Lec. 17. Equilibrium Calculations
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Shaka, A.J.
 UCI Chem 1B is the second quarter of General Chemistry and covers the following topics: properties of gases, liquids, solids; changes of state; properties of solutions; stoichiometry; thermochemistry; and thermodynamics.
 Inactive neurons are targeted by your immune system for complete destruction. This is just another variation on the theme of use it or lose it. Whichever neurons are just sitting around not pulling any weight, they eventually get tagged with a protein. And that protein tells the microglial cells to get rid of that stinker. It's not doing anything. And then you lose that one and all its connections. And that could be a physical mechanism for why even somebody who's extremely good at a musical instrument and doesn't do it for years comes back and they feel very rusty. They can't play with the same rapidity and style and verve that they had. Because when they stop practicing and those neurons become idle, your body says, well, look at all this metabolic load we've got here. 15 watts steaming out of this skull. And this guy here is taking energy and food and oxygen and nutrients and blah, blah, blah and not doing anything. We can prove he hasn't fired in years. He's gone. And perhaps when that mechanism goes astray or goes out of whack when you get older, your immune system starts getting rid of even parts of your brain that you would prefer to keep. And then you suffer cognitive loss. And the second is a dual thing, the observation that Europeans are putting on weight, well, as Gomer Pyle would have said, surprise, surprise, as they've become more like us, driving around everywhere. And the interesting connection here is that you are like a Prius. If you sit down for 20 minutes, it's over. You're on battery power. Your biochemistry detects in 20 minutes that you're bone idle and it shuts down. And why should it keep metabolizing fat and be ready to run and do things when you aren't doing anything? This is a self-defense mechanism so that you don't run out of energy. However, if you sit down for extended periods and then you eat, you pile it on. Your biochemistry can't believe how fortunate things are. Guess what? I'm like the king of France. I'm sitting down, I'm doing nothing, and I'm getting food handed to me. That never occurred, naturally. You had to climb a tree to get some sour apple. You had to chase something down. You had to at least run after the bugs you were eating because we ate plenty of them. They were much safer to eat than big animals because a big animal might injure you with its great big antlers. And if you got injured, you would often die of infection. If you know that, then you can explain why people who are every day on the treadmill, which I wouldn't recommend at all, are still heavy. They're the arts. You know what happens? On the way out, 300 calories of something or other, there's three miles down the tube right there. Did you do three miles on the treadmill? Maybe not. Then sit down in your car and then let nature do its work. To defeat that, interestingly, you don't have to be an avid exerciser, but just don't sit down for longer than 20 minutes. Always get up. Do a few deep knee bangs. You'll fool your system. It'll say, who? It'll keep the fat oxidizing enzymes going, and you'll be able to maintain your weight much more effortlessly than by making a big push, being tired, and then the rest of the day, you sit around. When you aren't keeping track, if you hadn't gone to the gym, you would have walked around a lot more. You actually defeat yourself because you pay attention to this big thing like the payoff in Vegas. I'm on the treadmill. I'm doing it. Then the rest of the day, when you do nothing, you don't even pay any attention. You say, well, I'm tired. I want to sit down, and I'm hungry, so I want to eat. Then you go backwards. Let's go on now and talk about equilibrium. First of all, equilibrium is dynamic. We talked about that last time. Carbon dioxide is the brown haze over Los Angeles. A lot of it comes from cars. It was much, much worse before the days of catalytic converters, which attempt to remove NL from the exhaust bloom. It has an odd number of electrons, which makes it very reactive, and it can dimerize to make N204, which is colorless. If we have a bulb then of NO2, we have a forward reaction, which I've just decided to put N204 as the reactant, and it falls apart into 2NO2, the revert, and then these sometimes recombine. At a certain temperature and pressure, there will be an equilibrium concentration of these two different molecules in this container. We know by now that if we press on the container, if we shrink it by putting pressure on it, that we will favor the reactant because the reactant is less moles of gas. If we heat it, if it gets hot in the afternoon, then we will favor this. We'll get much more of that coming up. Smog, of course, is much more than just NO2. It's a whole witch's brew of stuff that has to do with photochemistry, and that's why it's much worse in the afternoon. That's why the time to run is 5 a.m. because that's when the air is the best and all the idiots driving their cars are not up yet because they're still shaking off the night before. You're much safer when you cross the road as well because you won't have some car. I guess some people have actually been hit on the sidewalk by someone in a car because they're fiddling around doing something else and they jump the curb and then they smash into you. That's really unfortunate, but it does happen. Well let's suppose we looked in this bulb and we tag, we make an imaginary tag. We put a tag on one NO2 molecule and we follow that guy. Problem is this is very hard to do because molecules are so tiny we can't usually see them. We can imagine it. If we did that, we'd find that part of the time it spends its time as NO2 gas and part of the time it spends its time as N204 as the dimer hooked up with somebody else. The bonds are made and broken and remade all the time. In fact, if we follow the percentage of time that this guy is either NO2 or N204, we'd find that was directly related to what the equilibrium concentration was. There's a connection between time averages and averages of other properties. That study of so-called statistical mechanics is a highly interesting thing to engage in in a more advanced course. Let's have a look. Suppose then you don't believe this. You say you told us you should rely on evidence. What you've said is plausible, but what's the evidence that that actually happened? How could we either confirm it or rule it out? The answer is we could do so by employing isotopes. In this case, stable isotopes. We could use nitrogen 15, which is an isotope of nitrogen rather than nitrogen 14. It has pretty much the same bonding characteristics and everything else as nitrogen 14, but it has an extra neutron in the nucleus. The electrons are pretty much the same. We could use oxygen 18, which they use in some medical procedures. We could make some molecules that have N15. First we make a batch of molecules, N15 and oxygen 16, the abundant isotope. Then we make another bunch that are N14 and O18 in another container. Now we mix the containers. If after we mix them, we find by, for example, mass spectrometry, we find that there's an O18 and 14 hooked to an N15, O16, then we know that they rearrange, that they randomly assorted. If, on the other hand, when we prepare this guy and we only find this guy and we only find that guy in the mixture, that's all we find, then we know that they stayed put. Basically, they didn't rearrange at all. When we do this experiment with this particular molecule, we find that they rearrange all the time. In fact, sort of the presence of both the monomer and the dimer indicates that they have to break, otherwise we wouldn't have both. If you do the same experiment with carbon dioxide and you prepare some carbon dioxide with carbon 13, O16, and then carbon 12, O18, and you mix those together at normal temperature and pressure, there's no exchange. You never find any oxygen moving off one and onto the other. If you go to extremely high temperature, then you can make it happen, but that's extremely high temperature where those bonds can actually break. If you do the same thing with water in water, you find that the hydrogens hop around all the time. They fall off, they come on, they go every which way. They're exchanging quite rapidly in liquid water. Let's have a look at why this bond is so weak. Actually the Lewis structure is kind of a good way to start and look at a bond. Let's then write Lewis structures for NO2 and N204. Remember what we have to do. We have to count the valence electrons. We have to know how the atoms are bonded because it says NO2. That's going to be N in the middle and two oxygens sticking off like spokes. N20, which is nitrous oxide, has N then N then O. It's different. Let's draw them. There's five valence electrons for nitrogen. There's six for oxygen, but there's two oxygens, so I have 17. This is very rare because whenever you have an odd number of electrons, that means that you have one unpaired electron somewhere. Those guys are called free radicals. They usually cause mischief because they're quite reactive. If you ever breathe any NO2 and I have done it by mistake, you never forget it because your lungs tighten up and say that's enough of that and you feel like you need oxygen for the next couple of hours, depending how concentrated it was. Don't try to clean up any nitric acid because it won't work. It will produce NO2 instead and the penny will disappear. I tried to make an octet around each oxygen and I can do that, but in order to do it, I have to have a minus charge on this oxygen because remember we count one electron for the bond, then one, two, three, four, five, six. We count seven electrons, but oxygen's group six. This has a formal negative charge. Here we count one electron here, one, two, three, four, five, six. This has no formal charge, so we put nothing. The nitrogen group five, we count one, two, three, four. He's got a positive charge. Then we can play the usual game of flipping this guy in and flipping this guy out like playing dominoes and the negative charge moves back and forth. The actual structure of NO2 is 50% this and 50% that. It's one structure, but we can't write a proper Lewis structure for it like that. With Lewis structures, when we have more than one resonance structure, what we're saying is the real guy is a mule, but a mule is not a proper species because it's a mixture of a horse and a donkey. What we try to do to make it look like a mule is we flash a horse 50% of the time and a donkey 50% of the time. If we flash them very fast, we get kind of an average, and that way we can draw a proper Lewis structure for each one. This is not meant to mean that this is literally happening. This is just meant to mean there's a percentage of each in the real structure, which is an average, which we don't want to draw because then we have to draw dotted lines and funny things and we don't know where the electrons are. I think you can probably see that there's an electron here and an electron here. I think you can guess how N2O4 is made and I'll let you work out N2O4. Then I'll probably ask you about it on Tuesday to see if you did it. I think you can see why it's a weak bond because when I make it, I have to put those two formal positive charges together and those formal positive charges hate each other and they're closed. That bond is quite weak. That means we can break it apart fairly easily. It's an abnormally weak chemical bond that allows this thing to happen. We can take a kinetic view of equilibrium and I'm going to go through quite a lot of mathematics here just so you've seen it, not because I want you to reproduce it, but so that you've seen it so when you see it again, it'll be familiar. The chemical reactions can only happen when atoms or molecules are close enough to make a bond. That basically means the molecules have to collide. They first have to hit. We can figure out how often things hit in a gas. They have to collide and if we look at our kinetic theory of gases, we know how many are going how fast and we can figure out how many times they hit per second and in a gas it is a very, very, very large number of hits at normal pressures. We have the kinetic theory of gases and we can figure out how often the atoms or molecules are going to collide. If we heat things up, two things happen. They're going to collide more often because everybody's going faster, so it's as if time got shrunk and when they do collide more often, they're going to hit harder as well so there's more chance that they hit so hard that something gets knocked off and then you get a reaction initiated. That's why most reactions speed up in both directions, not just one, but in both directions when we heat things up. We'll see this in the summer if you're going to continue on in this series in Chem 1c with the Arrhenius equation. Let's have a look. Suppose that we are isomerizing cis-butene to trans-butene. We're taking something like this and we're isomerizing it like that, which means we have to formally break the double bond momentarily, slam into it, get this guy to twist and around. We know from before that usually the transform when you've got two things sticking out is slightly more favorable because when you're like this, these guys tend to repel each other a little bit and it puts a little bit of extra energy there that gets released when we go like that. That doesn't mean it's easy to do, but it can happen. If we have a forward rate constant, which we call little k for forward, of 5.59 times 10 to the minus 3 per second and the reverse reaction has a slower rate constant of 4.41 times 10 to the minus 3 per second and we begin with a mole of cis-butene in a one-liter flask, what will be the concentration of the two isomers at equilibrium? The forward reaction runs at a rate that depends on the product I meant of the rate constant and the concentration of cis-butene. It says however many I've got this many per second are going to tend to go this way, just like a conveyor belt. That makes trans-butene and then the trans-butene reacts backward but at a slower rate, more sluggish and that reverse reaction depletes trans-butene. And the one weapon that we have in this time view of equilibrium is that at equilibrium, whatever the amount of trans-butene is has to be fixed because that's what equilibrium means, the concentration's not changing. And that means that the rate of gain plus the rate of loss must equal to zero. So if the money in my bank account every month is the same, that means my income and what I'm spending match. That's the only way it can be the same. Let me say that again. The only way it can be the same is if the amount you get and the amount you spend match. If they're even slightly off, you eventually go into a black hole. If on the other hand you have a slight surplus, every month you end up in a white hole up here and that's the correct way to operate, of course, much less stressful. Now let's write the rate of gain. The rate of gain depends on the forward plus how much cis-butene is there at equilibrium. And the rate of loss depends on the reverse, but it's negative. I write plus, but I write the rate of loss being a negative number is minus whatever the rate of trans disappearing is times whatever the concentration is at equilibrium. And if it is at equilibrium, that should be zero. And if I solve this, I find that the ratio of the forward rate constant to the reverse rate constant is nothing other than the equilibrium concentration of trans-butene, the product, divided by the concentration of cis-butene. And that means I can get a number for it. So trans-over cis is forward, which is 5.59 times 10 to the minus 3 divided by 4.41 times 10 to the minus 3. 10 to the minus 3's go away and I get 1.27. Aha, but the concentration of the products at equilibrium divided by the reactants at equilibrium, that's none other than k. And so we have a number for k. K in this case is 1.27. Not a very favorable reaction, but more favorable to make trans than cis. This is a general conclusion. Equilibrium, so I had to abbreviate it, can be viewed as a detailed balance between all the ways that things are made and all the ways that things are destroyed. And at equilibrium, they're being made just as fast as they're being destroyed. It's not that nothing's happening, plenty's happening, but it's all a wash in the end. If you change one of the rate constants, for example, you have two kinds of cells that have to do with your bones, osteoblasts and osteoclasts. And one of them tends to make your bones bigger and stronger and the other one tends to make your bones disappear, get weaker and smaller. If you go up into space and float around in the space station, or you try to go to Mars, which wouldn't be a good idea, because there's no gravity, the osteoblasts say, guess what? There's no reason on earth to pile on bone. I'm not feeling anything here. The osteoclasts, however, are always running. And that's the rule, is that the machinery that's tearing you down is always running. So that the minute you stop building yourself up, you go to a lower state. That's again adaptive because you don't want to be trying to maintain gigantic bones and muscles that you don't need. Where are you going to find the stuff to eat to support all that if you're running around in the wild? So if you don't need it, you're going to just get rid of it, just like the brain. And that's why when they come back from the space station, they're basically on a gurney or in a wheelchair because they're so weak, their muscles didn't have anything to stand on. I don't weigh anything. I'm sitting down all the time. Plus, my blood doesn't even have to work to get back from my calves up to my heart. So my heart turns into this little pathetic thing. And I come back down to gravity. And I feel absolutely miserable. I'm weak, bones like balsa wood and so forth. And I expect that if you sent a human to Mars, they wouldn't even be able to get out of the spacecraft if they were the only person. Because when we take people out, we wheel them around and rehab them so that they can get back to normal function. My view is we might send a robot to Mars, which is made out of metal and doesn't have any active, degradatory mechanism. And that would be fine because that's not a living thing in the strict sense. So just like water, if we have a bunch of interconnected cisterns, we'll only be in equilibrium when the water level's at the same in all of them. So a system, no matter how complex it is, will only be at equilibrium when the rate of production and the rate of destruction of all the molecules in there is equal. Some of them may be destroyed and produced quickly. Others may be destroyed and produced slowly. It just depends. And in general, we can write this as a differential equation. And this is what I want you to see, but not necessarily digest today. So let's look at the rate of change. This is how the water level's changing. Is it going up or is it going down? What you read, the concentration of trans dt is just the rate of change. And because I don't want to write concentration of trans everywhere, I'll just write that as TR, and I'll write concentration of cis as C. Then we can get a differential equation. The rate of change of the trans depends on the rate of production, kf times C, and the rate of destruction minus kr, the reverse reaction times trans. And we can simplify this equation because in this case, I said we started with one mole, and so there's always one mole of material in there. It's either cis or trans. And that means we can substitute for C1 minus TR. So now we have a differential equation that involves the derivative of TR with respect to time and a bunch of stuff that has TR in it. This is called an ordinary differential equation. It's ordinary because this is a regular D and not the funny D. Yes? So this would only be if you were going from the cis to the trans estimate rate. Yeah, I'm talking about the rate of production of the product. And it's made by the reactant making the product, and it's subtracted by the product going backwards and making the reactant. You can solve these with Mathematica or Wolff from Alpha. It'll solve them for you. But until you learn how to solve them yourself, at least a little, the solution may not mean much. It just looks like a bunch of gobbledygook. So the problem is, often you can get the answer and you haven't learned anything. And that's something to remember when you say what's the answer. If somebody tells you the answer, you haven't learned anything. If you get the answer yourself, you have learned something. Let's put in the numbers that we have. We have those rate constants, so we have now a numerical equation. The rate of change of the concentration of trans with respect to time is equal to 5.59 times 10 to the minus 3, that's 3 times 1. And then these guys I put together, minus 1 times 10 to the minus 2. We solve this by separating variables. If you can separate variables, you can solve the differential equation instantly. And what separating variables means is get everything with the DTR and TR on one side, and then get everything with T and DT on the other side. This equation is easy because this side doesn't have anything that has T in it. If I had another term here plus T squared or something awful like that, then that would be much, much, much harder for us to solve. Because now I've got something else going on. It's as if not only am I trying to have it come to equilibrium, but I'm pushing one of the pots up, I'm lifting it up, and then I'm asking, okay, what's the water level? Then that's much more complicated to do. We can do that too. And when you're modeling weather or concentration of chemicals in the atmosphere, you have to solve lots of differential equations like this, and some of them can be very, very tricky to solve. Usually they're done numerically by giant computers that don't care what the equation is, they just integrate it the way a graphing calculator doesn't care what the function is, it just graphs it. Okay, so let's try separating variables. I look at this equation, and I want everything with TR on the left and everything with T or DT on the right. And I can pretty much see how to do that. I can just divide both sides by this guy on the right and multiply both sides by DT, and I get this great big equation here. DTR divided by 5.5 times 10 to the minus 3 minus 1 times 10 to the minus 2 TR equals DT. Good. Now I can solve that because this side only depends on TR, so I can integrate it. This side only depends on time, so I can integrate it. That's why I have to separate the variables. And so I put an integral on each side, and I say, okay, I'm going to integrate from the concentration of trans at some time I'm going to call zero, which is when I start the experiment maybe, or start the stopwatch. And I'm going to integrate up to some concentration of trans at time T. And then over here I'm going to integrate time from zero to T. Strictly speaking, this is sloppy notation because usually the variable you're integrating over and the limits, you don't like to use the same letter because sometimes it's confusing. If I were really doing this correctly, I should call this DX and integrate it from zero to T, but it doesn't matter because we can get the answer. The right hand side is just T minus zero because the integral of DT is T. And I evaluate it at the top limit and the bottom limit. And I subtract them so I get T minus zero is T. That's this side. And the right hand side, we can either hand over to integrals.com. If we're lazy or we can make a substitution, we can call U equals the denominator, 5.59 times 10 minus 3 minus 0.01 TR. And if I do that, which I'll do just for fun, we can set up the integral. And this is what you would probably do if you were doing it by hand. I have this new variable U. It's 5.59 times 10 minus 3 minus 0.01 TR. And I take its derivative so I know what DU is. And that the derivative of a constant is zero and the derivative of TR is just DTR. I end up with minus 1 times 10 to the minus 2 DTR. And then I solve this for DTR. DTR is minus 100 DU. And at T equals zero, TR is TR zero. And therefore, U at time zero is 5.59 times 10 to the minus 3 minus 1 times 10 to the minus 2 TR at time zero. And the upper limit is just the same thing, U at time T is this minus this times the concentration of transit time T. And then I put all those into the equation so I get something I can solve. When I put all these in, this minus 1 and this 10 to the minus 2 comes outside. And now I'm integrating DU over U and I know that's the natural log because now I have it in a form I can recognize it. The other side is T and that means the log, natural log of U at time T minus the natural log of U at time zero is equal to, I brought this thing over here, minus 1 times 10 to the minus 2 T. And I'll just remark as long as you make the variable X and you hand this thing like that to integrals.com, it'll give you the answer straight away. The variable has to be X for that particular website. And we know when we subtract logs, it's the same as dividing the arguments so we can clean this, tidy this up. We now have this. The natural log of U at time T divided by U at time zero is equal to minus 1 times 10 to the minus 2 times T. And then if I take the exponential function to get rid of the log, I end up with the following answer. Our new variable U at time T is equal to our variable U at time zero times E to the minus T over 100. The reason why the exponential function is so important is that lots of differential equations depend on how much stuff you've got. Compound interest, for example, is an exponential equation. You just plunk the money in and you leave it and you get 8% interest and you end up extremely rich in 30 years because it keeps going up and up and up. And as there's more money in the account, the interest is more. And then there's even more and then the interest is even more, et cetera, et cetera. On the other hand, if you borrow money, then you're going down like an exponential, especially if you keep borrowing more and more money. I'm amused that the credit card offers I get to get rid of my debt by transferring the debt to someone else. So you notice you didn't get rid of anything, but now you're paying the interest to them. That's what they're interested in. If you're a good risk, they want to get the 17% or whatever they're charging or even 29%, I guess, if you're unlucky and miss a payment and don't read the fine print. Okay, then now we don't like this. This is a nice equation, but it has u. It doesn't have the concentration of trans in it. So I have to go back and I have to substitute in the whole thing to obtain the final answer. I put in all this stuff for u at time t and I put in all the other stuff for u at time zero and now I have an equation where I have to solve now for transit time t, but basically I'm there and I can predict now from any starting place how much trans I'm going to have at any time. And here's the final result. The amount of transit time t is.559. Not surprisingly, that's the equilibrium concentration and you notice it doesn't have any time on it. That means as time goes on, it's going to settle down to this number. And then whatever the difference between where I start and the equilibrium amount and then multiply by e to the minus t over 100, when you multiply anything by e to a negative exponential, it gets smaller and smaller and smaller. So this part here dies out and I end up with the equilibrium concentration. We had an assumption here that we started at 1.0, but that was my first initial thing where I got rid of c. If now that we have more confidence, we could certainly go back and we could say, look, let the total amount of stuff be c naught or anything and we could do it again and then we'd have a completely general answer to this particular problem. Okay, just to emphasize, will you have to do this? No, not now and certainly not under the pressure of an exam. However, yes, eventually you will have to do things like this and much, much harder things because if you can't do things like this and much, much harder things, then you're down there with everybody else who also can't do that. If you can do something that other people can't do, you tend to get a job that they can't get. That's how that works. They have requirements. When you do this kind of computer programming, are you familiar with this process? Have you ever done this and so forth? If you have, you're a good candidate. Even if you do use modern software, you can get mixed up if you don't understand how the equations are behaving and what you're supposed to be trying to get. Most people view mathematics as something scary or difficult, but in fact, mathematics is the quickest, cleanest shortcut. That's why we make mathematical models of the universe because mathematics problems are much easier to solve than vague problems and words. Whenever somebody says something about the future in words and they don't have any mathematics behind it, then you have to ask how they're going to get it to happen. I will create 250,000 jobs. How? What's the equations? Show me the curve that goes up and levels off at 250,000. Why'd you pick that number? Why isn't it 253,000 or 225,000? What exactly are you going to do to create jobs out of thin air if there's no work to be done or no people who can do the work that's required? Gets much, much more complicated than to implement something like that. Let's plot our results. Once we've got an equation, we can see exactly how many molecules we're going to create. Let's suppose we start with no trans at all. It's all 100% cis. How long does it take to get to equilibrium? Well, we've got an exact solution, so we plot it. So I started with 100% 1.0 of cis and 0.0 of trans, and I just put those into the equations and I plotted it versus time here in seconds. I can see that after about 300 seconds or so, it settled down. That lets me know how long I have to wait in the lab before I make a measurement. If I want to know the equilibrium concentration as a function of temperature or pressure for this system if these rate constants are correct, then I have to change the temperature and then I have to go get a cup of coffee because I have to wait 300 seconds at least before I can know that it isn't moving around anymore, settle down. Chemical engineers really like to know exactly how long they have to wait. They're bunging in chemicals and they want something to come out. How long do I have to treat the wastewater? How long does it have to cycle around in there before it's clean enough or it's as clean as it's ever going to get? And then I get rid of it. It's very important to know these kinds of things. Sometimes the only way you can know is by careful measurement because you don't know these rate constants very well. If we start with no cis isomer, about how long is it? So now I've got the opposite. I start with 100% trans and some of it starts immediately going backwards because Q is bigger than K so it starts making cis and it takes about the same amount of time, about 300 seconds in, it's leveled off. And I would get the same conclusion no matter what I started with because these are the two extremes. If I'm lucky and I started with the equilibrium concentration, they both flat line all the way across. So here's the conclusion. The time it takes to establish equilibrium depends on the size of the individual rate constants more than the initial conditions. If the rate constants are fast, that means it's going to settle down very, very quickly. It means instantly I'll have equilibrium. If the rate constants are very slow, I'm going to have to wait forever to let equilibrium establish. On the other hand, the final concentrations don't depend on whether the rate constants are both big or both small. The final concentrations only depend on the ratio of the rate constants. How fast is this guy shoveling stuff in and how fast is the other guy shoveling stuff back? That ratio is going to tell how much I've got at equilibrium. And these same conclusions apply to processes like biological fermentation, which they're using to produce things like fuel and drugs. I'm growing yeast. I want to heat it up maybe to get the reaction to go better, growth, and other reactions that we'd like to be able to control. The equilibrium result of a bunch of fresh raspberries is the raspberries spoiled. They mold. They start smelling bad and so forth. And if you leave them out on the counter, that happens very quickly. If you put them in the refrigerator, the same thing happens. But you usually don't leave them there so long that you notice it happening. It just happens much, much more slowly. If you take food and put it in a freezer, it also spoils, but it spoils much, much more slowly. If you put it in a deep freezer or liquid nitrogen, it spoils very, very, very, very slowly. And so it keeps a very long time. And for each 10 degrees Fahrenheit you go down, you can keep the food in there a lot longer. Assuming it can tolerate being frozen, sometimes if you freeze things, they change. So frozen raspberries aren't the same as fresh because of the ice crystals rupturing things inside the fruit. Okay. Now let's talk about something which we will definitely have on Tuesday, which is we're going to add reactions. And we know that when we add reactions, any state function, big H, any big capital letter, big S, big G, they all add. Let's then talk about, suppose we have this reaction. This reaction in the 60s was a big problem because coal has a lot of other stuff in it, in particular sulfur. And high sulfur coal is a lot cheaper than low sulfur coal because of this reaction, basically. If we burn the coal to produce power and we have sulfur in it, then we also make SO2. So just like the C in the coal, the carbon makes CO2, the sulfur makes SO2. And if delta G for that reaction is delta G1, and I have another reaction where I take SO2 and it goes up into the air and it reacts over time with oxygen in the air out of the smokestack, then it makes SO3. SO3 is called sulfuric acid anhydride. If I add water to it, I get sulfuric acid. When a lot of coal plants in Ohio were burning a lot of coal and there were no emissions controls, the SO3 would drift over Canada, up north, and then mix with water and rain down on all their nice forests. And that was the so-called acid rain problem. The trees would defoliate because the pH of the ground was changing. And also, if you just spray dilute sulfuric acid onto the leaves of plants or pine needles, they don't like it. It's too acidic for them. So the trees wouldn't grow. And in Canada, of course, they're doing a lot of lumbering. They have a lot of people cutting down trees, turning them into planks and selling them. And if their trees don't grow, that business is dead. And so there was quite a tug of war. You've got to stop burning this stuff. That's your problem. But that drifts up there. You find another kind of tree. Well, eventually they said, well, OK, we'll stop this SO2 going up. But that cost a ton of money. Now you have to put something on the smokestack that captures the SO2. And then you have to take it out. And you have to keep it clean. And that costs a lot of money. And where am I going to put it? What am I going to do with it? Can I sell this material for any purpose? Or is it just junk that's dangerous and toxic that I've got to somehow figure out what I'm going to do with it? It depends. Well, if we add those reactions up, and we just imagine doing it in one step, then the SO2s go away here. And I have a net reaction, sulfur, elemental sulfur. And 3 halves of molecular oxygen gives SO3. Delta G for this reaction, whatever it is, is just the sum of this one and the sum of this one. We just add them up. If we turn a reaction around, we change the sign of delta G because we changed the initial and final state. And if we carry out these two reactions at the same temperature, then I can multiply by minus 1 and divide by RT. And I have, instead of this equation, minus delta G over RT is equal to, I'm sorry, minus delta G1 over RT plus minus delta G2 over RT. And that's of interest because that means since delta G over RT, minus delta G over RT is log K, that means that log K adds up. So log K for the total reaction is log K1 plus log K2. But I know how logs work. I can always combine them. If I add them, I multiply them and take the log of the total. If I divide them, I subtract them. And then if the log of K is equal to the log of K1, K2, that means that K is equal to K1, K2. Therefore if we add two chemical reactions, we multiply the equilibrium constants. Let me say that one more time. If we add two chemical reactions, we multiply the equilibrium constants for the reaction. The reason why this is not just academic is because one of these reactions might be unfavorable. It might be one we want to run, but it's not going anywhere because K is tiny. But if we couple it to another reaction that is favorable, then we can drive the whole log to go to where we want. And that's what chemists always do. Chemists run a ton of unfavorable reactions as do all living things. They assemble these gigantic molecules that are very, very low entropy, and they do so by running a ton of other reactions that are favorable, like burning up glucose with oxygen to make CO2. That's a favorable reaction. So as long as I can eat and I can run that favorable reaction, then I can run a lot of unfavorable reactions like get smarter. But if I can't eat anything, I can't get any smarter. And if I eat too much, I won't get smart either because then the blood ends up in your stomach and your brain needs blood to work. So before the exam, you don't just eat a ton of stuff because then you'll fall asleep in the exam. That's the natural result. You look at a lion that's hungry. How does it look? Right? Right alert. That's the way you want to be for an exam. What's the problem? You look at a lion that's just finished a giraffe. It's a dead, right? It's asleep. It's digesting all that food, all the bloods in its stomach, doing all that biochemistry. You only want to do one thing at a time. So you come into the exam hungry, and then after the exam, you eat and relax. But don't get the order wrong because if you do, you won't do as well. Okay, let's finish up by talking about Le Chatelier's principle. And Le Chatelier's principle is just that if we have a system at equilibrium and we change something, how will the equilibrium shift? Or will the equilibrium shift at all? While Le Chatelier worked this out and made the following general statement, the equilibrium will always shift in the direction that tends to counteract whatever the change is. For example, and we worked this out in some detail, if we increase the pressure in the Haber reaction to make ammonia, then we saw that we made a lot more ammonia, the right hand side, because making more ammonia minimizes the number of moles of gas and counteracts the increased pressure. When I squeeze on a chemical system, it tries to get smaller if it can. And if it can get smaller, it does. And in this case, that's our lever to get more product out of the reaction. Okay, so let's consider a reaction to synthesize Fos genes. Don't try this at home. Very, very, very reactive guy, this guy. Let's take molecular chlorine and carbon monoxide and make Fos gene, COCl2. It has a carbon, double bond oxygen, and two chlorine sticking out like a triangle. And let's consider the effect of the following changes. Let's suppose we have the reaction at equilibrium. That'll be the effect of each of the following on the concentration of carbon monoxide. One, I inject additional chlorine. Two, I add argon gas and I make the total pressure higher. Total pressure is higher because I've pumped in argon. But argon doesn't react with the other guys. Three, I just compress the mixture by increasing the pressure. Or four, I decompress the mixture by lowering the pressure. Okay, let's go through them then one by one. First we'll start with injecting additional chlorine. If I add additional chlorine, I'm going to increase the denominator in Q because Q has the concentration of Fos gene and in the denominator the concentration of chlorine times the concentration of carbon monoxide. Q is K at equilibrium and K doesn't change. Therefore if I add more chlorine in the denominator, I'm going to have to get less CO. Otherwise it's not going to work and I'm going to get more product. And this is another clue. Suppose I have two reactants. One of them is extremely expensive and the other one is cheap as dirt. And I want to use up the expensive reagent so I don't waste it. Then what I do is I just pile in the dirt until I drive it over to the other side. And if the dirt cheap reagent is a gas, it's great because I can pile it in, get the product, isolate the product, and then vent the gas off, assuming it's harmless. That's another trick. Two, adding a dirt gas does not shift the equilibrium. Why not? I just said if we increase the pressure, we shift the equilibrium if the number of moles of gas of products and reactants is different. And now I'm saying I'm increasing the pressure but nothing's happening. The answer is argon is not in Q, the reaction quotient. If I don't change the reaction quotient, I can't change any of the concentrations of the things. So although the total pressure in the container is higher because I didn't actually change the reaction quotient, I just added something else that doesn't react. Nothing changes. So there's no change in equilibrium if I add an inert gas. If on the other hand I squeeze on it like this and compress it, then I'm going to get more product because there's again Q, if I work it out with the moles of gas, there's going to be a P standard over P left over. And so if I increase the pressure, I'm going to decrease the number of moles of CO. And so not surprisingly, if I expand it, then I'm going to shift it and get more CO because some of this phosgene will actually react backwards and make both CO and CO2. And I'll get more CO. And that's how we keep track of it. So here's the summary. If I change the concentration of something in the chemical equation, not something else, but something that participates in Q, the reaction quotient, then I'm going to shift the equilibrium, but I'm not going to change the equilibrium constant. If I put pressure on a gas, I'm going to shift the equilibrium potentially if the number of moles of gas is different, but I'm not going to change the equilibrium constant. If I change the volume, I'm going to shift the equilibrium, but not change the equilibrium constant. If I shift the temperature, ah, log K minus delta G over RT, I know then that the equilibrium is going to shift and the equilibrium constant is going to shift. And if I add a catalyst to make the equilibrium come more quickly, a catalyst, strictly speaking, this is never true, but a catalyst, a true catalyst is unchanged. So if it starts on the reactant side, catalyst plus blah, blah, blah, blah, blah, and then on the product side is products plus the same catalyst, the catalyst cancels out because whatever its concentration is, it's in the numerator and the denominator. So if it's a true catalyst, it can't shift the equilibrium. It just makes the equilibrium happen more quickly. Really what happens is the catalyst gets dinged over time because our reagents aren't 100% pure and if we've got a tiny, tiny amount of some bad guy in there, it wrecks the catalyst. That's why the oil companies had to get rid of all the sulfur in the gasoline because it was reacting with the platinum and palladium and the catalytic converter and making it fail. And there was a big battle. You guys make a better catalytic converter. You guys get the sulfur out of the gas. We can't make a better catalytic converter. So usually the oil companies and the auto companies are on the same page, but not always. Whenever one's going to have to try to do something very hard that's going to cost a ton of money, then they try to say, well, the other guy should do something very hard that costs a ton of money. Instead, the result of that is that we have gas that doesn't have as much sulfur in it. Okay. We'll finish up here with a few subscripts on K. There's lots of little subscripts on K and they're all Ks. They're all ratios of products and reactants and they're just meant to be descriptive. So if I have K sub A, that means this reaction. That means an acid dissociates to give H plus and the anion. And here's the formula for K. And if I have KB, that refers to a base. I have a base. It pulls a proton off water and makes hydroxide. I could have a neutral here like ammonia plus water gives ammonia and myon plus hydroxide. That's why they tend to put ammonia into things like Windex. Okay. Because it gives the right amount. You say, why don't they put Drano in Windex and make hydroxide that way? And the answer is it will leave a residue on the glass, number one, because it won't evaporate like ammonia. And number two, it's usually too strong. And if I make it too basic, I'll actually start affecting the glass. I'll actually start polishing away the glass so my glass window is getting thinner and thinner as I keep cleaning it over time. And that's a bad strategy to do that. Okay. So at this time, what we'll do is work some problems, okay, which will be good ones to know.
 Inactive neurons are targeted by your immune system for complete destruction.  This is just another variation on the theme of use it or lose it.  Whichever neurons are just sitting around not pulling any weight, they eventually get  tagged with a protein.  And that protein tells the microglial cells to get rid of that stinker.  It's not doing anything.  And then you lose that one and all its connections.  And that could be a physical mechanism for why even somebody who's extremely good at  a musical instrument and doesn't do it for years comes back and they feel very rusty.  They can't play with the same rapidity and style and verve that they had.  Because when they stop practicing and those neurons become idle, your body says, well,  look at all this metabolic load we've got here.  15 watts steaming out of this skull.  And this guy here is taking energy and food and oxygen and nutrients and blah, blah, blah  and not doing anything.  We can prove he hasn't fired in years.  He's gone.  And perhaps when that mechanism goes astray or goes out of whack when you get older, your  immune system starts getting rid of even parts of your brain that you would prefer to keep.  And then you suffer cognitive loss.  And the second is a dual thing, the observation that Europeans are putting on weight, well,  as Gomer Pyle would have said, surprise, surprise, as they've become more like us, driving around  everywhere.  And the interesting connection here is that you are like a Prius.  If you sit down for 20 minutes, it's over.  You're on battery power.  Your biochemistry detects in 20 minutes that you're bone idle and it shuts down.  And why should it keep metabolizing fat and be ready to run and do things when you aren't  doing anything?  This is a self-defense mechanism so that you don't run out of energy.  However, if you sit down for extended periods and then you eat, you pile it on.  Your biochemistry can't believe how fortunate things are.  Guess what?  I'm like the king of France.  I'm sitting down, I'm doing nothing, and I'm getting food handed to me.  That never occurred, naturally.  You had to climb a tree to get some sour apple.  You had to chase something down.  You had to at least run after the bugs you were eating because we ate plenty of them.  They were much safer to eat than big animals because a big animal might injure you with  its great big antlers.  And if you got injured, you would often die of infection.  If you know that, then you can explain why people who are every day on the treadmill,  which I wouldn't recommend at all, are still heavy.  They're the arts.  You know what happens?  On the way out, 300 calories of something or other, there's three miles down the tube  right there.  Did you do three miles on the treadmill?  Maybe not.  Then sit down in your car and then let nature do its work.  To defeat that, interestingly, you don't have to be an avid exerciser, but just don't sit  down for longer than 20 minutes.  Always get up.  Do a few deep knee bangs.  You'll fool your system.  It'll say, who?  It'll keep the fat oxidizing enzymes going, and you'll be able to maintain your weight  much more effortlessly than by making a big push, being tired, and then the rest of the  day, you sit around.  When you aren't keeping track, if you hadn't gone to the gym, you would have walked around  a lot more.  You actually defeat yourself because you pay attention to this big thing like the payoff  in Vegas.  I'm on the treadmill.  I'm doing it.  Then the rest of the day, when you do nothing, you don't even pay any attention.  You say, well, I'm tired.  I want to sit down, and I'm hungry, so I want to eat.  Then you go backwards.  Let's go on now and talk about equilibrium.  First of all, equilibrium is dynamic.  We talked about that last time.  Carbon dioxide is the brown haze over Los Angeles.  A lot of it comes from cars.  It was much, much worse before the days of catalytic converters, which attempt to remove  NL from the exhaust bloom.  It has an odd number of electrons, which makes it very reactive, and it can dimerize to make  N204, which is colorless.  If we have a bulb then of NO2, we have a forward reaction, which I've just decided to put N204  as the reactant, and it falls apart into 2NO2, the revert, and then these sometimes recombine.  At a certain temperature and pressure, there will be an equilibrium concentration of these  two different molecules in this container.  We know by now that if we press on the container, if we shrink it by putting pressure on it,  that we will favor the reactant because the reactant is less moles of gas.  If we heat it, if it gets hot in the afternoon, then we will favor this.  We'll get much more of that coming up.  Smog, of course, is much more than just NO2.  It's a whole witch's brew of stuff that has to do with photochemistry, and that's why  it's much worse in the afternoon.  That's why the time to run is 5 a.m. because that's when the air is the best and all the  idiots driving their cars are not up yet because they're still shaking off the night before.  You're much safer when you cross the road as well because you won't have some car.  I guess some people have actually been hit on the sidewalk by someone in a car because  they're fiddling around doing something else and they jump the curb and then they smash  into you.  That's really unfortunate, but it does happen.  Well let's suppose we looked in this bulb and we tag, we make an imaginary tag.  We put a tag on one NO2 molecule and we follow that guy.  Problem is this is very hard to do because molecules are so tiny we can't usually see  them.  We can imagine it.  If we did that, we'd find that part of the time it spends its time as NO2 gas and part  of the time it spends its time as N204 as the dimer hooked up with somebody else.  The bonds are made and broken and remade all the time.  In fact, if we follow the percentage of time that this guy is either NO2 or N204, we'd  find that was directly related to what the equilibrium concentration was.  There's a connection between time averages and averages of other properties.  That study of so-called statistical mechanics is a highly interesting thing to engage in  in a more advanced course.  Let's have a look.  Suppose then you don't believe this.  You say you told us you should rely on evidence.  What you've said is plausible, but what's the evidence that that actually happened?  How could we either confirm it or rule it out?  The answer is we could do so by employing isotopes.  In this case, stable isotopes.  We could use nitrogen 15, which is an isotope of nitrogen rather than nitrogen 14.  It has pretty much the same bonding characteristics and everything else as nitrogen 14, but it  has an extra neutron in the nucleus.  The electrons are pretty much the same.  We could use oxygen 18, which they use in some medical procedures.  We could make some molecules that have N15.  First we make a batch of molecules, N15 and oxygen 16, the abundant isotope.  Then we make another bunch that are N14 and O18 in another container.  Now we mix the containers.  If after we mix them, we find by, for example, mass spectrometry, we find that there's an  O18 and 14 hooked to an N15, O16, then we know that they rearrange, that they randomly  assorted.  If, on the other hand, when we prepare this guy and we only find this guy and we only find  that guy in the mixture, that's all we find, then we know that they stayed put.  Basically, they didn't rearrange at all.  When we do this experiment with this particular molecule, we find that they rearrange all  the time.  In fact, sort of the presence of both the monomer and the dimer indicates that they have to  break, otherwise we wouldn't have both.  If you do the same experiment with carbon dioxide and you prepare some carbon dioxide  with carbon 13, O16, and then carbon 12, O18, and you mix those together at normal temperature  and pressure, there's no exchange.  You never find any oxygen moving off one and onto the other.  If you go to extremely high temperature, then you can make it happen, but that's extremely  high temperature where those bonds can actually break.  If you do the same thing with water in water, you find that the hydrogens hop around all  the time.  They fall off, they come on, they go every which way.  They're exchanging quite rapidly in liquid water.  Let's have a look at why this bond is so weak.  Actually the Lewis structure is kind of a good way to start and look at a bond.  Let's then write Lewis structures for NO2 and N204.  Remember what we have to do.  We have to count the valence electrons.  We have to know how the atoms are bonded because it says NO2.  That's going to be N in the middle and two oxygens sticking off like spokes.  N20, which is nitrous oxide, has N then N then O. It's different.  Let's draw them.  There's five valence electrons for nitrogen.  There's six for oxygen, but there's two oxygens, so I have 17.  This is very rare because whenever you have an odd number of electrons, that means that  you have one unpaired electron somewhere.  Those guys are called free radicals.  They usually cause mischief because they're quite reactive.  If you ever breathe any NO2 and I have done it by mistake, you never forget it because  your lungs tighten up and say that's enough of that and you feel like you need oxygen  for the next couple of hours, depending how concentrated it was.  Don't try to clean up any nitric acid because it won't work.  It will produce NO2 instead and the penny will disappear.  I tried to make an octet around each oxygen and I can do that, but in order to do it,  I have to have a minus charge on this oxygen because remember we count one electron for  the bond, then one, two, three, four, five, six.  We count seven electrons, but oxygen's group six.  This has a formal negative charge.  Here we count one electron here, one, two, three, four, five, six.  This has no formal charge, so we put nothing.  The nitrogen group five, we count one, two, three, four.  He's got a positive charge.  Then we can play the usual game of flipping this guy in and flipping this guy out like  playing dominoes and the negative charge moves back and forth.  The actual structure of NO2 is 50% this and 50% that.  It's one structure, but we can't write a proper Lewis structure for it like that.  With Lewis structures, when we have more than one resonance structure, what we're saying  is the real guy is a mule, but a mule is not a proper species because it's a mixture of  a horse and a donkey.  What we try to do to make it look like a mule is we flash a horse 50% of the time and a  donkey 50% of the time.  If we flash them very fast, we get kind of an average, and that way we can draw a proper  Lewis structure for each one.  This is not meant to mean that this is literally happening.  This is just meant to mean there's a percentage of each in the real structure, which is an  average, which we don't want to draw because then we have to draw dotted lines and funny  things and we don't know where the electrons are.  I think you can probably see that there's an electron here and an electron here.  I think you can guess how N2O4 is made and I'll let you work out N2O4.  Then I'll probably ask you about it on Tuesday to see if you did it.  I think you can see why it's a weak bond because when I make it, I have to put those two formal  positive charges together and those formal positive charges hate each other and they're  closed.  That bond is quite weak.  That means we can break it apart fairly easily.  It's an abnormally weak chemical bond that allows this thing to happen.  We can take a kinetic view of equilibrium and I'm going to go through quite a lot of mathematics  here just so you've seen it, not because I want you to reproduce it, but so that you've  seen it so when you see it again, it'll be familiar.  The chemical reactions can only happen when atoms or molecules are close enough to make  a bond.  That basically means the molecules have to collide.  They first have to hit.  We can figure out how often things hit in a gas.  They have to collide and if we look at our kinetic theory of gases, we know how many  are going how fast and we can figure out how many times they hit per second and in a gas  it is a very, very, very large number of hits at normal pressures.  We have the kinetic theory of gases and we can figure out how often the atoms or molecules  are going to collide.  If we heat things up, two things happen.  They're going to collide more often because everybody's going faster, so it's as if time  got shrunk and when they do collide more often, they're going to hit harder as well  so there's more chance that they hit so hard that something gets knocked off and then you  get a reaction initiated.  That's why most reactions speed up in both directions, not just one, but in both directions  when we heat things up.  We'll see this in the summer if you're going to continue on in this series in Chem 1c with  the Arrhenius equation.  Let's have a look.  Suppose that we are isomerizing cis-butene to trans-butene.  We're taking something like this and we're isomerizing it like that, which means we have  to formally break the double bond momentarily, slam into it, get this guy to twist and around.  We know from before that usually the transform when you've got two things sticking out is  slightly more favorable because when you're like this, these guys tend to repel each other  a little bit and it puts a little bit of extra energy there that gets released when we go  like that.  That doesn't mean it's easy to do, but it can happen.  If we have a forward rate constant, which we call little k for forward, of 5.59 times  10 to the minus 3 per second and the reverse reaction has a slower rate constant of 4.41  times 10 to the minus 3 per second and we begin with a mole of cis-butene in a one-liter  flask, what will be the concentration of the two isomers at equilibrium?  The forward reaction runs at a rate that depends on the product I meant of the rate constant  and the concentration of cis-butene.  It says however many I've got this many per second are going to tend to go this way, just  like a conveyor belt.  That makes trans-butene and then the trans-butene reacts backward but at a slower rate, more  sluggish and that reverse reaction depletes trans-butene.  And the one weapon that we have in this time view of equilibrium is that at equilibrium,  whatever the amount of trans-butene is has to be fixed because that's what equilibrium  means, the concentration's not changing.  And that means that the rate of gain plus the rate of loss must equal to zero.  So if the money in my bank account every month is the same, that means my income and what  I'm spending match.  That's the only way it can be the same.  Let me say that again.  The only way it can be the same is if the amount you get and the amount you spend match.  If they're even slightly off, you eventually go into a black hole.  If on the other hand you have a slight surplus, every month you end up in a white hole up  here and that's the correct way to operate, of course, much less stressful.  Now let's write the rate of gain.  The rate of gain depends on the forward plus how much cis-butene is there at equilibrium.  And the rate of loss depends on the reverse, but it's negative.  I write plus, but I write the rate of loss being a negative number is minus whatever  the rate of trans disappearing is times whatever the concentration is at equilibrium.  And if it is at equilibrium, that should be zero.  And if I solve this, I find that the ratio of the forward rate constant to the reverse  rate constant is nothing other than the equilibrium concentration of trans-butene, the product,  divided by the concentration of cis-butene.  And that means I can get a number for it.  So trans-over cis is forward, which is 5.59 times 10 to the minus 3 divided by 4.41 times  10 to the minus 3.  10 to the minus 3's go away and I get 1.27.  Aha, but the concentration of the products at equilibrium divided by the reactants at  equilibrium, that's none other than k.  And so we have a number for k.  K in this case is 1.27.  Not a very favorable reaction, but more favorable to make trans than cis.  This is a general conclusion.  Equilibrium, so I had to abbreviate it, can be viewed as a detailed balance between all  the ways that things are made and all the ways that things are destroyed.  And at equilibrium, they're being made just as fast as they're being destroyed.  It's not that nothing's happening, plenty's happening, but it's all a wash in the end.  If you change one of the rate constants, for example, you have two kinds of cells that  have to do with your bones, osteoblasts and osteoclasts.  And one of them tends to make your bones bigger and stronger and the other one tends to make  your bones disappear, get weaker and smaller.  If you go up into space and float around in the space station, or you try to go to Mars,  which wouldn't be a good idea, because there's no gravity, the osteoblasts say, guess what?  There's no reason on earth to pile on bone.  I'm not feeling anything here.  The osteoclasts, however, are always running.  And that's the rule, is that the machinery that's tearing you down is always running.  So that the minute you stop building yourself up, you go to a lower state.  That's again adaptive because you don't want to be trying to maintain gigantic bones and  muscles that you don't need.  Where are you going to find the stuff to eat to support all that if you're running around  in the wild?  So if you don't need it, you're going to just get rid of it, just like the brain.  And that's why when they come back from the space station, they're basically on a gurney  or in a wheelchair because they're so weak, their muscles didn't have anything to stand  on.  I don't weigh anything.  I'm sitting down all the time.  Plus, my blood doesn't even have to work to get back from my calves up to my heart.  So my heart turns into this little pathetic thing.  And I come back down to gravity.  And I feel absolutely miserable.  I'm weak, bones like balsa wood and so forth.  And I expect that if you sent a human to Mars, they wouldn't even be able to get out of the  spacecraft if they were the only person.  Because when we take people out, we wheel them around and rehab them so that they can  get back to normal function.  My view is we might send a robot to Mars, which is made out of metal and doesn't have  any active, degradatory mechanism.  And that would be fine because that's not a living thing in the strict sense.  So just like water, if we have a bunch of interconnected cisterns, we'll only be in equilibrium  when the water level's at the same in all of them.  So a system, no matter how complex it is, will only be at equilibrium when the rate  of production and the rate of destruction of all the molecules in there is equal.  Some of them may be destroyed and produced quickly.  Others may be destroyed and produced slowly.  It just depends.  And in general, we can write this as a differential equation.  And this is what I want you to see, but not necessarily digest today.  So let's look at the rate of change.  This is how the water level's changing.  Is it going up or is it going down?  What you read, the concentration of trans dt is just the rate of change.  And because I don't want to write concentration of trans everywhere, I'll just write that  as TR, and I'll write concentration of cis as C. Then we can get a differential equation.  The rate of change of the trans depends on the rate of production, kf times C, and the  rate of destruction minus kr, the reverse reaction times trans.  And we can simplify this equation because in this case, I said we started with one mole,  and so there's always one mole of material in there.  It's either cis or trans.  And that means we can substitute for C1 minus TR.  So now we have a differential equation that involves the derivative of TR with respect  to time and a bunch of stuff that has TR in it.  This is called an ordinary differential equation.  It's ordinary because this is a regular D and not the funny D. Yes?  So this would only be if you were going from the cis to the trans estimate rate.  Yeah, I'm talking about the rate of production of the product.  And it's made by the reactant making the product, and it's subtracted by the product  going backwards and making the reactant.  You can solve these with Mathematica or Wolff from Alpha.  It'll solve them for you.  But until you learn how to solve them yourself, at least a little, the solution may not mean  much.  It just looks like a bunch of gobbledygook.  So the problem is, often you can get the answer and you haven't learned anything.  And that's something to remember when you say what's the answer.  If somebody tells you the answer, you haven't learned anything.  If you get the answer yourself, you have learned something.  Let's put in the numbers that we have.  We have those rate constants, so we have now a numerical equation.  The rate of change of the concentration of trans with respect to time is equal to 5.59  times 10 to the minus 3, that's 3 times 1.  And then these guys I put together, minus 1 times 10 to the minus 2.  We solve this by separating variables.  If you can separate variables, you can solve the differential equation instantly.  And what separating variables means is get everything with the DTR and TR on one side,  and then get everything with T and DT on the other side.  This equation is easy because this side doesn't have anything that has T in it.  If I had another term here plus T squared or something awful like that, then that would  be much, much, much harder for us to solve.  Because now I've got something else going on.  It's as if not only am I trying to have it come to equilibrium, but I'm pushing one of  the pots up, I'm lifting it up, and then I'm asking, okay, what's the water level?  Then that's much more complicated to do.  We can do that too.  And when you're modeling weather or concentration of chemicals in the atmosphere, you have to  solve lots of differential equations like this, and some of them can be very, very tricky  to solve.  Usually they're done numerically by giant computers that don't care what the equation  is, they just integrate it the way a graphing calculator doesn't care what the function is,  it just graphs it.  Okay, so let's try separating variables.  I look at this equation, and I want everything with TR on the left and everything with T  or DT on the right.  And I can pretty much see how to do that.  I can just divide both sides by this guy on the right and multiply both sides by DT, and  I get this great big equation here.  DTR divided by 5.5 times 10 to the minus 3 minus 1 times 10 to the minus 2 TR equals  DT.  Good.  Now I can solve that because this side only depends on TR, so I can integrate it.  This side only depends on time, so I can integrate it.  That's why I have to separate the variables.  And so I put an integral on each side, and I say, okay, I'm going to integrate from the  concentration of trans at some time I'm going to call zero, which is when I start the experiment  maybe, or start the stopwatch.  And I'm going to integrate up to some concentration of trans at time T. And then over here I'm  going to integrate time from zero to T. Strictly speaking, this is sloppy notation because usually  the variable you're integrating over and the limits, you don't like to use the same letter  because sometimes it's confusing.  If I were really doing this correctly, I should call this DX and integrate it from zero to  T, but it doesn't matter because we can get the answer.  The right hand side is just T minus zero because the integral of DT is T. And I evaluate it  at the top limit and the bottom limit.  And I subtract them so I get T minus zero is T. That's this side.  And the right hand side, we can either hand over to integrals.com.  If we're lazy or we can make a substitution, we can call U equals the denominator, 5.59  times 10 minus 3 minus 0.01 TR.  And if I do that, which I'll do just for fun, we can set up the integral.  And this is what you would probably do if you were doing it by hand.  I have this new variable U. It's 5.59 times 10 minus 3 minus 0.01 TR.  And I take its derivative so I know what DU is.  And that the derivative of a constant is zero and the derivative of TR is just DTR.  I end up with minus 1 times 10 to the minus 2 DTR.  And then I solve this for DTR. DTR is minus 100 DU.  And at T equals zero, TR is TR zero.  And therefore, U at time zero is 5.59 times 10 to the minus 3 minus 1 times 10 to the  minus 2 TR at time zero.  And the upper limit is just the same thing, U at time T is this minus this times the concentration  of transit time T. And then I put all those into the equation so I get something I can  solve.  When I put all these in, this minus 1 and this 10 to the minus 2 comes outside.  And now I'm integrating DU over U and I know that's the natural log because now I have  it in a form I can recognize it.  The other side is T and that means the log, natural log of U at time T minus the natural  log of U at time zero is equal to, I brought this thing over here, minus 1 times 10 to  the minus 2 T. And I'll just remark as long as you make the variable X and you hand this  thing like that to integrals.com, it'll give you the answer straight away.  The variable has to be X for that particular website.  And we know when we subtract logs, it's the same as dividing the arguments so we can clean  this, tidy this up.  We now have this.  The natural log of U at time T divided by U at time zero is equal to minus 1 times 10  to the minus 2 times T.  And then if I take the exponential function to get rid of the log, I end up with the following  answer.  Our new variable U at time T is equal to our variable U at time zero times E to the minus  T over 100.  The reason why the exponential function is so important is that lots of differential  equations depend on how much stuff you've got.  Compound interest, for example, is an exponential equation.  You just plunk the money in and you leave it and you get 8% interest and you end up  extremely rich in 30 years because it keeps going up and up and up.  And as there's more money in the account, the interest is more.  And then there's even more and then the interest is even more, et cetera, et cetera.  On the other hand, if you borrow money, then you're going down like an exponential, especially  if you keep borrowing more and more money.  I'm amused that the credit card offers I get to get rid of my debt by transferring the  debt to someone else.  So you notice you didn't get rid of anything, but now you're paying the interest to them.  That's what they're interested in.  If you're a good risk, they want to get the 17% or whatever they're charging or even  29%, I guess, if you're unlucky and miss a payment and don't read the fine print.  Okay, then now we don't like this.  This is a nice equation, but it has u.  It doesn't have the concentration of trans in it.  So I have to go back and I have to substitute in the whole thing to obtain the final answer.  I put in all this stuff for u at time t and I put in all the other stuff for u at time  zero and now I have an equation where I have to solve now for transit time t, but basically  I'm there and I can predict now from any starting place how much trans I'm going to have at  any time.  And here's the final result.  The amount of transit time t is.559.  Not surprisingly, that's the equilibrium concentration and you notice it doesn't have any time on  it.  That means as time goes on, it's going to settle down to this number.  And then whatever the difference between where I start and the equilibrium amount and then  multiply by e to the minus t over 100, when you multiply anything by e to a negative exponential,  it gets smaller and smaller and smaller.  So this part here dies out and I end up with the equilibrium concentration.  We had an assumption here that we started at 1.0, but that was my first initial thing  where I got rid of c.  If now that we have more confidence, we could certainly go back and we could say, look, let  the total amount of stuff be c naught or anything and we could do it again and then we'd have  a completely general answer to this particular problem.  Okay, just to emphasize, will you have to do this?  No, not now and certainly not under the pressure of an exam.  However, yes, eventually you will have to do things like this and much, much harder things  because if you can't do things like this and much, much harder things, then you're down  there with everybody else who also can't do that.  If you can do something that other people can't do, you tend to get a job that they  can't get.  That's how that works.  They have requirements.  When you do this kind of computer programming, are you familiar with this process?  Have you ever done this and so forth?  If you have, you're a good candidate.  Even if you do use modern software, you can get mixed up if you don't understand how the  equations are behaving and what you're supposed to be trying to get.  Most people view mathematics as something scary or difficult, but in fact, mathematics  is the quickest, cleanest shortcut.  That's why we make mathematical models of the universe because mathematics problems  are much easier to solve than vague problems and words.  Whenever somebody says something about the future in words and they don't have any mathematics  behind it, then you have to ask how they're going to get it to happen.  I will create 250,000 jobs.  How?  What's the equations?  Show me the curve that goes up and levels off at 250,000.  Why'd you pick that number?  Why isn't it 253,000 or 225,000?  What exactly are you going to do to create jobs out of thin air if there's no work to  be done or no people who can do the work that's required?  Gets much, much more complicated than to implement something like that.  Let's plot our results.  Once we've got an equation, we can see exactly how many molecules we're going to create.  Let's suppose we start with no trans at all.  It's all 100% cis.  How long does it take to get to equilibrium?  Well, we've got an exact solution, so we plot it.  So I started with 100% 1.0 of cis and 0.0 of trans, and I just put those into the equations  and I plotted it versus time here in seconds.  I can see that after about 300 seconds or so, it settled down.  That lets me know how long I have to wait in the lab before I make a measurement.  If I want to know the equilibrium concentration as a function of temperature or pressure for  this system if these rate constants are correct, then I have to change the temperature and  then I have to go get a cup of coffee because I have to wait 300 seconds at least before  I can know that it isn't moving around anymore, settle down.  Chemical engineers really like to know exactly how long they have to wait.  They're bunging in chemicals and they want something to come out.  How long do I have to treat the wastewater?  How long does it have to cycle around in there before it's clean enough or it's as clean  as it's ever going to get?  And then I get rid of it.  It's very important to know these kinds of things.  Sometimes the only way you can know is by careful measurement because you don't know  these rate constants very well.  If we start with no cis isomer, about how long is it?  So now I've got the opposite.  I start with 100% trans and some of it starts immediately going backwards because Q is bigger  than K so it starts making cis and it takes about the same amount of time, about 300 seconds  in, it's leveled off.  And I would get the same conclusion no matter what I started with because these are the two  extremes.  If I'm lucky and I started with the equilibrium concentration, they both flat line all the  way across.  So here's the conclusion.  The time it takes to establish equilibrium depends on the size of the individual rate  constants more than the initial conditions.  If the rate constants are fast, that means it's going to settle down very, very quickly.  It means instantly I'll have equilibrium.  If the rate constants are very slow, I'm going to have to wait forever to let equilibrium  establish.  On the other hand, the final concentrations don't depend on whether the rate constants  are both big or both small.  The final concentrations only depend on the ratio of the rate constants.  How fast is this guy shoveling stuff in and how fast is the other guy shoveling stuff  back?  That ratio is going to tell how much I've got at equilibrium.  And these same conclusions apply to processes like biological fermentation, which they're  using to produce things like fuel and drugs.  I'm growing yeast.  I want to heat it up maybe to get the reaction to go better, growth, and other reactions that  we'd like to be able to control.  The equilibrium result of a bunch of fresh raspberries is the raspberries spoiled.  They mold.  They start smelling bad and so forth.  And if you leave them out on the counter, that happens very quickly.  If you put them in the refrigerator, the same thing happens.  But you usually don't leave them there so long that you notice it happening.  It just happens much, much more slowly.  If you take food and put it in a freezer, it also spoils, but it spoils much, much more  slowly.  If you put it in a deep freezer or liquid nitrogen, it spoils very, very, very, very slowly.  And so it keeps a very long time.  And for each 10 degrees Fahrenheit you go down, you can keep the food in there a lot  longer.  Assuming it can tolerate being frozen, sometimes if you freeze things, they change.  So frozen raspberries aren't the same as fresh because of the ice crystals rupturing things  inside the fruit.  Okay.  Now let's talk about something which we will definitely have on Tuesday, which is we're  going to add reactions.  And we know that when we add reactions, any state function, big H, any big capital letter,  big S, big G, they all add.  Let's then talk about, suppose we have this reaction.  This reaction in the 60s was a big problem because coal has a lot of other stuff in it,  in particular sulfur.  And high sulfur coal is a lot cheaper than low sulfur coal because of this reaction, basically.  If we burn the coal to produce power and we have sulfur in it, then we also make SO2.  So just like the C in the coal, the carbon makes CO2, the sulfur makes SO2.  And if delta G for that reaction is delta G1, and I have another reaction where I take  SO2 and it goes up into the air and it reacts over time with oxygen in the air out of the  smokestack, then it makes SO3.  SO3 is called sulfuric acid anhydride.  If I add water to it, I get sulfuric acid.  When a lot of coal plants in Ohio were burning a lot of coal and there were no emissions  controls, the SO3 would drift over Canada, up north, and then mix with water and rain  down on all their nice forests.  And that was the so-called acid rain problem.  The trees would defoliate because the pH of the ground was changing.  And also, if you just spray dilute sulfuric acid onto the leaves of plants or pine needles,  they don't like it.  It's too acidic for them.  So the trees wouldn't grow.  And in Canada, of course, they're doing a lot of lumbering.  They have a lot of people cutting down trees, turning them into planks and selling them.  And if their trees don't grow, that business is dead.  And so there was quite a tug of war.  You've got to stop burning this stuff.  That's your problem.  But that drifts up there.  You find another kind of tree.  Well, eventually they said, well, OK, we'll stop this SO2 going up.  But that cost a ton of money.  Now you have to put something on the smokestack that captures the SO2.  And then you have to take it out.  And you have to keep it clean.  And that costs a lot of money.  And where am I going to put it?  What am I going to do with it?  Can I sell this material for any purpose?  Or is it just junk that's dangerous and toxic that I've got to somehow figure out what I'm  going to do with it?  It depends.  Well, if we add those reactions up, and we just imagine doing it in one step, then the  SO2s go away here.  And I have a net reaction, sulfur, elemental sulfur.  And 3 halves of molecular oxygen gives SO3.  Delta G for this reaction, whatever it is, is just the sum of this one and the sum of  this one.  We just add them up.  If we turn a reaction around, we change the sign of delta G because we changed the initial  and final state.  And if we carry out these two reactions at the same temperature, then I can multiply by  minus 1 and divide by RT.  And I have, instead of this equation, minus delta G over RT is equal to, I'm sorry, minus  delta G1 over RT plus minus delta G2 over RT.  And that's of interest because that means since delta G over RT, minus delta G over  RT is log K, that means that log K adds up.  So log K for the total reaction is log K1 plus log K2.  But I know how logs work.  I can always combine them.  If I add them, I multiply them and take the log of the total.  If I divide them, I subtract them.  And then if the log of K is equal to the log of K1, K2, that means that K is equal to K1,  K2.  Therefore if we add two chemical reactions, we multiply the equilibrium constants.  Let me say that one more time.  If we add two chemical reactions, we multiply the equilibrium constants for the reaction.  The reason why this is not just academic is because one of these reactions might be unfavorable.  It might be one we want to run, but it's not going anywhere because K is tiny.  But if we couple it to another reaction that is favorable, then we can drive the whole  log to go to where we want.  And that's what chemists always do.  Chemists run a ton of unfavorable reactions as do all living things.  They assemble these gigantic molecules that are very, very low entropy, and they do so  by running a ton of other reactions that are favorable, like burning up glucose with oxygen  to make CO2.  That's a favorable reaction.  So as long as I can eat and I can run that favorable reaction, then I can run a lot of  unfavorable reactions like get smarter.  But if I can't eat anything, I can't get any smarter.  And if I eat too much, I won't get smart either because then the blood ends up in your stomach  and your brain needs blood to work.  So before the exam, you don't just eat a ton of stuff because then you'll fall asleep  in the exam.  That's the natural result.  You look at a lion that's hungry.  How does it look?  Right?  Right alert.  That's the way you want to be for an exam.  What's the problem?  You look at a lion that's just finished a giraffe.  It's a dead, right?  It's asleep.  It's digesting all that food, all the bloods in its stomach, doing all that biochemistry.  You only want to do one thing at a time.  So you come into the exam hungry, and then after the exam, you eat and relax.  But don't get the order wrong because if you do, you won't do as well.  Okay, let's finish up by talking about Le Chatelier's principle.  And Le Chatelier's principle is just that if we have a system at equilibrium and we change  something, how will the equilibrium shift?  Or will the equilibrium shift at all?  While Le Chatelier worked this out and made the following general statement, the equilibrium  will always shift in the direction that tends to counteract whatever the change is.  For example, and we worked this out in some detail, if we increase the pressure in the  Haber reaction to make ammonia, then we saw that we made a lot more ammonia, the right  hand side, because making more ammonia minimizes the number of moles of gas and counteracts  the increased pressure.  When I squeeze on a chemical system, it tries to get smaller if it can.  And if it can get smaller, it does.  And in this case, that's our lever to get more product out of the reaction.  Okay, so let's consider a reaction to synthesize Fos genes.  Don't try this at home.  Very, very, very reactive guy, this guy.  Let's take molecular chlorine and carbon monoxide and make Fos gene, COCl2.  It has a carbon, double bond oxygen, and two chlorine sticking out like a triangle.  And let's consider the effect of the following changes.  Let's suppose we have the reaction at equilibrium.  That'll be the effect of each of the following on the concentration of carbon monoxide.  One, I inject additional chlorine.  Two, I add argon gas and I make the total pressure higher.  Total pressure is higher because I've pumped in argon.  But argon doesn't react with the other guys.  Three, I just compress the mixture by increasing the pressure.  Or four, I decompress the mixture by lowering the pressure.  Okay, let's go through them then one by one.  First we'll start with injecting additional chlorine.  If I add additional chlorine, I'm going to increase the denominator in Q because Q has  the concentration of Fos gene and in the denominator the concentration of chlorine times the concentration  of carbon monoxide.  Q is K at equilibrium and K doesn't change.  Therefore if I add more chlorine in the denominator, I'm going to have to get less CO.  Otherwise it's not going to work and I'm going to get more product.  And this is another clue.  Suppose I have two reactants.  One of them is extremely expensive and the other one is cheap as dirt.  And I want to use up the expensive reagent so I don't waste it.  Then what I do is I just pile in the dirt until I drive it over to the other side.  And if the dirt cheap reagent is a gas, it's great because I can pile it in, get the product,  isolate the product, and then vent the gas off, assuming it's harmless.  That's another trick.  Two, adding a dirt gas does not shift the equilibrium.  Why not?  I just said if we increase the pressure, we shift the equilibrium if the number of moles  of gas of products and reactants is different.  And now I'm saying I'm increasing the pressure but nothing's happening.  The answer is argon is not in Q, the reaction quotient.  If I don't change the reaction quotient, I can't change any of the concentrations of  the things.  So although the total pressure in the container is higher because I didn't actually change  the reaction quotient, I just added something else that doesn't react.  Nothing changes.  So there's no change in equilibrium if I add an inert gas.  If on the other hand I squeeze on it like this and compress it, then I'm going to get  more product because there's again Q, if I work it out with the moles of gas, there's  going to be a P standard over P left over.  And so if I increase the pressure, I'm going to decrease the number of moles of CO.  And so not surprisingly, if I expand it, then I'm going to shift it and get more CO because  some of this phosgene will actually react backwards and make both CO and CO2.  And I'll get more CO.  And that's how we keep track of it.  So here's the summary.  If I change the concentration of something in the chemical equation, not something else,  but something that participates in Q, the reaction quotient, then I'm going to shift  the equilibrium, but I'm not going to change the equilibrium constant.  If I put pressure on a gas, I'm going to shift the equilibrium potentially if the number  of moles of gas is different, but I'm not going to change the equilibrium constant.  If I change the volume, I'm going to shift the equilibrium, but not change the equilibrium  constant.  If I shift the temperature, ah, log K minus delta G over RT, I know then that the equilibrium  is going to shift and the equilibrium constant is going to shift.  And if I add a catalyst to make the equilibrium come more quickly, a catalyst, strictly speaking,  this is never true, but a catalyst, a true catalyst is unchanged.  So if it starts on the reactant side, catalyst plus blah, blah, blah, blah, blah, and then  on the product side is products plus the same catalyst, the catalyst cancels out because  whatever its concentration is, it's in the numerator and the denominator.  So if it's a true catalyst, it can't shift the equilibrium.  It just makes the equilibrium happen more quickly.  Really what happens is the catalyst gets dinged over time because our reagents aren't 100%  pure and if we've got a tiny, tiny amount of some bad guy in there, it wrecks the catalyst.  That's why the oil companies had to get rid of all the sulfur in the gasoline because  it was reacting with the platinum and palladium and the catalytic converter and making it  fail.  And there was a big battle.  You guys make a better catalytic converter.  You guys get the sulfur out of the gas.  We can't make a better catalytic converter.  So usually the oil companies and the auto companies are on the same page, but not always.  Whenever one's going to have to try to do something very hard that's going to cost a  ton of money, then they try to say, well, the other guy should do something very hard  that costs a ton of money.  Instead, the result of that is that we have gas that doesn't have as much sulfur in it.  Okay.  We'll finish up here with a few subscripts on K. There's lots of little subscripts on  K and they're all Ks.  They're all ratios of products and reactants and they're just meant to be descriptive.  So if I have K sub A, that means this reaction.  That means an acid dissociates to give H plus and the anion.  And here's the formula for K. And if I have KB, that refers to a base.  I have a base.  It pulls a proton off water and makes hydroxide.  I could have a neutral here like ammonia plus water gives ammonia and myon plus hydroxide.  That's why they tend to put ammonia into things like Windex.  Okay.  Because it gives the right amount.  You say, why don't they put Drano in Windex and make hydroxide that way?  And the answer is it will leave a residue on the glass, number one, because it won't  evaporate like ammonia.  And number two, it's usually too strong.  And if I make it too basic, I'll actually start affecting the glass.  I'll actually start polishing away the glass so my glass window is getting thinner and  thinner as I keep cleaning it over time.  And that's a bad strategy to do that.  Okay.  So at this time, what we'll do is work some problems, okay, which will be good ones to  know.
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