 We'll have the grades posted, the scores posted soon and the exams are being scanned until you get them back. If there are any mistakes on your exam, just send me an email, caps, error, on exam, midterm two, and then what it is. And then I have a copy so I can look at it and fix it. And then we'll post all the updated scores for midterm one and midterm two. But look, don't sit on midterm two and come to me after I'm out of town at the end of the quarter and say, there's a goof on midterm two and I waited two weeks to find out about it. Go over it yourself, add up the scores, make sure there's no mistakes on your exam. If there are, we can fix them. If you wait till later, it's much more difficult. As a kid, I loved reading stories. I particularly liked reading the Greek myths about my purported ancestors and I liked reading about all the pagan gods because they seemed like bigger versions of what I wanted to be. They were awful, but they'd get in fights with each other like I did with my sister and I wished I could turn her into a spider some days. And I guess I sort of liked Hercules because I liked the idea of being strong. But my favorite character was Antaeus. Antaeus, as the myth goes, was the son of Mother Earth and another Greek god and he was a wrestler, but here's the metaphor. Antaeus could never be beaten because every time you threw him down to the ground and he touched Mother Earth, he came back stronger. That's the metaphor that you need sometimes in life. Things don't always go your way. They don't always go your way the first time. They may not go your way for a long time. But the difference is whether you give up or you come back stronger. That makes a world of difference in the long run. And with that in mind, I want to kind of go over a post-mortem on some of the problems that I graded that I saw mistakes on. I toyed with the idea of the following. If on the final you have those distributions of speeds and you draw it wrong a third time, you just flunk. You just post your three bad graphs as evidence that you learn nothing at all. Don't let that happen. That's a two-second problem and I'm going to go over it now. So make sure you get it right. It's not that important, but it's kind of a metaphor for actually paying attention and learning what you're supposed to do. First I want to start, though, with some corrections. Some of you figured out on your own that I made a goof when I analyzed the energy density of octane and hydrogen and the goof is big enough. It was a simple mistake that I'll show you. But like the Mars orbiter, it has a big impact. Hydrogen's not nearly as bad as what I sketched out in problem 41. So let's first fix my mistake. And here was the problem. I said, suppose we want to transition to a hydrogen economy. What volume of hydrogen gas at 1,000 psi and 25 Celsius is required to produce the same heat, and the enthalpy of combustion is 25 gallons of octane and then you get the density of octane and its standard enthalpy of combustion, 5,500 kilojoules per mole exothermic. And here's what went wrong. Right here in bold. What I originally put made no sense, but I put it, the units canceled out, so that was okay. But I got the ratio wrong. It's one mole is 114 grams, not 114 moles is one gram. And by putting the 114 on the top, I got a factor of 12,000 wrong. And that makes a huge difference in how practical you might assess hydrogen to be. So let's go through it now and get it right. So with this ratio correct, that one kilo mole is 114 kilograms, then we have 582 moles of octane in 25 gallons. And if we work that out, we get 3.2 million kilojoules. But I think before we had a much, much bigger number. With this 3.2 million, we have to figure how much we would get from hydrogen. For hydrogen, we get 241.8 kilojoules per mole, just from the enthalpy of formation of water vapor. And remember I said I was using water vapor. It doesn't matter as long as you use the same thing. And that means that when you convert it to the moles of hydrogen you need, you get 1.32 times 10 to the 4, or 13,000 moles of hydrogen. That's quite a lot, but it's not a million moles. Sorry, I have a duplicate. And then we had to figure out the volume. And now the volume comes out to be something much more reasonable because the number of moles is not so huge. We get 4,762 liters. And therefore the hydrogen gas tank that you'd need would have to be about 50 times larger than the petroleum gas tank. But it wouldn't have to be as big as the car or me set on three sides like a giant cube. And that makes quite a bit of difference. In addition we don't have to burn the hydrogen. If we use the hydrogen in a fuel cell then we can extract the energy from the hydrogen much more efficiently because we don't make heat. An internal combustion engine works by exploding and making heat and expansion. And whenever you make heat you take a bath because once you make heat you can't extract all the energy back out of the heat. You've made something very disordered and there's only so much order you can milk back out of it. That's what the second law says. But if you don't make heat if you make work first by converting the hydrogen in a fuel cell to electricity that can be far more efficient so we might I doubt we'll get a factor of 50 out of that but we'll get something out of that and that means that it might be possible. The big problem right now with the hydrogen is that the best kinds of fuel cells use platinum as a catalyst to make the fuel cell go. And because they use platinum that's like paving the roads with gold. It's extremely expensive to make cars that have appreciable amounts of platinum as part of the car. And that's where chemists have to come in and find a better solution to a fuel cell that only uses cheap stuff and that doesn't need to run at elevated temperature. The other problem with a fuel cell is to get the rate to be good enough to heat it up and that heat has to come from somewhere probably from your car battery but you still lose by having to heat that thing up all the time in order to run it. It has to be quite hot to get an appreciable rate and right now there doesn't seem to be a solution for that to have a fuel cell that runs more like a battery just sits there at room temperature and runs and produces electric power for your car to use. Okay, now let's go over a few of the exam. There were still errors on the plots of the relative numbers of atoms versus speed for argon and helium at 25 Celsius. We didn't grade on the exact shapes of the curve but there are zero atoms moving at zero speed because moving at zero speed means you're a solid and there's no solid in there so the curve starts at zero. It then comes up and then it has to die out again to zero. It can't keep going up because if it keeps going up you're implying that there's a ton of energy from somewhere to make those molecules keep going faster and faster and there isn't. It's at 25 Celsius. It has a finite amount of energy that depends on the temperature so we wouldn't want that. Let's have a look then. This was slide 62 I think before. They all start at zero. They all hit a maximum somewhere and the heavier you are the slower your most probable speed is. The most probable speed is the highest point on the curve. I think in the problem I said I was around 1,100 for helium, right? So you draw from zero to 1,100 and then you draw something that goes back down and you can draw it roughly symmetric. It's not quite symmetric but roughly the distance from here to here and the distance from here to where it goes back down to almost zero is about the same. Now you could have drawn me a very accurate one for argon. Here's argon here because you know that the most probable speed goes like the square root of the molar mass. So if you know the most probable speed of helium four grams per mole is 1,100 you can take the ratio of the square roots of the molar masses and figure out exactly where this is and you can draw beautiful things like that. What you don't want to do is draw argon as having a peak out here somewhere to the right because that implies that the heavier guy is moving faster and that can't possibly be true. If you just think about collisions between things the heavier guy doesn't get knocked around nearly as much because he's heavier. Three light guys hit him and he's still not moving very fast. Okay? Alrighty. Let's go on. Well, I just wanted to say again that it was Maxwell who worked out that this particular distribution maximizes the entropy so it's the most probable and we have all the masses. If you want to practice just take the function this is basically got v squared and then e to the minus some number times v squared. So just plot x squared times e to the minus ax squared and look at it and you'll see it starts at zero it goes up and then the exponential the Gaussian function kills it keeps it from going too fast. This one was a weak point of the papers I graded one person got this right I didn't grade every single paper but I did grade a lot of them because I like to see what people are doing and only one person of the ones I graded about fifty got it right. Let's then make sure that if this problem makes an on-core appearance that we get it right. We were supposed to estimate the enthalpy of combustion of propane using mean bond enthalpies. Okay. I tried to indicate by writing H3C and then dash CH2 and then dash CH3 what the structure would be rather than just writing C3H8 which is a little vague but the actual structure if you write it out looks exactly like this in a two-dimensional schematic form of course this is a three-dimensional molecule and the bond angles are not 90 degrees they're tetrahedral angles and they form a 3D structure and it can move, it can move around okay like spaghetti but for the purpose of analyzing it all we need is this we count 1, 2, 3, 4, 5, 6, 7, 8 CH bonds though we have to break and 2 CC bonds that we have to break and then we're asked to get delta H for some reaction now here's the thing whenever you write a state function delta G, delta H, whatever if you're referring to a chemical reaction write down the reaction that gives you kind of a fixed point to refer to what reaction am I talking about here many people just figured out how much energy it took to bust this thing apart but that's not what I'm interested in I'm interested in the enthalpy of combustion which means I'm reacting this with oxygen and producing CO2 and H2O so there's a lot more to it than just breaking this thing apart into atoms that's not the reaction we're talking about if you don't write down the reaction though you can kind of forget what you're talking about yourself and you start computing the wrong thing and I think that may have happened for some people so they did a right calculation but for the wrong reaction that's the upshot of it so delta H of combustion refers to a reaction always write down the reaction and then balance it atoms don't appear and disappear in chemistry they just rearrange that's all they do the nuclei in chemistry except for nuclear chemistry are fixed and it's just the electrons that move around and rearrange all the letters in the jumble we did are fixed it's just that one of them makes a word and the other is nonsense so we start here by saying right we're burning C3H8 propane with oxygen and we're producing CO2 and H2O and that's what complete combustion means we have three carbons here and this product side the only molecule with carbon is CO2 so we have to put a 3 there we have 8 hydrogens the only molecule with hydrogen on this side is H2O so we have to put a 4 there and then we count up the number of oxygens we've got 6 and 4 is 10 oxygen we have to put a 5 there and so now we want to break apart the reactants and the products into atoms and then figure out which side wins first let's estimate how much energy it takes to break the reactants formally into atoms we know we aren't ever going to do this but remember the beauty of state functions is the path doesn't matter so the path can be fantasy because the path doesn't matter it doesn't have to be something we can really do it can just be anything that we can figure so let's figure it let's break this apart into 3 carbon atoms in the gas phase 8 hydrogen atoms in the gas phase and 10 oxygen atoms in the gas phase well I have to break 8 CH bonds 2 CC bonds and 5 oxygen double bonds and you are given all those on the front and I think I've used the same numbers 8 times 412 plus 2 times 348 that's the carbon plus 5 times 496 to break these apart this reaction to make this go on paper requires 6472 kilojoules per mole it takes a ton of energy to break all those bonds apart ok I have that number now what I want to do is go to the other side and if I've balanced it I'm going to have the same atoms here but I have to count up different kinds of bonds because here I have OH bonds and here I didn't have any OH bonds and here I have the strong CO double bond and I didn't have any of those over here so we have to take a look at how those pan out and see what we get for this so let's do that here we have to again count up the bonds each CO2 has two CO double bonds so we're going to have six of those each of these has two OH single bonds so we're going to have eight of those and we get the same atoms the question is whether it takes more energy to break the products apart to atoms or the reactants apart to atoms and because of the very strong carbon-oxygen double bond this number 743 here is the winner in this game we get 6 times 743 plus 8 and it turns out OH bonds are pretty good too 463 so we get 8162 kilojoules per mole and now I'm going to go over this very carefully because I had a lot of questions is it reactants minus products or why is it different for this or that and what I want to do here is just go over this so that you can't get screwed up trying to remember if it's reactants minus products or this minus that is hopeless what you want to do is say I've got to add up these reactions somehow and however I add them up it's got to total up to the reaction I want not some other reaction out in left field once I've got them so that they add up to what I want it'll be absolutely clear which one's negative and which one's not and there's nothing to remember the less you have to remember the better off you are especially as you get older because your memory will not be as good if your only strategy to learn things is to somehow remember things you're going to be hopeless you have to just learn how to figure them out you have to remember something of course but you don't have to remember the details now the reaction that we want is this one and you see that the CO2 is on the right hand side here whereas the CO2 is on the left hand side here on the previous slide the propane was on the left hand side so the propane thing going to atoms is fine plus whatever it was this one however I've got to turn it around backwards otherwise I won't have CO2 on the right and therefore I have to change the sign of this one I don't give a toss which is products or reactants here I'm just saying there's an easy way to see what it has to be and that's the way to remember it not by saying it's that and for this is that so we have to turn around this product atomization this imaginary reaction we're forced to turn it around and when we turn a reaction around when the initial and final state swap the sign of the state function for this reaction also swaps because that's just the right minus the left and if I interchange them it changes the sign therefore this one is going to become negative simply by the fact that I have to turn this reaction around to get the one I want so let's do that let's turn that let's leave this one as it is we figured that one 6472 let's turn this one around and now we see that all the atoms in the way all these imaginary things go away and when we add up the two chemical equations we get the reaction we want it's balanced the way we want and now we see we're going to take 6472 minus 81 62 the way I think of it is I'm always adding but I had to change the sign of this one from what I figured because I figured it the other way around so I write them as plus and then I change the sign of any reaction that I turn around I don't even think of subtracting I always add everything but some numbers are negative and there's the right answer 1690 kilojoules per mole kilojoules per mole yes I'm going to go over that on the next few slides it depends whether you're talking about mean bond enthalpies whether you're going through this imaginary intermediate where you've broken all the bonds or not all we're doing we aren't saying that this reaction actually occurs nobody is proposing that we can do this in the lab but the beauty is it doesn't matter if it happens or not as long as we've got numbers in other words if I go to the top amount Everest and then come back and I come back 10 feet less I know I'm 10 feet higher than where I was before and it doesn't matter that I didn't actually go to Everest okay in fact based on recent news reports you don't want to go to Everest unless you want to end up with a very short life here's my thought then it's not worth trying to remember with trying to remember which gets the negative sign because it'll be obvious when you just add them up if you cancel out all the things either it'll be the reaction you want or it'll be some mumbo-jumbo that you don't want and then you have to figure out which ones you turn around but you just have to find reactions that add up to the ones that you want now we could also do this problem a different way a more accurate way we could use the heats of formation remember the bond enthalpies are an average and it depends what kind of compounds were included in the average what I get the average score for this exam which was pretty good actually I think I never look at the average but just marking the papers would be much higher for this class than it would be if I just gave it to every registered voter and since I don't know exactly which compounds they've included in the CH and so forth I expect this to only be approximate not not accurate now let's try it another way then using heats of formation let's try another imaginary path to get to what we want suppose we were given the same problem but we had the heats of formation of propane CO2 and H2O and remember the heat of formation the enthalpy of formation of molecular oxygen is zero because the heats of formation are from elements in their standard states made into the molecule the heat that we would get if we did that imaginary reaction but these can be measured in many cases and so they're much more accurate for the heat of formation of propane now I've got a different reaction I don't have gas phase atoms I have elements in their standard state at 298.15 kelvin and the elements then are three carbon in the form of graphite as a solid I should probably put an s, a graphite here but graphite is a solid plus four molecular hydrogens not hydrogen atoms molecular hydrogen is hydrogen in the standard state at 298.15 kelvin this hypothetical reaction if I could carry it out and make propane this way which I cannot gives an enthalpy of minus 104.6 kilojoules per mole how do I know that I have to look that up I don't know it and on an exam of course you'd have to be given all the heats of formation of every single compound in a chemical equation otherwise you can't do the problem and I can do the same thing with CO2 I can imagine taking three graphites and three molecular oxygens and making three carbon dioxide molecules in the standard state at 298 and since I've made three I have to take three times delta H of formation because delta H of formation of CO2 is per mole of CO2 and I'm making three moles so I got three times as much that's minus 393.5 I think we've seen that figure before and I have to multiply that by three I keep this sign and then the last thing I have to make is water and to make water I take four hydrogens as molecular hydrogen and two molecular oxygens and I make four water molecules and I have to look up the delta H of formation of water and I have to multiply by four and I have four times minus 241.8 now if I want if I were to say well what about the oxygen in the combustion reaction that's a reagent how would I make oxygen from elements in the standard state I'd make it from oxygen I'd start with O2 however many I need and I'd make O2 I'd do nothing and so delta H for that is zero because both sides are the same it has to be zero so then I realize that I don't need that one for this approach using formation because I'm using the elements not imagining going to atoms which are not the same as the elements and if I use then this approach here's the difference the products that we need are already on the ground that we need are already on the right hand side because we made them from the elements we made CO2 and H2O on the right hand side but the reactant propane was also on the right hand side and therefore I have to switch this around because I want this guy on the left hand side that's all chemical equation around and I changed this from minus put a minus sign in front of it because I've turned it around and before it was minus 104.6 therefore for this reaction written this way it's plus 104.6 and I put the plus sign just to emphasize it and therefore for this reaction that we want to do we have propane and then these we already had on the right and this is zero we take we add this equation and the other one so we take 104.6 plus 3 times minus 393.5 for this guy plus 4 times minus 241.8 for the water we get minus 204.2 kJ per mole which you see is pretty far off from the other one more than 10% off I would trust this number much more now keep in mind when you do this you're imagining doing the combustion at 298.15 if you're actually burning propane in a flame you have to be careful because the temperature is not 298.15 in a flame it's potentially far higher than that and you would have to correct for that by knowing all the heat capacities of all the products and reactants and then that would be much more detailed to actually figure it okay so just always write down the reaction you want put it together any way you want like tinker toys pull the things apart depending what information you have if you have delta H of formation for all the things involved always use that because it's more accurate if you don't guesstimate with the bond enthalpy you'll be in the ball part but you can't assume that that's going to be an accurate figure okay so here's the summary know what reaction delta H is referring to by whatever the definition is for bond enthalpy it's breaking it apart into atoms in the gas phase for formation it's using the elements in their standard states which are different then just add up the reactions until you get the reaction there and usually to make things cancel out you have to turn at least one reaction somewhere backwards and if you reverse a chemical equation you're just changing the initial and final state the other way around and therefore you have to reverse the algebraic sign of the state function because it is a state function it's a difference between final and initial if you swap them it changes sign and finally keep in mind that mean bond enthalpies are not going to be as accurate as using the enthalpies of formation heats of formation because they're averages and we're making specific molecules if we know the specific molecules we've got the data okay now then there was this bonus problem which I thought would be a gimme I thought I'd put the ball on the green about a foot away from the cup and all you had to do is tap it in for 20 points but what some of you did is you said, gee there's a ball on the green I think what I'll do is I'll stick this wooden stick in the ground, put the ball on it so I can really whack it and then I'll just hit it off into the wood somewhere and you can still actually complete the hole that way but the problem is it takes a lot longer in this problem we didn't give the temperature because the temperature doesn't matter, that was supposed to be a hint recall here's what we had we had these guys connected series of flasks we have argon, this is one version argon, one liter, one atmosphere, here it is helium one liter, half an atmosphere, here it is argon, two liters, one atmosphere here it is and of course we assume they don't react, these are noble gases they aren't going to react but you have to be careful if you have real chemicals here if they react, that's a nightmare because then the temperature will change for sure and the number of moles will change for sure and you have to be very careful but there was nothing like that here and what the question asked about was the partial pressure of each gas and the total pressure after the stop cocks or the valves are opened and at first blush this looks a little bit tricky but here's the thing you just look at each gas on its own and pretend the other ones aren't there you just pretend they aren't there because in terms of the helium they aren't there they're ideal gases the particles have no size in terms of heat argon here when I open it this is a vacuum and so is that and the argon just spreads out through the whole thing well I know p and v here and looks like v changed to four times so I know p end of story it takes one second let's do it we don't need to know the temperature each gas is independent of the others and we're assuming they're ideal but that's safe and none of them have any any excluded volume we're assuming the volume of the atoms is zero so all we do is just use Boyle's law for each gas separately we figure the pressure and then add up the pressures to get the total pressure therefore for each individual gas it's as simple as p1v1 equals p2v2 I need to know what p1v1 is but that's written on the diagram and I know v2 is four times v1 the argon starts at one atmosphere in one liter and ends up in a total of four liters so the argon is 0.25 atmospheres done the helium starts out in one liter and 0.5 atmospheres it divides by four the partial pressure of helium is 0.125 and the neon starts out at two liters in one atmosphere it goes to four liters therefore it ends up as half an atmosphere I add up to get the total a half and a quarter and an eight and I get 0.875 atmospheres very very quick I think I've got that on a slide but I'll get the slides in order for when they're posted on the video there was another one here osmotic pressure we had some strontium nitrate an imaginary pollutant from a farm the nitrate is probably not imaginary but I don't think anybody is putting strontium on their crops the key here is that strontium nitrate written like this it all dissolves we get one strontium ion and we get two nitrate anions so we get one strontium cation and two nitrate anions and therefore we get three moles of particles for this problem so we take the molarity that we're given 0.1, 0.08 whatever it was we multiply by three because I is equal to three for this problem and then we use pi the osmotic pressure as the moles per liter times RT T was 300 kelvin and then we get the answer and again it's quite quick and now I have the slide that was out of order with the answers here but I'll get it in the right order for the video this is exactly what I said and this is the total pressure 0.875 ok so I hope you can see that this wasn't as hard as it might have seemed on the day alright let's get back to free energy so recall the second law of thermodynamics asserts that for any spontaneous process in an isolated system the entropy change is non-negative at equilibrium the entropy change is zero but that's a special case or if we do a reversible if we do a reversible reaction then delta S is zero that's why it's reversible if we don't make any kind of additional entropy in the surroundings or in the system then we can go backward because in a certain sense entropy changes chart which way is forward in time since energy is conserved we can't look at energy and decide which way we're going forward but if I see a film that's running backward it's very easy for me to tell that it's running backward because there's somebody at the bottom of the pool there and suddenly magically there's all these swirls of water and the water kicks them up and into the air and back on the diving board and I say hey this film's running backwards I've never seen that happen but water spontaneously in a pool obviously the water has enough energy to do that but because the water molecules are not working together they can't do it on their own they can't just suddenly decide to hurl me out of the pool yes there's some chance 10 to the minus 130 that could happen sometime but it's not large enough that I'm ever going to see it in the lifetime of the universe and so if I see something like that I can tell immediately if a film's running backwards by the way things are happening what you're sensing is that it's very unnatural for entropy to decrease and when you see that you just automatically assume that the film's running backward and therefore it's really entropy that we want to watch out for when we burn things and create heat to generate power or do other things what's happening is that we're creating entropy energy's conserved we're turning chemical energy into heat but heat is much less useful than chemical energy so we're increasing the entropy of the universe and when we do that we can't go back and well in every actual case the total change is irreversible because reversible is kind of a special case for us to think about real reactions are irreversible why does entropy increase it just increases because it's the most likely outcome and atoms don't have brains it's the same reason that if you throw two dice you get seven a lot more often than you get two it doesn't mean the dice are loaded or the dice are trying to do anything you just throw them and it turns out there's many more ways to make seven one and six, two and five three and four then there are to make two you've got to have both dice come up one and one and you've got two snake eyes and you just lost your bet or if they both come up six you've got box cars you lost your bet that's not very likely but that does happen sometimes but if I have 10 to the 23 dice and I just throw them all there's going to be a certain range of numbers that it's going to be very likely yet and it's going to be very unlikely that all 10 to the 23 dice come up one that's extremely unlikely and therefore when we talk about large numbers of particles it's overwhelmingly probable that the entropy is going to increase and that's because molecular motion is random just like heat entropy is going to increase to the most likely state there's no rhyme or reason it's just probability in fact, Boltzmann expressed entropy in terms of the probability that a particular microstate could arise and this is in the book but we talked about this when we were looking at molecules dissolving all gas molecules on one side of a two-sided container that's like throwing snake eyes over and over that would have very low entropy that's where you start when the valves close because you've got them trapped but once you open it and they can decide what to do on their own they're never going to go back to one side when you've got large numbers of particles never going to happen but whatever you did there is irreversible when you open a tank of gas and let it get out that's it, it's out on the other hand equally distributed roughly would have a very high entropy and on Boltzmann's Tombstone is his famous formula that entropy is proportional to the natural log there's a reason why it's the natural log and not just the thing itself number of ways that you can actually make this state you can just think of W as the number of ways if you take a course in statistical mechanics then you'll see this formula or if you look at professor Tobias's license plate you'll see this formula as well okay, the universe whatever it is is an isolated system because when we say universe there's nothing outside there's no surrounding and therefore for any spontaneous process in this universe the entropy changes, the universe is greater than zero the sun doing nuclear fusion and creating heat and spewing off all the solar winds and everything else is creating huge amounts of entropy but when we're talking about the lab we aren't interested in the universe we're interested in our system and therefore we focus on what we're doing our chemical reaction in a flask and everything else the heat bath the stirring bar everything else is the surroundings and we focus on the system because that's what we want to study we don't want to try to study the whole universe it's too big this is what all living organisms do they create entropy in the surroundings so that they can order themselves if you look at your own physical being where do you think the lowest entropy is in your brain highly ordered amazing no way that you can take gas molecules and snap your fingers and you get a human brain and in order to keep that going which burns about 15 watts continuously we chuck out heat we're hot we're producing heat all the time and we're eating all kinds of fuel and we're burning it up and we're producing gases CO2 comes out in our breath and so forth and if we don't have a source of energy we die and as a society if we don't have a source of energy we die we're just at equilibrium we're in trouble we have to have a source of energy we have to have some way to pull that tractor through the field we have to have some to eat for ourselves and so forth and that's because the only way we can survive is by creating more entropy in the environment therefore the more things you have the more energy you you go through the more entropy you create that's the way it has to be there's no way around that whatsoever Delta S, recall it's just a measure of how much extra chaos we're creating and therefore and I think I gave you the analogy that in a library if you create a loud sentence everybody notices because it's so quiet in the library and likewise if I have a very ordered system at low temperature and I inject heat it does a lot of damage the entropy goes up like crazy but if I have a very disordered system that's already chaotic at high temperature and I inject a little heat it doesn't do much additional disordering therefore the change in entropy is so low and so we have to take a ratio of the heat that we add to the temperature that already exists that's the measure of the entropy and to be careful we have to add the heat reversibly because heat itself member is not a state function so I can't just say I add some heat I need to specify the path so that the heat is accountable and reversible is the nicest path so I add the heat at constant temperature then I can figure the entropy that I created otherwise it's much harder and for an exothermic reaction the reason most exothermic reactions are favorable like our metabolism and burning fuel whatever is that they disordered the surroundings because we add positive heat to the surroundings divided by the temperature of the surroundings and the positive heat we add to the surroundings at constant pressure is just delta H of the surroundings and that had to come from somewhere so that's minus delta H of the system therefore when delta H itself is negative minus delta H is positive and the entropy of the surroundings is positive the entropy of the surroundings has increased broadly speaking why exothermic reactions are favorable and then recall that this is how we derive why we are interested in delta G we don't want to take into account the surroundings at all so we divide the universe into the surroundings in the system and delta H of the surroundings we substitute delta S of the surroundings we substitute minus delta H over T that's the entropy we created by having a reaction add or subtract heat from the surroundings plus whatever happened to the system did the system itself become more ordered or disordered we multiply through the whole thing we substitute minus T we get this since delta S of the universe is greater than zero that means that minus T delta S of the universe is less than zero and that means that this thing which is the same is less than zero and because this thing occurs so often we give it a special symbol delta G so at constant temperature and pressure delta G decreases on hill G goes lower and lower and lower until G goes to the lowest possibility and then it sits there forever therefore we can say for any spontaneous process at constant temperature and pressure the free energy decreases though we write delta G is delta H minus T delta S and we forget about putting surroundings in system that means the system the whole point of delta G is we don't want to take into account the surroundings at all a reaction that dumps heat into the surroundings is favorable because it has a big negative delta H so I have a big negative number here it looks like delta G is going to be negative if in addition to dumping heat into the surroundings the system becomes more disordered for example I produce a gas solid then it's going to be wildly favorable and that's why burning a chunk of wood is wildly favorable because I produce heat into the surroundings and I produce CO2 and stuff from the wood from a solid I turn it into a gas see you later off it goes this order increases like crazy that reaction is very very favorable and humans figured that out what we do mostly is we burn stuff because that works this kind of reaction then that is exothermic and disorders the system will be spontaneous at all temperatures that doesn't mean that it will be fast it just means that it's eventually going to happen that if I leave a chunk of wood out in air eventually it's going to all turn to CO2 and H2O it's not going to help me much if I don't actually light it on fire because it's going to happen too slow but if I look at old books I notice that the pages are turning yellow they're sort of looking like I put them in the oven they're burning they're just burning up very very slowly and if I leave them a very long time they'll just turn into ash it's going to be a very long time but that's why preserving old manuscripts is tricky because they can be very fragile and they've been oxidized over time the only way I can stop that is by storing the manuscript in an atmosphere that's got nitrogen say no oxygen then I'd say I'm safe because there's no oxygen then the reaction can't happen since combustion requires oxygen so I could do that if I've got a reaction that absorbs heat from the surroundings a reaction that when I run it it gets cold so it absorbs heat from the surroundings it has a positive delta H and that positive delta H means that delta G is going to be positive unless something else makes it negative if the system becomes more disordered then delta S is positive and that means that minus T delta S is still negative and so it's a tug of war between delta H which is positive and minus T delta S which is negative and that means that if I make T it's going to happen that means that any reaction that absorbs heat and the system becomes more disordered will be favorable at high enough temperatures the example that I gave you was dissolving ammonium nitrate and water it gets cold but it still happens and this reason is part of the reason why your clothes usually get cleaner if you use warm or hot water because the kinds of things you're trying to dissolve don't want to actually dissolve that could be an endothermic process to try to get oily gunk off your clothes to go into solution and water the water is making all these hydrogen bonds and saying no keep out keep out and so to overcome that you heat it up and if you heat it in hotter and hotter water sometimes it works better it just depends what the reaction is and so forth for blood you don't want to put it in hot water because then that sets the protein like frying an egg and then the stain never comes out so if you have blood on something you wash it in cold water very gently to get it out and you don't just dunk it in very hot water because then the stain sets and that's it turns brown pretty much okay now if we have a reaction that dumps heat into the surroundings but the system becomes more ordered than minus T delta S is positive an example of this I haven't thought of a chemical equation but a phase change when I have water and it freezes to ice it dumps heat into the surroundings people find that confusing because they're saying well you're making ice isn't it getting colder the way to think about it is this if I have a chunk of ice and I want to melt it into water at the same temperature do I have to put in energy the answer is yes I have to put in energy to melt the ice therefore when the water freezes that same energy comes out into the surroundings okay in this case the reaction will be spontaneous at low enough temperatures and of course we know that ice only freezes at low enough temperatures if the temperature is not low enough the ice doesn't freeze often times if you've got citrus crops and you're worried about a hard freeze if you've got water you spray the trees with a ton of water because then a lot of the cold air will waste its time freezing the ice and it'll freeze the ice first before it gets any colder and hopefully by then it's dawn and you haven't frozen the leaves solid and so you haven't killed the plant and killed your harvest for the year solids always form at low enough temperatures because they're the most ordered form of matter with the lowest entropy we can make solids form just by cooling things down lower and lower and lower we can make solid nitrogen for example if we want by cooling it enough and that's this principle here we just lower the temperature if we have finally a reaction that absorbs heat from the surroundings and has a positive delta H and the system becomes more ordered that's a no starter this kind of reaction won't be spontaneous at any temperature so we cannot do a reaction like this and let it happen by itself though if we have some kind of chemical equilibrium that has this this characteristic that the system's becoming more ordered and it also has a positive delta H that means if we try to run a reaction like this with a positive delta G we aren't going to get much product we're always going to have a ton of reactant and a little product and what chemists do is they couple the reaction they want to run with another reaction that's wildly favorable like burning coal I can make any reaction run in the lab I can electrolyze water with electricity and produce H2 and O2 or whatever I want to do run things backwards as long as I have a power supply and the power supply is plugged into the wall and the wall somewhere is plugged into a power plant and the power plant I assure you is running a wildly favorable chemical reaction to generate power of some type burning natural gas oil, coal, nuclear you name it it's doing something like that and together if I include the power plant and what I'm doing with all delta G is negative and so I can synthesize very complicated molecules very highly ordered things by expanding enough energy and that's exactly what we do when you grow up and develop from just a cell that's sitting there and dividing and turn into you that takes a lot of energy if you figure out how much you eat over all those years etc etc etc it's a lot and all that's wildly favorable because you metabolize you produce CO2 and then your body takes things and intricately makes gigantic molecules that are very unlikely to occur spontaneously but they can only happen as long as you find something to eat they won't happen if you don't cannot we'll stop there and we'll pick this up next time