Lecture 01. Chemical Equilibrium Pt. 1.
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Arasasingham, Ramesh D.
UCI Chem 1C General Chemistry (Spring 2013) Lec 01. General Chemistry - Chemical Equilibrium - Part 1 Instructor: Ramesh D. Arasasingham, Ph.D. UCI Chem 1C is the third and final quarter of General Chemistry series and covers the following topics: equilibria, aqueous acid-base equilibria, solubility equilibria, oxidation reduction reactions, electrochemistry; kinetics; special topics. 
 Good morning. This is the third quarter of the general chemistry sequence, so I don't think I need an introduction. But welcome to this class. I see a lot of familiar faces. And we are going to pick up where we stopped last quarter. But before I begin, I want to kind of remind you that my primary mode of communicating with you will be through the class website. And so it is important that you check the class website often, particularly the one that says announcements, because I will be putting material there and reminders and so on. So it's important that before you come to class, sometimes I will put some notes there, I will put some extra problems there. It is important for you to check that as often as you can and print out the material that's there. Okay? So I am not, because you've already taken this class, you know, these classes before, and this is the third quarter, I'm not going to go into the details of housekeeping, all right? The dates when you'll have your midterm exams, how the grade distributions work out, all that information is on the class website and you can read it. So I'm not going to go through those details with you, because you already know what it is. However, I do want to remind you that if you want to do well in this class, there are three elements to this class and you all know what those elements are. The first is coming to lectures. As all of you know, it is important to come to lectures because as I say before, I'm your tour guide and if you don't know where your tour guide is taking you, obviously you're not going to know how to prepare for the exam, all right? Textbooks are very often written as encyclopedias, all right? So they have lots of information. It's impossible for you to kind of navigate through all the information that's provided in a general chemistry textbook, all right? So I'm going to emphasize things in class. This quarter, the material that we are covering actually involves a lot of numerical problem solving. Now we moved on. We looked at the fundamentals of bonding. Now we're moving on to looking at equilibria and later on we're going to look at electrochemistry, we're going to look at kinetics and so on. And the topics in 1c are very heavily weighted towards numerical problem solving. And so I'm going to be working lots of examples in class because one of the ways in which you teach problem solving skills is to model how experts solve problems, how do they approach problems, how do they solve these problems and particularly very challenging problems, all right? So I'm going to work lots and lots of examples in class. And so it's important that you come to class because you're going to see me work all these different types of problems and show you how to approach and solve those problems. And so if you don't come to lecture, you're going to miss out on that, okay? So lectures are important, but as all of you know, lectures has a disadvantage in that you're very passive. I'm doing all the talking and you're seated there just listening to me. And in that passive mode, your mind can wander off. Already it's wandered off, you know, and it'll come back. And so what you're doing is periodically you're paying attention and that's not very good, all right? But the reality is that in any large scale instruction like this, we have to get through the material and the reality is that the lecture mode of instruction is not going to change, all right, when you have such big lecture halls. So in the lecture, very often you're going to be passive, but I'm going to get you involved. I'm going to ask you to come prepared. I'm going to ask you to throw questions at you. I'm going to expect you to give me answers and sometimes I'm going to make you do the problem in class, all right? So come prepared. So that's the lecture part. But we also have the second element of this class, which is discussions. And because you're so passive during lectures, in discussions, I expect you to take control of the discussions. Discussions are where you are going to do the work, you're going to work the problems. The TAs are going to ask you to come forward. It's okay to make mistakes. Discussions are where you make mistakes and learn from your mistakes. So the TAs are going to make you come up and answer questions and so on. So the discussion is the place where you are going to be really engaged and you have to come prepared to discussions and you have to actively participate in discussions. So we don't want anybody just sitting in the back and copying the answers because that's not really what we want to do. We want you to be involved, okay? So discussions are where you're going to be very active. Every discussion, as you know, the prior week, I will be putting up a worksheet and those worksheets have representative problems. Again, because it's so heavily weighted towards numerical problems, the content that we're covering right now, and so I will give you lots of different problems and give you opportunities to practice as well, okay? So it will be a good idea if you could try the discussion problems prior to going to discussion so that, you know, when you go to discussions, you can actually ask a lot of questions. The TAs are going to ask you to come forward and work the problems on the board and then people are going to point out why it's wrong and so on. And so if you come prepared, it's very helpful, okay? So the second element to this course is discussions. And as all of you know, attendance and discussion is mandatory. We will mark attendance in discussion. And of course, the third element is homework, all right? All three elements are important. And for this class, you have online homework. All of you have seen the online homework and have tried it before. And the homework that we are going to be using is catalyst. So each one of you will be getting an individualized homework. So your homework is different from the others in your class. And remember, the goal is mastery. The goal is not to just complete the homework for the sake of completing the homework. It's not a one-shot deal. You can do the homework as many times as you wish. Because every time you restart it, it's going to give you a fresh set of problems, all right? And the goal is that you achieve mastery. So you can work it four times, five times. As long as you achieve mastery and recall that, you have to achieve 75% or more on your homework, all right? If you score 75% or more on your homework, that counts as getting 100% on your homework, all right? So once again, you have an opportunity to practice what you've learned using the online homework. But I want to remind you that online homework is an automated system. And there's only so much computers can do. And so where online homework really works is where you get a lot of practice. In other words, practice of fundamental basic stuff or some stuff that you have to learn by rote before you can actually work more complex problems. So the analogy that I'm going to give you is what I've given you before. And it's like, you know, multiplications. You know, you need to learn at some point that two times three is six and three times four is 12 and so on. And online homework is very good at getting you to practice that. But it is not very good at giving you a very complex problem where you're bringing in multiple concepts together, putting them together, and solving a complicated problem where you have to synthesize the information and use it in creative ways, all right? However, if you want to get to that state, you have to get past this algorithmic practice type problems where you have gone through just the basic stuff and you understand what it means when you say you're calculating equilibrium constant, you have to know what it means when you are looking at the law of mass actions. Those are all the fundamental stuff that we expect you to know if you're going to solve a more complex equilibrium problem. So the online homework is very good at getting you through the fundamental stuff and making sure you know the basic stuff. But it is not very good at giving you very complex problems that you have to synthesize the information. In order to do that, I'm also going to assign a few end of chapter problems from your textbook, all right? However, the textbook problems you do not need to turn it in, but I don't want to overload you with work, so I will choose a few examples of problems that I would like you to work from the end of chapter problems in your textbook. And the solutions to those problems are available in the back of your textbook. We've also placed the complete solutions manual on reserve at the library. So if you can't solve a problem and you can't come see me during my office hours or can't see the TAs during the office hours, you can actually go to the library and it's at the front desk. It's on reserve. You can just, you know, take a look at it and if you want to copy, you can just go down the hallway and use one of those copy machines to copy that particular page if you wanted to. All right? So there are three elements to this course and I liken it to a three-legged stool. If you choose to sit on a two-legged stool and you choose to not do homework or you choose not to go to discussion, what do you think is going to happen? It's not going to be very stable, all right? So if you want to do well in this class, if you want to benefit from this class, all three elements are important because all three are integrated with each other. Each one reinforces the other and I rely on you to go to discussions. I rely on you to carry online homework and I rely on you to come prepared to class. Is that clear to everybody? So that's all that I'm going to talk about the mechanics of this class, all right? Now before I proceed and start on the material for today, is there anything that you want to ask me? Any questions? All right, so let's go ahead and start and we're going to begin by reviewing a little bit of the material that we covered last quarter. Now the theme for this quarter is actually continuing chemical equilibria. So we're going to go on to looking at acid-base equilibria which happens to be chapter 11 from your textbook. However, it is important that you know chapter 10 really well. Chapter 10 deals with the fundamentals of chemical equilibrium. So it means that you have to have a good understanding of the basic principles that we've looked at last quarter because if you understand those principles, now it's a matter of applying them to different scenarios. So if you understand what chemical equilibrium is, now we're going to take different kinds of chemical equilibrium and look at them in greater detail. And the first one that we're going to start by looking at is looking at acid-base equilibria, okay? But before doing that, I wanted to kind of review some fundamental aspects of the material that we covered last quarter and I want to take one particular example. And the way I'm going to kind of review the material is actually to work a problem. And we're going to work on a complex problem and those of you who took my class last time know what I'm talking about because I want to show you different strategies of solving the problem. And I'm going to take a problem that I did work last quarter but is a really good example. And for the benefit of those who were not in my class last quarter, I wanted to go through this example and we're going to kind of look at all the main points as we go along. So I have put this problem on the class website. So I really don't want you to copy the problem. I want you to pay more attention to the strategies we're using to solve this problem. For some reason, this is not focusing very well today. So we're going to take an equilibrium problem and we're going to use the sort of general strategies that we use to solve an equilibrium problem. I apologize for this. I do not know why this thing is acting up today. There we are. Right. Keep your fingers crossed. So let's take a look at this problem. So you're given an equilibrium. All right. And we're told that at 35 degrees Celsius, KSC for this equilibrium is given to you. Calculate the concentrations of all the species at equilibrium for the following original mixtures. All right. So if you're going to solve an equilibrium problem, where is the starting point? What do you look at? Where's the most obvious thing you need to look at? The balanced equation. We start with the balanced equation. Never assume that any balanced equation provided to you is actually balanced. Always check because there can be typos. And sometimes people give you the equation without balancing it. And if it's unbalanced, you need to know that it's unbalanced. Now it turns out this equation is balanced. So we start by writing down the balanced equation and the physical states. And as a rule of thumb, I always suggest that you write the value of the equilibrium constant as well right next to it as a reminder to you what the magnitude of that equilibrium constant is. And it turns out this is 1.6 times 10 to the negative 5. Now all of you know that the equilibrium constant is called the law of mass action or it's the equilibrium constant expression. And how do you write the law of mass action? It's always defined as what? The concentrations of the products raised to the part of the coefficient divided by the concentration of the reactants raised to the part of the coefficient. So the bottom line is that when you look at the law of mass action or the equilibrium expression, the products are the numerator and the reactants are the denominator. And so essentially it gives you roughly the ratio of products to reactants. So the bottom line is it's not important to me that you know how to write the law of mass action. I tell you guys that any eighth grader can write that. If you are given a balanced equation, writing the law of mass action or the equilibrium expression, anyone can do it. It doesn't require any prior knowledge. What's more important to me is that you understand what that number tells you. Remember the equilibrium constant gives you the ratio, approximate ratio of products to reactants where the products are the numerator and the reactants are the denominator. So what K, the numerical value of K tells you is what the equilibrium composition looks like. It tells you what does that equilibrium mixture look like. It tells you whether that mixture has lots of products and very little reactants. So if your K value is much greater than 1, so let's say it's a value of the order of a thousand. If K is the value of the order of a thousand, what does that tell you about the equilibrium mixture? Lots of products, very little reactants. So it gives you an idea of the composition of the mixture. And that is the most important take home message. Because the equilibrium constant gives you an idea when equilibrium is established, what the mixture of reactants and products look like. Got it? So now let's look at this. This number is 10 to the negative 5. Is that a number? Now as a rough gauge, if your products and reactants were a 50-50 mixture, what should the value of K come out to be? Approximately 1. So if you're at equilibrium, when equilibrium is established, if you have a 50-50 mixture, it would be approximately 1. If the number is far greater than 1, we know that the mixture now should have lots of products and very little reactants. Now this is 10 to the negative 5, so this number is much, much smaller than 1. So what does that tell you about that equilibrium mixture? Lots of reactants, very little products. So that means this reaction took place only to a small amount. Do you see that? It reacted just a little bit and then established equilibrium. And we use the term extent of the reaction to describe that. And so if you want to describe this in terms of the extent of the reaction, does this reaction react to a large extent before establishing equilibrium or did it go to a small extent before establishing an equilibrium? Small extent. Do you see that? And that's important because we're going to use that idea. So one of the things that you get a feel for is by looking at the magnitude of k and then if that number is really small, you know that it only reacted to a small extent. If that number is really large, so instead of 10 to the negative 5, if it's 10 to the power 5, what does that mean? It reacted to a large extent. So what that means is like you have almost 99% product and 1% reactant. Do you see that? Roughly I'm using just approximate numbers but it means that it reacted to a large extent and if it's 10 to the positive 5, it's large extent, 99% products, about 1% reactants. If it's 10 to the negative 5, what does that tell you? You have only 1% product and you have 99% of unreacted reactants. Do you understand that? So that's the first important take home message and that's why you write the k value right next to the equilibrium so that when you look at the magnitude of that number, you have an idea of to what extent that reaction took place. So now we start by writing down all the information that's provided to us. So we're given the concentrations of NOCl. So we're told 2 moles of pure NOCl in a 2 liter flask. So since we're given moles, the units that are provided are moles and volume, now we're going to think of this in terms of concentration because remember in equilibrium expressions or law of mass actions, you can either give it in terms of concentrations or if they're all gases, you can give it in terms of pressure because pressure of gases is directly proportional to concentration as well and that's why you can give them in two different units. As long as they're gases, they're all gases, then you can give it in terms of atmospheres or you can give it in terms of concentration. But it turns out here, we're given moles and volume and as all of you know, what is the definition of concentration? It's moles divided by volume. So this is moles and this is volume in liters and therefore the unit is what? Moles per liter. So we're looking at this gas, we're given the number of moles of each one of those gases or in this case there's only one reactant and there's only one substance there to begin with and so what would the concentration come out to be? It's 2 moles divided by 2 liters and therefore my initial concentrations to begin with. Alright, so my initial concentration is moles per liter or what is moles per liter synonymous with? Molarity. Okay, you know that molarity and moles per liter are the same thing and it's a unit of concentration and therefore my initial would be before any reaction takes place. It's at time zero. Before any reaction starts at time zero is the initial concentration. So before any reaction takes place, we know that the concentration of NOCl is 2 moles divided by 2 liters so what does that come out to be? 1 mole per liter and zero. Okay, we have no products to begin with. Now we're going to allow the system to go to equilibrium but remember these are reversible reactions and therefore the reaction can take place in the forward direction or it could take place in the reverse direction. So now you have to figure out is the reaction when it goes to forming an equilibrium, is this reaction proceeding in the forward direction or is it proceeding in the reverse direction? And all of you can recognize that we have no products when you have an equilibrium you have to have mixtures of reactants and products. So if there are zero products what is the obvious direction in which the reaction is going to take place? Forward. Okay, there's another way you can do this and you learned about that last time. You can calculate what? Q. Q is the reaction quotient which is the ratio of products to reactants before any reaction takes place. What is Q here? If your products are zero what is Q? Zero. So there's only one way Q can go. From zero it's got to go to 10 to the negative 5. Do you understand that? And that means it has to go towards products as well. So if you're given the amounts of reactants and products then you calculate Q and you have to figure out which direction the reaction will proceed. In this case it's going to proceed in the forward direction. Alright? And so if it's proceeding in the forward direction then we're going to say that we're, let's say it is x is the amount that's used up. We know that it's moving in the forward direction therefore the concentration of this has to go down and the concentration of the products need to go up when equilibrium is established. And if it's going to proceed in the forward direction then some of this is going to be consumed and some of these are going to form. So let's say the amount that's consumed is 2x because the coefficient here is 2 and because in this balanced equation we know that the ratios are maintained for every two moles that's consumed how many moles a product have formed? Two moles of this and one mole of that. So that ratio is always maintained. That's the ratio in which they react. So for every two moles that's consumed you're going to have two moles of NO being formed and one mole of Cl2 being formed. So if 2x is being consumed what does that tell me about the amount of NO that will be formed? Plus 2x because you know that we put a plus sign to remind us that it's being formed and over here the ratio is 1 so that should be plus x. So that means at equilibrium what I would have is 1 minus 2x this would be 2xx. Does that make sense everybody? So when equilibrium is established if you want to know the concentrations of reactants and products that expression describes it and if we can calculate what the value of x is then we can solve for this and we can write an algebraic expression solve for x and figure out what is the concentration of NOCl at equilibrium, what is the concentration of NO, what is the concentration of Cl2. So in order to find out what x is we need an algebraic expression. We need to relate x to something else that the problem provides. If they didn't give us any additional information we could not solve this problem. And what is that additional information we're going to make use of? It's the equilibrium constant. So we can turn to the law of mass action and let me move on, let's take that out. So now we're going to move on to writing the law of mass action which is Kc. So we can say Kc is the concentration of the products raised to the power of the coefficient times Cl2 raised to the power of 1 divided by NOCl raised to the power of 2. So this is the law of mass action or this is the law of chemical equilibrium. And if we know that Kc is 1.6 times 10 to the negative 5 moles per liter. Remember equilibrium constants do have units, alright? And those units will depend on the law of mass action. So this is the K value and this equals NO would be 2x squared. That's the concentration of NO times x divided by 1 minus 2x squared. So that gives me this is 4x squared times x. That gives me 4x cubed 1 minus 2x. Alright? Now one of the biggest headaches of solving equilibrium problems as all of you know is the algebra. I mean up to this point it's pretty straightforward. But now you have to be able to solve for x, alright? And sometimes this can be very complicated. Now if you had a good calculator that could solve this, well it's good. But remember on an exam you don't have the luxury of having you know sophisticated program that can solve the algebraic equation for you. All you have is a basic scientific calculator. And you can see that if you're using just a basic scientific calculator this is going to be hard to solve. So one of the challenges that you're going to have when you solve these equilibrium problems is this mantra. And what's that mantra? Keep the math simple. Alright? If you can keep the math simple the chances that you'll make careless mistakes gets better. And it turns out that your answer should be the same as if you were to solve this, you know, go through the whole process of solving for this. So here one of the things, one way in which we keep the math simple is we make intelligent approximations. Alright? And so let's look at this equation. In this equilibrium if you look at on a scale, remember we said that k is 10 to the negative 5. And so if k is 10 to the negative 5 we said that because that number is so small we know that the reaction only proceeds to a small extent. Alright? If it proceeds to a small extent what does that tell you about x? Remember it only reacts so that 1% of reaction goes to product. 99% remains unreactive. So what does that tell you about x? Will that x be large or is it going to be small? Small. Okay? So if you have a scale and on this scale we have 100% reactants here. On this scale we have 100% products over here. This is initial. When we start this is we have only 100% reactants. If the reaction proceeded 100% if it goes to completion then what's going to happen? We have zero reactants, 100% products. If when equilibrium is established you're somewhere here what does that mean? Now when equilibrium is established right in the center is a 50-50 mixture. So that means k has to be approximately 1 there. Do you see that? We know this k is 10 to the negative 5. So what does that tell me about kc? kc when equilibrium is established and if you look at the ratio of products to reactants you know that you have lots of unreacted reactants and very little products. So if you look at this whole range of looking at the ratios of reactants to products you can have 100% reactant that means no reaction has taken place. You can have 100% product that means it's an example where the reaction went to completion. It didn't establish equilibrium it just went to completion. If it's right in the middle what happens when equilibrium is established you have a 50-50 mixture. If it's on the right hand side of this if it's somewhere here what does that tell you? You're going to have 75% product and 25% reactant. Do you see that? Alright so we're over here and so this is where equilibrium is established. So that means that it reacted only to a small extent and that's what x gives you. Alright the extent of the reaction is the amount that reacted and we know that that x is going to be really small because it only reacted to a small extent. Now because this x is really small because k is 10 to the negative 5 now that gives us a way to make an approximation because we know that it only reacted to a small extent and therefore to keep my math simple now I'm going to make a smart approximation and my approximation is that 1 minus 2x. Remember I want to simplify this expression. Alright and so if you start with one molar concentration and it reacts only to a small extent we're going to say that 2x is so small. 2x is so small compared to one. Compared to one molar we can neglect it. So this is an assumption that you're making. It's an approximation. It's very subjective. Do you understand that? Anybody can make any approximations. Alright we're going to start by making the approximations but then we have to go validate and make sure that approximation is actually valid or not. Okay so we're going to make this assumption and can you see that when we make this approximation it considerably simplifies my math because now I can say 1.6 times 7 to the negative 5 moles per liter equals 4x cubed divided by 1. Now what would have turned out to be a horrible algebraic equation to solve now just boils down to a very simple step. Alright now all I have to do is divide this number by 4 and take the root cube of that and so if you work that out it comes out to be, so if you solve for x, x comes out to be 1.6 times 10 to the negative 2 molar. Remember x is concentration so the unit for that should be multiple liter or molarity. Alright so I made this very subjective approximation and anybody can make any approximations you understand that. Alright but what do I need to do now? I actually have to go back and make sure that my approximation is valid. Now what was my approximation? Go back and ask yourself what is your approximation? 2x is small compared to 1. So we need to find out is 2x really small compared to 1. So my validity will be, so I'm going to check the validity of my approximation. And my validity is that what I'm saying is 2x is really small compared to 1. So I'm going to take the ratio of the 2 and find out what percent is 2x compared to 1. Alright so once again you have to go back, it's not x, I'm not saying x is small compared to 1 in that approximation I'm saying 2x is small compared to 1 and therefore I have to compare 2x with 1. Do you get it? And so I would go back and I'd say 2 times 1.6 times 10 to the negative 2 molar divided by 1 molar times 100 and if you work that out that comes out to be 3.2 percent. So what that tells us is that if you compare 2x to 1, 2x is 3.2 percent of 1. Alright so now you have to ask yourself is that good enough? And so we have this purely arbitrary rule and we call it the 5 percent rule. If it is approximately 5 percent or less we know that it's a really small number and that it can be neglected. Alright and so because this falls below this arbitrary 5 percent that we picked. Alright now we know that the approximation is valid. Okay and we know that 2x is indeed very small compared to 1. Alright so we say approximation is valid. Now let's say you made this approximation and it came out to be 25 percent. So what does that mean? That you can't make that approximation. Alright and that means that you have to go back and you have nothing, you just have to grit your teeth, sit down and solve that algebraic equation. Do you understand that? There's nothing you can do about it. Alright it turns out in this case you can. And where does that clue lie? Where does the clue lie when you can make those approximations? Kc. If Kc is really small, if it's 10 to the negative 4, 10 to the negative 5, 10 to the negative 6 those are really really small numbers for K. That means the extent of the reaction has to be really small and therefore you can make that approximation. Now we need to calculate the concentrations and so we go back. Now the problem is still not complete because now we need to figure out what the concentrations of NoCl is and the concentration of NoCl is 1 minus 2x which is 1 minus 3.2 times 10 to the negative 2 molar and that comes out to be.97 molar. No, concentration of No is 2x which is 3.2 times 10 to the negative 2 molar and then the concentration of Cl2 is x which is 1.6 times 10 to the negative 2 molar. Alright now what's interesting is if you did not make this approximation and if you had solved that algebraic equation you would end up with the same answer. Alright and so it makes it much easier if you make this approximation and therefore solving that problem gets easier. Alright now let's go to the second part of this problem. In the second part of this problem so can I remove this how you guys done with this? Okay great so let's go back to part 2 and now we're told things have changed a little bit. Now we have a different scenario where we have 3 moles of No and 1 mole of Cl2 in a 1 liter flask. Alright so we're going to do the same thing so we're going to go back. We start by writing the balanced equation. Make sure you look at the physical states because the physical states give you an idea of whether it's a homogeneous system or a heterogeneous system and the law of mass action does depend on whether it's a homogeneous system or a heterogeneous system. These are all in the gaseous states so are they homogeneous or are they heterogeneous? They're homogeneous. So once again, right KC to remind you that this is a very small number. Now if we write down the initial concentrations in moles per liter of molarity they tell us that we have 3 moles of No in a 1 liter flask so what is the concentration of No? 3 moles per liter and I have 1 mole per liter of Cl2. I'm not putting the units there because I'm putting that within parenthesis there and it reminds us that we're in units of concentration. And we don't have any reactants so what does that mean? You have 0. So in solving an equilibrium problem this is what you have to begin with. So what is the first question you ask? Remember equilibria are reversible reactions. They can proceed in the forward direction or the reverse direction. In this case which direction is that reaction proceeding? Reverse. So now we have no reactants so we know that if equilibrium is established we have to have some reactant and therefore we know so I'm going to put this arrow to remind me that that's the direction in which the reaction is going to take place. If the reaction is taking place in that direction then if we're looking at the relative change it's proceeding in this direction so what would this come out to be? So tell me what would I put here? Minus. Minus what? 2x. Here minus x and on this side plus 2x because now the reaction is proceeding in the reverse direction so products are going to be consumed to form reactants because products are being consumed you put the negative sign to show that your concentration is decreasing on the product side and the concentration is increasing on the reactant side. Okay therefore when equilibrium is established this would be 2x this would be 3 minus 2x and this would be 1 minus x. Everybody agrees with me? So now we turn to the law of mass action we know kc equals 1.6 times 10 to the negative 5 moles per liter equals again the form in which this equilibrium is written is that these are the products these are the reactants. Alright so the law of mass action would be NO squared over cl2 divided by NO cl squared which is NO would be 3 minus 2x squared times 1 minus x divided by 2x squared. Got it? So now ask yourselves can I make an approximation here? Now the reaction is proceeding in the reverse direction what does k tell me that at equilibrium what would my composition look like? At equilibrium k tends to negative 5 so regardless of whether the reaction proceeding in the forward direction or the reverse direction what would my equilibrium composition look like? Lots of products or very little products? Very little. Can you see that? Because when equilibrium is established my numerator products have to be small so I have very little products and lots of reactants. So if you have very little products and lots of reactants what would the magnitude of this x come out to be? Large or small? Large. So now this x is really big. So if x is really big can I say that 2x is small compared to 3? No that 2x is really going to be close to 3. Do you understand that? So here I can't make the approximation. Alright? And if I can't make the approximation then I have to sit there and solve this algebraic equation. Unless you have a computer that can solve this equation you won't be able to solve this equation you know in a classroom. Do you understand that? And so like I said before we have a problem. Alright? And therefore I'm just going to throw up my arms and say I can't do it. Alright? I can't solve this equation. Alright? But remember part of solving equilibrium problems is coming up with clever strategies. Alright? And the strategy we're going to use here is that once again let's go back to our scale. Remember we said if you have 100% reactants and over here you have 100% products and for this equilibrium we know that kc is right here. Alright? But we started with 100% product. Alright? So our problem was that we start here and we proceed this way and here x is large. So no approximation. And that creates a problem for us. Alright? So the other alternative is that we know that whenever the reaction proceeds in the forward direction x is small. When it proceeds in the reverse direction x is going to be large. Alright? So instead of making the reaction go this way what I'm going to do is I'm going to make it go 100% this way. Alright? So I'm going to let all of the reactants go to product and then I'm going to let it go in the forward direction. You see that? So by making it go 100% this way and then making it go here this x is going to be small. Alright? So our strategy is now to keep the math simple. If I make the reaction go in the forward direction now I can make the approximation and say x is really small. Alright? I'm going to call that y because I don't want you to get confused with the x up there because we're using x everywhere. So let's say that I'm going to call this y. Okay? So we have 100% reactant here, 100% product there. So how do I set about doing this? So I'm going to start with 2 NOCl gas giving me 2 NO gas plus CL2 gas. Kc is 1.6 times 10 to the negative 5 moles per liter. So I'm going to start with my initial to begin with what I have is 1 and 3 and 0. Now my general strategy is going to be that I'm going to make it go 100%. So I'm not going to let it go to equilibrium, I'm just going to go let it go 100%. So when I let it go 100% I have a new initial. Now I'm changing things around and my new initial would be when this reacts 100% to give me products. So if it's reacting 100% this is a limiting reagent problem. It's a reaction that goes to completion. So if it's a limiting reagent problem we look at these two reactants and figure out which is the excess reactant which is the limiting reactant. Now if you take CL2 remember the reactant 1 to 2 ratio. If all of this was consumed how much of this would it need? 2. Can you see that? If it only uses up 2 how much is left over? 1. So that means all of this is used up and this will react only 2 moles of this but it has 3. So this is the excess reagent and then if one of this reacts with 2 how much product would it form on this side? 2. Alright so now my change would be now I'm making the reaction go in the forward direction. So now it's proceeding. Now that I've shifted it all to 100% in order to establish equilibrium I know that it has to proceed in the forward direction. Alright if it's proceeding in the forward direction I know this would be negative 2y. This would be plus 2y. This would be y. I'm just using y so that you don't confuse it with the x from the previous example where we couldn't solve the equation. Okay so therefore at equilibrium the concentrations would be 2 minus 2y. This would be 1 plus 2y and that would be y. Alright so now I can say kc equals 1.6 times 10 to the negative 5 moles per liter equals the concentration of NO squared times the concentration of Cl2 divided by the concentration of NOCl squared which gives me 1 plus 2y squared times y divided by 2 minus 2y squared. But now I can simplify the math because now I can make an approximation and I can say my approximation is that 1 plus 2y is 1. In other words 2y is so small it can be neglected. Alright and 2 minus 2y will equal 2 as well. Alright so now I can see that kc equals 1.6 times 10 to the negative 5 moles per liter which equals 1 squared times y divided by 2 squared which is y over 4. Alright can you see now my math has got to be very simple. And so if I solve for x, x comes out to be 6.4 times 10 to the negative 5 moles per liter. So now by taking this approach I've simplified the math considerably. Alright and can you guys at home check the validity. Alright what is my validity here? I'm saying 2x is small compared to 1 or 2y. So check validity and then can you guys at home figure out what the concentration of NOCl is, what the concentration of NO is and what the concentration of Cl2 is. Alright we'll just check answers at the beginning of class next time.
 Good morning. This is the third quarter of the general chemistry sequence, so I don't  think I need an introduction. But welcome to this class. I see a lot of familiar faces.  And we are going to pick up where we stopped last quarter. But before I begin, I want to  kind of remind you that my primary mode of communicating with you will be through the  class website. And so it is important that you check the class website often, particularly  the one that says announcements, because I will be putting material there and reminders  and so on. So it's important that before you come to class, sometimes I will put some notes  there, I will put some extra problems there. It is important for you to check that as often  as you can and print out the material that's there. Okay? So I am not, because you've already  taken this class, you know, these classes before, and this is the third quarter, I'm  not going to go into the details of housekeeping, all right? The dates when you'll have your  midterm exams, how the grade distributions work out, all that information is on the class  website and you can read it. So I'm not going to go through those details with you, because  you already know what it is. However, I do want to remind you that if you want to do  well in this class, there are three elements to this class and you all know what those  elements are. The first is coming to lectures. As all of you know, it is important to come  to lectures because as I say before, I'm your tour guide and if you don't know where your  tour guide is taking you, obviously you're not going to know how to prepare for the exam,  all right? Textbooks are very often written as encyclopedias, all right? So they have lots  of information. It's impossible for you to kind of navigate through all the information  that's provided in a general chemistry textbook, all right? So I'm going to emphasize things  in class. This quarter, the material that we are covering actually involves a lot of  numerical problem solving. Now we moved on. We looked at the fundamentals of bonding.  Now we're moving on to looking at equilibria and later on we're going to look at electrochemistry,  we're going to look at kinetics and so on. And the topics in 1c are very heavily weighted  towards numerical problem solving. And so I'm going to be working lots of examples in  class because one of the ways in which you teach problem solving skills is to model  how experts solve problems, how do they approach problems, how do they solve these problems  and particularly very challenging problems, all right? So I'm going to work lots and lots  of examples in class. And so it's important that you come to class because you're going  to see me work all these different types of problems and show you how to approach and  solve those problems. And so if you don't come to lecture, you're going to miss out  on that, okay? So lectures are important, but as all of you know, lectures has a disadvantage  in that you're very passive. I'm doing all the talking and you're seated there just  listening to me. And in that passive mode, your mind can wander off. Already it's wandered  off, you know, and it'll come back. And so what you're doing is periodically you're  paying attention and that's not very good, all right? But the reality is that in any  large scale instruction like this, we have to get through the material and the reality  is that the lecture mode of instruction is not going to change, all right, when you have  such big lecture halls. So in the lecture, very often you're going to be passive, but  I'm going to get you involved. I'm going to ask you to come prepared. I'm going to ask  you to throw questions at you. I'm going to expect you to give me answers and sometimes  I'm going to make you do the problem in class, all right? So come prepared. So that's the  lecture part. But we also have the second element of this class, which is discussions.  And because you're so passive during lectures, in discussions, I expect you to take control  of the discussions. Discussions are where you are going to do the work, you're going  to work the problems. The TAs are going to ask you to come forward. It's okay to make  mistakes. Discussions are where you make mistakes and learn from your mistakes. So the TAs are  going to make you come up and answer questions and so on. So the discussion is the place  where you are going to be really engaged and you have to come prepared to discussions and  you have to actively participate in discussions. So we don't want anybody just sitting in the  back and copying the answers because that's not really what we want to do. We want you  to be involved, okay? So discussions are where you're going to be very active. Every discussion,  as you know, the prior week, I will be putting up a worksheet and those worksheets have representative  problems. Again, because it's so heavily weighted towards numerical problems, the content  that we're covering right now, and so I will give you lots of different problems and give  you opportunities to practice as well, okay? So it will be a good idea if you could try  the discussion problems prior to going to discussion so that, you know, when you go  to discussions, you can actually ask a lot of questions. The TAs are going to ask you  to come forward and work the problems on the board and then people are going to point out  why it's wrong and so on. And so if you come prepared, it's very helpful, okay? So the  second element to this course is discussions. And as all of you know, attendance and discussion  is mandatory. We will mark attendance in discussion. And of course, the third element is homework,  all right? All three elements are important. And for this class, you have online homework.  All of you have seen the online homework and have tried it before. And the homework that  we are going to be using is catalyst. So each one of you will be getting an individualized  homework. So your homework is different from the others in your class. And remember, the  goal is mastery. The goal is not to just complete the homework for the sake of completing the  homework. It's not a one-shot deal. You can do the homework as many times as you wish.  Because every time you restart it, it's going to give you a fresh set of problems, all right?  And the goal is that you achieve mastery. So you can work it four times, five times.  As long as you achieve mastery and recall that, you have to achieve 75% or more on your homework,  all right? If you score 75% or more on your homework, that counts as getting 100% on your  homework, all right? So once again, you have an opportunity to practice what you've learned  using the online homework. But I want to remind you that online homework is an automated system.  And there's only so much computers can do. And so where online homework really works  is where you get a lot of practice. In other words, practice of fundamental basic stuff  or some stuff that you have to learn by rote before you can actually work more complex  problems. So the analogy that I'm going to give you is what I've given you before. And  it's like, you know, multiplications. You know, you need to learn at some point that  two times three is six and three times four is 12 and so on. And online homework is very  good at getting you to practice that. But it is not very good at giving you a very complex  problem where you're bringing in multiple concepts together, putting them together, and  solving a complicated problem where you have to synthesize the information and use it in  creative ways, all right? However, if you want to get to that state, you have to get  past this algorithmic practice type problems where you have gone through just the basic  stuff and you understand what it means when you say you're calculating equilibrium constant,  you have to know what it means when you are looking at the law of mass actions. Those  are all the fundamental stuff that we expect you to know if you're going to solve a more  complex equilibrium problem. So the online homework is very good at getting you through  the fundamental stuff and making sure you know the basic stuff. But it is not very good  at giving you very complex problems that you have to synthesize the information. In order  to do that, I'm also going to assign a few end of chapter problems from your textbook,  all right? However, the textbook problems you do not need to turn it in, but I don't  want to overload you with work, so I will choose a few examples of problems that I would like  you to work from the end of chapter problems in your textbook. And the solutions to those  problems are available in the back of your textbook. We've also placed the complete solutions  manual on reserve at the library. So if you can't solve a problem and you can't come see  me during my office hours or can't see the TAs during the office hours, you can actually  go to the library and it's at the front desk. It's on reserve. You can just, you know, take  a look at it and if you want to copy, you can just go down the hallway and use one of  those copy machines to copy that particular page if you wanted to. All right? So there  are three elements to this course and I liken it to a three-legged stool. If you choose  to sit on a two-legged stool and you choose to not do homework or you choose not to go  to discussion, what do you think is going to happen? It's not going to be very stable,  all right? So if you want to do well in this class, if you want to benefit from this class,  all three elements are important because all three are integrated with each other. Each  one reinforces the other and I rely on you to go to discussions. I rely on you to carry  online homework and I rely on you to come prepared to class. Is that clear to everybody?  So that's all that I'm going to talk about the mechanics of this class, all right? Now  before I proceed and start on the material for today, is there anything that you want  to ask me? Any questions? All right, so let's go ahead and start and we're going to begin  by reviewing a little bit of the material that we covered last quarter. Now the theme  for this quarter is actually continuing chemical equilibria. So we're going to go on to looking  at acid-base equilibria which happens to be chapter 11 from your textbook. However, it  is important that you know chapter 10 really well. Chapter 10 deals with the fundamentals  of chemical equilibrium. So it means that you have to have a good understanding of the  basic principles that we've looked at last quarter because if you understand those principles,  now it's a matter of applying them to different scenarios. So if you understand what chemical  equilibrium is, now we're going to take different kinds of chemical equilibrium and look at  them in greater detail. And the first one that we're going to start by looking at is  looking at acid-base equilibria, okay? But before doing that, I wanted to kind of review  some fundamental aspects of the material that we covered last quarter and I want to take  one particular example. And the way I'm going to kind of review the material is actually  to work a problem. And we're going to work on a complex problem and those of you who  took my class last time know what I'm talking about because I want to show you different  strategies of solving the problem. And I'm going to take a problem that I did work last  quarter but is a really good example. And for the benefit of those who were not in my class  last quarter, I wanted to go through this example and we're going to kind of look at  all the main points as we go along. So I have put this problem on the class website. So I  really don't want you to copy the problem. I want you to pay more attention to the strategies  we're using to solve this problem. For some reason, this is not focusing very well today.  So we're going to take an equilibrium problem and we're going to use the sort of general  strategies that we use to solve an equilibrium problem. I apologize for this. I do not know  why this thing is acting up today. There we are. Right. Keep your fingers crossed. So  let's take a look at this problem. So you're given an equilibrium. All right. And we're  told that at 35 degrees Celsius, KSC for this equilibrium is given to you. Calculate the  concentrations of all the species at equilibrium for the following original mixtures. All right.  So if you're going to solve an equilibrium problem, where is the starting point? What  do you look at? Where's the most obvious thing you need to look at? The balanced equation.  We start with the balanced equation. Never assume that any balanced equation provided  to you is actually balanced. Always check because there can be typos. And sometimes  people give you the equation without balancing it. And if it's unbalanced, you need to know  that it's unbalanced. Now it turns out this equation is balanced. So we start by writing  down the balanced equation and the physical states. And as a rule of thumb, I always suggest  that you write the value of the equilibrium constant as well right next to it as a reminder  to you what the magnitude of that equilibrium constant is. And it turns out this is 1.6 times  10 to the negative 5. Now all of you know that the equilibrium constant is called the  law of mass action or it's the equilibrium constant expression. And how do you write  the law of mass action? It's always defined as what? The concentrations of the products  raised to the part of the coefficient divided by the concentration of the reactants raised  to the part of the coefficient. So the bottom line is that when you look at the law of mass  action or the equilibrium expression, the products are the numerator and the reactants  are the denominator. And so essentially it gives you roughly the ratio of products to  reactants. So the bottom line is it's not important to me that you know how to write  the law of mass action. I tell you guys that any eighth grader can write that. If you are  given a balanced equation, writing the law of mass action or the equilibrium expression,  anyone can do it. It doesn't require any prior knowledge. What's more important to me is  that you understand what that number tells you. Remember the equilibrium constant gives  you the ratio, approximate ratio of products to reactants where the products are the numerator  and the reactants are the denominator. So what K, the numerical value of K tells you  is what the equilibrium composition looks like. It tells you what does that equilibrium  mixture look like. It tells you whether that mixture has lots of products and very little  reactants. So if your K value is much greater than 1, so let's say it's a value of the order  of a thousand. If K is the value of the order of a thousand, what does that tell you about  the equilibrium mixture? Lots of products, very little reactants. So it gives you an  idea of the composition of the mixture. And that is the most important take home message.  Because the equilibrium constant gives you an idea when equilibrium is established, what  the mixture of reactants and products look like. Got it? So now let's look at this. This  number is 10 to the negative 5. Is that a number? Now as a rough gauge, if your products  and reactants were a 50-50 mixture, what should the value of K come out to be? Approximately  1. So if you're at equilibrium, when equilibrium is established, if you have a 50-50 mixture,  it would be approximately 1. If the number is far greater than 1, we know that the mixture  now should have lots of products and very little reactants. Now this is 10 to the negative  5, so this number is much, much smaller than 1. So what does that tell you about that equilibrium  mixture? Lots of reactants, very little products. So that means this reaction took place only  to a small amount. Do you see that? It reacted just a little bit and then established equilibrium.  And we use the term extent of the reaction to describe that. And so if you want to describe  this in terms of the extent of the reaction, does this reaction react to a large extent  before establishing equilibrium or did it go to a small extent before establishing an  equilibrium? Small extent. Do you see that? And that's important because we're going to  use that idea. So one of the things that you get a feel for is by looking at the magnitude  of k and then if that number is really small, you know that it only reacted to a small extent.  If that number is really large, so instead of 10 to the negative 5, if it's 10 to the  power 5, what does that mean? It reacted to a large extent. So what that means is like  you have almost 99% product and 1% reactant. Do you see that? Roughly I'm using just approximate  numbers but it means that it reacted to a large extent and if it's 10 to the positive  5, it's large extent, 99% products, about 1% reactants. If it's 10 to the negative  5, what does that tell you? You have only 1% product and you have 99% of unreacted reactants.  Do you understand that? So that's the first important take home message and that's why  you write the k value right next to the equilibrium so that when you look at the magnitude of  that number, you have an idea of to what extent that reaction took place. So now we start  by writing down all the information that's provided to us. So we're given the concentrations  of NOCl. So we're told 2 moles of pure NOCl in a 2 liter flask. So since we're given  moles, the units that are provided are moles and volume, now we're going to think of this  in terms of concentration because remember in equilibrium expressions or law of mass  actions, you can either give it in terms of concentrations or if they're all gases, you  can give it in terms of pressure because pressure of gases is directly proportional to concentration  as well and that's why you can give them in two different units. As long as they're gases,  they're all gases, then you can give it in terms of atmospheres or you can give it in  terms of concentration. But it turns out here, we're given moles and volume and as all of  you know, what is the definition of concentration? It's moles divided by volume. So this is moles  and this is volume in liters and therefore the unit is what? Moles per liter. So we're  looking at this gas, we're given the number of moles of each one of those gases or in  this case there's only one reactant and there's only one substance there to begin with and  so what would the concentration come out to be? It's 2 moles divided by 2 liters and therefore  my initial concentrations to begin with. Alright, so my initial concentration is moles per liter  or what is moles per liter synonymous with? Molarity. Okay, you know that molarity and  moles per liter are the same thing and it's a unit of concentration and therefore my initial  would be before any reaction takes place. It's at time zero. Before any reaction starts  at time zero is the initial concentration. So before any reaction takes place, we know  that the concentration of NOCl is 2 moles divided by 2 liters so what does that come  out to be? 1 mole per liter and zero. Okay, we have no products to begin with. Now we're  going to allow the system to go to equilibrium but remember these are reversible reactions  and therefore the reaction can take place in the forward direction or it could take place  in the reverse direction. So now you have to figure out is the reaction when it goes  to forming an equilibrium, is this reaction proceeding in the forward direction or is  it proceeding in the reverse direction? And all of you can recognize that we have no products  when you have an equilibrium you have to have mixtures of reactants and products. So if  there are zero products what is the obvious direction in which the reaction is going  to take place? Forward. Okay, there's another way you can do this and you learned about  that last time. You can calculate what? Q. Q is the reaction quotient which is the ratio  of products to reactants before any reaction takes place. What is Q here? If your products  are zero what is Q? Zero. So there's only one way Q can go. From zero it's got to go  to 10 to the negative 5. Do you understand that? And that means it has to go towards  products as well. So if you're given the amounts of reactants and products then you  calculate Q and you have to figure out which direction the reaction will proceed. In this  case it's going to proceed in the forward direction. Alright? And so if it's proceeding  in the forward direction then we're going to say that we're, let's say it is x is the  amount that's used up. We know that it's moving in the forward direction therefore the concentration  of this has to go down and the concentration of the products need to go up when equilibrium  is established. And if it's going to proceed in the forward direction then some of this  is going to be consumed and some of these are going to form. So let's say the amount  that's consumed is 2x because the coefficient here is 2 and because in this balanced equation  we know that the ratios are maintained for every two moles that's consumed how many  moles a product have formed? Two moles of this and one mole of that. So that ratio is always  maintained. That's the ratio in which they react. So for every two moles that's consumed  you're going to have two moles of NO being formed and one mole of Cl2 being formed. So  if 2x is being consumed what does that tell me about the amount of NO that will be formed?  Plus 2x because you know that we put a plus sign to remind us that it's being formed and  over here the ratio is 1 so that should be plus x. So that means at equilibrium what  I would have is 1 minus 2x this would be 2xx. Does that make sense everybody? So when equilibrium  is established if you want to know the concentrations of reactants and products that expression  describes it and if we can calculate what the value of x is then we can solve for this  and we can write an algebraic expression solve for x and figure out what is the concentration  of NOCl at equilibrium, what is the concentration of NO, what is the concentration of Cl2. So  in order to find out what x is we need an algebraic expression. We need to relate x  to something else that the problem provides. If they didn't give us any additional information  we could not solve this problem. And what is that additional information we're going  to make use of? It's the equilibrium constant. So we can turn to the law of mass action and  let me move on, let's take that out. So now we're going to move on to writing the law  of mass action which is Kc. So we can say Kc is the concentration of the products raised  to the power of the coefficient times Cl2 raised to the power of 1 divided by NOCl raised  to the power of 2. So this is the law of mass action or this is the law of chemical equilibrium.  And if we know that Kc is 1.6 times 10 to the negative 5 moles per liter. Remember equilibrium  constants do have units, alright? And those units will depend on the law of mass action.  So this is the K value and this equals NO would be 2x squared. That's the concentration  of NO times x divided by 1 minus 2x squared. So that gives me this is 4x squared times  x. That gives me 4x cubed 1 minus 2x. Alright? Now one of the biggest headaches of solving  equilibrium problems as all of you know is the algebra. I mean up to this point it's  pretty straightforward. But now you have to be able to solve for x, alright? And sometimes  this can be very complicated. Now if you had a good calculator that could solve this,  well it's good. But remember on an exam you don't have the luxury of having you know sophisticated  program that can solve the algebraic equation for you. All you have is a basic scientific  calculator. And you can see that if you're using just a basic scientific calculator this  is going to be hard to solve. So one of the challenges that you're going to have when  you solve these equilibrium problems is this mantra. And what's that mantra? Keep the math  simple. Alright? If you can keep the math simple the chances that you'll make careless  mistakes gets better. And it turns out that your answer should be the same as if you were  to solve this, you know, go through the whole process of solving for this. So here one of  the things, one way in which we keep the math simple is we make intelligent approximations.  Alright? And so let's look at this equation. In this equilibrium if you look at on a scale,  remember we said that k is 10 to the negative 5. And so if k is 10 to the negative 5 we  said that because that number is so small we know that the reaction only proceeds to a  small extent. Alright? If it proceeds to a small extent what does that tell you about  x? Remember it only reacts so that 1% of reaction goes to product. 99% remains unreactive. So  what does that tell you about x? Will that x be large or is it going to be small? Small.  Okay? So if you have a scale and on this scale we have 100% reactants here. On this scale  we have 100% products over here. This is initial. When we start this is we have only 100% reactants.  If the reaction proceeded 100% if it goes to completion then what's going to happen?  We have zero reactants, 100% products. If when equilibrium is established you're somewhere  here what does that mean? Now when equilibrium is established right in the center is a 50-50  mixture. So that means k has to be approximately 1 there. Do you see that? We know this k is  10 to the negative 5. So what does that tell me about kc? kc when equilibrium is established  and if you look at the ratio of products to reactants you know that you have lots of unreacted  reactants and very little products. So if you look at this whole range of looking at  the ratios of reactants to products you can have 100% reactant that means no reaction has  taken place. You can have 100% product that means it's an example where the reaction went  to completion. It didn't establish equilibrium it just went to completion. If it's right  in the middle what happens when equilibrium is established you have a 50-50 mixture. If  it's on the right hand side of this if it's somewhere here what does that tell you? You're  going to have 75% product and 25% reactant. Do you see that? Alright so we're over here  and so this is where equilibrium is established. So that means that it reacted only to a small  extent and that's what x gives you. Alright the extent of the reaction is the amount that  reacted and we know that that x is going to be really small because it only reacted to  a small extent. Now because this x is really small because k is 10 to the negative 5 now  that gives us a way to make an approximation because we know that it only reacted to a  small extent and therefore to keep my math simple now I'm going to make a smart approximation  and my approximation is that 1 minus 2x. Remember I want to simplify this expression. Alright  and so if you start with one molar concentration and it reacts only to a small extent we're  going to say that 2x is so small. 2x is so small compared to one. Compared to one molar  we can neglect it. So this is an assumption that you're making. It's an approximation.  It's very subjective. Do you understand that? Anybody can make any approximations. Alright  we're going to start by making the approximations but then we have to go validate and make sure  that approximation is actually valid or not. Okay so we're going to make this assumption  and can you see that when we make this approximation it considerably simplifies my math because  now I can say 1.6 times 7 to the negative 5 moles per liter equals 4x cubed divided  by 1. Now what would have turned out to be a horrible algebraic equation to solve now  just boils down to a very simple step. Alright now all I have to do is divide this number  by 4 and take the root cube of that and so if you work that out it comes out to be, so  if you solve for x, x comes out to be 1.6 times 10 to the negative 2 molar. Remember x is  concentration so the unit for that should be multiple liter or molarity. Alright so I  made this very subjective approximation and anybody can make any approximations you understand  that. Alright but what do I need to do now? I actually have to go back and make sure that  my approximation is valid. Now what was my approximation? Go back and ask yourself what  is your approximation? 2x is small compared to 1. So we need to find out is 2x really  small compared to 1. So my validity will be, so I'm going to check the validity of my approximation.  And my validity is that what I'm saying is 2x is really small compared to 1. So I'm going  to take the ratio of the 2 and find out what percent is 2x compared to 1. Alright so once  again you have to go back, it's not x, I'm not saying x is small compared to 1 in that  approximation I'm saying 2x is small compared to 1 and therefore I have to compare 2x with  1. Do you get it? And so I would go back and I'd say 2 times 1.6 times 10 to the negative  2 molar divided by 1 molar times 100 and if you work that out that comes out to be 3.2  percent. So what that tells us is that if you compare 2x to 1, 2x is 3.2 percent of  1. Alright so now you have to ask yourself is that good enough? And so we have this purely  arbitrary rule and we call it the 5 percent rule. If it is approximately 5 percent or  less we know that it's a really small number and that it can be neglected. Alright and  so because this falls below this arbitrary 5 percent that we picked. Alright now we know  that the approximation is valid. Okay and we know that 2x is indeed very small compared  to 1. Alright so we say approximation is valid. Now let's say you made this approximation  and it came out to be 25 percent. So what does that mean? That you can't make that approximation.  Alright and that means that you have to go back and you have nothing, you just have to  grit your teeth, sit down and solve that algebraic equation. Do you understand that? There's  nothing you can do about it. Alright it turns out in this case you can. And where does that  clue lie? Where does the clue lie when you can make those approximations? Kc. If Kc is  really small, if it's 10 to the negative 4, 10 to the negative 5, 10 to the negative  6 those are really really small numbers for K. That means the extent of the reaction has  to be really small and therefore you can make that approximation. Now we need to calculate  the concentrations and so we go back. Now the problem is still not complete because now  we need to figure out what the concentrations of NoCl is and the concentration of NoCl is  1 minus 2x which is 1 minus 3.2 times 10 to the negative 2 molar and that comes out to  be.97 molar. No, concentration of No is 2x which is 3.2 times 10 to the negative 2 molar  and then the concentration of Cl2 is x which is 1.6 times 10 to the negative 2 molar. Alright  now what's interesting is if you did not make this approximation and if you had solved that  algebraic equation you would end up with the same answer. Alright and so it makes it much  easier if you make this approximation and therefore solving that problem gets easier.  Alright now let's go to the second part of this problem. In the second part of this problem  so can I remove this how you guys done with this? Okay great so let's go back to part  2 and now we're told things have changed a little bit. Now we have a different scenario  where we have 3 moles of No and 1 mole of Cl2 in a 1 liter flask. Alright so we're going  to do the same thing so we're going to go back. We start by writing the balanced equation.  Make sure you look at the physical states because the physical states give you an idea  of whether it's a homogeneous system or a heterogeneous system and the law of mass action  does depend on whether it's a homogeneous system or a heterogeneous system. These are  all in the gaseous states so are they homogeneous or are they heterogeneous? They're homogeneous.  So once again, right KC to remind you that this is a very small number. Now if we write  down the initial concentrations in moles per liter of molarity they tell us that we have  3 moles of No in a 1 liter flask so what is the concentration of No? 3 moles per liter  and I have 1 mole per liter of Cl2. I'm not putting the units there because I'm putting  that within parenthesis there and it reminds us that we're in units of concentration. And  we don't have any reactants so what does that mean? You have 0. So in solving an equilibrium  problem this is what you have to begin with. So what is the first question you ask? Remember  equilibria are reversible reactions. They can proceed in the forward direction or the  reverse direction. In this case which direction is that reaction proceeding? Reverse. So now  we have no reactants so we know that if equilibrium is established we have to have some reactant  and therefore we know so I'm going to put this arrow to remind me that that's the direction  in which the reaction is going to take place. If the reaction is taking place in that direction  then if we're looking at the relative change it's proceeding in this direction so what  would this come out to be? So tell me what would I put here? Minus. Minus what? 2x. Here  minus x and on this side plus 2x because now the reaction is proceeding in the reverse  direction so products are going to be consumed to form reactants because products are being  consumed you put the negative sign to show that your concentration is decreasing on the  product side and the concentration is increasing on the reactant side. Okay therefore when  equilibrium is established this would be 2x this would be 3 minus 2x and this would be  1 minus x. Everybody agrees with me? So now we turn to the law of mass action we know  kc equals 1.6 times 10 to the negative 5 moles per liter equals again the form in which this  equilibrium is written is that these are the products these are the reactants. Alright so  the law of mass action would be NO squared over cl2 divided by NO cl squared which is  NO would be 3 minus 2x squared times 1 minus x divided by 2x squared. Got it? So now ask  yourselves can I make an approximation here? Now the reaction is proceeding in the reverse  direction what does k tell me that at equilibrium what would my composition look like? At equilibrium  k tends to negative 5 so regardless of whether the reaction proceeding in the forward direction  or the reverse direction what would my equilibrium composition look like? Lots of products or  very little products? Very little. Can you see that? Because when equilibrium is established  my numerator products have to be small so I have very little products and lots of reactants.  So if you have very little products and lots of reactants what would the magnitude of this  x come out to be? Large or small? Large. So now this x is really big. So if x is really  big can I say that 2x is small compared to 3? No that 2x is really going to be close  to 3. Do you understand that? So here I can't make the approximation. Alright? And if I  can't make the approximation then I have to sit there and solve this algebraic equation.  Unless you have a computer that can solve this equation you won't be able to solve this  equation you know in a classroom. Do you understand that? And so like I said before we have a  problem. Alright? And therefore I'm just going to throw up my arms and say I can't do it.  Alright? I can't solve this equation. Alright? But remember part of solving equilibrium problems  is coming up with clever strategies. Alright? And the strategy we're going to use here is  that once again let's go back to our scale. Remember we said if you have 100% reactants  and over here you have 100% products and for this equilibrium we know that kc is right  here. Alright? But we started with 100% product. Alright? So our problem was that we start  here and we proceed this way and here x is large. So no approximation. And that creates  a problem for us. Alright? So the other alternative is that we know that whenever the reaction  proceeds in the forward direction x is small. When it proceeds in the reverse direction  x is going to be large. Alright? So instead of making the reaction go this way what I'm  going to do is I'm going to make it go 100% this way. Alright? So I'm going to let all  of the reactants go to product and then I'm going to let it go in the forward direction.  You see that? So by making it go 100% this way and then making it go here this x is going  to be small. Alright? So our strategy is now to keep the math simple. If I make the reaction  go in the forward direction now I can make the approximation and say x is really small.  Alright? I'm going to call that y because I don't want you to get confused with the  x up there because we're using x everywhere. So let's say that I'm going to call this y.  Okay? So we have 100% reactant here, 100% product there. So how do I set about doing  this? So I'm going to start with 2 NOCl gas giving me 2 NO gas plus CL2 gas. Kc is 1.6  times 10 to the negative 5 moles per liter. So I'm going to start with my initial to  begin with what I have is 1 and 3 and 0. Now my general strategy is going to be that I'm  going to make it go 100%. So I'm not going to let it go to equilibrium, I'm just going  to go let it go 100%. So when I let it go 100% I have a new initial. Now I'm changing  things around and my new initial would be when this reacts 100% to give me products.  So if it's reacting 100% this is a limiting reagent problem. It's a reaction that goes  to completion. So if it's a limiting reagent problem we look at these two reactants and  figure out which is the excess reactant which is the limiting reactant. Now if you take CL2  remember the reactant 1 to 2 ratio. If all of this was consumed how much of this would  it need? 2. Can you see that? If it only uses up 2 how much is left over? 1. So that means  all of this is used up and this will react only 2 moles of this but it has 3. So this  is the excess reagent and then if one of this reacts with 2 how much product would it form  on this side? 2. Alright so now my change would be now I'm making the reaction go in  the forward direction. So now it's proceeding. Now that I've shifted it all to 100% in order  to establish equilibrium I know that it has to proceed in the forward direction. Alright  if it's proceeding in the forward direction I know this would be negative 2y. This would  be plus 2y. This would be y. I'm just using y so that you don't confuse it with the x  from the previous example where we couldn't solve the equation. Okay so therefore at equilibrium  the concentrations would be 2 minus 2y. This would be 1 plus 2y and that would be y. Alright  so now I can say kc equals 1.6 times 10 to the negative 5 moles per liter equals the  concentration of NO squared times the concentration of Cl2 divided by the concentration of NOCl  squared which gives me 1 plus 2y squared times y divided by 2 minus 2y squared. But now I  can simplify the math because now I can make an approximation and I can say my approximation  is that 1 plus 2y is 1. In other words 2y is so small it can be neglected. Alright and  2 minus 2y will equal 2 as well. Alright so now I can see that kc equals 1.6 times 10  to the negative 5 moles per liter which equals 1 squared times y divided by 2 squared which  is y over 4. Alright can you see now my math has got to be very simple. And so if I solve  for x, x comes out to be 6.4 times 10 to the negative 5 moles per liter. So now by taking  this approach I've simplified the math considerably. Alright and can you guys at home check the  validity. Alright what is my validity here? I'm saying 2x is small compared to 1 or 2y.  So check validity and then can you guys at home figure out what the concentration of  NOCl is, what the concentration of NO is and what the concentration of Cl2 is. Alright  we'll just check answers at the beginning of class next time.
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