Just like that. >> All right. Good afternoon. >> Reminder that when you enter this room 48 hours from now, it'll be time to see how much of what we've been talking about recently has made sense to you. And I remind you that we will have that review session this evening aged 15 to 20. Brown lab it Yaakov for today, I want to do a little bit more with Dynamic Equilibrium. Chapter 15, talk about how to solve a couple of problems. They're mostly, I want to introduce a technique you will find useful for solving problems in chapter 15. And if there's time left at the end, I'd be willing to address any questions anybody might have about Friday's exam and the time we have left. And we might have a little bit of time left because the fellow who was supposed to be IRR, the beginning of class today to talk to you about a study abroad opportunity has had to postpone until Monday, so it'll be here Monday to see what he has to you instead. >> So just to refresh your memory a little bit about where we were last time, we used this reaction referred to as the Haber process. >> Afterwards, discover as an example of an equilibrium and also as an example of how to go about finding an equilibrium constant. That is, you can write an equilibrium expression war reaction quotient looking something like this, with concentrations of products in the numerator, concentrations of reactants are the denominator, everything raised to the exponent that matches up with its coefficient in the balanced equation. And then at equilibrium, that is going to be equal to the equilibrium constant. So one way of evaluating an equilibrium constant is to measure the concentrations of everything once the system has come to a state of equilibrium. And then enter the numbers and go ahead and evaluate it as a numerical value. Now that's pretty straightforward and that's actually kind of like what you're gonna be doing in next week's laboratory experiment, which has to do with the concept of finding an equilibrium constant. The somewhat more challenging problem is, if you already know the equilibrium constant, can you figure out what the concentration of everything is going to be once the system comes to a state of equilibrium. A lot of people have already measured equilibrium constants for Many chemical reactions, you can frequently look up equilibrium constants. But the question then becomes, can you use that information to find the concentrations of everything at a state of equilibrium. For those of you who are following along in the lecture notes, we have the two problems today. The first one appears on page 35 and the second on page 36. So here's the problem for page 35. Let's consider this equilibrium which involves phosphorus penta chloride, PCl5, decomposing the form phosphorus trichloride, PCl3, and chlorine gas. And in fact, all three of these substances happened to be gases. That turns out to be important for reasons that we'll talk about later. But the point is, the equilibrium constant for this reaction has been measured a value of 7.30 times ten to the negative two. Alright, suppose you have an empty one liter box. Into that box, you inject one mole of PCl5. And then you let this reaction take place until it reaches a state of dynamic chemical equilibrium. At that point, what are the concentrations of PCl5, PCl3, and Cl2? That's what you're trying to figure out. Now, the technique that I want to introduce for solving problems like this is explained in your textbook on one thing that your textbook tends to do, which you may have noticed last semester. They have these two page spreads and they have a starting point down here where the green dot is. And then you follow the arrows around, see what's going on. But the point is what the two-page spread on pages 6-8 forward 685, is constructing ICE tables to solve equilibrium problems. Here's what we mean by that. Well, first of all, before we get around to what an ICE table is, how would you write the equilibrium expression for this equilibrium? In other words, we're going to create a fraction that is equal to this equilibrium constant. What's the numerator of the fraction going to look like? >> What's the denominator of the fraction that'll look like refer back to last time if you need a hint. Yes, That's correct. >> Basic idea is products over reactants, stuff on the right side, over stuff on the left side. >> So our products are PCl3 and Cl2 are only react, it is PCl5. >> And so the equilibrium expression or the reaction quotient looks like this. And at equilibrium, this ratio of concentrations should have a value of 7.3 times ten to the negative two. >> Now, here's what we mean by setting up an ICE table. >> The purpose of the ICE table is to provide a visual aid that will hopefully help you think about conceptually what's going on as we go from the initial state right after we inject or one mole of PCl5 to the final or equilibrium state. Once this reaction has come to a state of equilibrium, it's called an ICE table because you're gonna make three columns. >> I, C, E stands for Initial Change equilibrium. >> And the idea of an ICE table is to monitor what happens to the concentrations of all three things involved in the equilibrium as we go from the initial state before any chemical reaction has taken place to the final or equilibrium state. So let's think about the initial column for a moment. Notice what it says here. We have a one liter box into which we're going to inject one mole of PCl5 gas. Now the gas, of course, fills up the entire volume of the box, so it's going to occupy a volume of one liter. By when we say initial column, we assume that this reaction has not taken place yet, so just PCl5. So what's the initial concentration of PCl5? >> Yes. >> One mole per liter. We injected one mole into the box. The box is a volume of one liter. These concentrations are all in moles per liter. So one mole divided by one leader. The initial concentration of PCl5 is one moment. What's the initial concentration of PCl3? Yeah, right. We didn't inject any BCL3, we just injected PCl5. And we're assuming that this reaction has not taken place yet. So the initial concentration of PCl3 is 0. How about the initial concentration of Cl2? Did we inject any Cl2? So what's the initial concentration of CO2? >> 0. >> So the initial column looks like this. >> Any questions about where the initial column numbers came from? >> Okay. >> Now the change column means these concentrations aren't going to change as we go from the initial state to the final equilibrium state. Specifically, some of the PCL5 is going to break apart and form some PCl3 and some Cl2. How much breaks apart? >> We don't know. >> So we're going to call that quantity x change column, looks like this. The concentration of PCl5, it's going to go down by some amount. The minus in front of the x means that concentration is going down. >> The other two concentrations are going to go up by the same amount x. >> The point is they can't go down there starting from 0, they can only go up. As some of this breaks apart, we make some of this and some of this. So these two concentrations go up, this concentration goes down. And since all of the coefficients in the balanced equation r one, that means for every molecule of this that breaks apart, you get one molecule of this, one molecule this. >> So this goes down by x. >> These other G, We're going to go up by x. >> Still making sense, okay? >> The equilibrium column is just the sum of the first two columns. >> So for PCl5, one plus negative x is one minus x. For the other 20 plus x is just x. >> And the id is after you construct this ice table. >> When you have all of these final concentrations are equilibrium concentrations in terms of x, then you just insert these things in terms of x back into this equation and then do whatever arithmetic is necessary to solve for the value of x. We'll do that in just a moment before I do just want to give people a chance to finish with this slide question because think about what's going on here. We injected one mole of PCl5 into the box, nothing else but what this PCl5 do, it decomposes deformed PCl3 and Cl2. In other words, look up here at the balanced equation. So the point is, since some of this is going to break apart and form this, at this, this concentration must be going. Now these other two, you have no choice but to go up because they started from 0. The point is, where's this material coming from? >> Some of this breaking up before these two guys, okay, makes sense to everyone. >> Okay? >> They might need more time, but this slide, Alright, so Here's the equilibrium expression again. >> Concentration of PCl3 times concentration of Cl2 divided by the concentration of PCl5. That's equal to the equilibrium constant, which is 7.30 times ten to the negative two at equilibrium. But going back to the ICE table again, Concentration of PCl5 and equilibrium is one minus x. The other two concentrations are both x. So when we insert that into the table, we get x times x in the numerator and one minus x in the denominator. And again, our goal is to do whenever we have to do to solve for x. So a little bit of algebra here. X times x is of course x squared. >> You get x squared by itself. >> I multiply both sides by the denominator. So now we have 7.3 times ten to the negative two times the quantity one minus x on the other side. On the next line, I just multiply through to expand this 7.30 times ten to the negative two times 1.307 times. So that they can do minus 7.30 times ten to the negative two times x And then finally, to get to this next line, I just subtracted the entire right side from both sides, which means the right side is now 0. And here's what the other side looks like. If you do all that and there's a reason that we're doing all this. Take a look at this equation. What kind of equation is this? Presumably something you have seen before, hopefully prior to coming to the University of Delaware. >> What kind of equation is this? >> Yeah, this equation takes the form a x squared plus bx plus c equals 0, which is the basic form of the quadratic equation. >> So to solve this problem, we need to solve a quadratic equation. >> Here's an aside. >> How many of you were sitting in your high school math class, whether we're talking about quadratic equations at the person next to you leaned over and said, We're never going to need this. Why are we studying this, that app and your assignment over spring break is to go back and find year-old math teacher and apologize because the quadratic equation is good for stuff. And here's one thing it's good for. Who remembers the solution to the quadratic equation in terms of a, B, and C? >> Yes, very well. >> What I'm saying is, given that this is the form of the quadratic equation, given the coefficients a, b, and c. What is the solution to the quadratic equation x equals what? >> In terms of a, B, and C. >> Okay? In other words, it was pounded into your brain. >> So for those for whom it was not pounded into their brain, that's what today's handout is all about. >> You didn't pick up today's handout. You can do so on the way out. But in line one of the handout, it just shows you the form of the quadratic equation and the solution to it in terms of a, b, and c, which are the coefficients of x squared x and the unit step as we go. Now in this particular quadratic equation, what is the value of a one? A is the coefficient of x squared. We don't have a coefficient for x squared, so a is just one. >> What's the value of B in this particular quadratic equation? What is the value of big yep, 7.30 times ten to the negative two. >> And what's the value of C, though it is not the same thing. >> This is usually written as a x squared plus bx plus c. But since we have a subtraction sign here, That means the value of c is negative 7.3 times ten to the negative two. >> Now the point is, if you know what the values of a, B, and C are, you can solve the quadratic equation. >> The rest of this handout just walks you through the arithmetic of actually doing that. Some calculators can be programmed to solve quadratic equations. And you plug in the values of a, b, and c phi. >> You can do it that way for that matter. >> What the heck buddy, hey Siri, what's the solution to this quadratic equation? If a is this, b is that. See you that my word, I haven't tried it, but it might work. Point is, you should in principle be able to solve the quadratic equation as shown on the handout. And if you do, you get a value for X of 0.236. Now, just as an aside, when I asked a moment ago, what was the solution for the quadratic equation? You said correctly, negative b plus or minus the square root of blah, blah, blah, plus or minus means that there are usually two solutions to the quadratic equation. >> However, I've only included plus, not minus on the handout. >> And the reason for that is that most of the time when you're doing a problem like this one, the other solution to x will be a negative number. But the problem is what x is going to represent here are concentrations. >> Is there such a thing as a negative concentration, though there is not. >> So even though you might find two solutions to the particular quadratic equation you're trying to solve. It's probably the case that only one of them will have any physical meaning. The point is, once you know what x is, you go back to the ice. Did in fact, let me just briefly take you back to the ICE table. Once you know what x is, then you go back to the ICE table and say, okay, whatever x is, that's the chlorine concentration, that's also the PCl3 concentration. And one minus x is the PCL5 concentration. In other words, once you have a numerical value for x, go back to the ICE table, look at the equilibrium column, enter that value for x wherever you see it, and evaluate. And if you do that in this case, you should get that the concentration of PCl3 and the concentration of chlorine are both 0.236 lowered. And the concentration of PCl5, which is one minus x, turns out to be 0.76 forward. So that's the answer to the question. The question wanted to know what was the concentration of everything? Once you come to a state of equilibrium, and you can always check yourself to be sure that you did it right. How can you check yourself based on the information that you have on the screen? >> Yeah. >> If you did it right, you take these numbers and enter them right here and evaluate this, you should get 7.3 times ten to the negative two. And if you do that for these numbers, that will work out to be the case. So you can always check yourself at the end to make sure you did it right. >> So the main point for right now, your textbook talks about how to make an ICE table. >> An ICE table like this is just a visual aid to help you solve the problem. The real work of actually solving the problem might come down doing something like solving a quadratic equation or whatever. But the point is, once you have everything in terms of x and the ICE table, your goal is to solve for x by any means necessary questions about this problem. >> Yes, what I did was to leave this x square right where it was and subtract everything else that leaves me with 0 on the right side of the equal sign. >> And then these other terms become the negative of themselves on the other side. >> Okay? Yep. Yes, yes. >> Before we do that other problem, any of the questions about this one. >> Okay. Lets do that other page 36 of electronics, same kind of question, but now we're looking at a different reaction which is in equilibrium. >> And it has a different value for the equilibrium constant, 1.8 times ten to the negative six, an empty one liter box into which we inject one mole of NO2 at equilibrium. What are the concentrations of NO2, NO, and O2 according to this equilibrium? Alright, let's start where we did with the previous problem. How would you write the equilibrium expression for this equilibrium? >> In other words, what would the numerator of the fraction, luckily, the denominator of the fraction look like, Yes, you're close. >> It is products over reactants. So the eta o, the 02 are going to be in the numerator and the NO2 is going to be the denominator. >> But what about the coefficients? >> What happens to them? >> They don't cancel out. >> Remembers what happens to the coefficients when you write the equilibrium expression, they become the exponents. >> So equilibrium expression looks like this. >> No and O2 are the products. They go on top, but NO has a coefficient of two. That means the concentration of NO is going to have an exponent of two. That's concentration of NO squared. And then in the denominator we have NO2, which is our only reacted, but it has a coefficient of two also. So its concentration term is also squared. >> In the denominator. >> So remember, whenever you are writing the equilibrium expression, the coefficients in the balanced equation become exponents when you write the equilibrium expression. Okay? That's equal to the equilibrium constant, which is 1.8 times ten to the negative six. And to help us sort all this out, we're going to put together another ICE table. So remember what it says in the problem? We take an empty one liter box, we inject one mole of NO2. >> That's it. >> We assume that this reaction has not yet taken place. >> What is the initial concentration of NO2 Sagan? One molar. >> Yeah, it's moles per liter, which is molar. >> But yeah, again, one mole divided by one liter. It's gotta be one more. What's the initial concentration of NO, 0? Because this reaction has taken place yet. The initial concentration of O2 also 0. >> The initial column looks like this change column. >> Now this is going to be a little bit different from the previous problem, but I think you'll see why we know that as this comes to equilibrium, the sum of this breaks down and these concentrations are gonna go up. But look at the coefficients. >> Suppose the concentration of O2 goes up by x. What will the concentration of NO go up by? Yeah, 2x because its coefficient is two. >> Therefore, what will the concentration of NO2 go down by negative 2x? >> Yep. >> In other words, I'll just show you the rest of the slide in the previous example, the change column is pretty straightforward because all of the coefficients in the balanced equation, we're just one here, they're not. But the numbers that you see in the chain of column should reflect the stoichiometry of the reaction. In question. Two moles of NO2 break down to form two moles of NO and one mole of O2. So the coefficient for the change of the O2 concentration is one for the change in the concentration of NO, it's two for the change in the concentration of NO2, it's minus two goes down. >> This goes up for positives and negatives, and the numbers match up with the coefficients in the balanced equation. >> Still making sense. >> Yeah. >> Ok. >> This is where you just have to read what the problem says and see what's going on. >> Now in this case, it just says we injected one mole of NO2. >> That's it. >> Ok? >> Now suppose it was something like we add a plus B yields C plus D. And we injected one mole of a to moles of B. Well then you're going to have several things ABCD your column. And for a, the initial is going to be one. For B the initial is going to be too, because you objected to moles. So the point is you're just going to read what the problem says, see what you did. But the point is the initial column just assumes that the reaction hasn't taken place yet. Just whatever you eject, that's it. >> But nothing has happened yet. >> Anyway, the equilibrium column here, just like before, add up the first two columns and you get equilibrium column. So for O2, 0 plus x is x. For n, 0 plus 2x is 2x ever NO2 one plus negative 2x is one minus 2x molar. And then as before, our goal is to take these guys, plug them in up here and solve for x by any means necessary before we do that. But anyway, like more time with this slide. >> Alright, let's see what happens at equilibrium. >> Q, which is the reaction quotient. >> The fraction is equal to k Q, which is the numerical value of the equilibrium constant. >> And going back to the ICE table, concentration of O2 is X, concentration of NO is 2x, concentration of NO2 is one minus 2x. >> We, we put all of that into the equilibrium expression and it looks something like this. >> Now we could do a ton of algebra here, but let me just make the process a little bit simpler for you to X quantity squared is four X squared. Multiply that by x and you get 4X cubed. So the numerator becomes 4X cubed. And then you can play whatever games you want with the denominator. The point is, if you eventually do all that algebra and rearrange this thing and write it out, you're going to have what's called a cubic equation, which looks sort of like a quadratic equation, except it's got an x cubed term in there somewhere. So anybody here know the solution to the cubic equation. If not, that's okay, I don't either. And therefore Chatelier's first rule kicks in. This is not to be confused with Le Chatelier's Principle. Le Chatelier's principle would shift to the left, shift to the right. We're talking about that on Monday Chatelier's first rule is if I don't know, you don't have to. So we're not going to solve the cubic equation, but we can still solve this problem because this number right here, the equilibrium constant, large number or small number, theta, Yeah, that's a very small number. You said small. >> You were holding up fingers like this little teeny number, very small number. >> And what that tells you is by the time this reaction is done, you're going to have mostly reactants, not very much of the products. In other words, what that tells you is that x is going to be a very small number. We can still solve this problem because we can make an assumption here. The assumption is that since the equilibrium constant is very small, and for our purposes in Chem 102, I'm defining very small as less than one times ten to the negative four, which this number is. Then x will also be very small, so much so that subtracting x, or perhaps even subtracting 2x one isn't going to make very much difference. The point is, if we can ignore this term and the denominator here. And the denominator basically comes down to one squared, which is just one. And so basically this isn't even a fraction anymore. In short, this simplifies down to four x cubed. And now we don't have to solve a cubic equation. The worst thing you have to do to solve this problem. Let's take a cube root plosive. You should be able to take cube roots and your calculators. If you don't know how to do that, bring me your calculator will show you how. But the point is to solve this problem at this point, all you have to do is divide both sides by four and take a cube root. >> So divide 1.8 times ten to the negative six by four. >> You get 4.5 times to the negative seven. And then take the cube root of 4.5 times ten to the negative seven. That turns out to be 7.7 times ten to the negative third, that's x. So now we go back to the ICE table and we enter this numerical value for x in the equilibrium column. Wherever we see it, x represents the concentration of O2. >> Okay? >> 7.7 times ten to the negative third molar concentration of NO is 2x. Multiply this number by two, we get 1.5 times ten to the negative two lower. Finally, concentration of NO2 was one minus 2x e. Subtract this number from one, you get 0.98. And how can you check yourself to be sure you did it right? After you check to be sure these answers are correct. >> Yes. >> If we did this right, they should be able to take these numbers, put it back into the equilibrium expression up here, the arithmetic, and we should get 1.8 times ten to the negative six. The other thing that is also worth doing whenever you make a simplifying assumption like this, check yourself at the end to make sure your assumption was correct. The assumption that we made here is that since X was going to be very small, subtracting 2x from one wasn't going to make very much of a difference. The denominator should come out pretty close to what did it? Yeah, 0.98 is pretty close to one off by maybe about 2%. That's not too bad. Sound reasonable? So here is the take home message for today. A technique that is sometimes helpful for solving equilibrium problems like this. Construct an ICE table. An ICE table with a visual aid to help you think in terms of what happens as you go from the initial state to the equilibrium state. Then you can take these values, plug them back into the equilibrium expression, and solve for x by any means necessary. Now sometimes that might involve doing something like solving a quadratic equation. But sometimes, especially when you have a very small number for the equilibrium constant, you can make a simplifying assumption to ignore the x term in the denominator. And that makes life a whole lot simpler than trying to do half a ton of algebra. >> And as always, you can check yourself when you're done by taking the answers you get and putting them back in here. >> It's a technique. Hopefully you will find it useful as we move forward. We'll stop there for today, but I will hang out for the rest of the class, period. If people have questions they want to have answered right now about Friday's exam and again, 815 tonight, Brown lab to 20. I'll be there to answer any other questions you might have. >> Look how can we get your chapter? But everybody gets caught. I'd rather go for a package of old stuff if you've already got picked me up all your data. Yeah, totally an opportunity to impress her that he was a wonderful book, by the way, what we eat. And I thought, well, yeah, we're not going to be asking you a question. Just say what's the definition then again, if we use a term like what? >> You did something like know what the energy of activation or this reaction, you don't know what that term you're probably going to have a hard time answering that went out. >> So he didn't know what but we're probably not going to ask you very often are right out of my my point is we should be grateful or whatever. Don't drive yourself crazy trying to reclaim. I think through many weird things up there in my mind. I thought wow. Actually, I remember when I first okay. A practice exam. That is why it is one of the wonderful work that you have to look at the constancy of the table. I've tried them. Okay. >> Now, the problem says, well, the normal freezing point and they both involve looking water. There's water boils at a 100. >> And here are the KDF to K_p value for B. >> It's urea. >> They told you about the way though its molecular weight is about this many grams of IRI. >> And we figure out how many moles. >> And then 66.7 milliliters of water. And I want you walk is loud footballs going, right? >> But this is milliliters. >> Uh, what, what are the density of one? So 66.7 billion years, if they re going to kilograms, you can think you figure out the molarity, molality, pretty straightforwardly. Same thing up here, but the only part that makes part a tricky. >> It's so important and growing more than sodium chloride because that's where the AI, that's where AI comes into play, the blame game. And we just calculated the area. But you're like, well, I back it was World War II to calculate allowing this ball of white light to solids We're getting late one by one. Mm-hm. Mm-hm. Yeah. Mm-hm. Whoa, whoa, whoa, whoa, whoa, whoa, whoa. Oh, oh, oh, oh, oh, oh, oh, oh, oh, oh,

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From Dana Chatellier March 04, 2020

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Video Created by UD Capture Classroom Recording in Brown 206 on 2020-03-04 12:20:00.

Appears In: CHEM-102 Chatellier

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