Lec. 16. Equilibrium Constants: Temperature and Pressure
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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.
 I heard that the world is coming to an end this year, so I guess all our problems are over. Notice how easy it is when the world is coming to an end. You can forget about long term planning. Party on. Of course all of our worlds do come to an end and sometimes unpredictably. So it's kind of important from time to time to imagine what you would do if you were going to die today for sure. Who would you call? What grudges would you drop? What would you say to your parents? Your friends? Anybody who's had an influence on your life. And then my advice would be to go ahead and say some of that even though you don't think you're going to die. There's no point in waiting. Many people wait too long and then the other person dies and you never get a chance to tell them how much you appreciated something they did because you kept putting it off. So don't put things off forever. We all have a shelf life. Unfortunately we're thermodynamically unstable. And that means we're doomed. Although hopefully not too quickly. But eventually we are doomed. But for now let's imagine we're going to live long enough to take the final. Get a grave. That type of thing. And so it would be a good idea to plan for that. And with that in mind let's get on with things. Let's talk about equilibrium constants. I'll pick up where I left off, we'll put the problem up again. And we were actually trying to figure out if we knew the equilibrium constant for the synthesis of ammonia at room temperature based on the tables. Could we figure out what the yield would be at some higher temperature where we have to run the reaction? And the answer is we can do that for sure but we have to make some approximations unless we make some measurements. Here's what we have. We had delta G of formation of ammonia is minus 16.6 kilojoules per mole. Negative. What would you say if somebody came to you and said, hey, I found something interesting in the table. Here's a molecule that has a positive delta G of formation. Most of them are negative but there are a few in the table. I'll give you an example, hydrazine N2H4, methylhydrazine I think is used in rocket fuels sometimes. If you were given that information what would you say? Since the molecule exists. Have you got any ideas? I'll tell you what I would say. Number one, although hydrazine has nitrogen and hydrogen like ammonia, I can't make hydrazine and high yield from nitrogen and hydrogen because it has a positive delta G so it has a tiny K. I could probably make a little bit of it but not much. Second, hydrazine must be kinetically stable like a human being. Hydrazine must be doomed eventually to fall apart at least into hydrogen and nitrogen because it's unstable with respect to the elements it was made from. So it's going to fall downhill eventually. Anything like that usually you want to handle extremely carefully. Those are a lot of things like explosives. They have a big delta G and then boom when they detonate you get them going to products with a lot of energy. But ammonia doesn't have that problem and we had figured out that the balanced reaction for one mole of ammonia because delta G refers to one mole of ammonia is one half molecular nitrogen plus three halfs molecular hydrogen gives one mole of ammonia. If I double a reaction, I double delta G. That's very important. When delta G a formation is listed it's per mole. If I make two moles of ammonia delta G is twice as big just like two kilograms of ammonia is twice the mass of one kilogram of ammonia. Delta G behaves that way. We have a formation, excuse me, an equation that relates K to delta G and there it is. We just put in the numbers and we make sure we have the right units. We get K is E to the plus 6.5963 and that's about 730. That's favorable. That means we're going to get mostly ammonia. Now at the other temperatures we need to make some assumptions. First we have to make the assumption because we don't know the heat capacities of the products or reactants that delta H and delta S don't depend themselves too much on temperature. The entropy is different than the enthalpy because there's no entropy of formation. The entropy is referenced to absolute zero temperature and any kind of molecule or atom at finite temperature has some disorder. It's moving, it's tumbling around, energy is distributed in the molecule in different ways. The entropy says the energy is going to randomize to the maximum extent possible because that's simply the most likely thing that's going to happen. If we use our equation for the state function of delta G, delta G equals delta H minus t delta S and the relationship between log of K and delta G, we can set this up at two different temperatures and then find the value of log K for the second temperature if we know the value of log K at the first temperature. That's how we have to do it. Here we go. Each temperature, I'm going to write log and I'm going to put in parentheses the temperature I'm calling it just to keep track. Log of K at temperature T1 is equal to minus delta G standard at temperature T1 over RT1 and then I'm going to put in delta H and delta S. I'm going to do the same thing for temperature T2 except I'm going to put in T2. Then since this is equal to this and this is equal to this, then log K T1 minus log K T2 is equal to the right hand side here minus the other right hand side. I've written that all out here. Log K at temperature T1 minus log K at temperature T2 factored out R minus 1 over R delta H at T1 over T1 minus delta H over T2 over T2 plus, does this add a minus up here, 1 over R times delta S at T1 minus delta S at T2. Now we have to make an assumption. The assumption we make is that delta H and delta S themselves don't depend much on temperature. Whereas we know that's not quite true but we're going to assume it's true because we don't know the heat capacities. We assume delta H at T1 is equal to delta H at T2 and we'll just call that delta H and the same for delta S then conveniently because this delta S and this delta S are the same, this term goes away. So we have the natural log of K at temperature T1 minus natural log K at temperature T2 is now equal to minus delta H over R times 1 over T1 minus 1 over T2. This not surprisingly is very, very similar to the Clausius-Clapeyron equation. The only difference is that we had the vapor pressure. Well it turns out that if I have an equilibrium between a pure liquid and a vapor pressure at some pressure, the equilibrium constant is just the vapor pressure. So this is Clausius-Clapeyron equations, just a special case of this more powerful equation. And then because I'm subtracting the logs I can tidy this up and say the natural log of K at temperature T1 divided by K at temperature T2, the natural log of that whole thing is equal to this. And therefore what I need to know in addition to delta G at one temperature so I can get log K is I need to know delta H for the reaction at some temperature and if I assume it doesn't depend on temperature then 298 will do. So I look that up and I recall that delta H of formation for ammonia for one mole is minus 46,300. It's exothermic joules per mole. So I put that in. Log of K at temperature T1 divided by K at T2 minus 46,300 over R in joules 1 over 298 minus 1 over 598. I put the 0.15 but I didn't want to have this drop over a line because it makes it hard to read. It didn't fit and I don't want to make it tiny because it's hard to read. But when I did the calculation I put 298.15 and I find that log of K at temperature T1 divided by K at T2 is equal to plus 9.367 and T1 was 298 because I'm keeping things in order. While we know what K at temperature T1 is, so K at temperature T1 divided by K at temperature T2, the exponential function of the log is just the thing and the exponential function of this plus 9 is e to the plus 9.367 and if I solve that using K of T1 is 730 I get K of T2 is equal to 730 times e to the minus 9.367 and so I get a measly 6.24 times 10 to the minus 2. So you can see that my yield went from wildly favorable, like 730 to 6 percent per K by cranking the temperature up. That's how a lot of things work and that's why in part cranking the temperature up on the planet just a little bit can shift the equilibrium concentrations of things like a giant teeter totter with unimaginable consequences. Or we can imagine them but we wouldn't want to live through them necessarily. It's not just that it'll be hotter, it'll be totally different too and sometimes you don't know all the equilibria going on in the ocean and so you get surprised. Okay, at the highest temperature it's less favorable so if we go up to 798 that's where they run it remember about 500 Celsius. Then we find with delta H minus 46,300 joules per mole we do the same thing and now we find that log K at T1 divided by K at T2 is equal to plus 11.7 and the bad thing is that it's an exponential function and so now when I multiply 730 by e to the minus 11.7 I get less than a percent I get 6.05 times 10 to the minus 3. I'll just remark here that the assumption that we've made that delta H and delta S themselves don't change much with temperature is not going to be accurate when you change the temperature hundreds and hundreds of degrees or a thousand degrees. That's too far off for that assumption to be any good but if you change a reaction from 25 Celsius to 35 Celsius that's probably going to be a very good assumption and sometimes if you're in the lab and you're optimizing a reaction if you just heat it or just cool it a little bit more use a different solvent that has a higher boiling point when you reflux it you suddenly get much higher yield because of this exponential function you can go from 50% to 99% pretty readily and that's just much much more useful than if you have a reaction that does that kind of thing. Okay, some observations. The equilibrium constant depends on the exponential function of the free energy. Therefore small differences in free energy lead to large changes in the equilibrium constant. That's why for a lot of reactions if I see that delta G is some big number of kilojoules per mole and negative I can just pretty much assume 100% yield because of this exponential function. Likewise temperature because it's in the denominator of the exponential can shift the equilibrium greatly as we saw in the last problem. Now let's try to shift the equilibrium back in our favor. We were carrying this out at pressure of one atmosphere and we weren't really asking how much ammonia we got versus how much hydrogen we started with. Since the most expensive reagent in the synthesis, nitrogen is pretty much free from the air. So we'll do it with respect to the hydrogen. But first let's just look at this shifting. So let's just take a hypothetical reaction. A is in equilibrium with B. But A and B have different free energies. The question is what should the difference in free energy be to get 90% yield of B? What should it be to get 99% yield and what should it be to get 99.9% yield which would be essentially considered quantitative? And it turns out here it works in our favor. So let's assume we run the reaction at 298 and let's just take delta G of this reaction A goes to B, plug in 298 and figure out what we need. Well we want to get 90% of B at equilibrium and that means we have 10% of A and so 0.9 over 0.1 is 9. That's K, K is 9. And therefore if we take the natural log, if we solve this, excuse, yeah, if we take the natural log and solve this for delta G, we get delta G for the reaction should be minus RT log K, K is 9. We put in R in joules. We put in 298.15 Kelvin and we put in the log of 9 and we find that we only need minus 5.45 kilojoules per mole, which is hardly anything. For 99% yield, the only difference now is we have 99% of B at equilibrium and that means we have 1% of A left over, so 0.99 over 0.01 is 99. The log of 99 is fortunately, by the way, logs work. They never get that big, even if the numbers are big. We get delta G for this case is minus RT log 99. It only goes to minus 11.39 kilojoules per mole. So it's just twice. And for 99.9, B at equilibrium is 0.999, A at equilibrium is 0.001, so K is 999, but again, log 999 doesn't get that big and when I multiply it by minus RT, I get negative 17.12 kilojoules per mole. For typical reactions that you're likely to pencil out that involve burning a hydrocarbon and getting CO2 and H2O, delta G for that reaction is going to be huge and negative. And so we can pretty much consider that there's nothing left if, in fact, the material is pure. What's usually left when you burn tons and tons of stuff is the small impurities that aren't written in your chemical equation, sulfur, mercury, cadmium, other things that don't burn uranium, thorium, radon, other things, and all those come out in the fly ash if you burn coal. And if you want to burn it cleanly, then you have to figure out what you're going to do with all those things. For the longest time, we didn't care and we let all the mercury in the fly ash flow into the ocean and that's why fish may have higher mercury content. Now I think fish always have some mercury content, but I don't think we helped that much. Okay. Now let's go back to the Haber process and let's try to squeeze the product. We're forced to heat it up so we know K is small, but we aren't forced to run an atmospheric pressure. And because there's more moles of gas in the reactants than there are in the products, I made a comment without saying why, that we could put pressure on it and force the reactants to go over to products. Now we're going to do a calculation to show how that actually works. Let's take our same reaction and let's figure out what the equilibrium constant Kp is in terms of partial pressure. The reason I'm using Kp is because all these are gases and that's the natural thing to use. And even though these are non-integer stoichiometric coefficients, when I write the equilibrium constant, I always raise the concentration to the stoichiometric power. If it's one-half, then it's the square root of the concentration. If it's three-halves, it's the concentration of the three-halves and so forth. It's whatever the actual number is. Doesn't matter if it's not integer. And of course, non-integer powers don't scare us anymore because we are not going to ever try to solve the equation analytically. We're just going to guess the answer. And we're so good at guessing by this point that we're going to get it right. So let's have a look. This looks a little bit messy, but we'll go through it step by step by step. Kp, the definition of Kp, is its products, partial pressure of products divided by the standard pressure because Kp has no units. Divided by the partial pressure of hydrogen divided by the standard pressure raised to the three-halves power because it had a three-halves in the chemical equation. Multiplied by the partial pressure of nitrogen divided by the standard and raised to the one-half power. And then I can use Dalton's law. Dalton's law says the partial pressure of an ideal gas is equal to the mole fraction times the total pressure. And here I'm using p to be the total pressure. Therefore if I look on the diagram and it says p is 300 atmospheres, then that's the p I'm going to put in there for the total pressure. And likewise I can do the same thing here, mole fraction of hydrogen times the total pressure and the mole fraction of nitrogen times the total pressure raised to the powers. And then I'm going to pull out the mole fractions and separate out the mole fraction part from the p part. Now I've got this leading term which is all the mole fractions which is going to be what I'm going to call the yield. And then I've got all these pressure terms which I don't care so much about. But it seems like the wind's blowing from behind because pressure to the three-halves times pressure to the one-half, I add the x-bonus, that's just pressure squared. I have p over p naught squared on the bottom, p over p naught on the top. If I simplify that, I end up with p naught over p. Now we can sort of see how it works. When we change the pressure, Kp doesn't change because Kp is related to delta G standard. And delta G standard is at the standard pressure of one atmosphere. Therefore when I change the pressure, Kp doesn't change. When I change the temperature, Kp does change. All things can change things. But here's the thing. If Kp doesn't change, let's suppose we've got our miserable yield of 0.6% here for K. And now I crank this guy up. This guy was 1 before. But I crank p up to some huge number like 300. So this is now 1 over 300. Well that means that this thing here with the x's has to get bigger because p made this term here get smaller. And the only way this guy can get bigger is if the mole fraction of ammonia increases like crazy. But that's what I just said is that because I've got more gases as reactants, if I squeeze down on it, I'm going to make much more products. So these guys will go down and this guy will go up. And the harder I push on it, the higher yield I get. Why don't I just push on it to 10,000 atmospheres then? The answer is that costs too much money. Because if I want to squeeze that hard, I've got to make a chemical plant where all the pipes are that thick. So they can take that pressure and that costs me a ton of money. And therefore it's more economical for me to recycle the reactants that I can't quite get into products, bleed off the ammonia by condensing it and cooling it, and then just go around again at the same pressure. That's going to be more economical than trying to run one reaction and get it to go all the way. That's just economics of designing a plant. And economics are important because if the fertilizer costs 10 times as much because I'm doing something in a crazy way, then I can't afford to eat. OK. So let's actually do this calculation. This is going to be very long, unfortunately, so I hope you can follow it. But we'll do it step by step by step. Here's the problem. What pressure would it take at 598, the intermediate temperature, to boost the yield of ammonia compared to the residual hydrogen by 100 times compared to whatever yield we got at atmospheric pressure? So problem 54, we calculated K, but we didn't calculate the equilibrium concentrations. And we have to assume that the reaction reaches equilibrium. In a real chemical process, people don't really care if it reaches equilibrium. They just care if they get a pretty good yield of product, especially if they're going to recycle the reactants. And let's assume that we start with a 1 to 3 nitrogen to hydrogen ratio, the stoichiometric ratio for ammonia. That'll just simplify our calculations. But of course, again, when you're running a plant, you may not have that condition. And you may actually try to game the system a little bit by putting in much more concentration of something that's a cheaper reagent to drive the equilibrium to the right. So you may actually use much more nitrogen than hydrogen as you go to try to convert all the hydrogen over to ammonia. I don't actually know how they do it in practice. OK, let's first figure out what the ratio of the mole fractions of ammonia to hydrogen would be at one atmosphere. That's going to take us a little bit of calculation. And then next, let's increase whatever this ratio turns out to be by a factor of 100 and then figure out what P naught over P should be. P naught's one atmosphere, so that'll tell us what P should be in atmospheres. We're talking to a chemical engineer. We may convert it to PSI so they know that they may have things in terms of pounds per square inch because that's how a lot of things are quoted. We have to proceed step by step. We use the partial pressures, then Dalton's law, and to convert to mole fractions. And we have to be careful to include the powers of 3 half and 1 half. And I've just rewritten that here so that we can see it again. We have k at 598. We got that from before. And after everything drops out, we have the mole fraction of ammonia in the numerator and in the denominator, we have the mole fraction of nitrogen to the 1 half power and the mole fraction of hydrogen to the 3 half power. And then we have this term left over, P standard over P. If we have the same number of moles of gas on both sides, it's pretty easy to see that the powers of P here are going to be the same on the top and bottom. And in that case, if you have the same number of moles of gas on both sides, squeezing it won't do anything much because it won't change it. And so therefore, you don't squeeze it under those conditions. We're lucky that we're taking more moles of gas when we make ammonia because that gives us a lever to push on. And let's, for the sake of argument, start with 1 half times some number. It turns out it won't matter what it is, but I'll call it n naught. Let's start with 1 half n naught moles of nitrogen. And we have three times as much hydrogen. Therefore we have 3 halves n naught moles of hydrogen initially before we actually conduct the reaction. So we have no nH3. This is our initial condition. The number of moles of nitrogen starting out is 3 halves n naught. The number of moles of hydrogen is 1 half n naught. And the number of moles of ammonia is zero. And now we let a fraction alpha of the reactants react. And for some reason, which I don't actually know myself, whenever chemists deal with gases, the fraction that reacts is alpha, and whenever they deal with solutions, the molarity that reacts is x. But anyway, it doesn't matter what you call it. It's an unknown, and we're going to try to solve for the fraction alpha that gives products based on the value of k and a very, very messy equation which would intimidate everybody if they thought they had to solve it exactly because it could be potentially quite different to, quite difficult to solve. Now this fraction alpha is a number we're going to try to get. It has no units, and it's between zero and one. If alpha is one, then all of it reacts to ammonia. That's 100% yield. If alpha is zero, that's what we started with here. And that's what I've written. Whenever I set up an equation like this, I always put in alpha or x or whatever it is, the range it can be because I want to know where I should guess. In chemistry, all these things are positive numbers, and they aren't imaginary numbers or anything else funny, and that helps you a lot. If you have to guess in two dimensions, it gets very, very much harder to do. You may need a computer to do that. And I see that it all works out. It makes sense when alpha is equal to one, there's no reactants left over. It's all products. When alpha is zero, there's no products. It's all reactants. And I see that they're going away in the right way. And that's because I put the three halves and the one half here and not some crazy thing here trying to multiply that by three halves, which is a very, very confusing and bad strategy. So let's figure out the partial pressure, or let's figure out the mole fraction so we can get the partial pressure. To figure out the mole fraction, we have to add up the total number of moles of reactants and products. We now know if a fraction alpha reacts how many moles of nitrogen, hydrogen, and ammonia we've got, now let's add them up. So we add up the number of moles of nitrogen plus the number of moles of hydrogen plus the number of moles of ammonia, the three halves and the one half add up to two. The one minus alpha was in both of these guys. So I get two times quantity one minus alpha times n naught. And this guy here was alpha times n naught. And if I expand this out, I get two minus two alpha plus alpha, so I get two minus alpha times n naught. That's the total number of moles. When alpha is zero, there's two moles because there's three half moles of hydrogen and one half mole of nitrogen. When alpha is one, there's one mole because there's one mole of ammonia. So again, I check, put in zero and one, and I make sure it makes sense that I haven't done some sort of logical error in setting it up. And then I've got this equation here for the total number of moles. The mole fraction is just the number of moles of each of these guys divided by the total number of moles. And I just very patiently work it out. I don't try to shortcut it. But for the mole fraction of nitrogen, I get three halves times one minus alpha times n naught divided by two minus alpha times n naught. And you see n naught goes away. It doesn't matter because we're talking mole fractions, so it has to go away. And therefore, the mole fraction of nitrogen is three halves times one minus alpha divided by two minus alpha. And I don't sweat about that any further at this point. And just leave it like that. Then I move on to hydrogen. Sorry. I'll fix this. This was meant to be hydrogen because hydrogen has the three halves. I'll fix this before I post these slides. This is meant to be the mole fraction of hydrogen, not nitrogen. And likewise, this one is also backwards. But fortunately, it won't matter since they're multiplied by each other. So I think the rest of the calculation's OK. And for hydrogen, we should have a three halves instead of a one half, but it's basically the same thing. We get a number and we get one minus alpha divided by two minus alpha. And then the mole fraction of ammonia is, again, the number of moles of ammonia divided by the total. And both of them depend on how far the reaction goes. The top part is alpha n naught. The bottom part is two minus alpha times n naught. It's an i of alpha over two minus alpha. And the thing I want to emphasize is that this is a very simple reaction. And yet I could potentially have alpha to the fifth power here to solve. And that's why I don't advise trying to solve things analytically. There is a formula for a quadratic for the roots. A x squared plus B x plus C. There's a formula. There's a formula for a cubic. It's pretty big. There's a formula for a quadratic. A x to the fourth plus B x cubed plus C x squared plus D x. That formula is 28 pages long. Says that here are the four roots. X1 equals ABC raised to all these power. And then amazingly for a Quintic equation, it was shown quite a long time ago that there is no formula for a general Quintic equation. It's not that we can't figure it out. It's that it doesn't exist in terms of raising things to a power, multiplying things, adding, subtracting, and dividing. And those are the weapons we have in arithmetic. The fact that it suddenly at five becomes different is very mysterious to a mathematician and fascinating. And as you go beyond five, none of them have formulas that you can write down. But you can always guess the answer for the one you want. And we don't want all the funny roots that are negative and imaginary. We just want the positive real number that's the pressure or mole fraction of the gas. So we don't have to be intimidated by that. And I'll just remark here, mole fractions don't depend on n-naught, of course. So if your mole fraction depends on n-naught, you did something wrong. I usually put n-naught in because I like to see it go away. But a shortcut is to just say n-naught's one and leave it out. It's a little risky because if you make an arithmetic mistake, you won't catch it. Okay, now let's substitute in those horrible things for the mole fractions. And we get this, my goodness. Gigantic equation. At the start of the quarter, this would have been so intimidating. Imagine if I had given you this with no explanation and said solve it for alpha. You'd be gone for two weeks and when you came back you'd be looking terrible. You wouldn't have any coffee in your apartment either. But now this is a breeze because you know alpha's between 0 and 1 and you're going to guess the answer. And I'll show you how we guess the answer fairly quickly. It doesn't matter how intimidating it is because they're all the same. We don't care if we can write down the exact solution. Now the first thing here is that when you've got fractions within fractions, they are extremely confusing and prone to error. You have to be very careful. If I write 3, line, 4, line, same size, 5. I don't know whether I mean 3 fifths, 3 fourths divided by 5, or 3 divided by 4 fifths. And unlike multiplication, division is not associative. I have to know the order that I'm doing the operations. And therefore you have to get used to writing great big lines here and then tiny ones here to indicate which is done first. And I like to put these things in braces to say look do this first before you do it. And I do not keep these things in here for very long. I write it down and then the fastest thing I do is try to get rid of it. And the way I get rid of it is never by dividing. Because division is prone to error. For the reasons I just said, I get rid of it by multiplying like crazy by stuff until I clear out all the junk out of the stable there. So I have to find some way to clear things out, especially denominators in the denominator. I don't like those. I'm going to focus on denominators in the denominator. And I'm going to clear those guys out of there as fast as I can by multiplying by the appropriate thing. And if the talk gets messy, so be it. So here let's clear out the 2 minus alpha squared and let's clear out the numbers. Here's what we had originally. I've just tidied it up a bit. 3 halves to the 3 halves, 1 half to the 1 half, 1 minus alpha over 2 minus alpha to the 3 halves plus 1 half which is squared. I see this thing down here. I multiply the top and the bottom by 2 minus alpha squared to get rid of this guy. I see these numbers here with fractions in them. I don't like them. I multiply the top and the bottom by 2 to the 1 half. I just turn it upside down, keep the power. And I multiply the top and the bottom by 2 thirds to the 3 halves. That clears out this, this, and that. And finally I can simplify my result. I get a lot of spinach on the front, square it at 2, 2 thirds to the 3 halves power. Then I have the alpha I had here. I had a 2 minus alpha here so when I multiply by 2 minus alpha squared, I pick up a 2 minus alpha. The other stuff's gone on the bottom. I have 1 minus alpha quantity squared and then I have my P naught over P. So it's not so bad but this is not so easy either. My goal now is given a value of Kp, solve for alpha, assuming that P is equal to P standard. In other words, solve for the mole fraction of ammonia versus hydrogen at 1 atmosphere pressure and 598.15 Kelvin. And I go straight to putting in the numbers. 0.7698 is the square root of 2 times 2 thirds to the 3 halves power. Kp at 598 was 0.0624, near enough. P is equal to P standard so that's 1. And now I have an equation with alpha in it and a target to hit that doesn't depend on alpha. And that's the conditions for guessing and furthermore I have a range on alpha between 0 and 1. If I know nothing at all, I can play a game of cut in half. Guess 0.5, if it's too big guess 0.25 and then subdivide. That's called the bisection method. You don't have to actually do it that way because there are faster ways to guess. But let's look. When alpha is 0, this whole side here is 0. So alpha equals 0 is too small. When alpha is equal to 1 or near 1, this thing blows up to a huge number because 100% yield means K is infinity. And K is nowhere near infinity. K is small so I know alpha is going to be small. And based on that guess, I'm going to go ahead and guess some values of alpha. And if I don't know anything, I'm going to put in 0. I get 0. My target value is 0.0624 and I say that's too small. Now the question is what should I put in? And one trick is if K is small, put in K. Just try it. You know a number, 0.0624, it's pretty small. Put it in. See what you get. That's what I did. I put in 0.0624 and out popped 0.10588. I don't believe all those digits but I like to write them down just to see what they are. My target is still 0.0624. Now it's too big. Now I went up by 0 to 0.6 and the function went up from 0 to 0.1. Therefore I'm going to guess something about half way in here. This is what I picked, 0.31, about half and sure enough this is 0.05. Now it's looking pretty good because now I can see how it's behaving. I want 0.0624, now it's too small. But now I can see about how much I need to bump this up to get this to go from 0.05 to 0.06. It's about 0.04 so that's what I guessed. In bingo I get 0.06449, little too big. I want 0.624. But now I'm very close and I've got two estimates here. So I can zoom in pretty well. And I guess 0.038 and I get 0.06202, just a wee bit small. And now I have two estimates here. I know it's between here and I know it's closer to this one. And I guess 0.0382 and bingo to at least three digits it's on. So I'm going to quit here and say look this is close enough. The value of alpha that's between 0 and 1 and that's real that solves this equation is 0.0382. I think that's really 0.0382 is even closer. So that's the value of alpha. Now we have to go back and figure out what the ratio of the mole fraction of ammonia to hydrogen is. But we have that because we have a value of alpha. So we take the ratio of those two and if we simplify it, the mole fraction of ammonia is alpha over 2 minus alpha. The mole fraction of hydrogen, this might be a mistake again. I'll fix it, sorry. One half, it should be three halves, times one minus alpha over two minus alpha. And so instead of two, I should have two thirds here, two thirds alpha. Okay, but let's just follow it through assuming we get this. If I put in the value I got, then the ratio of the product to the major expensive reactant is 0.07948. I got about 8% yield based on how much hydrogen I used. Now if I want to increase the yield, this is actually with respect to nitrogen. It's worse with respect to hydrogen. Anyway, to increase the ratio by 100, we just say, well, we want this number instead of to be 0.07, we want it to be 100 times bigger. So this number, the ratio should be 7.4948. This will be another number when I fix it. And if you solve that, that's easy to solve for alpha. That's about 0.8. And so I'll let you figure out if alpha is 0.8 and you know Kp and you know the formula for it, it's simple to figure out what the pressure should be. And you'll easily see that it'll be a couple of hundred atmospheres and that's exactly what they use. And so that explains exactly how they're thinking. Okay, sorry for the confusion between nitrogen and hydrogen. And I guess I knew I was going to be confused because I said I personally find these Kp problems to be the ones that require the most care. They're the easiest ones to get wrong. They're the toughest ones to figure out because you end up with a lot of fractions. The equations that we have to solve in the end, if we've got a big reaction, the equations are absolutely intimidating. They have things to all kinds of powers. Yes? How many? Don't copy down something wrong because that's not going to help you. Which slide? This one? This one? This is just the equation to solve for alpha. This part's okay. Because luckily the juxtaposition cancels out. So this is the numerical equation to solve for alpha. Okay, so we'll go back here. But it doesn't matter how intimidating the equation is because you know alpha's between zero and one. So it's fairly easy to get it. Guessing is the most efficient way as long as you only have to obtain one answer. If you have to obtain things versus a variable, you don't want to be guessing over and over. Although if you don't change the pressure much, your last guess is a pretty good estimate for the next guess. So you can kind of creep along like a snake and you're close and then you just guess the next value. And you watch the yield go up as you turn the pressure up so you know alpha's going to get bigger. And you just guess a little bigger and you zoom in on it and then you can plot it. If you've got a graphing calculator, this is the thing to do is to graph it. But don't graph it between zero and one because usually at one it blows up to infinity. And your graphing calculator just shows you nothing because it gets too big and it can't scale it to fit on the display. Therefore I usually graph it between 10% and 90%. So it's not too big. And if it's not in the range I need there, then I zoom in here where I need it and graph it from that range. And that gives me an estimate of what value of alpha will hit the target that I want to get for Kp. So a graphing calculator can help speed things up a lot. Or if you have a spreadsheet, obviously. Okay. Equilibrium is dynamic. And that means that even though at equilibrium the concentrations of reactants and products don't change, that does not mean that the reactions have stopped. They have not stopped. They are continuing to occur. And what it means is that they're balancing out. I've got two jugs of water and one reaction is pumping water into one side and one reaction is pumping water into the other side and they're canceling out. So that the level of water is staying fixed. But if I look in detail, it's chaos. Things are moving around all over the place all the time because the reactions can still occur and they do. Ammonia, some of it goes backwards to make hydrogen and nitrogen. And some nitrogen and hydrogen then goes forward. If I take any chemical reaction and look at it microscopically, then it can certainly in principle run backwards because that conserves energy too. There's no difference. And to assume, even though we'd like to, to assume that an unlikely reaction running backwards, a disfavored reaction, never happens. It's too unlikely. Entropy steps in and says, no, no, no. That's too unlikely. The unlikely thing has to happen sometimes. You have to get hit by lightning occasionally. That happens. And then in fact, if you look at how many lightning strikes there are on the planet all the time, it's amazing that you haven't been hit yet. But of course, people who enter the lottery don't seem to realize that their chance of winning the lottery is about a million times less than getting hit by lightning. And yet they aren't worried about getting hit by lightning, but they somehow think they're going to win the lottery. Humans have a bias for positive things. If you give people bad news, they say, the first reaction is, I don't think that's going to happen. That's a natural human reaction. If you give them good news and a chance, they say, I'm in. I can win. I can fantasize what I'm going to do with the money, too. But then when you enter and play, you lose, of course, because they don't build those giant casinos by losing a ton of money. And in fact, it's interesting psychologically if you study why people get addicted to gambling, when you lose, there's never any remark. The machine doesn't say, you lost. You're going broke. Watch out. It's just quiet. And then when you win, even a quarter, ding, ding, ding, ding. And when you think back, you only remember winning, because those are the only memorable moments. You talk to people who go to Las Vegas. Everybody won. Nobody says, usually, I got cleaned out. First of all, it makes you seem like a sucker. Nobody wants to actually broadcast that. But second of all, you may actually be misremembering, because you actually remember the big deal when you win. They have that psychology figured out. That's how they make money. Okay. Let's talk quickly about homogeneous reactions. It's a simple term. You can think of milk. They homogenize milk so that the cream doesn't separate out on the top. And homogeneous means that it's all in one phase, like all gases or a single solution. If you've got a homogeneous equilibrium, the reactions never go 100%. And the reason why they never go 100% is that it turns out, even if the products are wildly favorable, it's always even more favorable to contaminate the products with a little bit of the reactants. And the reason why is that when you contaminate the products with a little bit of reactants, they get to mix. And mixing is entropically favorable. So rather than being a pure thing which has too low entropy, the products will spontaneously give a little bit of reactants so they can mix with them, because they're all in one phase. And then that turns out to be the most favorable outcome. Sometimes, of course, there's hardly any reactants left if the reaction's very, very favorable. But in principle, there's always a little. On the other hand, if you have heterogeneous equilibria, so you have more than one phase, like a solid turning into a liquid, a phase transition, or a gas reaction coming down on a solid surface, then because there are two phases, these reactions can go 100%. And the big difference is that by the definition of phase, phases don't mix. And if there's no chance to mix, then there's no penalty in making 100% of the product. Because you can't mix with the other guy anyway because he's a different phase than you are, and therefore you can go 100%. All phase, all type one phase transitions are like this. So for example, if the temperature is above zero Celsius and the pressure is one atmosphere, there is no ice, none, present at equilibrium. It's all liquid water because ice to water is a heterogeneous reaction. And if the temperature is below zero and the pressure is one atmosphere, there is no liquid water present at equilibrium. It's all ice, same reason. And likewise, if we have the temperature at exactly zero, we could have any amount of ice and water. We can't say what the concentration is. If we have the pressure exactly at the vapor pressure of water, then the container will have a certain concentration of vapor, but the amount of water in the bottom can be anything, as long as it's enough that it doesn't all evaporate. And if the pressure is higher than the vapor pressure, that means that if I had the thing in a piston and the pressure is higher than the vapor pressure, that the piston comes down right on the surface of the water. There's nothing left. It's 100% water once the vapor pressure exceeds the vapor pressure. As long as there's nothing else in there. And if it's exactly the vapor pressure, it depends how far I pull it up. I could have a lot of vapor and a little water or a lot of water and a little vapor. And therefore, heterogeneous equilibria, we don't have this fraction alpha that goes to this and that. It's usually all or nothing. And we decide based on looking at the reaction. Which way will a reaction go? Well, this is pretty obvious. If k is large, we're going to get products. If k is small, we're going to stick with reactants. And if k is 1, if delta G is 0, they're going to be present in roughly equal amounts, but not exactly equal because we could have 3 halves of hydrogen and 1 half of nitrogen, for example, and so they would be in a 3 to 1 ratio. If I have a reaction like this, 1 propane going back and forth to cyclo propane, you don't need to know what these are right now, but they're 3 carbon molecules. And let's suppose kc is 1 for this. Well, kc is the concentration of the products over the reactants. At equilibrium. And that's 1. Then if I just multiply through by the concentration of 1 propane, I come to the conclusion that the concentration of cyclo propane at equilibrium, whatever it is, is equal to the concentration of 1 propane. So it's 50-50. Now if k is near 1, you can think it's going to be a big mess, a big mixture of things. If k is huge, that means delta G is big and negative, it's going to be all products, and if k is tiny, it's going to be all reactants. You aren't going to get much material. Okay. Let's do one more. Let's try to decide if we have an initial condition, whether the reaction will go backwards or forwards from where we start. Because sometimes we have to figure out something. Somebody dumps something into a river by mistake. And we have to figure out if it's going to precipitate, or if it's going to dissolve and get in the ground water, and what we can do to lock it down, maybe so it doesn't get in the ground water, so we have time to clean it up. And we'd like to know how much of whatever we're going to add, we need to add, because we don't want to add too much. That usually costs money. And sometimes if we add too much, then a lot of it will be left over. Then we have to clean that up, because it's not supposed to be there in the first place. Well, we can decide which way it's going to shift fairly easy. If k, if q, the reaction quotient, wherever you start is bigger than k, that means you've got too many products. And that means that if you start at this condition with too many products, the reaction is actually going to go backwards. So the reaction shifts to the left, to the products. Excuse me, to the reactants. Chemists always write reactants on the left. If q is smaller than k, that means we haven't made enough products yet, then the reaction will run forwards until it makes enough products so that q comes up to the value of k. And once the water lines are equal, it's just going to sit there. And if q equals k, nothing further happens that we can see in terms of measuring concentration. They still run both ways. But there's no net change. Still lots of chemistry occurring, but no net change. And let's do one example. So at 1400 Kelvin, kc is 4.7. This is the first stage of the Haber process where you take methane and water. Those are two things you can get a hold of, and you heat them up, and you get carbon monoxide and hydrogen. You get three moles of hydrogen. And because you're making more moles of gas here, you predict that this reaction is going to be favorable, roughly speaking, if you heat it, because of the increased delta S. And that's why you heat it up. So you heat it up like crazy, you're making this. And if we know this value, it's 4.7, and we start out with this concentration, 0.035 molar H2O, 0.05 molar methane, 0.15 molar carbon monoxide, and 0.2 molar H2. Which direction is it going to run? Is it going to make us more of the hydrogen, or are we going to lose our hydrogen? Well, we just put in the reaction quotient, and we have to put in the hydrogen cubed, because there's a three, and then we put in the numbers, carbon monoxide, 0.15, hydrogen, 0.2 cubed, 0.035, 0.05, and we get Q is 0.69. Aha! But K is 4.7, so because Q is less than K, we're going to make more hydrogen if we run this reaction. We can start like this, and sit there, and we will get more hydrogen out than what we started with. If Q were bigger than K, then we would have too much of something there, and then it would run backwards. Okay, I think that's plenty for today.
 I heard that the world is coming to an end this year, so I guess all our problems are  over.  Notice how easy it is when the world is coming to an end.  You can forget about long term planning.  Party on.  Of course all of our worlds do come to an end and sometimes unpredictably.  So it's kind of important from time to time to imagine what you would do if you were going  to die today for sure.  Who would you call?  What grudges would you drop?  What would you say to your parents?  Your friends?  Anybody who's had an influence on your life.  And then my advice would be to go ahead and say some of that even though you don't think  you're going to die.  There's no point in waiting.  Many people wait too long and then the other person dies and you never get a chance to  tell them how much you appreciated something they did because you kept putting it off.  So don't put things off forever.  We all have a shelf life.  Unfortunately we're thermodynamically unstable.  And that means we're doomed.  Although hopefully not too quickly.  But eventually we are doomed.  But for now let's imagine we're going to live long enough to take the final.  Get a grave.  That type of thing.  And so it would be a good idea to plan for that.  And with that in mind let's get on with things.  Let's talk about equilibrium constants.  I'll pick up where I left off, we'll put the problem up again.  And we were actually trying to figure out if we knew the equilibrium constant for the  synthesis of ammonia at room temperature based on the tables.  Could we figure out what the yield would be at some higher temperature where we have to  run the reaction?  And the answer is we can do that for sure but we have to make some approximations unless  we make some measurements.  Here's what we have.  We had delta G of formation of ammonia is minus 16.6 kilojoules per mole.  Negative.  What would you say if somebody came to you and said, hey, I found something interesting  in the table.  Here's a molecule that has a positive delta G of formation.  Most of them are negative but there are a few in the table.  I'll give you an example, hydrazine N2H4, methylhydrazine I think is used in rocket  fuels sometimes.  If you were given that information what would you say?  Since the molecule exists.  Have you got any ideas?  I'll tell you what I would say.  Number one, although hydrazine has nitrogen and hydrogen like ammonia, I can't make hydrazine  and high yield from nitrogen and hydrogen because it has a positive delta G so it has  a tiny K.  I could probably make a little bit of it but not much.  Second, hydrazine must be kinetically stable like a human being.  Hydrazine must be doomed eventually to fall apart at least into hydrogen and nitrogen  because it's unstable with respect to the elements it was made from.  So it's going to fall downhill eventually.  Anything like that usually you want to handle extremely carefully.  Those are a lot of things like explosives.  They have a big delta G and then boom when they detonate you get them going to products  with a lot of energy.  But ammonia doesn't have that problem and we had figured out that the balanced reaction  for one mole of ammonia because delta G refers to one mole of ammonia is one half molecular  nitrogen plus three halfs molecular hydrogen gives one mole of ammonia.  If I double a reaction, I double delta G. That's very important.  When delta G a formation is listed it's per mole.  If I make two moles of ammonia delta G is twice as big just like two kilograms of ammonia  is twice the mass of one kilogram of ammonia.  Delta G behaves that way.  We have a formation, excuse me, an equation that relates K to delta G and there it is.  We just put in the numbers and we make sure we have the right units.  We get K is E to the plus 6.5963 and that's about 730.  That's favorable.  That means we're going to get mostly ammonia.  Now at the other temperatures we need to make some assumptions.  First we have to make the assumption because we don't know the heat capacities of the products  or reactants that delta H and delta S don't depend themselves too much on temperature.  The entropy is different than the enthalpy because there's no entropy of formation.  The entropy is referenced to absolute zero temperature and any kind of molecule or atom  at finite temperature has some disorder.  It's moving, it's tumbling around, energy is distributed in the molecule in different  ways.  The entropy says the energy is going to randomize to the maximum extent possible because that's  simply the most likely thing that's going to happen.  If we use our equation for the state function of delta G, delta G equals delta H minus t  delta S and the relationship between log of K and delta G, we can set this up at two different  temperatures and then find the value of log K for the second temperature if we know the  value of log K at the first temperature.  That's how we have to do it.  Here we go.  Each temperature, I'm going to write log and I'm going to put in parentheses the temperature  I'm calling it just to keep track.  Log of K at temperature T1 is equal to minus delta G standard at temperature T1 over RT1  and then I'm going to put in delta H and delta S. I'm going to do the same thing for temperature  T2 except I'm going to put in T2.  Then since this is equal to this and this is equal to this, then log K T1 minus log K  T2 is equal to the right hand side here minus the other right hand side.  I've written that all out here.  Log K at temperature T1 minus log K at temperature T2 factored out R minus 1 over R delta H  at T1 over T1 minus delta H over T2 over T2 plus, does this add a minus up here, 1 over  R times delta S at T1 minus delta S at T2.  Now we have to make an assumption.  The assumption we make is that delta H and delta S themselves don't depend much on temperature.  Whereas we know that's not quite true but we're going to assume it's true because we  don't know the heat capacities.  We assume delta H at T1 is equal to delta H at T2 and we'll just call that delta H and  the same for delta S then conveniently because this delta S and this delta S are the same,  this term goes away.  So we have the natural log of K at temperature T1 minus natural log K at temperature T2 is  now equal to minus delta H over R times 1 over T1 minus 1 over T2.  This not surprisingly is very, very similar to the Clausius-Clapeyron equation.  The only difference is that we had the vapor pressure.  Well it turns out that if I have an equilibrium between a pure liquid and a vapor pressure  at some pressure, the equilibrium constant is just the vapor pressure.  So this is Clausius-Clapeyron equations, just a special case of this more powerful equation.  And then because I'm subtracting the logs I can tidy this up and say the natural log  of K at temperature T1 divided by K at temperature T2, the natural log of that whole thing is  equal to this.  And therefore what I need to know in addition to delta G at one temperature so I can get  log K is I need to know delta H for the reaction at some temperature and if I assume it doesn't  depend on temperature then 298 will do.  So I look that up and I recall that delta H of formation for ammonia for one mole is  minus 46,300.  It's exothermic joules per mole.  So I put that in.  Log of K at temperature T1 divided by K at T2 minus 46,300 over R in joules 1 over 298  minus 1 over 598.  I put the 0.15 but I didn't want to have this drop over a line because it makes it hard  to read.  It didn't fit and I don't want to make it tiny because it's hard to read.  But when I did the calculation I put 298.15 and I find that log of K at temperature T1  divided by K at T2 is equal to plus 9.367 and T1 was 298 because I'm keeping things  in order.  While we know what K at temperature T1 is, so K at temperature T1 divided by K at temperature  T2, the exponential function of the log is just the thing and the exponential function  of this plus 9 is e to the plus 9.367 and if I solve that using K of T1 is 730 I get  K of T2 is equal to 730 times e to the minus 9.367 and so I get a measly 6.24 times 10  to the minus 2.  So you can see that my yield went from wildly favorable, like 730 to 6 percent per K by  cranking the temperature up.  That's how a lot of things work and that's why in part cranking the temperature up on  the planet just a little bit can shift the equilibrium concentrations of things like  a giant teeter totter with unimaginable consequences.  Or we can imagine them but we wouldn't want to live through them necessarily.  It's not just that it'll be hotter, it'll be totally different too and sometimes you  don't know all the equilibria going on in the ocean and so you get surprised.  Okay, at the highest temperature it's less favorable so if we go up to 798 that's where  they run it remember about 500 Celsius.  Then we find with delta H minus 46,300 joules per mole we do the same thing and now we find  that log K at T1 divided by K at T2 is equal to plus 11.7 and the bad thing is that it's  an exponential function and so now when I multiply 730 by e to the minus 11.7 I get  less than a percent I get 6.05 times 10 to the minus 3.  I'll just remark here that the assumption that we've made that delta H and delta S  themselves don't change much with temperature is not going to be accurate when you change  the temperature hundreds and hundreds of degrees or a thousand degrees.  That's too far off for that assumption to be any good but if you change a reaction from  25 Celsius to 35 Celsius that's probably going to be a very good assumption and sometimes  if you're in the lab and you're optimizing a reaction if you just heat it or just cool  it a little bit more use a different solvent that has a higher boiling point when you reflux  it you suddenly get much higher yield because of this exponential function you can go from  50% to 99% pretty readily and that's just much much more useful than if you have a reaction  that does that kind of thing.  Okay, some observations.  The equilibrium constant depends on the exponential function of the free energy.  Therefore small differences in free energy lead to large changes in the equilibrium constant.  That's why for a lot of reactions if I see that delta G is some big number of kilojoules  per mole and negative I can just pretty much assume 100% yield because of this exponential  function.  Likewise temperature because it's in the denominator of the exponential can shift the equilibrium  greatly as we saw in the last problem.  Now let's try to shift the equilibrium back in our favor.  We were carrying this out at pressure of one atmosphere and we weren't really asking how  much ammonia we got versus how much hydrogen we started with.  Since the most expensive reagent in the synthesis, nitrogen is pretty much free from the air.  So we'll do it with respect to the hydrogen.  But first let's just look at this shifting.  So let's just take a hypothetical reaction.  A is in equilibrium with B. But A and B have different free energies.  The question is what should the difference in free energy be to get 90% yield of B?  What should it be to get 99% yield and what should it be to get 99.9% yield which would  be essentially considered quantitative?  And it turns out here it works in our favor.  So let's assume we run the reaction at 298 and let's just take delta G of this reaction  A goes to B, plug in 298 and figure out what we need.  Well we want to get 90% of B at equilibrium and that means we have 10% of A and so 0.9  over 0.1 is 9.  That's K, K is 9.  And therefore if we take the natural log, if we solve this, excuse, yeah, if we take  the natural log and solve this for delta G, we get delta G for the reaction should be  minus RT log K, K is 9.  We put in R in joules.  We put in 298.15 Kelvin and we put in the log of 9 and we find that we only need minus  5.45 kilojoules per mole, which is hardly anything.  For 99% yield, the only difference now is we have 99% of B at equilibrium and that means  we have 1% of A left over, so 0.99 over 0.01 is 99.  The log of 99 is fortunately, by the way, logs work.  They never get that big, even if the numbers are big.  We get delta G for this case is minus RT log 99.  It only goes to minus 11.39 kilojoules per mole.  So it's just twice.  And for 99.9, B at equilibrium is 0.999, A at equilibrium is 0.001, so K is 999, but again,  log 999 doesn't get that big and when I multiply it by minus RT, I get negative 17.12 kilojoules  per mole.  For typical reactions that you're likely to pencil out that involve burning a hydrocarbon  and getting CO2 and H2O, delta G for that reaction is going to be huge and negative.  And so we can pretty much consider that there's nothing left if, in fact, the material is pure.  What's usually left when you burn tons and tons of stuff is the small impurities that  aren't written in your chemical equation, sulfur, mercury, cadmium, other things that  don't burn uranium, thorium, radon, other things, and all those come out in the fly  ash if you burn coal.  And if you want to burn it cleanly, then you have to figure out what you're going to do  with all those things.  For the longest time, we didn't care and we let all the mercury in the fly ash flow into  the ocean and that's why fish may have higher mercury content.  Now I think fish always have some mercury content, but I don't think we helped that  much.  Okay.  Now let's go back to the Haber process and let's try to squeeze the product.  We're forced to heat it up so we know K is small, but we aren't forced to run an atmospheric  pressure.  And because there's more moles of gas in the reactants than there are in the products,  I made a comment without saying why, that we could put pressure on it and force the reactants  to go over to products.  Now we're going to do a calculation to show how that actually works.  Let's take our same reaction and let's figure out what the equilibrium constant Kp is in  terms of partial pressure.  The reason I'm using Kp is because all these are gases and that's the natural thing to  use.  And even though these are non-integer stoichiometric coefficients, when I write the equilibrium  constant, I always raise the concentration to the stoichiometric power.  If it's one-half, then it's the square root of the concentration.  If it's three-halves, it's the concentration of the three-halves and so forth.  It's whatever the actual number is.  Doesn't matter if it's not integer.  And of course, non-integer powers don't scare us anymore because we are not going to ever  try to solve the equation analytically.  We're just going to guess the answer.  And we're so good at guessing by this point that we're going to get it right.  So let's have a look.  This looks a little bit messy, but we'll go through it step by step by step.  Kp, the definition of Kp, is its products, partial pressure of products divided by the  standard pressure because Kp has no units.  Divided by the partial pressure of hydrogen divided by the standard pressure raised to  the three-halves power because it had a three-halves in the chemical equation.  Multiplied by the partial pressure of nitrogen divided by the standard and raised to the  one-half power.  And then I can use Dalton's law.  Dalton's law says the partial pressure of an ideal gas is equal to the mole fraction  times the total pressure.  And here I'm using p to be the total pressure.  Therefore if I look on the diagram and it says p is 300 atmospheres, then that's the  p I'm going to put in there for the total pressure.  And likewise I can do the same thing here, mole fraction of hydrogen times the total  pressure and the mole fraction of nitrogen times the total pressure raised to the powers.  And then I'm going to pull out the mole fractions and separate out the mole fraction part from  the p part.  Now I've got this leading term which is all the mole fractions which is going to be what  I'm going to call the yield.  And then I've got all these pressure terms which I don't care so much about.  But it seems like the wind's blowing from behind because pressure to the three-halves  times pressure to the one-half, I add the x-bonus, that's just pressure squared.  I have p over p naught squared on the bottom, p over p naught on the top.  If I simplify that, I end up with p naught over p.  Now we can sort of see how it works.  When we change the pressure, Kp doesn't change because Kp is related to delta G standard.  And delta G standard is at the standard pressure of one atmosphere.  Therefore when I change the pressure, Kp doesn't change.  When I change the temperature, Kp does change.  All things can change things.  But here's the thing.  If Kp doesn't change, let's suppose we've got our miserable yield of 0.6% here for K.  And now I crank this guy up.  This guy was 1 before.  But I crank p up to some huge number like 300.  So this is now 1 over 300.  Well that means that this thing here with the x's has to get bigger because p made this  term here get smaller.  And the only way this guy can get bigger is if the mole fraction of ammonia increases  like crazy.  But that's what I just said is that because I've got more gases as reactants, if I squeeze  down on it, I'm going to make much more products.  So these guys will go down and this guy will go up.  And the harder I push on it, the higher yield I get.  Why don't I just push on it to 10,000 atmospheres then?  The answer is that costs too much money.  Because if I want to squeeze that hard, I've got to make a chemical plant where all the  pipes are that thick.  So they can take that pressure and that costs me a ton of money.  And therefore it's more economical for me to recycle the reactants that I can't quite  get into products, bleed off the ammonia by condensing it and cooling it, and then just  go around again at the same pressure.  That's going to be more economical than trying to run one reaction and get it to go all the  way.  That's just economics of designing a plant.  And economics are important because if the fertilizer costs 10 times as much because I'm  doing something in a crazy way, then I can't afford to eat.  OK.  So let's actually do this calculation.  This is going to be very long, unfortunately, so I hope you can follow it.  But we'll do it step by step by step.  Here's the problem.  What pressure would it take at 598, the intermediate temperature, to boost the yield of ammonia  compared to the residual hydrogen by 100 times compared to whatever yield we got at atmospheric  pressure?  So problem 54, we calculated K, but we didn't calculate the equilibrium concentrations.  And we have to assume that the reaction reaches equilibrium.  In a real chemical process, people don't really care if it reaches equilibrium.  They just care if they get a pretty good yield of product, especially if they're going to  recycle the reactants.  And let's assume that we start with a 1 to 3 nitrogen to hydrogen ratio, the stoichiometric  ratio for ammonia.  That'll just simplify our calculations.  But of course, again, when you're running a plant, you may not have that condition.  And you may actually try to game the system a little bit by putting in much more concentration  of something that's a cheaper reagent to drive the equilibrium to the right.  So you may actually use much more nitrogen than hydrogen as you go to try to convert  all the hydrogen over to ammonia.  I don't actually know how they do it in practice.  OK, let's first figure out what the ratio of the mole fractions of ammonia to hydrogen  would be at one atmosphere.  That's going to take us a little bit of calculation.  And then next, let's increase whatever this ratio turns out to be by a factor of 100 and  then figure out what P naught over P should be.  P naught's one atmosphere, so that'll tell us what P should be in atmospheres.  We're talking to a chemical engineer.  We may convert it to PSI so they know that they may have things in terms of pounds per  square inch because that's how a lot of things are quoted.  We have to proceed step by step.  We use the partial pressures, then Dalton's law, and to convert to mole fractions.  And we have to be careful to include the powers of 3 half and 1 half.  And I've just rewritten that here so that we can see it again.  We have k at 598.  We got that from before.  And after everything drops out, we have the mole fraction of ammonia in the numerator  and in the denominator, we have the mole fraction of nitrogen to the 1 half power and the mole  fraction of hydrogen to the 3 half power.  And then we have this term left over, P standard over P.  If we have the same number of moles of gas on both sides, it's pretty easy to see that  the powers of P here are going to be the same on the top and bottom.  And in that case, if you have the same number of moles of gas on both sides, squeezing it  won't do anything much because it won't change it.  And so therefore, you don't squeeze it under those conditions.  We're lucky that we're taking more moles of gas when we make ammonia because that gives  us a lever to push on.  And let's, for the sake of argument, start with 1 half times some number.  It turns out it won't matter what it is, but I'll call it n naught.  Let's start with 1 half n naught moles of nitrogen.  And we have three times as much hydrogen.  Therefore we have 3 halves n naught moles of hydrogen initially before we actually conduct  the reaction.  So we have no nH3.  This is our initial condition.  The number of moles of nitrogen starting out is 3 halves n naught.  The number of moles of hydrogen is 1 half n naught.  And the number of moles of ammonia is zero.  And now we let a fraction alpha of the reactants react.  And for some reason, which I don't actually know myself, whenever chemists deal with gases,  the fraction that reacts is alpha, and whenever they deal with solutions, the molarity that  reacts is x.  But anyway, it doesn't matter what you call it.  It's an unknown, and we're going to try to solve for the fraction alpha that gives products  based on the value of k and a very, very messy equation which would intimidate everybody  if they thought they had to solve it exactly because it could be potentially quite different  to, quite difficult to solve.  Now this fraction alpha is a number we're going to try to get.  It has no units, and it's between zero and one.  If alpha is one, then all of it reacts to ammonia.  That's 100% yield.  If alpha is zero, that's what we started with here.  And that's what I've written.  Whenever I set up an equation like this, I always put in alpha or x or whatever it is,  the range it can be because I want to know where I should guess.  In chemistry, all these things are positive numbers, and they aren't imaginary numbers  or anything else funny, and that helps you a lot.  If you have to guess in two dimensions, it gets very, very much harder to do.  You may need a computer to do that.  And I see that it all works out.  It makes sense when alpha is equal to one, there's no reactants left over.  It's all products.  When alpha is zero, there's no products.  It's all reactants.  And I see that they're going away in the right way.  And that's because I put the three halves and the one half here and not some crazy thing  here trying to multiply that by three halves, which is a very, very confusing and bad strategy.  So let's figure out the partial pressure, or let's figure out the mole fraction so we  can get the partial pressure.  To figure out the mole fraction, we have to add up the total number of moles of reactants  and products.  We now know if a fraction alpha reacts how many moles of nitrogen, hydrogen, and ammonia  we've got, now let's add them up.  So we add up the number of moles of nitrogen plus the number of moles of hydrogen plus  the number of moles of ammonia, the three halves and the one half add up to two.  The one minus alpha was in both of these guys.  So I get two times quantity one minus alpha times n naught.  And this guy here was alpha times n naught.  And if I expand this out, I get two minus two alpha plus alpha, so I get two minus alpha  times n naught.  That's the total number of moles.  When alpha is zero, there's two moles because there's three half moles of hydrogen and one  half mole of nitrogen.  When alpha is one, there's one mole because there's one mole of ammonia.  So again, I check, put in zero and one, and I make sure it makes sense that I haven't  done some sort of logical error in setting it up.  And then I've got this equation here for the total number of moles.  The mole fraction is just the number of moles of each of these guys divided by the total  number of moles.  And I just very patiently work it out.  I don't try to shortcut it.  But for the mole fraction of nitrogen, I get three halves times one minus alpha times  n naught divided by two minus alpha times n naught.  And you see n naught goes away.  It doesn't matter because we're talking mole fractions, so it has to go away.  And therefore, the mole fraction of nitrogen is three halves times one minus alpha divided  by two minus alpha.  And I don't sweat about that any further at this point.  And just leave it like that.  Then I move on to hydrogen.  Sorry.  I'll fix this.  This was meant to be hydrogen because hydrogen has the three halves.  I'll fix this before I post these slides.  This is meant to be the mole fraction of hydrogen, not nitrogen.  And likewise, this one is also backwards.  But fortunately, it won't matter since they're multiplied by each other.  So I think the rest of the calculation's OK.  And for hydrogen, we should have a three halves instead of a one half, but it's basically  the same thing.  We get a number and we get one minus alpha divided by two minus alpha.  And then the mole fraction of ammonia is, again, the number of moles of ammonia divided  by the total.  And both of them depend on how far the reaction goes.  The top part is alpha n naught.  The bottom part is two minus alpha times n naught.  It's an i of alpha over two minus alpha.  And the thing I want to emphasize is that this is a very simple reaction.  And yet I could potentially have alpha to the fifth power here to solve.  And that's why I don't advise trying to solve things analytically.  There is a formula for a quadratic for the roots.  A x squared plus B x plus C. There's a formula.  There's a formula for a cubic.  It's pretty big.  There's a formula for a quadratic.  A x to the fourth plus B x cubed plus C x squared plus D x.  That formula is 28 pages long.  Says that here are the four roots.  X1 equals ABC raised to all these power.  And then amazingly for a Quintic equation, it was shown quite a long time ago that there  is no formula for a general Quintic equation.  It's not that we can't figure it out.  It's that it doesn't exist in terms of raising things to a power, multiplying things, adding,  subtracting, and dividing.  And those are the weapons we have in arithmetic.  The fact that it suddenly at five becomes different is very mysterious to a mathematician  and fascinating.  And as you go beyond five, none of them have formulas that you can write down.  But you can always guess the answer for the one you want.  And we don't want all the funny roots that are negative and imaginary.  We just want the positive real number that's the pressure or mole fraction of the gas.  So we don't have to be intimidated by that.  And I'll just remark here, mole fractions don't depend on n-naught, of course.  So if your mole fraction depends on n-naught, you did something wrong.  I usually put n-naught in because I like to see it go away.  But a shortcut is to just say n-naught's one and leave it out.  It's a little risky because if you make an arithmetic mistake, you won't catch it.  Okay, now let's substitute in those horrible things for the mole fractions.  And we get this, my goodness.  Gigantic equation.  At the start of the quarter, this would have been so intimidating.  Imagine if I had given you this with no explanation and said solve it for alpha.  You'd be gone for two weeks and when you came back you'd be looking terrible.  You wouldn't have any coffee in your apartment either.  But now this is a breeze because you know alpha's between 0 and 1 and you're going to guess the answer.  And I'll show you how we guess the answer fairly quickly.  It doesn't matter how intimidating it is because they're all the same.  We don't care if we can write down the exact solution.  Now the first thing here is that when you've got fractions within fractions,  they are extremely confusing and prone to error.  You have to be very careful.  If I write 3, line, 4, line, same size, 5.  I don't know whether I mean 3 fifths, 3 fourths divided by 5, or 3 divided by 4 fifths.  And unlike multiplication, division is not associative.  I have to know the order that I'm doing the operations.  And therefore you have to get used to writing great big lines here and then tiny ones here  to indicate which is done first.  And I like to put these things in braces to say look do this first before you do it.  And I do not keep these things in here for very long.  I write it down and then the fastest thing I do is try to get rid of it.  And the way I get rid of it is never by dividing.  Because division is prone to error.  For the reasons I just said, I get rid of it by multiplying like crazy by stuff  until I clear out all the junk out of the stable there.  So I have to find some way to clear things out, especially denominators in the denominator.  I don't like those.  I'm going to focus on denominators in the denominator.  And I'm going to clear those guys out of there as fast as I can by multiplying by the appropriate thing.  And if the talk gets messy, so be it.  So here let's clear out the 2 minus alpha squared and let's clear out the numbers.  Here's what we had originally.  I've just tidied it up a bit.  3 halves to the 3 halves, 1 half to the 1 half, 1 minus alpha over 2 minus alpha  to the 3 halves plus 1 half which is squared.  I see this thing down here.  I multiply the top and the bottom by 2 minus alpha squared to get rid of this guy.  I see these numbers here with fractions in them. I don't like them.  I multiply the top and the bottom by 2 to the 1 half.  I just turn it upside down, keep the power.  And I multiply the top and the bottom by 2 thirds to the 3 halves.  That clears out this, this, and that.  And finally I can simplify my result.  I get a lot of spinach on the front, square it at 2, 2 thirds to the 3 halves power.  Then I have the alpha I had here.  I had a 2 minus alpha here so when I multiply by 2 minus alpha squared, I pick up a 2 minus alpha.  The other stuff's gone on the bottom.  I have 1 minus alpha quantity squared and then I have my P naught over P.  So it's not so bad but this is not so easy either.  My goal now is given a value of Kp, solve for alpha, assuming that P is equal to P standard.  In other words, solve for the mole fraction of ammonia versus hydrogen at 1 atmosphere pressure and 598.15 Kelvin.  And I go straight to putting in the numbers.  0.7698 is the square root of 2 times 2 thirds to the 3 halves power.  Kp at 598 was 0.0624, near enough.  P is equal to P standard so that's 1.  And now I have an equation with alpha in it and a target to hit that doesn't depend on alpha.  And that's the conditions for guessing and furthermore I have a range on alpha between 0 and 1.  If I know nothing at all, I can play a game of cut in half.  Guess 0.5, if it's too big guess 0.25 and then subdivide. That's called the bisection method.  You don't have to actually do it that way because there are faster ways to guess.  But let's look.  When alpha is 0, this whole side here is 0.  So alpha equals 0 is too small.  When alpha is equal to 1 or near 1, this thing blows up to a huge number  because 100% yield means K is infinity.  And K is nowhere near infinity.  K is small so I know alpha is going to be small.  And based on that guess, I'm going to go ahead and guess some values of alpha.  And if I don't know anything, I'm going to put in 0.  I get 0. My target value is 0.0624 and I say that's too small.  Now the question is what should I put in?  And one trick is if K is small, put in K.  Just try it.  You know a number, 0.0624, it's pretty small.  Put it in.  See what you get.  That's what I did.  I put in 0.0624 and out popped 0.10588.  I don't believe all those digits but I like to write them down just to see what they are.  My target is still 0.0624.  Now it's too big.  Now I went up by 0 to 0.6 and the function went up from 0 to 0.1.  Therefore I'm going to guess something about half way in here.  This is what I picked, 0.31, about half and sure enough this is 0.05.  Now it's looking pretty good because now I can see how it's behaving.  I want 0.0624, now it's too small.  But now I can see about how much I need to bump this up to get this to go from 0.05 to 0.06.  It's about 0.04 so that's what I guessed.  In bingo I get 0.06449, little too big.  I want 0.624.  But now I'm very close and I've got two estimates here.  So I can zoom in pretty well.  And I guess 0.038 and I get 0.06202, just a wee bit small.  And now I have two estimates here.  I know it's between here and I know it's closer to this one.  And I guess 0.0382 and bingo to at least three digits it's on.  So I'm going to quit here and say look this is close enough.  The value of alpha that's between 0 and 1 and that's real that solves this equation is 0.0382.  I think that's really 0.0382 is even closer.  So that's the value of alpha.  Now we have to go back and figure out what the ratio of the mole fraction of ammonia to hydrogen is.  But we have that because we have a value of alpha.  So we take the ratio of those two and if we simplify it,  the mole fraction of ammonia is alpha over 2 minus alpha.  The mole fraction of hydrogen, this might be a mistake again.  I'll fix it, sorry.  One half, it should be three halves, times one minus alpha over two minus alpha.  And so instead of two, I should have two thirds here, two thirds alpha.  Okay, but let's just follow it through assuming we get this.  If I put in the value I got, then the ratio of the product to the major expensive reactant is 0.07948.  I got about 8% yield based on how much hydrogen I used.  Now if I want to increase the yield, this is actually with respect to nitrogen.  It's worse with respect to hydrogen.  Anyway, to increase the ratio by 100, we just say,  well, we want this number instead of to be 0.07, we want it to be 100 times bigger.  So this number, the ratio should be 7.4948.  This will be another number when I fix it.  And if you solve that, that's easy to solve for alpha.  That's about 0.8.  And so I'll let you figure out if alpha is 0.8 and you know Kp and you know the formula for it,  it's simple to figure out what the pressure should be.  And you'll easily see that it'll be a couple of hundred atmospheres and that's exactly what they use.  And so that explains exactly how they're thinking.  Okay, sorry for the confusion between nitrogen and hydrogen.  And I guess I knew I was going to be confused because I said I personally find these Kp problems  to be the ones that require the most care.  They're the easiest ones to get wrong.  They're the toughest ones to figure out because you end up with a lot of fractions.  The equations that we have to solve in the end, if we've got a big reaction,  the equations are absolutely intimidating.  They have things to all kinds of powers.  Yes?  How many?  Don't copy down something wrong because that's not going to help you.  Which slide?  This one?  This one?  This is just the equation to solve for alpha.  This part's okay.  Because luckily the juxtaposition cancels out.  So this is the numerical equation to solve for alpha.  Okay, so we'll go back here.  But it doesn't matter how intimidating the equation is because you know alpha's between zero and one.  So it's fairly easy to get it.  Guessing is the most efficient way as long as you only have to obtain one answer.  If you have to obtain things versus a variable, you don't want to be guessing over and over.  Although if you don't change the pressure much, your last guess is a pretty good estimate for the next guess.  So you can kind of creep along like a snake and you're close and then you just guess the next value.  And you watch the yield go up as you turn the pressure up so you know alpha's going to get bigger.  And you just guess a little bigger and you zoom in on it and then you can plot it.  If you've got a graphing calculator, this is the thing to do is to graph it.  But don't graph it between zero and one because usually at one it blows up to infinity.  And your graphing calculator just shows you nothing because it gets too big and it can't scale it to fit on the display.  Therefore I usually graph it between 10% and 90%.  So it's not too big.  And if it's not in the range I need there, then I zoom in here where I need it and graph it from that range.  And that gives me an estimate of what value of alpha will hit the target that I want to get for Kp.  So a graphing calculator can help speed things up a lot.  Or if you have a spreadsheet, obviously.  Okay. Equilibrium is dynamic.  And that means that even though at equilibrium the concentrations of reactants and products don't change,  that does not mean that the reactions have stopped.  They have not stopped.  They are continuing to occur.  And what it means is that they're balancing out.  I've got two jugs of water and one reaction is pumping water into one side  and one reaction is pumping water into the other side and they're canceling out.  So that the level of water is staying fixed.  But if I look in detail, it's chaos.  Things are moving around all over the place all the time because the reactions can still occur and they do.  Ammonia, some of it goes backwards to make hydrogen and nitrogen.  And some nitrogen and hydrogen then goes forward.  If I take any chemical reaction and look at it microscopically,  then it can certainly in principle run backwards because that conserves energy too.  There's no difference.  And to assume, even though we'd like to, to assume that an unlikely reaction running backwards,  a disfavored reaction, never happens.  It's too unlikely.  Entropy steps in and says, no, no, no.  That's too unlikely.  The unlikely thing has to happen sometimes.  You have to get hit by lightning occasionally.  That happens.  And then in fact, if you look at how many lightning strikes there are on the planet all the time,  it's amazing that you haven't been hit yet.  But of course, people who enter the lottery don't seem to realize that their chance of winning the lottery  is about a million times less than getting hit by lightning.  And yet they aren't worried about getting hit by lightning,  but they somehow think they're going to win the lottery.  Humans have a bias for positive things.  If you give people bad news, they say, the first reaction is, I don't think that's going to happen.  That's a natural human reaction.  If you give them good news and a chance, they say, I'm in.  I can win.  I can fantasize what I'm going to do with the money, too.  But then when you enter and play, you lose, of course,  because they don't build those giant casinos by losing a ton of money.  And in fact, it's interesting psychologically if you study why people get addicted to gambling,  when you lose, there's never any remark.  The machine doesn't say, you lost. You're going broke.  Watch out. It's just quiet.  And then when you win, even a quarter, ding, ding, ding, ding.  And when you think back, you only remember winning,  because those are the only memorable moments.  You talk to people who go to Las Vegas. Everybody won.  Nobody says, usually, I got cleaned out.  First of all, it makes you seem like a sucker.  Nobody wants to actually broadcast that.  But second of all, you may actually be misremembering,  because you actually remember the big deal when you win.  They have that psychology figured out.  That's how they make money.  Okay. Let's talk quickly about homogeneous reactions.  It's a simple term.  You can think of milk.  They homogenize milk so that the cream doesn't separate out on the top.  And homogeneous means that it's all in one phase,  like all gases or a single solution.  If you've got a homogeneous equilibrium,  the reactions never go 100%.  And the reason why they never go 100%  is that it turns out, even if the products are wildly favorable,  it's always even more favorable to contaminate the products  with a little bit of the reactants.  And the reason why is that when you contaminate the products  with a little bit of reactants, they get to mix.  And mixing is entropically favorable.  So rather than being a pure thing which has too low entropy,  the products will spontaneously give a little bit of reactants  so they can mix with them, because they're all in one phase.  And then that turns out to be the most favorable outcome.  Sometimes, of course, there's hardly any reactants left  if the reaction's very, very favorable.  But in principle, there's always a little.  On the other hand, if you have heterogeneous equilibria,  so you have more than one phase,  like a solid turning into a liquid, a phase transition,  or a gas reaction coming down on a solid surface,  then because there are two phases, these reactions can go 100%.  And the big difference is that by the definition of phase,  phases don't mix.  And if there's no chance to mix,  then there's no penalty in making 100% of the product.  Because you can't mix with the other guy anyway  because he's a different phase than you are,  and therefore you can go 100%.  All phase, all type one phase transitions are like this.  So for example, if the temperature is above zero Celsius  and the pressure is one atmosphere,  there is no ice, none, present at equilibrium.  It's all liquid water because ice to water  is a heterogeneous reaction.  And if the temperature is below zero  and the pressure is one atmosphere,  there is no liquid water present at equilibrium.  It's all ice, same reason.  And likewise, if we have the temperature at exactly zero,  we could have any amount of ice and water.  We can't say what the concentration is.  If we have the pressure exactly at the vapor pressure of water,  then the container will have a certain concentration of vapor,  but the amount of water in the bottom can be anything,  as long as it's enough that it doesn't all evaporate.  And if the pressure is higher than the vapor pressure,  that means that if I had the thing in a piston  and the pressure is higher than the vapor pressure,  that the piston comes down right on the surface of the water.  There's nothing left.  It's 100% water once the vapor pressure exceeds the vapor pressure.  As long as there's nothing else in there.  And if it's exactly the vapor pressure,  it depends how far I pull it up.  I could have a lot of vapor and a little water  or a lot of water and a little vapor.  And therefore, heterogeneous equilibria,  we don't have this fraction alpha that goes to this and that.  It's usually all or nothing.  And we decide based on looking at the reaction.  Which way will a reaction go?  Well, this is pretty obvious.  If k is large, we're going to get products.  If k is small, we're going to stick with reactants.  And if k is 1, if delta G is 0,  they're going to be present in roughly equal amounts,  but not exactly equal because we could have 3 halves of hydrogen  and 1 half of nitrogen, for example,  and so they would be in a 3 to 1 ratio.  If I have a reaction like this, 1 propane going back and forth  to cyclo propane, you don't need to know what these are right now,  but they're 3 carbon molecules.  And let's suppose kc is 1 for this.  Well, kc is the concentration of the products over the reactants.  At equilibrium.  And that's 1.  Then if I just multiply through by the concentration of 1 propane,  I come to the conclusion that the concentration of cyclo propane  at equilibrium, whatever it is,  is equal to the concentration of 1 propane.  So it's 50-50.  Now if k is near 1, you can think it's going to be a big mess,  a big mixture of things.  If k is huge, that means delta G is big and negative,  it's going to be all products,  and if k is tiny, it's going to be all reactants.  You aren't going to get much material.  Okay.  Let's do one more.  Let's try to decide if we have an initial condition,  whether the reaction will go backwards or forwards from where we start.  Because sometimes we have to figure out something.  Somebody dumps something into a river by mistake.  And we have to figure out if it's going to precipitate,  or if it's going to dissolve and get in the ground water,  and what we can do to lock it down,  maybe so it doesn't get in the ground water,  so we have time to clean it up.  And we'd like to know how much of whatever we're going to add,  we need to add, because we don't want to add too much.  That usually costs money.  And sometimes if we add too much,  then a lot of it will be left over.  Then we have to clean that up,  because it's not supposed to be there in the first place.  Well, we can decide which way it's going to shift fairly easy.  If k, if q, the reaction quotient,  wherever you start is bigger than k,  that means you've got too many products.  And that means that if you start at this condition  with too many products,  the reaction is actually going to go backwards.  So the reaction shifts to the left, to the products.  Excuse me, to the reactants.  Chemists always write reactants on the left.  If q is smaller than k,  that means we haven't made enough products yet,  then the reaction will run forwards  until it makes enough products  so that q comes up to the value of k.  And once the water lines are equal,  it's just going to sit there.  And if q equals k,  nothing further happens  that we can see in terms of measuring concentration.  They still run both ways.  But there's no net change.  Still lots of chemistry occurring,  but no net change.  And let's do one example.  So at 1400 Kelvin, kc is 4.7.  This is the first stage of the Haber process  where you take methane and water.  Those are two things you can get a hold of,  and you heat them up, and you get carbon monoxide and hydrogen.  You get three moles of hydrogen.  And because you're making more moles of gas here,  you predict that this reaction is going to be favorable,  roughly speaking, if you heat it,  because of the increased delta S.  And that's why you heat it up.  So you heat it up like crazy, you're making this.  And if we know this value, it's 4.7,  and we start out with this concentration,  0.035 molar H2O, 0.05 molar methane,  0.15 molar carbon monoxide,  and 0.2 molar H2.  Which direction is it going to run?  Is it going to make us more of the hydrogen,  or are we going to lose our hydrogen?  Well, we just put in the reaction quotient,  and we have to put in the hydrogen cubed,  because there's a three,  and then we put in the numbers,  carbon monoxide, 0.15, hydrogen, 0.2 cubed,  0.035, 0.05, and we get Q is 0.69.  Aha! But K is 4.7,  so because Q is less than K,  we're going to make more hydrogen if we run this reaction.  We can start like this, and sit there,  and we will get more hydrogen out than what we started with.  If Q were bigger than K,  then we would have too much of something there,  and then it would run backwards.  Okay, I think that's plenty for today.
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