Lecture 16. Chemical Equilibrium, Part III
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Blake, Donald R.
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've received several emails from students asking me to think about changing their scores because of problems on their exam. At least a dozen, maybe two dozen, have failed to attach their exam. I refuse. I don't have access to your exam folks. I can't believe that somebody would send me an email and say, I deserve three more points on this problem, two more on this one, and one more on this one, and not include an exam. Like I'm just supposed to believe you. So you haven't told tomorrow, get me your exams, get me your questions. Anything after that, you're out of luck if there's a problem. Okay? All right. Just a little quick review. You seem to all, well most of you, many of you are having problems with trying to determine what the equilibrium concentrations for a particular reaction are when I give you the K. We're going to do a few of those today, but folks, you just got to practice, practice, practice. Okay? I can't do it all for you. All right? So number one, these notes were posted yesterday, express the equilibrium concentrations of all species in terms of the initial concentration and a single unknown X, which represents the change in concentration. Write the equilibrium constant expression in terms of equilibrium concentrations, go on the value of the equilibrium constant, solve for X. Having solve for X, calculate the equilibrium concentration for all species, and then treat units, I don't care about units. Okay? This is the one time that I'm happy with just a number. So once again, if I ask you what is the equilibrium constant for this, and I give you the concentrations at equilibrium, you can give me a number of 16. Okay? No units are needed. All right? 16 tells me where things lie in terms of the reaction scheme. Do this. Equilibrium constant for the decomposition of solid NH4Cl is Kp equals 0.0100. Folks, look at this question and tell me how many significant figures am I expecting you to give me? If I give you 4, let me say 3, 4 sig figs, 1, 2, 3, 4, okay, is what I'm expecting. If I hadn't put those 2 zeros there, then it would be 1 sig fig, okay? On the exam, folks, on the exam, if I gave you 4 sig figs, and you gave me an answer of with 3 sig figs, 4 sig figs, or 5 sig figs, you got it right. You got it right, okay? You could be off by 1. All right? But some of you gave me 7 sig figs for something I gave you 2. Okay? Yes? Okay, maybe it's 3. What is it? Let's take a poll. 3 or 4? 3 or 4, okay? Probably 3. But 3 or 4 would work on the exam, okay? I'm giving you some leeway. Do the problem, hurry up. Quit asking me about sig figs. I would be smart enough to have given you a 1 right there. Then there would have been 4, for sure. For those of you who just showed up, hurry up. This is the question. You're missing out if you're not doing it yourself with a calculator. So what's the first thing you're going to do here, folks? What's the first thing you should do? Write the equation. There's the equation. Notice, folks, that this is a solid. Okay? That's going to tell you something, I hope. Conclude. So you started with only this, right? Solid this is what you start with. You got none of these. What do you think? 3 or 4 sig figs? 3 or 4 sig figs. What do you think? 3 or 4. Everybody should be doing this problem, folks. So you started with this. This doesn't count in the equilibrium constant calculation. So the initial, you had this. None of this. How much of this are you going to have when you finish? You can end up with x. For every one of these that breaks up, you make one of these and one of these. So you end up with a little bit of this and a little bit of that. In the end, that's what you've got. So since I post these notes, it's not a big surprise what the answer is, right? How do you set this up now? Okay? It's going to be k sub p is equal to this, which is equal to this times this over 1. Okay? We give this solid to 1 or you just forget it. So it's going to be x squared is equal to 0.0100 or you solve for x and x is equal to 0.1. Just like I said, 3 significant figures. 3 I said. Okay. So x is equal to 0.1. Folks, I cannot make it any simpler than this. This is the most simple one. I give you so you don't have to worry about what's on this side and there is nothing to start with on the other side. So everything that goes over is x. Do this one. At a certain temperature, the equilibrium constant k sub c for the reaction Br2 goes to 2 Br is 1.1 times 10 to the minus 3. Okay? In this case though, I'm being a little bit trickier. You have some of both. You got some of both. The question is, is what you've got exactly what you want? Well, do a quick q calculation and find out. You don't have to, but that's how I do it. So you've got some of this and you started with that much. 0.063 molar. Okay? And it goes to this. And this you started with 0.012 molar. Okay? What's q? Anybody do a q? What did you get? 0.33? 0.33. Really? What did you get? Is it really 0.3? Do you make sure you square that side? Sounds kind of wrong. What did you get? 0.002. So 0.002 is smaller than 0.002. I thought it was going to, I mean, just my own gut. Okay? You've got about, you know, five times more of this than this, but that's kind of a small number. So I would've thought things were going to go in this direction. Okay? Which means q is bigger than k. Is it? Okay. Perfect. Good job. So then what's going to happen with this? Okay? What do you set this thing up with? So you have the initial concentrations. Okay? I gave you this, I gave you this. Okay? Now in this case, in this case folks, they didn't do a q. Okay? They just, they, and you don't have to. You don't have, I like to, I like to know which way the equation is going, or the reaction is going, but you don't have to. You can just say, I'm going to say that this is going to be a minus. I'm going to lose one of these. I'm going to make two of these. It absolutely doesn't matter. Okay? So you say, but you've got to have these be opposite. If this is a negative, then this is a positive. This is a two. If you say, no, I want this to be a plus x, then this is going to be a minus 2x. Okay? And it all comes out in the quadratic. You don't have to worry. So in this case, it's set up and that says, okay, we're going to say that x, we're going to say is, every one of these you lose, you make two of those. So that's the change. Here's the equilibrium. Okay? Set it up like this, and then you plug that in to your equation. Okay? Which is that? There you go. How many got that far? Okay, good. So you've got k sub c is equal to that whole mess right there. Okay? You solve for x. You expand it out. You set it equal to zero. We back to the old quadratic. Okay? Lucky us. So how many got this far? How many need more time to get that far? Okay. How many don't have a clue? Okay. I saw one hand. So multiply that thing out, add stuff, subtract stuff, put it together, and then set it equal to zero. Once it's set it equal to zero, then you've got your a, b, and c for the quadratic. Folks, if you get it this far on an exam, yes. What? That should be, well, no, no, no, no, no, because we said the other side was a minus. So what I said, it doesn't matter. Folks, if you make this a plus and this a minus or this a minus and this a plus, you're still going to get the right stuff here. Trust me. Trust me. You can say x is whatever you want. Okay? Whether it's coming this direction or that direction, it's going to then give you the opposite. Okay? Maybe we should do it two ways. Okay? Just do this. Okay, who's got an answer? Yes? Okay, so let me do this. Okay? Let's do this problem. Okay? Get an answer. Just do it. And then we'll do something else. So you've got an a, b, c right there. You need to, what I was saying was is that on an exam, if you say, you know, the quadratic just, I can't do it. Okay? Then as long as you say that a is four and b is that and c is that, you'll get some points. Okay? You're not going to get full credit. So you put in the quadratic, you get two answers. Okay? You get two answers. This one and this one. Okay? You can do it a couple ways. You can plug them both in if you want. Okay? So if you want, you can plug this one in, try and solve it, and then this one. The problem is, folks, is that if you look at this, that number right there, 0.0105, okay, that's a negative, right? If I multiply that by two, what do you get? You get 0.021, which is bigger than that. That's an imaginary number, folks, okay? You cannot do this. You cannot have a negative concentration. Okay? No negative. So this one is no good. That's the root that's no good. So this is the value you're going to put in there. Okay? So this is what x is equal to. In this case, x is equal to minus that. You plug it in, and then you come up with that. Look, I got this, so it's going to be, where have I got it? I hope I got one more to go. Okay, so this is what Br is. This is what Br2 is. Notice, folks, notice this got bigger. This started at 0.062, and it actually got bigger. So do me a favor. Do me a favor, okay? I'm going to go back. Okay, these are the answers. Look at the answer right there. The same value right there is the answer for Br2. Okay? Let's go back. Let's go back and start over. Okay? This time, and it's up to you guys to do it because I'm not going to do it, this time I want you to make that a plus x and that a minus 2x. Just do it. This is a plus x, this is a minus 2x. Do it. Do the whole problem again, and guess what? The sign is going to change. Okay? Yes? Absolutely are the products on top. Absolutely. It's always products over reactants. Yes? Really loud. Loud. Geez Louise. Shout. That means it's got to go this direction. Yes? If Q is bigger than K, then it's got to go this direction. Just do the problem. Geez Louise, folks, please just do it. I told you it doesn't matter. Just do it. Make this a plus x and make that a minus x and then this will show you that it doesn't matter. If Q is smaller than K, it goes that way. If Q is bigger than K, it goes that way. But I'm just saying I want you to see that it doesn't matter in this case whether you say that this is a minus x or a plus x. So just do it. Solve the problem again. I'm trying to help you folks, okay? Okay. So folks, in your equation, you should have this setup like this and this is going to be 0.012 minus 2x and this is going to be 0.063 plus x. Okay? Do it. Expand it. Solve for 0 and you'll end up with something like this. Okay? So who wants to tell me what do you get for this part when you switched them, like I said? Who's got that answer? Who's got an answer? Yes? 4x squared minus. So this is a minus and then a plus. So you plug those into the quadratic equation and out comes a number. Okay? And you see that when you have that number then, it's the same number but with a minus sign. Okay? Or a plus sign, whatever it was. So the values, so this folks just turns into a plus. This turns into a plus. You add that to this. You subtract that from that times 2 and you get the same darn numbers. So it doesn't matter as long as you get it right with the x and the 2x. Okay? As long as you keep track of the number of moles, it doesn't matter what you say. The quadratic is going to solve it for you. Yes? So it's not going to be that you need to have q. You don't need to have q. Yes? You don't need to have q. Now for me, I just kind of always like stuff to go from the left to the right. Or at least I like to know whether something is going to be eaten on one side or produced on the other. Okay? But you're right. You don't need to have q and if q is a calculation that's going to take you too long then don't do it. Okay? You can solve it with this being a plus x and this being 2 minus, minus 2x or the other way around. I like q. Okay? I like to know where things are headed before I do the problem but you don't have to. Okay? So do you understand this now? It doesn't matter. As long as you get the x plus or minus and that is a 2x. If it's just cubed, it would be a 3x. Alright. We're switching gears. How many of you heard of Le Chatelier? So what Le Chatelier discovered, hypothesized and discovered was that you can shift equilibrium by playing with products or reactants. And this is extremely important in industry. So look at this right here. N2 plus 3H2 goes to 2NH3. So if I said I want more NH3, tell me one thing I can do, just one to make that happen. Raise your hand. Yes? Wait, wait, wait. I want more of this. You say you want me to add more of that? Well, it looks like I want gold and I've got to put more gold in. That is not how equilibrium works. I want more of this. Okay? I can't add it. Yes? Less liquid. No. Try again. Look at equilibrium. What do I have to do to make this get more and more? Yes? Add to the reactant side. Yes, that's what we learned because this, folks, is going to be products over reactants. So the way to force things in this direction is to add more. Okay? So I add to this side, it forces things this way. Okay? Do we understand that? Okay. What is something else I could do? Yes? I could increase the volume and what would that do? That's not what I want. I want this. Decrease the volume. Okay? So the way the book, I don't know, decreasing volume, we have to understand, folks, that means increasing the pressure. Okay? You increase the pressure on something. Count how many moles. How many moles of gas are there on the left side? How many on the right side? Two. So in a situation where I'm going to be putting pressure on something, okay, I can force things to the side that has fewer moles. Okay? If I put pressure on it, the side with fewer moles is the side that is favored. So if I start pushing on this thing, I can force things to go this way. Okay? Let's say that this was an exothermic reaction. So we got like heat over here. Okay? So we need to write it almost like it's part of the reaction. Okay? So it's this, this, this, this, plus heat. Okay? Then what's another way that I can make this reaction come this way? So if heat is on the right side, that's almost like it would fit into the equation, you know, where it would be this squared on the top and then heat on the top divided by these guys on the bottom. Okay? So if I want to have it come this way more, what do I do if it gives off heat? Decrease the temperature. So stick it in ice water or something. Okay? So if you cool it down or let's say this has a, not a very high boiling point, what if we just stick the reaction in ice water? Okay? And this ends up liquefying. Okay? We can just keep things coming this direction. Okay? These gases do not liquefy at ice temperatures. So if you have something here, you can actually condense this out or you can just have it be colder if it's exothermic. So there's lots of things we can do. We can change the temperature. Okay? We can change the volume or the pressure or we can change the number of moles. Okay? I can double them out of N2 and I force things that way. Okay? So there are lots of things that we can tweak and L'achatelier was the one that sort of pioneered this work. Okay? Well, it's going to, if we add, now like this guy was saying, if we add NH3, what happens? Comes back this way. Okay? This is just a little cartoon to illustrate. So here we are at equilibrium for this reaction. N2 here, NH3 here, H2 there. All right? And we're flat versus time. There's no change in concentration so we're in equilibrium throughout this whole time. And then all of a sudden right here we dump in twice as much NH3. Dump in, yeah, about twice. Okay? And what happens? Okay? It gets dumped in but then it slowly goes part of it away. Okay? Why do these other ones grow? Why does the H2 grow three times faster than the N2? Three moles. That's right. Okay? And this comes down twice as fast as that goes up and two-thirds as fast as that goes up. Okay? Because that's got a two there. So this eventually finds equilibrium here. There all of the equilibrium right there. Those are the new equilibrium concentrations. Okay? That, folks, is when we change N. Okay? We change N and the K doesn't change. Okay? The K stays the same. That is not the case if you change temperature. Okay? You can change some other things and then the K changes. And that's why in problems it'll say specifically you know at a certain temperature this is what, this is the K. Because I can change the K based on temperature. But by adding, changing N, the K stays the same. All right, now, last night students had a real hard time with this. I'm not sure why, but look at the colors. NO2 is this rusty color that we see in Los Angeles in the morning. Okay? The low part of the atmosphere if we look north from Irvine. N2O is colorless. Okay? So this has color. This doesn't have color. So there are going to be some questions that are asked about is something going to become more, you know, darker. Okay? So this reaction right now, if, if I say that all we do to this, I got it in a bottle, all I do to it is I increase the pressure. Increase the pressure. Okay? Which way is this reaction going to go? To the right. Because look at this. It says two moles. This has one. One mole is easier to contain in a constant in a certain volume than two. Okay? So things are going to shift that way. Okay? So, so what we do is we've said that we're going to be going this direction. So is this flask going to get darker? Going through and change the life, I guess. Adolescents. Finally. Or lighter. Lighter. Okay? So if it goes to the right, it gets lighter. And this is a concept that last night, I couldn't get through to them. Okay? So in this case, folks, we got the dark going to the light. Unfortunately, the light is the product side. And that seems to be the hang up in the way these questions are asked. Okay? It's exothermic. What else can I do to this thing? What if I want it to go this direction? And it's exothermic. What do I got to do? Cool it down. Oh, exothermic. What am I saying? Heat it up. So it's exothermic. If I heat it up, it goes this direction, right? It's exo- wait a second. See, I was right. It's exothermic for the first time. Exothermic means it gives off energy. It gets hotter. So if I add heat, it comes back to the left. You guys, I trusted you. You said decrease. Yeah. Okay, let's start over. I'm very confused. I could have said that if you have exothermic and you add heat, it's going to decrease. It's going to move things this direction. Okay? That's what I thought I was saying. So that's if we change the temperature. If I add some of this, what happens? Okay? I can add clear stuff and make it get darker. So we know this. Q is equal to K at equilibrium, right? Whatever K is, it's eddy-clearant, it's equal to Q. So, and I wrote something else in on your notes, it's not there, but the volume is double. And I said, which means the pressure is halved, okay? By opening the valve, as the system approaches the equilibrium, the color intensity will, what's it going to be? This is the problem. So, we know that Q is equal to K at equilibrium. So we know that Q is equal to K here. What happens when we decrease the pressure of everything by a factor of two? Okay? Everything's by a factor of two and now Q is equal to 2K. Okay? Q is equal to, it's not as big as K, right? Is that correct? Q is twice as big as K. So what does that, what happens? What does that mean? Which way is the reaction going to go? It's going to go left or right? Okay, because Q is bigger, so things are going to head back this direction, which means an increase. Because this side is the dark side. That's the problem, okay? It's a little bit confusing. It actually comes back. We think of the reaction going forward. Anyway, the chatty class. Alright, here's a cartoon of that, okay? This is the same exact question. You got your N, you're at equilibrium? You got your NO and you got your N204 right here. Okay? You change the volume. This is going to shoot up. Then you go here because the pressure changed, the Q changed. Now all of a sudden we're at equilibrium again. Okay? So this got darker. Okay? So once again I would have preferred that they chose something that, I like things that as the reaction goes further, you got color. The opposite is, any questions about this? It seemed last night, maybe I just did a better job today, or you're asleep. Let's talk about something else that the chatelier pointed out. Some of you want to be doctors, okay? And a problem with carbon monoxide is that if you breathe it, it likes to bind to hemoglobin. Okay? And I think not so much anymore because cars are so well tuned today. Do not try this at home, okay? But people used to kill themselves by sticking a hose in the exhaust pipe of their car and then run the hose into the cab of the car and then roll up the windows and just fill the car with CO, well quite a bit of CO and they would have all this carbon monoxide bonded to the hemoglobin instead of oxygen and they would die. Okay? Cars now almost give off like no CO. So you'd just be sitting a long time with a headache. So I haven't told you a story for a while. And this has absolutely nothing to do with this class but I'm going to tell you anyway because I just reminded myself. So I grew up on a farm and my dad had cancer and a lot of problems but I went to my parents' house one day and I see my dad's pickup in the tree orchard with a hose in the exhaust pipe and the windows rolled up in the truck and I thought, geez dad, did you really, were things this bad? So I pulled off the driveway real quickly and I run to the truck and there's my dad inside listening to his country western music. Turns out he had the hose going into a gopher hole. He figured he was going to do this reaction on a gopher's and for about 30 seconds I was scared to death thinking I was going to find my dad blue in the pickup. In reality, he was fine. Nice story. So what does the Chalier tell us about this? Here's the reaction. This is what is happening in your body right now. Okay? You've got hemoglobin taking oxygen from your lungs to the rest of your body that we use to burn up glucose and stuff. Okay? We need that oxygen. Alright? If however you get near something that's giving off a lot of CO, turns out that the binding energy that this reaction is, that's a stronger bond there than there. So preferentially this will bond. Okay? Now cyanide is a real bad one right there. Cyanide bonds to hemoglobin like not quite irreversibly but really, really, really well. Okay? So that's how we kill people in gas chambers. Yeah, yeah. You just drop some hydrogen cyanide, some potassium cyanide in some water. Out comes cyanide, grabs onto the hemoglobin. It's true because this is almost irreversible when you have cyanide. Fortunately with CO it's not. Okay? So how would we try and help somebody who had had CO poisoning? Okay? Well look it. Here's the equilibrium right here. Okay? Products reactants. Alright? How would we force this reaction back that way? Okay? In order, how would we get rid of the CO and make it be free CO? Okay? Well look, we've talked about a variety of things. You can add, you can increase the N. Okay? You can make something higher concentrations on this side but we've also learned about pressure and we've learned about volume and we've learned about temperature. Alright? So the question is how would you do this? Would you increase the pressure of oxygen, cool off, or remove hemoglobin? That I think would require you to slit your wrist and bleed to death. That's the only way I know how to get rid of hemoglobin is to bleed a lot. Not a good idea. So I don't think that, that should even be on there. That gives you bad ideas. Cool off! You know you can't cool very much. You're 37 degrees right now and if you get to like even 35 or 34 you start getting kind of shaky. So that's not a good idea. So what is it, look at this. How do you increase, what do you do? Instead of breathing 21% oxygen you breathe more. Okay? 100%? Well you know I think you can do that for a few breaths. Maybe more. But back in the old days they used to take babies that were born just a little bit premature. This is back in the 50s and put them in like these things where they had high concentrations of oxygen to try and help them whatever and it made them blind. A lot of blind kids back in the 40s and 50s because of this high oxygen that they were given at birth. So I don't think you can just breathe, I got all kinds of bad stories today. I had two of them in my class. Two kids out of 30 in my class were blind because of that. So anyway. So we increase this, force things this way. Yes? What's that? If the moles are the same on both sides you mean the number of moles. Yes? No, it won't matter. That's right. Her question is is that if you have the same number of moles, okay if we're talking about gases here, if you've got the same number of moles on either side or then change of pressure there's nothing. What's that? Is that the same number of moles? Is this the same number of moles? Looks like the same number of moles. So would it increase the pressure and not do anything? That's right. Increasing the pressure overall but increasing the amount of this forces it this way. So we're changing the end and not the pressure plus you know it's not a good idea to get inside like a several atmospheres of pressure and be pressurized. Lots of bad examples here of things not to do. Okay. So yeah so all we're doing is changing N. Okay so once again I said you can change you know N, P, T or V. And in this case since you've got a whole body like this you don't really want to be changing T or P or V. At least if it's my body I don't want that happening. You can do whatever you want. Okay so Le Chatelier shows us how we can force this thing this way. That of course is assuming folks that this is really reversible. For CO let's say in this reaction the K is 10. Okay for this reaction K is equal to 10. What does that mean? That means things lie on this side right? Okay. Things lie on this side. Maybe it's not quite 10 but let's say it is 10 and by increasing this by a factor of 2 or 3 then we can shift things back this way and make the K go that direction. If instead for cyanide the K is 10,000 then adding a whole bunch of oxygen here really doesn't shift the equilibrium very much. So you're a goner. And cyanide smells like almonds so be careful when you smell almonds. Alright here's another one. The temperature for N2 is raised. What happens? The reaction is exothermic. So this means folks that it's NO2 goes to N2O4 plus heat. Exothermic. Heat on this side. So if I add heat which way is things going to shift? That direction. Okay. Which then means is it going to increase in color, stay the same or decrease? Increase. Yeah. Because it's going a lot. I'd like them to say are you going to go to reactants, not change or go to products? The whole color thing I think is confusing because once again your reactants are what are starting with color. Have you already answered this question? Perfect. So once again if you look at it the K with temperature as you increase temperature K is going to get smaller. Alright. If you have an exothermic K is smaller. If it's endothermic the more temperature you give it the better it likes it. Alright. So this is the exothermic and endothermic K versus temperature plot. And this is just to say that for this reaction we've been talking about not the N2O1 but the N2 plus H2 that in order to make this and folks this it's not the holy grail but for some reason in industry we like a lot of ammonia. Okay. We use ammonia and fertilizer and lots of other things. So making this is profitable particularly because this is free. Alright. So if you have some of this you get it somewhat purified. You have some hydrogen. You can make it go this way. The problem is K is very small. Okay. I think I said I didn't say oh it means okay. Anyway it's very small. So what do you do to make it go that direction? Okay. Look at this. We've got four moles of gas on this side and we've got two moles on this side. So what are we going to do? Increase the pressure. Okay. So it turns out that this is exactly what they do and when they do this in industry it's done at 500 degrees but look at that. 200 atmospheres. Huge pressure. And you know what they'd probably like to go to 300 or 400 but it's just too expensive. You'd have to have a vessel that would handle you know it wouldn't be worth it. So it's just not cost effective but if they had something that could go to that pressure they would. All right do this. So the equilibrium folks, these are the equilibrium concentrations. Two, two and one. And then I'm going to somehow suck all of the A out. All of it is just immediately gone. So it has to equilibrate. So the first thing you've got to do folks is figure out what is the K. Since this is an atmosphere this is K sub p. So now you have to set it. So you say okay fine I've got my K so now I've got to figure out what are my equilibrium because that's the question. What is the equilibrium concentrations right? What are the pressures after the system to equilibrate? Well look folks you had A going to B and C. I think that works yes. And then all of a sudden you yanked all the A out. All of the A is gone. So all you have is product. What's going to happen? Some of it's going to have to go from product back to reactant. It's got to go in this direction. So you set it up like that. You started with two you're going to lose X. You started with two you're going to lose X and you're going to make X. So those are the after equilibrium. So now you just go ahead and plug that in for case of p it was equal to so it's going to be 2X squared. Each of these is a 2 minus X squared over X. So it's this times this which is just squared over that. So X squared is going to equal this right there. So you should be doing this folks. Yes. Loud. They can't hear it. What are you saying? No. The pressure of A is zero. So you had five at equilibrium you had five atmospheres. You had two of B and two of C and one of A. Okay. And you're at equilibrium. Then you somehow sucked out the A. But you still have four atmospheres. So now it's got to re-equilibrate. And that's where the X comes in. So you lost some X from this. You lost some X from this. And you made X here. So this is the equation. Okay. You do this was four. You calculated that in the beginning. And now you've come up with this. And once again. These I'm going to tell you. These are the two values that you can have for X. Once you've done the quadratic. So I'll give you a second. Yes. Why is there this? Oh, that should be a PC. Sorry. Okay. So folks, once again. You got two answers for the quadratic. And one of them makes no sense at all. How can X be 7.5 when you only started with two? Okay. So this is some common sense that you've got to use. Okay. So the only X that makes sense is this one's about 0.5. So what it's saying is, is that you're going to subtract about 0.5 from two. All right. And you're going to have X of 0.5. So just to humor me. Plug that Vos values back into the original right at the very beginning. Say, okay. I'm going to say that the pressure of A is 0.53. The pressure of B and C is going to be 2 minus that. Plug it in and see if you don't get four. Okay. Always, if you have a chance, plug your values in and see if they give you back the right answer. Okay. You should get, if you put in 0.5, if you put, make that a 0.5 and make that a 1.1 and a 1.1, you're going to get very close to four. So once again, I don't know how they sucked this A out like that. But they did. And this is a nice question for an exam because I don't have to give you K. Okay. I don't have to give you K. You figured out K early on. And then I said, oh, now I'm going to take away something. So then you had to do all this kind of logical thinking quadratic. I like this kind of problem. That's all I'm going to say. Okay. So this folks is going to take us one minute. It's in your notes. It should be something that is second nature. Okay. That if you change the concentration of products, which means you're over here, which ways are going to shift equilibrium? This direction. You always go in the opposite direction of where you add stuff. Okay. Okay. If you decrease the concentration of products, which ways are you going to go? So I want to hear loud, left or right? Right. Okay. Increase the concentration of reactants. Right. Decrease the concentration of reactants. And you guys are so excited about this. Will the equilibrium concentration of NO2 increase or decrease or stay constant if we? So we're always looking at NO2. Okay. So here's the first question. I want to know what's going to happen to NO2. Increase or decrease if we increase NOCl. What's it going to do? That's right. How about if we increase NO? Increase NO. What's it going to do? Okay. Increase NO. Hopefully you'll say the opposite. Increase NO2Cl. What's it going to do? Okay. Good. All right. And this is just the same stuff. Okay. I'm not even going to go over this. I'm so bored. Okay. So study this on your exam. I mean on your stuff. So look at it. I will do this little thing at the bottom. Okay. So look once again. This is exothermic. Okay. It's exothermic. And once again, the products are colorless. So if it's exothermic and we add heat, which side is equal? We're even going to go on. This side or this side. Okay. If we add heat, we're adding something on the product side, shifts things this way. Right? This way, okay, is darker. If this has color, this doesn't. At room temperature, it looks this way. Look what happens when we make it colder. Colder is just the opposite of hot. So for something that's exothermic, if you cool it down, it forces things more this way. More this way means more into O4, less NO2, less color. Don't forget, folks, changing temperature changes the K equilibrium. Actually, changes the K. All right? By adding more of this, it doesn't change the K. That's like a good multiple choice question. All right. Now we're going to shift gears back to Delta G. So this is the beginning of the beginning of the end. Yes? Do you know why your body turns pale when you start doing the wearable? No. I have to think about that. I turn pale for lots of reasons. I was going to say something, but since we're being filmed, I better not. Okay, so back to Delta G. I love Delta G because it allows me to not have to worry about entropy. I love disorder, but I have a hard time knowing whether the whole world and universe is being disordered because of something I do. Delta G tells me, you got a negative Delta G, it's spontaneous. You got a positive Delta G, it's spontaneous in the other direction. Okay? Now Delta G is equal to products minus reactants, and it's zero in equilibrium. So we got this equation again. Okay? So Delta G, products minus reactants. Okay? We've got, I can't do this back here. So the products is this side. So we've got, the products is going to be two times the standard G plus RT log A, and then it's going to be minus the reactants. Okay? So you're going to get two more of these things up here. Okay? Now we haven't learned the Delta G is equal to minus RT log K, which is what's coming up here. Okay? Now, let me just tell you something. I told you about four weeks ago that you cannot take the log of a unit. It must be a number. Okay? So in order for us to do these things with K in here, okay, this is actually going to be a K at some point. Let me, look at this. Okay? The A here are what we call the activity. Okay? And I don't like to use activities, but I have to explain to you why we use this. Because if you look at this, you've got two moles here times three moles times one mole. You've got a total of four moles on the bottom and four moles on the top. And these are in atmospheres. So in theory, this would give you, if each one was at one atmosphere, this would give you, you know, one ATM squared divided by three ATM square, or by one ATM cubed and one ATM, you'd have units in your log. Okay? Can't do that. So we, somebody came up with this idea of activities. So it's actually, these activities are a pressure divided by a pressure. Okay? So you get rid of the units. So all you have is a value right there. So in this case, and this is all you really need to know, it's just pressure squared over pressure cubed times this, but these are now not units. Okay? We've removed the units. Okay? By using these A's. This is what you need to know. Okay? The delta G is equal to minus RT log K. Okay? So, folks, if I want something to be spontaneous, what does RT log K need to be? Just this part, not the minus. RT log K has to be what? If I want something to be spontaneous, it's got to be positive because we're subtracting it then, okay? So as long as an RT log K is a positive number, then it's spontaneous. Everybody take out your calculators, take them out, everybody. Find the LN button, looks a lot like that. LN, natural log. Okay? So what I'm saying is that as long as this part from here to here is positive, then this is spontaneous. T is always positive. T is Kelvin. Kelvin, the absolute zero is zero. That's just cold. We can't get there, but that is the value. There's no negative temperatures. It's not minus 50 degrees. It's minus 50 degrees in North Dakota because we use Fahrenheit. But in Kelvin, it's a balmy, you know, 220 degrees. Okay? So this is always positive, and we know R is positive. It's either 8.3 or 0.082, but it's always positive. So the only thing, folks, that determines what the sign of the RT log K is, is this. So I want you to find out what is the LN of one. Do it. For those of you who think you know or just believe that these other people are right, it's important for you to do that. Okay? So at equilibrium, at equilibrium, what is K? What is that equal to at equilibrium? When delta G is zero, K is equal to one. Okay? Now, punch in 0.9, log of 0.9. What is it? Negative point something. So a negative time is a negative gives me what? So if K is less than one, then delta G is positive. Type in 1.1. It's not a huge number. What is the log of 1.1? Point zero something, but it's positive. So that whole thing is positive, therefore delta G is a negative. Okay? So please understand. I'm sorry that I'm dumbing this down, but you have to understand that's what this whole thing is about. Okay? So folks, in an example I could give you, I could say what is the K of this reaction? Let's do it. What is the K of this reaction if delta G is equal to, no, let me say what is delta G? Make it easier. No, that's not what I want to do there. What is delta G if K is two? At room temperature, 298. What is delta G? First of all, you know it's going to be negative, right? You know because anything over one means that the log is a positive number. Anything less than one, it's a negative number. And at one, it's zero. So I can ask you a question as simple as that. I can give you the temperature, you know what R is, and I give you K, figure out G. Okay? Very simple. Or I can give you something more complicated, right? We also know that delta G is equal to minus, delta H minus T delta S. Do we remember that? Delta G is equal to delta H minus T delta S. So now I could say delta H minus T delta S is equal to minus RT log K. I can get fancy. Okay? These are those activities. Look at it in the book. Okay. Once again, we know that if delta G is less than zero, K is greater than one and spontaneous. Delta G is greater than zero, K is less than one and not spontaneous. And if delta G equals zero, K equals one. Delta G equals one. You know that I put a minus in that last night? It's weird. It didn't show up. Okay, this is a minus right here. It's weird. It was there, then it disappeared, now it's gone again. So okay, well anyway, there it is. I fixed it. Now, we also learned when we were doing the Clausius Clapperon, when we had a log of something, we had a log of P2 over P1 and we were solving for a P that the only way that you can get something out of the log is to take it to the E. Okay? So let's say I've got, I want to get rid of this log K. All right? If I say I want log K on one side, I can say delta G over RT minus delta G over RT is equal to log K. Do everybody see that? I can bring this to this side and I get delta G over minus RT is equal to log K. Then I can take it out of the E power and I get this. K is equal to this. Okay? So folks, there are going to be times I give you delta G and a temperature and you're going to give me K. You can do it through this way if you want. It doesn't matter, okay? This is a very, very handy equation to know. There will be something using this on the exam. Okay? What are delta G and not NK for this at 25 degrees C? Room temperature. Okay? At room temperature, folks, I have water. Okay? Is it boiling at room temperature? That means it's not spontaneous, which means what? The delta G must be greater than zero. Right? Let's see if I'm right. NK must be less than one. Oh, look at that. I'm incredible. Uh-oh. I warned you about this. The delta G is equal to delta H minus T delta S. So I can set that equal to that. Yeah. Gotta love that. All it is is a substitution, no magic. Just substitute one thing for another. All right? So if I want to have this be the log of K, I think this is going to be log K over 1 over T. Okay. What would that plot look like? Here is the log of K over 1 over T. Here's the log of K right there. Okay? I've got delta H here. I've got R. I've got all this stuff, but I'm plotting it versus 1 over T. Log of K versus 1 over temperature. Okay? Maybe this one would be better. Okay? You've got K over T. What happens to this? Can you predict what it's going to look like? If it's exothermic, then 1 over T, the log of K gets bigger. Okay? In reality, we know that if something's exothermic, that means there's energy given off on the product side that as we increase the temperature, we can shift the K smaller and smaller and smaller as we change temperature. Yeah, that's what this means. Okay? Sometimes plotting is 1 over T, you get a nice straight line. Now, folks, I should be able to give you any kind of a plot. This is 1 over T. I should be able to give you a delta H, and you should be able to plot that. Okay? That's endothermic. Oops. Okay? Once again. Endothermic, we know that K gets bigger as we heat things up. Okay? Put energy in on the left side, forces it to the right. It needs energy to go, and look at that. Look how the K gets bigger and bigger and bigger as we change the temperature. Plotting things 1 over T, oftentimes will give us a straight line. Okay? I think we should quit. All right, folks. There's one more lecture.
 I've received several emails from students asking me to think about changing their scores  because of problems on their exam. At least a dozen, maybe two dozen, have failed to attach  their exam. I refuse. I don't have access to your exam folks. I can't believe that somebody  would send me an email and say, I deserve three more points on this problem, two more  on this one, and one more on this one, and not include an exam. Like I'm just supposed  to believe you. So you haven't told tomorrow, get me your exams, get me your questions.  Anything after that, you're out of luck if there's a problem. Okay? All right. Just a  little quick review. You seem to all, well most of you, many of you are having problems  with trying to determine what the equilibrium concentrations for a particular reaction are  when I give you the K. We're going to do a few of those today, but folks, you just got  to practice, practice, practice. Okay? I can't do it all for you. All right? So number one,  these notes were posted yesterday, express the equilibrium concentrations of all species  in terms of the initial concentration and a single unknown X, which represents the change  in concentration. Write the equilibrium constant expression in terms of equilibrium concentrations,  go on the value of the equilibrium constant, solve for X. Having solve for X, calculate  the equilibrium concentration for all species, and then treat units, I don't care about  units. Okay? This is the one time that I'm happy with just a number. So once again, if  I ask you what is the equilibrium constant for this, and I give you the concentrations  at equilibrium, you can give me a number of 16. Okay? No units are needed. All right?  16 tells me where things lie in terms of the reaction scheme. Do this. Equilibrium constant  for the decomposition of solid NH4Cl is Kp equals 0.0100. Folks, look at this question  and tell me how many significant figures am I expecting you to give me? If I give you  4, let me say 3, 4 sig figs, 1, 2, 3, 4, okay, is what I'm expecting. If I hadn't put those  2 zeros there, then it would be 1 sig fig, okay? On the exam, folks, on the exam, if  I gave you 4 sig figs, and you gave me an answer of with 3 sig figs, 4 sig figs, or 5 sig figs,  you got it right. You got it right, okay? You could be off by 1. All right? But some  of you gave me 7 sig figs for something I gave you 2. Okay? Yes? Okay, maybe it's 3.  What is it? Let's take a poll. 3 or 4? 3 or 4, okay? Probably 3. But 3 or 4 would work  on the exam, okay? I'm giving you some leeway. Do the problem, hurry up. Quit asking me about  sig figs. I would be smart enough to have given you a 1 right there. Then there would have  been 4, for sure. For those of you who just showed up, hurry up. This is the question.  You're missing out if you're not doing it yourself with a calculator. So what's the  first thing you're going to do here, folks? What's the first thing you should do? Write  the equation. There's the equation. Notice, folks, that this is a solid. Okay? That's  going to tell you something, I hope. Conclude. So you started with only this, right? Solid  this is what you start with. You got none of these. What do you think? 3 or 4 sig figs?  3 or 4 sig figs. What do you think? 3 or 4. Everybody should be doing this problem,  folks. So you started with this. This doesn't count in the equilibrium constant calculation.  So the initial, you had this. None of this. How much of this are you going to have when  you finish? You can end up with x. For every one of these that breaks up, you make one  of these and one of these. So you end up with a little bit of this and a little bit of that.  In the end, that's what you've got. So since I post these notes, it's not a big surprise  what the answer is, right? How do you set this up now? Okay? It's going to be k sub p  is equal to this, which is equal to this times this over 1. Okay? We give this solid  to 1 or you just forget it. So it's going to be x squared is equal to 0.0100 or you  solve for x and x is equal to 0.1. Just like I said, 3 significant figures. 3 I said. Okay.  So x is equal to 0.1. Folks, I cannot make it any simpler than this. This is the most  simple one. I give you so you don't have to worry about what's on this side and there  is nothing to start with on the other side. So everything that goes over is x. Do this  one. At a certain temperature, the equilibrium constant k sub c for the reaction Br2 goes  to 2 Br is 1.1 times 10 to the minus 3. Okay? In this case though, I'm being a little bit  trickier. You have some of both. You got some of both. The question is, is what you've got  exactly what you want? Well, do a quick q calculation and find out. You don't have to,  but that's how I do it.  So you've got some of this and you started with that much. 0.063 molar. Okay? And it  goes to this. And this you started with 0.012 molar. Okay? What's q? Anybody do a q? What  did you get? 0.33? 0.33. Really? What did you get? Is it really 0.3? Do you make sure  you square that side? Sounds kind of wrong. What did you get? 0.002. So 0.002 is smaller  than 0.002. I thought it was going to, I mean, just my own gut. Okay? You've got about, you  know, five times more of this than this, but that's kind of a small number. So I would've  thought things were going to go in this direction. Okay? Which means q is bigger than k. Is it?  Okay. Perfect. Good job. So then what's going to happen with this? Okay? What do you set  this thing up with? So you have the initial concentrations. Okay? I gave you this, I gave  you this. Okay? Now in this case, in this case folks, they didn't do a q. Okay? They just,  they, and you don't have to. You don't have, I like to, I like to know which way the equation  is going, or the reaction is going, but you don't have to. You can just say, I'm going  to say that this is going to be a minus. I'm going to lose one of these. I'm going to make  two of these. It absolutely doesn't matter. Okay? So you say, but you've got to have these  be opposite. If this is a negative, then this is a positive. This is a two. If you say,  no, I want this to be a plus x, then this is going to be a minus 2x. Okay? And it all  comes out in the quadratic. You don't have to worry. So in this case, it's set up and  that says, okay, we're going to say that x, we're going to say is, every one of these  you lose, you make two of those. So that's the change. Here's the equilibrium. Okay?  Set it up like this, and then you plug that in to your equation. Okay? Which is that?  There you go. How many got that far? Okay, good. So you've got k sub c is equal to that  whole mess right there. Okay? You solve for x. You expand it out. You set it equal to  zero. We back to the old quadratic. Okay? Lucky us. So how many got this far? How many  need more time to get that far? Okay. How many don't have a clue? Okay. I saw one hand.  So multiply that thing out, add stuff, subtract stuff, put it together, and then set it equal  to zero. Once it's set it equal to zero, then you've got your a, b, and c for the quadratic.  Folks, if you get it this far on an exam, yes. What? That should be, well, no, no, no,  no, no, because we said the other side was a minus. So what I said, it doesn't matter.  Folks, if you make this a plus and this a minus or this a minus and this a plus, you're  still going to get the right stuff here. Trust me. Trust me. You can say x is whatever you  want. Okay? Whether it's coming this direction or that direction, it's going to then give  you the opposite. Okay? Maybe we should do it two ways. Okay? Just do this. Okay, who's  got an answer? Yes? Okay, so let me do this. Okay? Let's do this problem. Okay? Get an  answer. Just do it. And then we'll do something else. So you've got an a, b, c right there.  You need to, what I was saying was is that on an exam, if you say, you know, the quadratic  just, I can't do it. Okay? Then as long as you say that a is four and b is that and c  is that, you'll get some points. Okay? You're not going to get full credit. So you put in  the quadratic, you get two answers. Okay? You get two answers. This one and this one.  Okay? You can do it a couple ways. You can plug them both in if you want. Okay? So if  you want, you can plug this one in, try and solve it, and then this one. The problem is,  folks, is that if you look at this, that number right there, 0.0105, okay, that's a negative,  right? If I multiply that by two, what do you get? You get 0.021, which is bigger than  that. That's an imaginary number, folks, okay? You cannot do this. You cannot have a negative  concentration. Okay? No negative. So this one is no good. That's the root that's no  good. So this is the value you're going to put in there. Okay? So this is what x is equal  to. In this case, x is equal to minus that. You plug it in, and then you come up with  that. Look, I got this, so it's going to be, where have I got it? I hope I got one more  to go. Okay, so this is what Br is. This is what Br2 is. Notice, folks, notice this got  bigger. This started at 0.062, and it actually got bigger. So do me a favor. Do me a favor,  okay? I'm going to go back. Okay, these are the answers. Look at the answer right there.  The same value right there is the answer for Br2. Okay? Let's go back. Let's go back and  start over. Okay? This time, and it's up to you guys to do it because I'm not going to  do it, this time I want you to make that a plus x and that a minus 2x. Just do it. This  is a plus x, this is a minus 2x. Do it. Do the whole problem again, and guess what? The  sign is going to change. Okay? Yes? Absolutely are the products on top. Absolutely. It's  always products over reactants. Yes? Really loud. Loud. Geez Louise. Shout. That means  it's got to go this direction. Yes? If Q is bigger than K, then it's got to go this direction.  Just do the problem. Geez Louise, folks, please just do it. I told you it doesn't matter.  Just do it. Make this a plus x and make that a minus x and then this will show you that  it doesn't matter. If Q is smaller than K, it goes that way. If Q is bigger than K, it  goes that way. But I'm just saying I want you to see that it doesn't matter in this  case whether you say that this is a minus x or a plus x. So just do it. Solve the problem  again. I'm trying to help you folks, okay?  Okay.  So folks, in your equation, you should have this setup like this and this is going to  be 0.012 minus 2x and this is going to be 0.063 plus x. Okay? Do it. Expand it. Solve  for 0 and you'll end up with something like this. Okay? So who wants to tell me what do  you get for this part when you switched them, like I said? Who's got that answer? Who's  got an answer? Yes?  4x squared minus. So this is a minus and then a plus. So you plug those into the quadratic  equation and out comes a number. Okay? And you see that when you have that number then,  it's the same number but with a minus sign. Okay? Or a plus sign, whatever it was. So  the values, so this folks just turns into a plus. This turns into a plus. You add that  to this. You subtract that from that times 2 and you get the same darn numbers. So it  doesn't matter as long as you get it right with the x and the 2x. Okay? As long as you  keep track of the number of moles, it doesn't matter what you say. The quadratic is going  to solve it for you. Yes?  So it's not going to be that you need to have q.  You don't need to have q. Yes? You don't need to have q. Now for me, I just kind of  always like stuff to go from the left to the right. Or at least I like to know whether  something is going to be eaten on one side or produced on the other. Okay? But you're  right. You don't need to have q and if q is a calculation that's going to take you too  long then don't do it. Okay? You can solve it with this being a plus x and this being  2 minus, minus 2x or the other way around. I like q. Okay? I like to know where things  are headed before I do the problem but you don't have to. Okay? So do you understand  this now? It doesn't matter. As long as you get the x plus or minus and that is a 2x.  If it's just cubed, it would be a 3x. Alright. We're switching gears. How many of you heard  of Le Chatelier? So what Le Chatelier discovered, hypothesized and discovered was that you can  shift equilibrium by playing with products or reactants. And this is extremely important  in industry. So look at this right here. N2 plus 3H2 goes to 2NH3. So if I said I want  more NH3, tell me one thing I can do, just one to make that happen. Raise your hand.  Yes? Wait, wait, wait. I want more of this. You say you want me to add more of that? Well,  it looks like I want gold and I've got to put more gold in. That is not how equilibrium  works. I want more of this. Okay? I can't add it. Yes? Less liquid. No. Try again. Look  at equilibrium. What do I have to do to make this get more and more? Yes? Add to the reactant  side. Yes, that's what we learned because this, folks, is going to be products over reactants.  So the way to force things in this direction is to add more. Okay? So I add to this side,  it forces things this way. Okay? Do we understand that? Okay. What is something else I could  do? Yes? I could increase the volume and what would that do? That's not what I want. I want  this. Decrease the volume. Okay? So the way the book, I don't know, decreasing volume,  we have to understand, folks, that means increasing the pressure. Okay? You increase the pressure  on something. Count how many moles. How many moles of gas are there on the left side? How  many on the right side? Two. So in a situation where I'm going to be putting pressure on  something, okay, I can force things to the side that has fewer moles. Okay? If I put  pressure on it, the side with fewer moles is the side that is favored. So if I start  pushing on this thing, I can force things to go this way. Okay? Let's say that this  was an exothermic reaction. So we got like heat over here. Okay? So we need to write  it almost like it's part of the reaction. Okay? So it's this, this, this, this, plus  heat. Okay? Then what's another way that I can make this reaction come this way? So  if heat is on the right side, that's almost like it would fit into the equation, you know,  where it would be this squared on the top and then heat on the top divided by these  guys on the bottom. Okay? So if I want to have it come this way more, what do I do if  it gives off heat? Decrease the temperature. So stick it in ice water or something. Okay?  So if you cool it down or let's say this has a, not a very high boiling point, what if  we just stick the reaction in ice water? Okay? And this ends up liquefying. Okay? We can  just keep things coming this direction. Okay? These gases do not liquefy at ice temperatures.  So if you have something here, you can actually condense this out or you can just have it  be colder if it's exothermic. So there's lots of things we can do. We can change the temperature.  Okay? We can change the volume or the pressure or we can change the number of moles. Okay?  I can double them out of N2 and I force things that way. Okay? So there are lots of things  that we can tweak and L'achatelier was the one that sort of pioneered this work. Okay?  Well, it's going to, if we add, now like this guy was saying, if we add NH3, what happens?  Comes back this way. Okay? This is just a little cartoon to illustrate. So here we are at equilibrium  for this reaction. N2 here, NH3 here, H2 there. All right? And we're flat versus time. There's  no change in concentration so we're in equilibrium throughout this whole time. And then all of  a sudden right here we dump in twice as much NH3. Dump in, yeah, about twice. Okay? And  what happens? Okay? It gets dumped in but then it slowly goes part of it away. Okay?  Why do these other ones grow? Why does the H2 grow three times faster than the N2? Three  moles. That's right. Okay? And this comes down twice as fast as that goes up and two-thirds  as fast as that goes up. Okay? Because that's got a two there. So this eventually finds equilibrium  here. There all of the equilibrium right there. Those are the new equilibrium concentrations.  Okay? That, folks, is when we change N. Okay? We change N and the K doesn't change. Okay?  The K stays the same. That is not the case if you change temperature. Okay? You can change  some other things and then the K changes. And that's why in problems it'll say specifically  you know at a certain temperature this is what, this is the K. Because I can change the  K based on temperature. But by adding, changing N, the K stays the same. All right, now, last  night students had a real hard time with this. I'm not sure why, but look at the colors.  NO2 is this rusty color that we see in Los Angeles in the morning. Okay? The low part  of the atmosphere if we look north from Irvine. N2O is colorless. Okay? So this has color.  This doesn't have color. So there are going to be some questions that are asked about  is something going to become more, you know, darker. Okay? So this reaction right now,  if, if I say that all we do to this, I got it in a bottle, all I do to it is I increase  the pressure. Increase the pressure. Okay? Which way is this reaction going to go? To  the right. Because look at this. It says two moles. This has one. One mole is easier to  contain in a constant in a certain volume than two. Okay? So things are going to shift  that way. Okay? So, so what we do is we've said that we're going to be going this direction.  So is this flask going to get darker? Going through and change the life, I guess. Adolescents.  Finally. Or lighter. Lighter. Okay? So if it goes to the right, it gets lighter. And  this is a concept that last night, I couldn't get through to them. Okay? So in this case,  folks, we got the dark going to the light. Unfortunately, the light is the product side.  And that seems to be the hang up in the way these questions are asked. Okay? It's exothermic.  What else can I do to this thing? What if I want it to go this direction? And it's exothermic.  What do I got to do? Cool it down. Oh, exothermic. What am I saying? Heat it up. So it's exothermic.  If I heat it up, it goes this direction, right? It's exo- wait a second. See, I was right.  It's exothermic for the first time. Exothermic means it gives off energy. It gets hotter.  So if I add heat, it comes back to the left. You guys, I trusted you. You said decrease.  Yeah. Okay, let's start over. I'm very confused. I could have said that if you have exothermic  and you add heat, it's going to decrease. It's going to move things this direction. Okay?  That's what I thought I was saying. So that's if we change the temperature. If I add some  of this, what happens? Okay? I can add clear stuff and make it get darker. So we know this.  Q is equal to K at equilibrium, right? Whatever K is, it's eddy-clearant, it's equal to Q.  So, and I wrote something else in on your notes, it's not there, but the volume is double.  And I said, which means the pressure is halved, okay? By opening the valve, as the system approaches  the equilibrium, the color intensity will, what's it going to be? This is the problem.  So, we know that Q is equal to K at equilibrium. So we know that Q is equal to K here. What  happens when we decrease the pressure of everything by a factor of two? Okay? Everything's by a  factor of two and now Q is equal to 2K. Okay? Q is equal to, it's not as big as K, right?  Is that correct? Q is twice as big as K. So what does that, what happens? What does that  mean? Which way is the reaction going to go? It's going to go left or right? Okay, because  Q is bigger, so things are going to head back this direction, which means an increase. Because  this side is the dark side. That's the problem, okay? It's a little bit confusing. It actually  comes back. We think of the reaction going forward. Anyway, the chatty class. Alright,  here's a cartoon of that, okay? This is the same exact question. You got your N, you're  at equilibrium? You got your NO and you got your N204 right here. Okay? You change the  volume. This is going to shoot up. Then you go here because the pressure changed, the  Q changed. Now all of a sudden we're at equilibrium again. Okay? So this got darker. Okay? So once  again I would have preferred that they chose something that, I like things that as the  reaction goes further, you got color. The opposite is, any questions about this? It  seemed last night, maybe I just did a better job today, or you're asleep. Let's talk about  something else that the chatelier pointed out. Some of you want to be doctors, okay?  And a problem with carbon monoxide is that if you breathe it, it likes to bind to hemoglobin.  Okay? And I think not so much anymore because cars are so well tuned today. Do not try this  at home, okay? But people used to kill themselves by sticking a hose in the exhaust pipe of their  car and then run the hose into the cab of the car and then roll up the windows and just  fill the car with CO, well quite a bit of CO and they would have all this carbon monoxide  bonded to the hemoglobin instead of oxygen and they would die. Okay? Cars now almost give  off like no CO. So you'd just be sitting a long time with a headache. So I haven't told  you a story for a while. And this has absolutely nothing to do with this class but I'm going  to tell you anyway because I just reminded myself. So I grew up on a farm and my dad  had cancer and a lot of problems but I went to my parents' house one day and I see my  dad's pickup in the tree orchard with a hose in the exhaust pipe and the windows rolled  up in the truck and I thought, geez dad, did you really, were things this bad? So I pulled  off the driveway real quickly and I run to the truck and there's my dad inside listening  to his country western music. Turns out he had the hose going into a gopher hole. He  figured he was going to do this reaction on a gopher's and for about 30 seconds I was  scared to death thinking I was going to find my dad blue in the pickup. In reality, he  was fine. Nice story. So what does the Chalier tell us about this? Here's the reaction.  This is what is happening in your body right now. Okay? You've got hemoglobin taking oxygen  from your lungs to the rest of your body that we use to burn up glucose and stuff. Okay?  We need that oxygen. Alright? If however you get near something that's giving off a lot  of CO, turns out that the binding energy that this reaction is, that's a stronger bond  there than there. So preferentially this will bond. Okay? Now cyanide is a real bad one  right there. Cyanide bonds to hemoglobin like not quite irreversibly but really, really,  really well. Okay? So that's how we kill people in gas chambers. Yeah, yeah. You just drop  some hydrogen cyanide, some potassium cyanide in some water. Out comes cyanide, grabs onto  the hemoglobin. It's true because this is almost irreversible when you have cyanide.  Fortunately with CO it's not. Okay? So how would we try and help somebody who had had  CO poisoning? Okay? Well look it. Here's the equilibrium right here. Okay? Products reactants.  Alright? How would we force this reaction back that way? Okay? In order, how would we get  rid of the CO and make it be free CO? Okay? Well look, we've talked about a variety of  things. You can add, you can increase the N. Okay? You can make something higher concentrations  on this side but we've also learned about pressure and we've learned about volume and  we've learned about temperature. Alright? So the question is how would you do this? Would  you increase the pressure of oxygen, cool off, or remove hemoglobin? That I think would  require you to slit your wrist and bleed to death. That's the only way I know how to get  rid of hemoglobin is to bleed a lot. Not a good idea. So I don't think that, that should  even be on there. That gives you bad ideas. Cool off! You know you can't cool very much.  You're 37 degrees right now and if you get to like even 35 or 34 you start getting kind  of shaky. So that's not a good idea. So what is it, look at this. How do you increase,  what do you do? Instead of breathing 21% oxygen you breathe more. Okay? 100%? Well you know  I think you can do that for a few breaths. Maybe more. But back in the old days they  used to take babies that were born just a little bit premature. This is back in the  50s and put them in like these things where they had high concentrations of oxygen to  try and help them whatever and it made them blind. A lot of blind kids back in the 40s  and 50s because of this high oxygen that they were given at birth. So I don't think you  can just breathe, I got all kinds of bad stories today. I had two of them in my class. Two kids  out of 30 in my class were blind because of that. So anyway. So we increase this, force  things this way. Yes? What's that? If the moles are the same on both sides you mean  the number of moles. Yes? No, it won't matter. That's right. Her question is is that if you  have the same number of moles, okay if we're talking about gases here, if you've got the  same number of moles on either side or then change of pressure there's nothing. What's  that? Is that the same number of moles? Is this the same number of moles? Looks like the  same number of moles. So would it increase the pressure and not do anything? That's right.  Increasing the pressure overall but increasing the amount of this forces it this way. So  we're changing the end and not the pressure plus you know it's not a good idea to get  inside like a several atmospheres of pressure and be pressurized. Lots of bad examples here  of things not to do. Okay. So yeah so all we're doing is changing N. Okay so once again  I said you can change you know N, P, T or V. And in this case since you've got a whole  body like this you don't really want to be changing T or P or V. At least if it's my  body I don't want that happening. You can do whatever you want. Okay so Le Chatelier  shows us how we can force this thing this way. That of course is assuming folks that  this is really reversible. For CO let's say in this reaction the K is 10. Okay for this  reaction K is equal to 10. What does that mean? That means things lie on this side right?  Okay. Things lie on this side. Maybe it's not quite 10 but let's say it is 10 and by  increasing this by a factor of 2 or 3 then we can shift things back this way and make  the K go that direction. If instead for cyanide the K is 10,000 then adding a whole bunch  of oxygen here really doesn't shift the equilibrium very much. So you're a goner. And cyanide  smells like almonds so be careful when you smell almonds. Alright here's another one.  The temperature for N2 is raised. What happens? The reaction is exothermic. So this means  folks that it's NO2 goes to N2O4 plus heat. Exothermic. Heat on this side. So if I add  heat which way is things going to shift? That direction. Okay. Which then means is it going  to increase in color, stay the same or decrease? Increase. Yeah. Because it's going a lot.  I'd like them to say are you going to go to reactants, not change or go to products?  The whole color thing I think is confusing because once again your reactants are what  are starting with color. Have you already answered this question? Perfect. So once again  if you look at it the K with temperature as you increase temperature K is going to get  smaller. Alright. If you have an exothermic K is smaller. If it's endothermic the more  temperature you give it the better it likes it. Alright. So this is the exothermic and  endothermic K versus temperature plot. And this is just to say that for this reaction  we've been talking about not the N2O1 but the N2 plus H2 that in order to make this  and folks this it's not the holy grail but for some reason in industry we like a lot  of ammonia. Okay. We use ammonia and fertilizer and lots of other things. So making this is  profitable particularly because this is free. Alright. So if you have some of this you get  it somewhat purified. You have some hydrogen. You can make it go this way. The problem is  K is very small. Okay. I think I said I didn't say oh it means okay. Anyway it's very small.  So what do you do to make it go that direction? Okay. Look at this. We've got four moles of  gas on this side and we've got two moles on this side. So what are we going to do? Increase  the pressure. Okay. So it turns out that this is exactly what they do and when they do this  in industry it's done at 500 degrees but look at that. 200 atmospheres. Huge pressure. And  you know what they'd probably like to go to 300 or 400 but it's just too expensive. You'd  have to have a vessel that would handle you know it wouldn't be worth it. So it's just  not cost effective but if they had something that could go to that pressure they would.  All right do this. So the equilibrium folks, these are the equilibrium concentrations. Two,  two and one. And then I'm going to somehow suck all of the A out. All of it is just immediately  gone. So it has to equilibrate. So the first thing you've got to do folks is figure out  what is the K. Since this is an atmosphere this is K sub p. So now you have to set it.  So you say okay fine I've got my K so now I've got to figure out what are my equilibrium  because that's the question. What is the equilibrium concentrations right? What are the pressures  after the system to equilibrate? Well look folks you had A going to B and C. I think that  works yes. And then all of a sudden you yanked all the A out. All of the A is gone. So all  you have is product. What's going to happen? Some of it's going to have to go from product  back to reactant. It's got to go in this direction. So you set it up like that. You started with  two you're going to lose X. You started with two you're going to lose X and you're going  to make X. So those are the after equilibrium. So now you just go ahead and plug that in  for case of p it was equal to so it's going to be 2X squared. Each of these is a 2 minus  X squared over X. So it's this times this which is just squared over that. So X squared  is going to equal this right there. So you should be doing this folks. Yes. Loud. They  can't hear it. What are you saying? No. The pressure of A is zero. So you had five at  equilibrium you had five atmospheres. You had two of B and two of C and one of A. Okay.  And you're at equilibrium. Then you somehow sucked out the A. But you still have four  atmospheres. So now it's got to re-equilibrate. And that's where the X comes in. So you lost  some X from this. You lost some X from this. And you made X here. So this is the equation.  Okay. You do this was four. You calculated that in the beginning. And now you've come  up with this. And once again. These I'm going to tell you. These are the two values that  you can have for X. Once you've done the quadratic. So I'll give you a second.  Yes. Why is there this? Oh, that should be a PC. Sorry.  Okay. So folks, once again. You got two answers for the quadratic. And one of them makes no  sense at all. How can X be 7.5 when you only started with two? Okay. So this is some common  sense that you've got to use. Okay. So the only X that makes sense is this one's about  0.5. So what it's saying is, is that you're going to subtract about 0.5 from two. All  right. And you're going to have X of 0.5. So just to humor me. Plug that Vos values  back into the original right at the very beginning. Say, okay. I'm going to say that the pressure  of A is 0.53. The pressure of B and C is going to be 2 minus that. Plug it in and see if  you don't get four. Okay. Always, if you have a chance, plug your values in and see if they  give you back the right answer. Okay. You should get, if you put in 0.5, if you put,  make that a 0.5 and make that a 1.1 and a 1.1, you're going to get very close to four.  So once again, I don't know how they sucked this A out like that. But they did. And this  is a nice question for an exam because I don't have to give you K. Okay. I don't have to  give you K. You figured out K early on. And then I said, oh, now I'm going to take away  something. So then you had to do all this kind of logical thinking quadratic. I like  this kind of problem. That's all I'm going to say. Okay. So this folks is going to take  us one minute. It's in your notes. It should be something that is second nature. Okay. That  if you change the concentration of products, which means you're over here, which ways are  going to shift equilibrium? This direction. You always go in the opposite direction of  where you add stuff. Okay. Okay. If you decrease the concentration of products, which ways  are you going to go? So I want to hear loud, left or right? Right. Okay. Increase the concentration  of reactants. Right. Decrease the concentration of reactants. And you guys are so excited  about this. Will the equilibrium concentration of NO2 increase or decrease or stay constant  if we? So we're always looking at NO2. Okay. So here's the first question. I want to know  what's going to happen to NO2. Increase or decrease if we increase NOCl. What's it going  to do? That's right. How about if we increase NO? Increase NO. What's it going to do? Okay.  Increase NO. Hopefully you'll say the opposite. Increase NO2Cl. What's it going to do? Okay.  Good. All right. And this is just the same stuff. Okay. I'm not even going to go over  this. I'm so bored. Okay. So study this on your exam. I mean on your stuff. So look at  it. I will do this little thing at the bottom. Okay. So look once again. This is exothermic.  Okay. It's exothermic. And once again, the products are colorless. So if it's exothermic  and we add heat, which side is equal? We're even going to go on. This side or this side.  Okay. If we add heat, we're adding something on the product side, shifts things this way.  Right? This way, okay, is darker. If this has color, this doesn't. At room temperature,  it looks this way. Look what happens when we make it colder. Colder is just the opposite  of hot. So for something that's exothermic, if you cool it down, it forces things more  this way. More this way means more into O4, less NO2, less color. Don't forget, folks,  changing temperature changes the K equilibrium. Actually, changes the K. All right? By adding  more of this, it doesn't change the K. That's like a good multiple choice question. All  right. Now we're going to shift gears back to Delta G. So this is the beginning of the  beginning of the end. Yes?  Do you know why your body turns pale when you start doing the wearable?  No. I have to think about that. I turn pale for lots of reasons. I was going to say something,  but since we're being filmed, I better not. Okay, so back to Delta G. I love Delta G because  it allows me to not have to worry about entropy. I love disorder, but I have a hard time knowing  whether the whole world and universe is being disordered because of something I do. Delta  G tells me, you got a negative Delta G, it's spontaneous. You got a positive Delta G, it's  spontaneous in the other direction. Okay? Now Delta G is equal to products minus reactants,  and it's zero in equilibrium. So we got this equation again. Okay? So Delta G, products  minus reactants. Okay? We've got, I can't do this back here. So the products is this  side. So we've got, the products is going to be two times the standard G plus RT log  A, and then it's going to be minus the reactants. Okay? So you're going to get two more of these  things up here. Okay? Now we haven't learned the Delta G is equal to minus RT log K, which  is what's coming up here. Okay? Now, let me just tell you something. I told you about  four weeks ago that you cannot take the log of a unit. It must be a number. Okay? So in  order for us to do these things with K in here, okay, this is actually going to be a  K at some point. Let me, look at this. Okay? The A here are what we call the activity.  Okay? And I don't like to use activities, but I have to explain to you why we use this.  Because if you look at this, you've got two moles here times three moles times one mole.  You've got a total of four moles on the bottom and four moles on the top. And these are in  atmospheres. So in theory, this would give you, if each one was at one atmosphere, this  would give you, you know, one ATM squared divided by three ATM square, or by one ATM  cubed and one ATM, you'd have units in your log. Okay? Can't do that. So we, somebody  came up with this idea of activities. So it's actually, these activities are a pressure  divided by a pressure. Okay? So you get rid of the units. So all you have is a value right  there. So in this case, and this is all you really need to know, it's just pressure squared  over pressure cubed times this, but these are now not units. Okay? We've removed the  units. Okay? By using these A's. This is what you need to know. Okay? The delta G is equal  to minus RT log K. Okay? So, folks, if I want something to be spontaneous, what does RT  log K need to be? Just this part, not the minus. RT log K has to be what? If I want something  to be spontaneous, it's got to be positive because we're subtracting it then, okay? So  as long as an RT log K is a positive number, then it's spontaneous. Everybody take out  your calculators, take them out, everybody. Find the LN button, looks a lot like that.  LN, natural log. Okay? So what I'm saying is that as long as this part from here to here  is positive, then this is spontaneous. T is always positive. T is Kelvin. Kelvin, the  absolute zero is zero. That's just cold. We can't get there, but that is the value. There's  no negative temperatures. It's not minus 50 degrees. It's minus 50 degrees in North  Dakota because we use Fahrenheit. But in Kelvin, it's a balmy, you know, 220 degrees.  Okay? So this is always positive, and we know R is positive. It's either 8.3 or 0.082,  but it's always positive. So the only thing, folks, that determines what the sign of the  RT log K is, is this. So I want you to find out what is the LN of one. Do it. For those  of you who think you know or just believe that these other people are right, it's important  for you to do that. Okay? So at equilibrium, at equilibrium, what is K? What is that equal  to at equilibrium? When delta G is zero, K is equal to one. Okay? Now, punch in 0.9,  log of 0.9. What is it? Negative point something. So a negative time is a negative gives me  what? So if K is less than one, then delta G is positive. Type in 1.1. It's not a huge  number. What is the log of 1.1? Point zero something, but it's positive. So that whole  thing is positive, therefore delta G is a negative. Okay? So please understand. I'm  sorry that I'm dumbing this down, but you have to understand that's what this whole  thing is about. Okay? So folks, in an example I could give you, I could say what is the  K of this reaction? Let's do it. What is the K of this reaction if delta G is equal to,  no, let me say what is delta G? Make it easier. No, that's not what I want to do there. What  is delta G if K is two? At room temperature, 298. What is delta G? First of all, you know  it's going to be negative, right? You know because anything over one means that the log  is a positive number. Anything less than one, it's a negative number. And at one, it's  zero. So I can ask you a question as simple as that. I can give you the temperature, you  know what R is, and I give you K, figure out G. Okay? Very simple. Or I can give you something  more complicated, right? We also know that delta G is equal to minus, delta H minus T  delta S. Do we remember that? Delta G is equal to delta H minus T delta S. So now I could  say delta H minus T delta S is equal to minus RT log K. I can get fancy. Okay? These are  those activities. Look at it in the book. Okay. Once again, we know that if delta G is less  than zero, K is greater than one and spontaneous. Delta G is greater than zero, K is less than  one and not spontaneous. And if delta G equals zero, K equals one. Delta G equals one. You  know that I put a minus in that last night? It's weird. It didn't show up. Okay, this  is a minus right here. It's weird. It was there, then it disappeared, now it's gone again.  So okay, well anyway, there it is. I fixed it. Now, we also learned when we were doing  the Clausius Clapperon, when we had a log of something, we had a log of P2 over P1 and  we were solving for a P that the only way that you can get something out of the log  is to take it to the E. Okay? So let's say I've got, I want to get rid of this log K.  All right? If I say I want log K on one side, I can say delta G over RT minus delta G over  RT is equal to log K. Do everybody see that? I can bring this to this side and I get delta  G over minus RT is equal to log K. Then I can take it out of the E power and I get this.  K is equal to this. Okay? So folks, there are going to be times I give you delta G and a  temperature and you're going to give me K. You can do it through this way if you want.  It doesn't matter, okay? This is a very, very handy equation to know. There will be something  using this on the exam. Okay? What are delta G and not NK for this at 25 degrees C? Room  temperature. Okay? At room temperature, folks, I have water. Okay? Is it boiling at room  temperature? That means it's not spontaneous, which means what? The delta G must be greater  than zero. Right? Let's see if I'm right. NK must be less than one. Oh, look at that.  I'm incredible. Uh-oh. I warned you about this. The delta G is equal to delta H minus  T delta S. So I can set that equal to that. Yeah. Gotta love that. All it is is a substitution,  no magic. Just substitute one thing for another. All right? So if I want to have this be the  log of K, I think this is going to be log K over 1 over T. Okay. What would that plot  look like? Here is the log of K over 1 over T. Here's the log of K right there. Okay?  I've got delta H here. I've got R. I've got all this stuff, but I'm plotting it versus  1 over T. Log of K versus 1 over temperature. Okay? Maybe this one would be better. Okay?  You've got K over T. What happens to this? Can you predict what it's going to look like?  If it's exothermic, then 1 over T, the log of K gets bigger. Okay? In reality, we know  that if something's exothermic, that means there's energy given off on the product side  that as we increase the temperature, we can shift the K smaller and smaller and smaller  as we change temperature. Yeah, that's what this means. Okay? Sometimes plotting is 1  over T, you get a nice straight line. Now, folks, I should be able to give you any kind  of a plot. This is 1 over T. I should be able to give you a delta H, and you should be able  to plot that. Okay? That's endothermic. Oops. Okay? Once again. Endothermic, we know that  K gets bigger as we heat things up. Okay? Put energy in on the left side, forces it to the  right. It needs energy to go, and look at that. Look how the K gets bigger and bigger  and bigger as we change the temperature. Plotting things 1 over T, oftentimes will give us a  straight line. Okay? I think we should quit. All right, folks. There's one more lecture.
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