# 14.5 Integrated Rate Laws and Half Lives

### Chad's General Chemistry Videos

- 19.1 Oxidation Reduction Reactions and Oxidation States
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- 19.3 Galvanic Cells
- 19.4 Standard Cell Potentials aka emf or Voltage
- 19.5 Nonstandard Cell Potentials the Nernst Equation
- 19.6 Reduction Potentials and the Relationship between Cell Potential, Delta G, and the Equilibrium Constant
- 19.7 Electrolytic Cells
- 19.8 Electrolysis Calculations

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##### Video Transcript

All right. All right. Last, last topic in kinetics we'll talk about integrated rate laws. Here they are: zero order, first order, second order. Depending on- depending on what order it is, it follows a different mathematical equation. So that's the first part. These are going to be given to you on the front of your exam. And they'll even be labeled as zero, first, and second. In fact they may be given in a couple of different forms, but here they are. So notice I've given you them in slightly different forms of what's on your sheet I did that for reason as we'll see in a minute. But the big thing these are used for, is they allow you to determine concentration changes over time. Like let's say I told you reactant is zero order, and you have this much now, and the rate constant is whatever, how much will you have in 10 minutes? Okay great. I gave you the time, 10 minutes. I gave you the rate constant, K. I said you had this much initially. The only thing you don't know is him, and it's just plug and chug and solve. But that's what the integrated rate laws are for. They allow you to determine concentration changes over time. Why do they call them integrated rate laws? Because we derived these from calculus, or at least using calculus. Good times. Is calc a prereq for this class? Aren't you happy you don't have to know how to derive these? You just got to know how to use them. Plug and chug. It's algebra. First of all I got to tell you what order a reactant is. That way you know which equation to use. But after that, I got to give you everything except what I'm asking you to solve for. Okay great. So the other thing here. So what is the slope intercept equation of a line? Y equals mx plus B. So we're going to use that because you also have to know some graphical relationships involving stuff like this. Stuff you probably did in your crystal violet lab, FYI. So and, y equals mx plus B. I wrote these up here in the way I did to make them match up with the equation for a line. No matter what order I have, guess what I'm plotting on the x-axis. I'm going to put time there regardless of order. Time is always going to go there. Notice if I want to get super tricky, guess what I put there instead of a lowercase T for time. I'd put a capital T, which represents not time but temperature. Be careful. I saw that show up last year. I was like really, really? Wasn't in Cabirac's class. Bu just FYI. It's a lowercase T for time, not temperature. So if I had a zero order reactant and I want to get a straight line, what do I need to plot on the y-axis then? Concentration of A. And so here, if I'm looking at a reactant at zero order, I'll just put the concentration of the reactant on the y-axis. And so in this case for zero order what would the slope, M, be equal to? Negative K. Negative K. And so it tells me it's not just K, it's negative K, it's a downhill slope. It's a negative slope. Now let's say that rather than having a zero order reactant, I had a first-order reactant. Having a first-order reactant, would I have gotten a straight line for this graph? No I probably would have gotten something more along the lines of a curve had it been first-order and I made this graph. If it's first-order this graph wouldn't give a straight line. And so notice if I don't get a straight line here- if I give you this graph and there's not a straight line, what do you know? Actually you don't know what it is, you just know what it isn't. You'd know it wasn't zero order. So if I give you a curved line, it's not zero order. Could be first. Could be second, I don't know. But at least I know it's not zero. But again, in this case, I gave you a straight line. And so if you see a straight line for the concentration versus time graph, that's zero order. So let's look at the first-order plot as well. Still plotting time on the x-axis. What do we need to plot on the y-axis now to get a straight line? Now it's got to be the natural log of A on the y-axis. So in this case, what's my slope going to be equal to for the first-order plot? Also negative K again. So again another negative slope. It's a downhill slope. And last but not least, second-order. Still going to put time on the x-axis. What do I need to put on the y-axis to get a straight line for a second-order reactant? 1 over A. What's another way of writing 1 over A? A to the negative 1. Same diff. Same diff. So what's my slope for a second-order plot? Positive K. This is the only one that should have an uphill slope, not a downhill slope. And if I want to ask you a tricky question, I might give you a question and these are the answer choices. And I might say "Which of the following corresponds to a second-order reactant?" And you'd be like, "Sweet! That's the one." That's choice A, and that's wrong. That's choice B and that's wrong. But choice C that looks second-order. But guess what I put up there for choice D. Just to make this super tricky, I put that up there. Does that correspond to a second-order reaction? Well I've plotted the right things and I did get a straight line. But the slope should not be negative. The slope really should be positive. And so in this case, that's the tricky wrong answer I'm trying to get you to pick. Correct. For zero order and first order, the plot should come out to have a negative slope. But for second order it should come out to have a positive slope. And again this is not something you have to memorize, because these equations on the front of the exam. Look at what's multiplied by T. Whatever is multiplied by T is your slope. Here it's negative K. Negative slope. Here it's negative K. Here it's just positive K. And that's- yeah if I really had a second-order reactant, this is the straight line graph. Had I tried to graph ln(A) versus time, it would not have given me a straight line. It would have totally given me a curve. Don't worry about that. That doesn't matter. That's not a question they're ever going to ask you. So you should just know, is it a straight line? Yes or no. Okay. Next thing we'll talk about, half-life. What is a half-life? What is a half-life? A set amount of time. What is it the amount of time for? Yeah. It's where your concentration is dropped in half of what it originally was. And the amount of time that takes? We just call it the half-life. Usually what do we use to symbolize the half-life? That's the symbol we give. It's not just t, it's t^1/2. It's a specific set amount of time for a given reaction. And it's just the amount of time it takes for your concentration of your reactant to get cut in half. So if I started out with, you know, a 10 molar concentration of a some reactant. How much would be left after one half-life? Five molar. Sweet. How much would be left after another half-life? 2 and a half molar. And so on and so forth. You just keep dividing by two for every half-life you go through. Okay. And here's the deal, there are equations for the half-lives of all three of these. And those are also going to be given to you. But we're going to talk about one of them in particular, because one of them is more important than the other two. So you may get questions involving zero-order half-life or a second-order half-life, but more than likely the most common one we talk about is the first-order half-life. And the reason we talk about it, is it's special. Turns out this guy, the half-life is equal to 0.693 over K. And so what is the only thing the half-life time depends on? The rate constant. That is it. And the key is this if you look at the half-life equations for zero and second-order, they depend on A initial. And so depending on how much A initial you have, how much the initial concentration of the reactant is, the half light's going to be a different time. Well if I start out with this much A initial and I go through a half-life, what's going to happen to its concentration? It's going to get cut in half. And then for the next half-life I have a new initial concentration, and the next half-life is going to be a totally different amount of time. But for first-order, the half-life time doesn't change. As the concentration of a gets used up, the half-life time doesn't change though, because it's not dependent upon the concentration of A at all. The concentration of your reactant. So it's constant. And that's why we most commonly talk about it. So if you notice, so if I tell you that a reactant is first-order, and I want to look at how the concentration changes over time, then what am I using? To do my calculations? The integrated rate law. So if I want you to solve for that final concentration what do I have to give you? I have to give you the initial concentration, the amount of time that's passed, and the rate constant K. Well sometimes we don't give you the rate constant K, we give you the half-life instead. But if I give you the half-life, what should you be able to calculate? The K value, as well. So sometimes instead of giving you the K value, like we should, we give you the half-life. But you should know that that allows you to calculate out the K value. You guys cool with that? Okay. So we talk about half lives all the time however, we don't always talk about like a fourth life or an eighth life or anything like this. But every once while it shows up on an exam. So let me show you how to derive this half life, that way you know how to drive any other life that gets presented to you as well. So if we look at this first order integrated rate law and rearrange it just a little bit, and I'm going to take ln(A) still, but I'm going to subtract ln(Ao) to the other side, so I get minus ln(A) initial here. And that's equal to negative KT. But when you subtract logs, you can combine them under one log term. And what does that look like? Yeah. Whatever gets subtracted ends up in the denominator. And so I end up with ln(A) over Ao, one log term equals, negative KT. Now in this case if A initial was 10 molar, after a half-life, what would be the concentration of A? 5. Notice you didn't have to plug that into an equation, you just divided by 2, right? What if I started out with like, you know, 10 grams of something. How many grams would be left after a half-life? 5 grams. So what if I, you know, had a hundred percent of something. What percent would be left after a half-life? 50 percent. So you can do this in molarities, grams, percentages, whatever. So here's the deal. Whether it's 5 grams to 10 grams, 5 molar to 10 molar, 50% over 100%, this ratio after half-life always comes out to 1/2. And so specifically for a half-life, this equation turns into ln(1/2) equals negative K. But it's not just t anymore, it is specifically, the half-life time, T one-half, in the equation now. Guess what the natural log of 1/2 is? It's negative 0.693. And when you divide by K, so that's why the equation looks the way it does. So t 1/2 equals 0.693 over K. Well what if instead of a half-life, I wanted like a 1/8 life. What would you do? What fraction of your sample should be left after a 1/8 life? 1/8. So instead of plugging in a 1/2, plug in 1/8. And instead of t 1/2, it would be t 1/8. Same thing. What if I had a 1/3 life? 1/3 and t 1/3 instead. You could solve for it. Cool. So I just want to show you where the half-life equation was derived from. That way in case you get asked anything else, you can figure it out. If they ask for the second order to zero order, their equations are a little more complex and it'll have A not showing up in it one way or the other. But that equation is still going to be on the front of your exam. You just use that equation instead. I'm just again going through this one because this one is much more likely. Cool. By the way you, those are also much more difficult, you probably won't get asked anything other than something about half-life for zero order and second order. But first order, because we take the time to derive this for you, we might ask you something a little trickier as well. So you can also do this one other way. Let's look at this for a sec. Let's say I told you that K equaled 0.00693. I'll make this nice. If K was 0.00693 for a first order process, that's the unit, then my question first off is what is t 1/2? What would be the half-life for a reaction that had that K value being first-order? We'll plug it in. 0.693 over that K value gets you a hundred, somebody said. And because it was s to the minus one, your half-life time comes out in seconds. Cool. See, so you got half-life. Now we said you could plug in like 1/8 life and get the eight lifetime and stuff like that. And you totally can. But you can go about- as long as it's something that's divisible by two, you can do something else as well. So if you start with all of your sample let's say. All of it. We'll call that one. What fraction would be left after one half-life? A half. What fraction would be left after another half-life in this case? Another hundred seconds. A fourth. What fraction will be left after another hundred seconds? Cool. How long is an eighth life? In this case then. How long is it? Three half-lives. So if the half-life is 100 seconds, an eighth life would be 300 seconds. How long would a 1/4 life be? 200 seconds. So on and so forth. So as long as it's some multiple of two, a half-life, a fourth life, an eighth life, a sixteenth life, you could just use the equation get the half-life. And then if it's a fourth life, double it. If it's an eighth life, triple it. Whatever. Do it that route as well.

In this lesson you will learn:

### The zero, first, and second order integrated rate laws

### How to calculate concentration changes over time using integrated rate laws

### How to perform calculations involving half-lives

### How to obtain linear plots for zero, first, and second order reactants