, here's where we're doing chemistry.  And this guy right here, we're open to the atmosphere  and the Gibbs function is going to tell us whether this blue stuff here is going  to react spontaneously.  All right, we don't need the par bomb.  Okay, so today and last Friday, we've taken the consideration,  we've taken the condition for spontaneous change for non-isolated systems.  We consider the total entropy change, system plus surroundings.  That's got to be greater than or equal to zero.  And from that, we derived all of these different conditions that apply  for these different constraints.  Volume and entropy, temperature and volume, pressure and entropy,  temperature and pressure.  We've got these four different.  And what I've told you today is look, these two are not super useful to us.  As chemists, these two are more useful.  And this one is way more useful than that.  All right, we derived these all.  We didn't have to assume anything.  Very proud of that.  It's hard.  Okay, now these conditions here also serve to tell us whether this system is proceeding  towards equilibrium.  It not only tells us whether the chemistry is spontaneous,  it'll tell us whether the system is proceeding towards equilibrium or not.  For example, what I'm plotting here is the Gibbs energy for some chemical process.  And on this axis, I have the reaction coordinate.  So this represents reactants right here and this represents products.  This is 100% products.  This is 100% reactants.  But as you move along the axis in this direction,  we're converting reactants into products.  That's what I mean by a reaction coordinate.  Sometimes we'll call this reaction coordinate X or chi.  Okay, reactants getting converted to products, very generic.  What does DG that should be P and T?  How did that happen?  Less than zero.  There it is, P and T. All right, so first of all, this difference here  between the reaction and product Gibbs functions,  that's the delta G reaction.  Makes sense.  Now, let's consider a process that starts right here and ends right here.  Or we can ask, is such a process going to be spontaneous or not?  Well, we have a criterion here if we just change that V to a T. All right,  we know DG should be less than zero.  Okay, and so we can say G final minus G initial is that DG final minus initial.  Is that going to be less than zero or greater than zero?  What do you think?  Yeah, it's a small number minus a bigger number,  and so that difference is going to be negative, isn't it?  All right, and so we would predict that's a spontaneous process.  Yes, DG at constant T and P is less than zero.  What about this guy?  Same conclusion.  What about this guy?  No, final minus initial is going to be a positive number now.  All right, so that's not going to be a spontaneous process going  from here to here.  No. All right, and what about that guy?  Yes, final minus initial is going to be negative again,  and so that should be spontaneous.  All right, so basically what we're concluding is that if you're over here,  we're going to go spontaneously downhill in this direction,  and if you're over here, you're going to go spontaneously downhill in this direction,  and that this minimum here in the Gibbs energy is going  to indicate the equilibrium position of this reaction.  It's the point where DG over DX,  where X is now my reaction coordinate, is equal to zero.  At that point, there's no more driving force  for spontaneous change.  We're at equilibrium.  Okay. Now, amongst these four thermodynamic potentials, U, H, A,  and G, G will be by far the most important to us.  Yes, yes, yes.  How does G depend on temperature?  All right, how does the Gibbs energy depend on temperature?  Well, that's a rather important thing for us to understand,  because as chemists, if we want to accelerate a reaction,  G is going to tell us whether the reaction is spontaneous or not.  All right, we want to understand how temperature will influence that spontaneity.  G is equal to H minus TS.  We know that, and so we can take the derivative immediately, DG, DT.  Even I can take this derivative.  I get minus S. Okay, and so what this tells us is two things.  First of all, since we know that S is always a positive quantity,  there's no such thing as negative entropy.  S is always a positive quantity.  All right, that tells us that G has to decrease  with increasing temperature, because that derivative is always going to be negative.  All right, that's kind of surprising.  The Gibbs energy is going to go down as the temperature goes up.  That's counterintuitive.  Don't all energies go up when you increase the temperature?  Not this one.  All right, the Gibbs energy goes down as you increase the temperature.  Not only that, but the rate of change of G with temperature is greatest  for systems having high entropy.  The higher the entropy, the greater the change in G is going to be with temperature.  Well, what kind of systems have high entropy?  Well, gases, my laser pointer's dying, gases have the highest entropy.  And so the rate of change of the free energy with temperature is going to be the highest,  then liquids, then solids.  Solids have the lowest entropy.  Okay, so this plot is right out of your chapter 16.  Gases, biggest slope, right?  Here's the Gibbs energy on this axis.  Here's temperature.  It's going down.  For every single one of these guys, it's going down, right?  But it's going down at a rate that depends on the state.  Gases show the largest decrease in Gibbs energy with temperature.  Liquids next, solids show the least.  Okay, so a couple things are, I mean, one thing that's surprising for sure is  that the Gibbs energy goes down with temperature.  All right, it's an unusual energy, isn't it, that goes down with temperature.  Okay, so we can evaluate this derivative.  And then we can go back to this equation right here, and we can just say,  we can solve for minus s.  All right, if I solve for minus s here, I'm going to get g minus h over t,  just solving for minus s in that equation right there.  Okay, so I've got dg over t at constant p is g minus h over t.  And then we can rearrange that, just split this into two terms, move g  over t to the left-hand side.  I don't know why we actually did that, because I don't think we need this result here.  All right, maybe we'll come back to this in a second.  But let's just look at this for a second.  This is the derivative of g over t.  I don't know, that has anything to do with you.  This is just the derivative of g over t.  If I use the quotient rule to evaluate this derivative, I've got 1 over t times the  derivative of g with respect to t, and I've got g times the derivative of t  with respect to t, right?  Two terms in my quotient rule expansion.  Okay, now, the derivative of 1 over t is just minus 1 over t squared, right?  Okay, and so that's that derivative right there.  And this guy, if we factor out 1 over t, so I'm going to pull the 1 over t out of both  of these two terms and put it right there, all right?  Now I've got this expression here, and that is just the entropy, right?  The derivative of the Gibbs energy with respect to t at constant p, that's the entropy.  Okay, and so I can plug that in to this expression right here.  I still got g over t, and then I can move, maybe that's why I did it.  No, all right, forget that.  This, all right, is just plugging in for g from that equation two slides ago.  Okay, and so this is s over t.  This is s over t, so we're going to get rid of the s over t.  We're just going to be left with h over t squared, all right?  And this is your equation 15.62b.  This is the Gibbs-Hammholtz equation, which is important because it allows us to measure h  by looking at the temperature dependence of g.  And the temperature dependence of g is something we're going to be able  to measure experimentally, all right?  And so we can get h directly from that using this Gibbs-Hammholtz equation.  Now, if this is a delta h and that's a delta g, this equation still is fine.  Okay, so now let's ask some questions about the Gibbs function.  We already asked a question about the temperature dependence.  We said the Gibbs function goes down with increasing temperature, surprising.  The rate at which it goes down depends on the entropy, all right?  Higher the entropy, the faster the temperature rate of change  of the Gibbs function.  What about pressure?  All right, we've got this expression here for dg.  And if we want to look at this at constant temperature, we can say dt is zero, all right?  And so that term is just going to go away.  We've got dg is Vdp.  And so to find out what the free energy, free energy, the Gibbs energy change is,  there's the Gibbs energy at p final minus the Gibbs energy at p initial.  I can just integrate this Vdp, all right, from initial to final pressure.  And if I know what that is, if these are molar quantities, of course, there's going  to be an m there, an m there, and an m there, all right?  If these are all molar quantities.  And so this equation is not super useful to us unless we know how this volume changes  with pressure, all right?  But one thing that's obvious is for phases like solids and liquids  that are essentially incompressible,  Vm is virtually constant, independent of pressure, all right?  There's not much compressibility of a solid phase or a liquid phase, right?  And so there isn't much pressure dependence of Vm.  And so we can write a simpler expression.  If we can pull Vm into the front of this integral sign, right,  because it's constant, then the integral just turns into this.  And we've got an extremely simple expression that allows us  to evaluate the pressure dependence on the Gibbs energy, right?  It's just the molar volume times the change in pressure.  So if you look, once again, this is the Gibbs energy on this vertical axis here.  And this is the pressure on this horizontal axis, right?  And for liquids and solids, you get virtually a horizontal line.  Because they're incompressible, right?  Their volume doesn't depend on pressure very strongly.  Interestingly, as the pressure gets higher, the Gibbs function goes up a little bit.  All right?  With gases, there's a much stronger effect.  Gibbs energies of gas depend strongly on the pressure.  And you might expect them to because gases are far from incompressible.  They're highly compressible.  All right?  So their molar volume is highly dependent on pressure.  Consequently, their Gibbs energy is highly dependent on pressure.  In fact, their Gibbs energy goes up with increasing pressure.  Now, we can actually figure out what this is for ideal gases very readily.  All right?  We can just substitute for Vm from the ideal gas equation.  Move the RT out front.  All right?  So that's going to move out front.  We've got 1 over P. So we're just going to log P final over P initial.  That is the equation that describes the change in the Gibbs energy for an ideal gas  as a function of pressure.  All right?  So change the pressure.  We have very simple equation.  Probably should go on your equations page for quiz five.  What is with this projector?  What is with my laser pointer?  Too cheap to buy new batteries for it.  Please, please.  OK.  So this is what the volume is doing as a function of pressure for an ideal gas.  It's following this purple line here.  OK?  And so if we want to evaluate this integral,  we're going to be integrating from some initial pressure  to some final pressure.  This is the area underneath this curve.  So this is the Gibbs energy.  All right?  And it's obvious that as we make PF higher and higher and higher,  this integral is going to get bigger, right?  And so it's obvious that the Gibbs energy is going to go up.  Just based on that.  Now we can define a standard molar Gibbs free energy.  All right?  We're constantly, there shouldn't be the word free in here.  We're trying to get rid of the word free.  Should be the standard molar Gibbs energy.  All right?  That's defined at a defined pressure, which is one bar.  All right?  That's how we define the standard Gibbs energy.  Or the standard entropy.  Or the standard anything.  If it says standard, it's one bar.  OK?  And so in this case, we can write this expression.  It just follows directly from this guy right here,  except that we've now defined a particular Gibbs energy that  applies when the pressure is one bar.  OK?  So this initial guy is now that guy.  OK?  So all this plot shows.  Here's the molar Gibbs energy, and here's the pressure.  And what we said earlier is that as I increase pf,  I'm going to make this integral larger.  And so that tells us the Gibbs energy's  got to go up with increasing pressure.  All right?  What I just showed on this slide right here  is that we're going to define a special Gibbs energy  at one bar.  And so that's what this becomes.  This becomes one bar.  Imagine this is one bar, and now we're integrating to higher  pressure from that.  So the same intuitive picture applies.  If we move this to higher pressure,  the integral is going to go up.  And that's why this plot is going up, up, up, up.  But you can see that it's got downward curvature.  It's got downward curvature because here there's  a big change in the Gibbs energy.  Smaller, smaller, smaller.  This has got upward curvature.  So if we can integrate this guy, increases in pressure  are going to have a progressively smaller and smaller  effect on the Gibbs energy.  And that's what we're seeing here.  That's why there's downward curvature of this guy.  All right?  So the Gibbs energy goes down as we increase the pressure  and up as we increase, sorry, goes down  as we increase the temperature and up  as we increase the pressure.  Is all of this confusing?  Yes.  Absolutely.  I mean, if you don't think so, you're just not paying attention.  OK.  Should we do some examples?  The change in the Gibbs energy of 25 grams of methanol,  mass density 0.791 grams per cubic centimeter, when the  pressure is increased isothermally from 100 kilopascals  to 100 megapascals, oh, I should say, calculate the change  in the Gibbs energy.  No verb.  Calculate the Gibbs energy.  When we subject this methanol to a change in pressure,  that's a change by a factor of 1,000.  It's a big change.  But so your first thought process should be,  it's a liquid, it's incompressible,  let's use the simplest equation we can.  And the simplest equation for any liquid or solid  is just V.  M times delta P. V. M times delta P. I immediately get the  change in the Gibbs energy.  This is what I did.  And so if I'm willing to give up these molar, forget molar,  I'm just going to calculate the difference.  It doesn't ask me for the change in the molar Gibbs energy,  it just asks me for the change in the Gibbs energy.  So here I need, I'm going to calculate the change in the  Gibbs energy.  The volume of my 25 grams of methanol, there's the density  and so I can calculate the volume.  That's 10 to the 5, that's 10 to the 8, so there's my factor  of 1,000 and so when I plug these numbers in I get 3.157  times 10 to the 9 cubic centimeters per cubic centimeter  pascals.  That's the delta G that I get.  I don't like these units but I can convert them later.  Now, oh, so I converted them right here.  3.157 times 10 to the 9 centimeters cubed pascals,  no I don't like that and so I can use the definition for,  here's the conversion factor for pascals to ATMs,  there's the conversion factor from cubic centimeters to  later, now I've got later atmospheres and so then I can use  the ratio between the two different Rs to do the unit  conversion to get joules.  Tidious, but when I do that I get 3.157 times 10 to the 3  joules, or roughly 3 kilojoules.  That's what the change in the Gibbs energy is going to be.  Now, when I looked up the answer in the key it had something  far fancier.  It used the isothermal compressibility of methanol,  which you can look up in the back of your book.  That's the isothermal compressibility, it's 1.26 times  10 to the minus 9 per pascal, well for goodness sakes.  We can go through and do it with the isothermal compressibility  and see how different our answers will be.  The isothermal compressibility is defined by equation 14.59,  it's 1 over V times the derivative of volume with  pressure constant temperature, and we can just linearize  this differential, final minus initial, because we know the  change in volume is going to be small.  Pressure difference is not small actually, but we're going  to linearize that also.  So we're going to make two approximations, we're going to  say the initial volume is just the final volume also, it's  not going to change much.  So that's going to become VI, so I know what that is.  And the final pressure is much, much larger than the  initial pressure, and so that difference right there I can  just approximate as the final pressure.  The initial pressure is just a rounding error on the final  pressure, literally.  So that's the equation I'm going to use.  And so if I just do a little algebra, split this into two  terms, now solve for VF, I get this expression right here,  which can be further simplified to give me this  expression right here, and then what I've got to do,  sadly, is integrate that, so I'm going to plug that in for  VM and do the integral.  And so that's what I did here.  I can calculate what this VI is, that's the mass,  that's the density, and here are the two derivatives I'm  going to do.  I'm going to actually pull this out front, and then I'm going  to take the derivative of dp, well that's pretty easy to do,  and then kappa t times the derivative of p, and so I can  run these two integrals.  Now this is just going to be 10 to the 8 minus 10 to the 5.  This guy is 1 half kappa tp squared, and when I plug the  integration limits into these two antiderivatives, this guy  ends up being almost 10 to the 8.  That's the rounding error that gets subtracted from him,  and these are the results for these other two terms.  And so what I calculate is 2.96 times 10 to the 3 joules,  which is also about 3 kilojoules, but if I do this  carefully considering the isothermal compressibility  of methanol, I get a slightly different answer.  It's different by one part in 30 roughly.  That's wrong, that's right, but pretty darn close to being  right.  So it depends on the precision that you need on the quiz.  That might be good enough, unless this is also an answer.  That would be cruel, wouldn't it?  One part in 30 difference in the answer?  No, I would never do that.  Everyone see how to do that?  Isothermal compressibility makes it a little bit more  complicated.  So the last thing to say today is that, and this may be  completely obvious to everybody, is that if you want to  know the standard Gibbs free energy for a reaction,  like this reaction right here, what you do is you look up  on a table, the Gibbs standard, that free shouldn't be  there again, the Gibbs energy of this stuff, the Gibbs  energy of this stuff, and the Gibbs energy of this stuff,  and the standard energy difference is that minus that  plus that.  I can look this up in tables.  Now, if I don't have a table of Gibbs energies,  well I've only got a table of entropies and enthalpies,  which is not terribly uncommon, you've got to use this  equation right here.  You can look up these enthalpies, you can look up these  entropies, and if you know the temperature at which the  reaction is happening, you can figure out whether the  reaction is spontaneous at that temperature or not.  You can look up the delta G. This is the delta G reaction.  So in the case of H, we've got this delta sub R H, what  that notation means is that R is the reaction, right?  This is the enthalpy change for the reaction.  This is the delta H formation.  That's what we're going to look up in the table.  What we want is we want to look up the delta H formation  for all the reactants and all the products.  Add up the reactant delta H formation, subtract, sorry,  add up the product delta H formations, subtract the  reactant delta H formations to get the delta H for the  reaction.  This zero means standard.  What does that mean?  Pressure equals one bar.  Temperature equals 298.16 K. What is that new?  It's just a stoichiometric coefficient in front of these.  So that would be one, one, and one in this case.  We need to do the same thing for the entropy.  Here's the enthalpy, here's the entropy.  We still get stoichiometric coefficients.  We're going to look up these standard entropies,  products minus reactants.  So we can do that.  Here's, I've done it by Golly.  That's for this guy right here, which is what?  Propionic acid.  And these two guys are for those two in no particular order.  Okay, so the delta H is just that, an evaluation of that,  all of those numbers.  Okay, and you can do the same thing for the entropy,  from a table of standard entropies.  You do exactly the same thing.  And the delta G then is just the delta H minus T times the  delta S, where we put plug in the T.  All right, we get 72.6 kilojoules per mole at the delta G.  We conclude that this reaction should be spontaneous  at this temperature.  Okay, I think that's all I got.  105.  Are there questions about this stuff?  I know it's not riveting.  All right, we'll do, hopefully, it's going to get more  interesting, I think.
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Okay, how are you guys doing? You got those mid-quarter daldrums. I know how you're feeling. Okay, this is a chapter 16 topic. We're really cranking away on chapter 16 here. Edging our way towards the end of thermodynamics, which I hope will come probably in the middle of next week. Now we posted the quiz scores this morning. This was the hardest quiz. It might have been too hard. First question was pretty easy. Didn't you think? So these guys got A's. These guys got B's. There's a few C's here. We usually don't have hardly any C's. Okay, quiz five is Friday. The first two are supposed to be easy. The first two problems. In this case, the second one was actually not so easy. It was the body temperature one. Okay, I also posted the key in case you want to look at that. So we won. And what's even sweeter is we beat SC. Very, very sweet. I'm always amused by the fighting anteater. The anteater is the most docile animal that we know of in the animal kingdom. But to dress him up for the athletic department, here he is fierce. There's no such thing as a fierce anteater. In nature, they're not known to be fierce. Okay, so we're going to review a little from Friday. Why does that say Wednesday? We're going to talk about how the Gibbs free energy varies with temperature and pressure. We'll do a couple of examples. Just sort of easy ones to ease into this subject. Okay, we'll do a bunch more examples on Wednesday. All right, so on Friday, not Wednesday, we said, look, there's three types of systems. Okay, and uniquely for this guy right here, it's an isolated system. There's no energy or matter exchanged with the environment, the surroundings. We don't have to consider anything except the system when we think about the spontaneity of processes that occur within it. Okay, it's blocked off to the surroundings. It doesn't even know about them. And so we can say any process that has an entropy, positive entropy change, is going to be a spontaneous process for an isolated system. We don't even have to think about the surroundings. They're not part of our thought process. But we don't have isolated systems in chemistry too often. They're almost always in communication with the environment. And so we have to consider open systems and closed systems as well. And in those cases, because there is communication with the surroundings, it's the total entropy, surroundings plus system that matters. Okay, in terms of figuring out if this process is spontaneous. Notice that the focus is totally on entropy. We're not saying anything about the energy. The energy can do anything it wants. We're only focusing attention on the entropy to understand whether these processes are spontaneous or not. Okay, so we've got to have this term in here for the surroundings. We didn't need it for the isolated system. Okay, so now we're just going to do a little algebra. We're going to move the surroundings over to the right hand side, put a negative sign in front of it. And then we're going to remember that dS is minus dQ over T. And so we can make that substitution for the surroundings right here. And then we're just going to remember that Q is a conserved quantity. In other words, if plus Q is heat entering the system, that heat had to come from the surroundings. And so the surroundings has to be minus dQ. There's conservation of Q. And then we have to think back and remember that dU is dW plus dQ. And so we can just solve for dQ in this expression. dU minus dW. And if we plug that guy in for dQ, we get this guy right here. And if we consider only pressure volume work, this dW is PDV. Okay, and so then we're going to multiply by T surroundings and move it over to the left hand side. So we get rid of this T surroundings now. Now it's over here. And we just have dU plus PDV. We're going to drop that cis subscript. If you don't see a subscript, just assume we're talking about the system. Okay, so this is the pink equation. It only took us about four steps to get there from conservation of entropy. Or not conservation of entropy from entropy dictating which process is spontaneous or not. We kept coming back to this pink equation on Friday in deriving different thermodynamic state functions from it. In fact, we showed that if the process occurs under conditions of constant volume and constant entropy, then it's the internal energy that tells us whether the process is spontaneous or not. And if instead the process occurs under conditions of constant pressure and entropy, why it's the enthalpy that's going to tell us whether the process is spontaneous or not. But unfortunately, we don't encounter these two sets of conditions very often. It's virtually never the case that the entropy is constant. If you're a chemist, you can ask, you know, how do you do that? How do you do an experiment with constant entropy? I don't know the answer. So when you're doing an experiment and you want to understand whether it's spontaneous or not, the chemistry that you're looking at, it's unlikely that you're going to be paying attention to these two variables to figure that out. They're not going to help guide your decision-making process in figuring out whether your chemistry is spontaneous or not. That's what we care about here. So we need some other state functions. And we talked about one, all right? The Helmholtz energy, all right? In chemistry, temperature is frequently constant but not the entropy, all right? Constant temperature, lots of constant temperature chemical processes that we can think about. So let's consider the case where Dt is zero and the volume is zero. We're going to need for two things to be zero. Otherwise, we're not going to end up with a state function. And so we'll also define a new state function A, which is going to be called the Helmholtz energy. It's going to be defined as the internal energy minus Ts. And so if we take the derivative now to get Da on the left-hand side, we're going to get Du and we're going to have Dts. And so we can split that into two terms, Dts minus Sdt. And then we can just solve for Du. And this expression, Du is going to be equal to Da plus Tds plus Sdt. And of course, the next thing that we do is we plug this thing into the pink equation, put this expression for Du into the pink equation, and then look for the terms that are going to cancel for us, all right? So we've got Da, Tds, Sdt, and our expression for pressure volume work from the pink equation. Okay, and the first thing that we notice is that we got Tds here. We got Tds here. Now these two Ts are different in principle. This is a T for the surroundings. This is a T for the system. But as we converge on equilibrium, these two temperatures will become very, very close. And under those conditions, we can expect these two terms to cancel for us. And then under conditions where we said Dt is 0 and Dv is 0, there's Dv. We can cancel out that term. And Dt, we can cancel out that term. All right, and we're just left with Da. There's nothing else left. And so it's going to be Da is less than 0. And so this Hemmholtz energy is going to be a state function that we can use to tell whether the chemistry that we care about is spontaneous or not when temperature and volume are held constant. And in the laboratory, we can enforce those limits. We can do an experiment at constant temperature, maintaining the volume constant, let the pressure do whatever it wants. Okay, we need one of these things. All right, it's got a defined volume and it's built like a tank and so even if the pressure changes a lot, we're going to enforce constant pressure. And in principle, it's the Hemmholtz energy that will tell us whether a reaction in this parbomb is going to be spontaneous or not. Right, we would want to, if we do an experiment in here, we would want to use the Hemmholtz energy to figure out whether it's spontaneous or not. Now, if you do undergraduate research, how many people have done undergraduate research? How many people have seen a parbomb? All right, a few of you, who do you guys work for? You all work for the same person? Oh. He's got a parbomb in his lab. Okay, so it's not completely impossible that you would use one of these things, right? They are in rather common use but I would say probably 99.9% of all the chemistry that we're likely to do is not going to be in a parbomb. All right, 99.9%. So, we need a different thermodynamic function. The Hemmholtz energy is fine but constant volume is inconvenient for us to use because we needed parbomb to do it in many cases. In chemistry, it's even more useful to make predictions about processes occurring at constant pressure and temperature because that's dead easy. Right, we live in an environment of quasi-constant pressure. Okay, and so we can do chemistry that's open to the environment and make predictions about whether it's spontaneous or not. To that, we're going to use something called the Gibbs energy. All right, we're going to define it as H minus Ts. Okay, we're going to do the same kind of algebra we did before, Dg is DH minus Dts and so we've got two terms here now. And then we're going to think back to Friday when we wrote an expression for DH. We said DH is Du plus PDV plus VDP. Okay, and so we can just plug that in for DH here. Now, we've got this long thing here that's equal to Dg. Okay, and so once again, we're just going to solve for Du. All right, put all of this other stuff on the right-hand side. And then once we've got Du, we're going to just plug it into the pink equation again. There it is. Put all of this stuff in for Du. We get this long thing here. Okay, and some of these terms are going to start canceling for us. As usual, PDV, PDV, right? Remember these two pressures are not in principle identical. That's the system pressure. That's the surroundings pressure. All right, but in the limit of equilibrium, they will be the same. All right, Tds, Tds, same idea there. And then because we're talking about G, we're going to make P constant. And so we're going to lose that guy. And Dt, so we're going to lose that guy. So everything cancels out here except Dg, which is going to be less than or equal to zero. And so that's going to be the state function that we're going to want a key on most of the time as chemists. All right, now if you're a physicist, if you're some other kind of scientist, these other state functions might be more important to you under other sets of conditions. But for chemists, it's all about the Gibbs free energy. The Gibbs energy, we're not supposed to call it the Gibbs free energy anymore. It's just the Gibbs energy. Okay, now I know that's tedious. But this is important, right? This is actually one of the more important things in thermodynamics that we need to be able to understand. All right, here's where we're doing chemistry. And this guy right here, we're open to the atmosphere and the Gibbs function is going to tell us whether this blue stuff here is going to react spontaneously. All right, we don't need the par bomb. Okay, so today and last Friday, we've taken the consideration, we've taken the condition for spontaneous change for non-isolated systems. We consider the total entropy change, system plus surroundings. That's got to be greater than or equal to zero. And from that, we derived all of these different conditions that apply for these different constraints. Volume and entropy, temperature and volume, pressure and entropy, temperature and pressure. We've got these four different. And what I've told you today is look, these two are not super useful to us. As chemists, these two are more useful. And this one is way more useful than that. All right, we derived these all. We didn't have to assume anything. Very proud of that. It's hard. Okay, now these conditions here also serve to tell us whether this system is proceeding towards equilibrium. It not only tells us whether the chemistry is spontaneous, it'll tell us whether the system is proceeding towards equilibrium or not. For example, what I'm plotting here is the Gibbs energy for some chemical process. And on this axis, I have the reaction coordinate. So this represents reactants right here and this represents products. This is 100% products. This is 100% reactants. But as you move along the axis in this direction, we're converting reactants into products. That's what I mean by a reaction coordinate. Sometimes we'll call this reaction coordinate X or chi. Okay, reactants getting converted to products, very generic. What does DG that should be P and T? How did that happen? Less than zero. There it is, P and T. All right, so first of all, this difference here between the reaction and product Gibbs functions, that's the delta G reaction. Makes sense. Now, let's consider a process that starts right here and ends right here. Or we can ask, is such a process going to be spontaneous or not? Well, we have a criterion here if we just change that V to a T. All right, we know DG should be less than zero. Okay, and so we can say G final minus G initial is that DG final minus initial. Is that going to be less than zero or greater than zero? What do you think? Yeah, it's a small number minus a bigger number, and so that difference is going to be negative, isn't it? All right, and so we would predict that's a spontaneous process. Yes, DG at constant T and P is less than zero. What about this guy? Same conclusion. What about this guy? No, final minus initial is going to be a positive number now. All right, so that's not going to be a spontaneous process going from here to here. No. All right, and what about that guy? Yes, final minus initial is going to be negative again, and so that should be spontaneous. All right, so basically what we're concluding is that if you're over here, we're going to go spontaneously downhill in this direction, and if you're over here, you're going to go spontaneously downhill in this direction, and that this minimum here in the Gibbs energy is going to indicate the equilibrium position of this reaction. It's the point where DG over DX, where X is now my reaction coordinate, is equal to zero. At that point, there's no more driving force for spontaneous change. We're at equilibrium. Okay. Now, amongst these four thermodynamic potentials, U, H, A, and G, G will be by far the most important to us. Yes, yes, yes. How does G depend on temperature? All right, how does the Gibbs energy depend on temperature? Well, that's a rather important thing for us to understand, because as chemists, if we want to accelerate a reaction, G is going to tell us whether the reaction is spontaneous or not. All right, we want to understand how temperature will influence that spontaneity. G is equal to H minus TS. We know that, and so we can take the derivative immediately, DG, DT. Even I can take this derivative. I get minus S. Okay, and so what this tells us is two things. First of all, since we know that S is always a positive quantity, there's no such thing as negative entropy. S is always a positive quantity. All right, that tells us that G has to decrease with increasing temperature, because that derivative is always going to be negative. All right, that's kind of surprising. The Gibbs energy is going to go down as the temperature goes up. That's counterintuitive. Don't all energies go up when you increase the temperature? Not this one. All right, the Gibbs energy goes down as you increase the temperature. Not only that, but the rate of change of G with temperature is greatest for systems having high entropy. The higher the entropy, the greater the change in G is going to be with temperature. Well, what kind of systems have high entropy? Well, gases, my laser pointer's dying, gases have the highest entropy. And so the rate of change of the free energy with temperature is going to be the highest, then liquids, then solids. Solids have the lowest entropy. Okay, so this plot is right out of your chapter 16. Gases, biggest slope, right? Here's the Gibbs energy on this axis. Here's temperature. It's going down. For every single one of these guys, it's going down, right? But it's going down at a rate that depends on the state. Gases show the largest decrease in Gibbs energy with temperature. Liquids next, solids show the least. Okay, so a couple things are, I mean, one thing that's surprising for sure is that the Gibbs energy goes down with temperature. All right, it's an unusual energy, isn't it, that goes down with temperature. Okay, so we can evaluate this derivative. And then we can go back to this equation right here, and we can just say, we can solve for minus s. All right, if I solve for minus s here, I'm going to get g minus h over t, just solving for minus s in that equation right there. Okay, so I've got dg over t at constant p is g minus h over t. And then we can rearrange that, just split this into two terms, move g over t to the left-hand side. I don't know why we actually did that, because I don't think we need this result here. All right, maybe we'll come back to this in a second. But let's just look at this for a second. This is the derivative of g over t. I don't know, that has anything to do with you. This is just the derivative of g over t. If I use the quotient rule to evaluate this derivative, I've got 1 over t times the derivative of g with respect to t, and I've got g times the derivative of t with respect to t, right? Two terms in my quotient rule expansion. Okay, now, the derivative of 1 over t is just minus 1 over t squared, right? Okay, and so that's that derivative right there. And this guy, if we factor out 1 over t, so I'm going to pull the 1 over t out of both of these two terms and put it right there, all right? Now I've got this expression here, and that is just the entropy, right? The derivative of the Gibbs energy with respect to t at constant p, that's the entropy. Okay, and so I can plug that in to this expression right here. I still got g over t, and then I can move, maybe that's why I did it. No, all right, forget that. This, all right, is just plugging in for g from that equation two slides ago. Okay, and so this is s over t. This is s over t, so we're going to get rid of the s over t. We're just going to be left with h over t squared, all right? And this is your equation 15.62b. This is the Gibbs-Hammholtz equation, which is important because it allows us to measure h by looking at the temperature dependence of g. And the temperature dependence of g is something we're going to be able to measure experimentally, all right? And so we can get h directly from that using this Gibbs-Hammholtz equation. Now, if this is a delta h and that's a delta g, this equation still is fine. Okay, so now let's ask some questions about the Gibbs function. We already asked a question about the temperature dependence. We said the Gibbs function goes down with increasing temperature, surprising. The rate at which it goes down depends on the entropy, all right? Higher the entropy, the faster the temperature rate of change of the Gibbs function. What about pressure? All right, we've got this expression here for dg. And if we want to look at this at constant temperature, we can say dt is zero, all right? And so that term is just going to go away. We've got dg is Vdp. And so to find out what the free energy, free energy, the Gibbs energy change is, there's the Gibbs energy at p final minus the Gibbs energy at p initial. I can just integrate this Vdp, all right, from initial to final pressure. And if I know what that is, if these are molar quantities, of course, there's going to be an m there, an m there, and an m there, all right? If these are all molar quantities. And so this equation is not super useful to us unless we know how this volume changes with pressure, all right? But one thing that's obvious is for phases like solids and liquids that are essentially incompressible, Vm is virtually constant, independent of pressure, all right? There's not much compressibility of a solid phase or a liquid phase, right? And so there isn't much pressure dependence of Vm. And so we can write a simpler expression. If we can pull Vm into the front of this integral sign, right, because it's constant, then the integral just turns into this. And we've got an extremely simple expression that allows us to evaluate the pressure dependence on the Gibbs energy, right? It's just the molar volume times the change in pressure. So if you look, once again, this is the Gibbs energy on this vertical axis here. And this is the pressure on this horizontal axis, right? And for liquids and solids, you get virtually a horizontal line. Because they're incompressible, right? Their volume doesn't depend on pressure very strongly. Interestingly, as the pressure gets higher, the Gibbs function goes up a little bit. All right? With gases, there's a much stronger effect. Gibbs energies of gas depend strongly on the pressure. And you might expect them to because gases are far from incompressible. They're highly compressible. All right? So their molar volume is highly dependent on pressure. Consequently, their Gibbs energy is highly dependent on pressure. In fact, their Gibbs energy goes up with increasing pressure. Now, we can actually figure out what this is for ideal gases very readily. All right? We can just substitute for Vm from the ideal gas equation. Move the RT out front. All right? So that's going to move out front. We've got 1 over P. So we're just going to log P final over P initial. That is the equation that describes the change in the Gibbs energy for an ideal gas as a function of pressure. All right? So change the pressure. We have very simple equation. Probably should go on your equations page for quiz five. What is with this projector? What is with my laser pointer? Too cheap to buy new batteries for it. Please, please. OK. So this is what the volume is doing as a function of pressure for an ideal gas. It's following this purple line here. OK? And so if we want to evaluate this integral, we're going to be integrating from some initial pressure to some final pressure. This is the area underneath this curve. So this is the Gibbs energy. All right? And it's obvious that as we make PF higher and higher and higher, this integral is going to get bigger, right? And so it's obvious that the Gibbs energy is going to go up. Just based on that. Now we can define a standard molar Gibbs free energy. All right? We're constantly, there shouldn't be the word free in here. We're trying to get rid of the word free. Should be the standard molar Gibbs energy. All right? That's defined at a defined pressure, which is one bar. All right? That's how we define the standard Gibbs energy. Or the standard entropy. Or the standard anything. If it says standard, it's one bar. OK? And so in this case, we can write this expression. It just follows directly from this guy right here, except that we've now defined a particular Gibbs energy that applies when the pressure is one bar. OK? So this initial guy is now that guy. OK? So all this plot shows. Here's the molar Gibbs energy, and here's the pressure. And what we said earlier is that as I increase pf, I'm going to make this integral larger. And so that tells us the Gibbs energy's got to go up with increasing pressure. All right? What I just showed on this slide right here is that we're going to define a special Gibbs energy at one bar. And so that's what this becomes. This becomes one bar. Imagine this is one bar, and now we're integrating to higher pressure from that. So the same intuitive picture applies. If we move this to higher pressure, the integral is going to go up. And that's why this plot is going up, up, up, up. But you can see that it's got downward curvature. It's got downward curvature because here there's a big change in the Gibbs energy. Smaller, smaller, smaller. This has got upward curvature. So if we can integrate this guy, increases in pressure are going to have a progressively smaller and smaller effect on the Gibbs energy. And that's what we're seeing here. That's why there's downward curvature of this guy. All right? So the Gibbs energy goes down as we increase the pressure and up as we increase, sorry, goes down as we increase the temperature and up as we increase the pressure. Is all of this confusing? Yes. Absolutely. I mean, if you don't think so, you're just not paying attention. OK. Should we do some examples? The change in the Gibbs energy of 25 grams of methanol, mass density 0.791 grams per cubic centimeter, when the pressure is increased isothermally from 100 kilopascals to 100 megapascals, oh, I should say, calculate the change in the Gibbs energy. No verb. Calculate the Gibbs energy. When we subject this methanol to a change in pressure, that's a change by a factor of 1,000. It's a big change. But so your first thought process should be, it's a liquid, it's incompressible, let's use the simplest equation we can. And the simplest equation for any liquid or solid is just V. M times delta P. V. M times delta P. I immediately get the change in the Gibbs energy. This is what I did. And so if I'm willing to give up these molar, forget molar, I'm just going to calculate the difference. It doesn't ask me for the change in the molar Gibbs energy, it just asks me for the change in the Gibbs energy. So here I need, I'm going to calculate the change in the Gibbs energy. The volume of my 25 grams of methanol, there's the density and so I can calculate the volume. That's 10 to the 5, that's 10 to the 8, so there's my factor of 1,000 and so when I plug these numbers in I get 3.157 times 10 to the 9 cubic centimeters per cubic centimeter pascals. That's the delta G that I get. I don't like these units but I can convert them later. Now, oh, so I converted them right here. 3.157 times 10 to the 9 centimeters cubed pascals, no I don't like that and so I can use the definition for, here's the conversion factor for pascals to ATMs, there's the conversion factor from cubic centimeters to later, now I've got later atmospheres and so then I can use the ratio between the two different Rs to do the unit conversion to get joules. Tidious, but when I do that I get 3.157 times 10 to the 3 joules, or roughly 3 kilojoules. That's what the change in the Gibbs energy is going to be. Now, when I looked up the answer in the key it had something far fancier. It used the isothermal compressibility of methanol, which you can look up in the back of your book. That's the isothermal compressibility, it's 1.26 times 10 to the minus 9 per pascal, well for goodness sakes. We can go through and do it with the isothermal compressibility and see how different our answers will be. The isothermal compressibility is defined by equation 14.59, it's 1 over V times the derivative of volume with pressure constant temperature, and we can just linearize this differential, final minus initial, because we know the change in volume is going to be small. Pressure difference is not small actually, but we're going to linearize that also. So we're going to make two approximations, we're going to say the initial volume is just the final volume also, it's not going to change much. So that's going to become VI, so I know what that is. And the final pressure is much, much larger than the initial pressure, and so that difference right there I can just approximate as the final pressure. The initial pressure is just a rounding error on the final pressure, literally. So that's the equation I'm going to use. And so if I just do a little algebra, split this into two terms, now solve for VF, I get this expression right here, which can be further simplified to give me this expression right here, and then what I've got to do, sadly, is integrate that, so I'm going to plug that in for VM and do the integral. And so that's what I did here. I can calculate what this VI is, that's the mass, that's the density, and here are the two derivatives I'm going to do. I'm going to actually pull this out front, and then I'm going to take the derivative of dp, well that's pretty easy to do, and then kappa t times the derivative of p, and so I can run these two integrals. Now this is just going to be 10 to the 8 minus 10 to the 5. This guy is 1 half kappa tp squared, and when I plug the integration limits into these two antiderivatives, this guy ends up being almost 10 to the 8. That's the rounding error that gets subtracted from him, and these are the results for these other two terms. And so what I calculate is 2.96 times 10 to the 3 joules, which is also about 3 kilojoules, but if I do this carefully considering the isothermal compressibility of methanol, I get a slightly different answer. It's different by one part in 30 roughly. That's wrong, that's right, but pretty darn close to being right. So it depends on the precision that you need on the quiz. That might be good enough, unless this is also an answer. That would be cruel, wouldn't it? One part in 30 difference in the answer? No, I would never do that. Everyone see how to do that? Isothermal compressibility makes it a little bit more complicated. So the last thing to say today is that, and this may be completely obvious to everybody, is that if you want to know the standard Gibbs free energy for a reaction, like this reaction right here, what you do is you look up on a table, the Gibbs standard, that free shouldn't be there again, the Gibbs energy of this stuff, the Gibbs energy of this stuff, and the Gibbs energy of this stuff, and the standard energy difference is that minus that plus that. I can look this up in tables. Now, if I don't have a table of Gibbs energies, well I've only got a table of entropies and enthalpies, which is not terribly uncommon, you've got to use this equation right here. You can look up these enthalpies, you can look up these entropies, and if you know the temperature at which the reaction is happening, you can figure out whether the reaction is spontaneous at that temperature or not. You can look up the delta G. This is the delta G reaction. So in the case of H, we've got this delta sub R H, what that notation means is that R is the reaction, right? This is the enthalpy change for the reaction. This is the delta H formation. That's what we're going to look up in the table. What we want is we want to look up the delta H formation for all the reactants and all the products. Add up the reactant delta H formation, subtract, sorry, add up the product delta H formations, subtract the reactant delta H formations to get the delta H for the reaction. This zero means standard. What does that mean? Pressure equals one bar. Temperature equals 298.16 K. What is that new? It's just a stoichiometric coefficient in front of these. So that would be one, one, and one in this case. We need to do the same thing for the entropy. Here's the enthalpy, here's the entropy. We still get stoichiometric coefficients. We're going to look up these standard entropies, products minus reactants. So we can do that. Here's, I've done it by Golly. That's for this guy right here, which is what? Propionic acid. And these two guys are for those two in no particular order. Okay, so the delta H is just that, an evaluation of that, all of those numbers. Okay, and you can do the same thing for the entropy, from a table of standard entropies. You do exactly the same thing. And the delta G then is just the delta H minus T times the delta S, where we put plug in the T. All right, we get 72.6 kilojoules per mole at the delta G. We conclude that this reaction should be spontaneous at this temperature. Okay, I think that's all I got. 105. Are there questions about this stuff? I know it's not riveting. All right, we'll do, hopefully, it's going to get more interesting, I think.
from it. In fact, we showed that if the process other stuff on the right-hand side. And then once the atmosphere and the Gibbs function is going to systems. We consider the total entropy change, Gibbs energy for some chemical process. And on going to indicate the equilibrium position of thermodynamic potentials, U, H, A, and G, G will are molar quantities, of course, there's going to should go on your equations page for quiz five. I'm going to calculate the change in the Gibbs never do that. Everyone see how to do that? enthalpy change for the reaction. This is the