In our last couple of lessons we learned about rate laws and we learned about rate constants and the orders and how to determine those orders from experimental data. This lesson is a little bit different in that we're gonna be talking about integrated rate laws. Integrated rate laws incorporate time. The others was just rate versus concentration. This will incorporate time. So we're gonna be able to see how the concentration changes as time goes by. This is kind of like the difference between saying this is how fast a train is traveling in miles per hour. Or, this is how much time it takes for the train to go from one point to some other point down the line. Our integrated rate laws are like that latter one. How much time does it take to go from this amount of reactant to having it drop down to this amount? Or, if I know I'm starting with this amount and I wanna know how much time it's gonna take to end up with another amount, we can calculate the time. We will be focusing on the simplest cases of 1st order overall reactions. Then we'll look at 2nd order overall reactions. Then we'll back up a little bit and spend a little bit of time talking about 0th order because there are reactions in which the concentration has no effect on the rate. This learning objective five is going to focus on the first order. So, in our lesson today, we're going to be focusing on how to derive the first order integrated rate law, and the interdependence between time and concentration for that. So for first-order kinetics, we know a couple of things already. If we had a generic reaction like this one, A turning into products, we could write this rate law, rate is equal to K times A, we know that simply because we've been told it is first order. If we back up a few more lessons and remember what we learned about integrated, no not integrated, but writing a rate expression for reaction, we could take that reaction up here and we could write this rate expression. We remember that the minus sign is because that's a reactant. And the reactant's concentration is decreasing as time goes by. We want the rate of reaction to be positive, so we put that minus sign in there. Because both this portion and this portion are both equal to the same thing, both equal to rate, we can set them equal to each other. This gives us this equation. Now, taking that equation and doing a little calculus will give us the integrated rate law. Before we look at that integrated rate law, let me say that calculus is not a prerequisite for this course, so you don't have to be able to derive this. But if you've had calculus and you take it and you look at it like this, because that would be a little bit more familiar symbols for you. And you integrate that, you will obtain this integrated rate law. And this is the integrated rate law for a first order reaction. And you need to commit it to memory. It's easy to remember if you can derive it using calculus, but that is the equation. The parts and pieces that we see here is the bottom, which is the concentration of your reactant, we are calling it A, at the beginning or at the initial point of the reaction. And up here would be the concentration of A, that reactant, at some time down the line. And that time would be that time that's in the equation when you solve it. Now I'm gonna do a little rearranging. The purpose for this rearranging is I want to get it into a nice, linear expression, something that we can plot a straight line using. The first thing I've done is, when you've got the natural log of a quotient as we see here, that can be translated as this portion. You subtract numerator and not denominator. Then I'm gonna do a little rearranging, to rearrange it into this form. Now, all I've done is I have moved the natural log of A over here, and I've grouped the minus k out. This is gonna give me the slope intercept form of a line, m stands for slope, b stands for intercept. And what this tells me is that this I could plot along the y-axis, t I could plot along the x-axis, and when I did that, I would obtain a straight line, and the slope of that line would be equal to -k. Because that's m and of course y intercept would be natural log of the initial concentration. So that's what that says. A plot of natural log of A, that would be the y-axis versus time, gives a straight line with slope equal to a -k. And here's a couple plots. A plot for first order kinetics of concentration versus time will not give you a straight line, but will give you this exponential decay. But if you take the natural log of that concentration, you suddenly have a straight line. The straight line will always point in the down direction, okay? It'll be a negative slope because if you take the negative of a negative slope, you will get a positive value for k, okay? So the slope is always gonna be negative for this plot. The slope is always gonna be equal to a -k, where k is the rate constant. Here is an actual reaction, a plot of natural log of A versus time. Time is not written there, versus time. And we did it in minutes instead of seconds, and that's okay. Looks like we've got four data plots that have been posted and the best straight line through those data points. Now, if you wanna know the slope of that line, what you would do is you would take two points on the line. Not two data points, but two points on the line, and then you would do a change in y over a change in x. Let me erase that dot because it's in the way now. So the slope of that line would be obtained by the change in y, that's from here to here, minus a change in x. Keep consistent with your data points, so that you're doing the same order. That the two y values, which are these two here, are plugged in. The two x values, which are these here and here, are plugged in and you will obtain a value for k, in this instance, of being 0.034 minutes to the -1. Like I said, it doesn't always have to be units of seconds. We could convert this to 1 over seconds if we wanted using that there's 60 seconds in a minute. But any time you have 1 over a time unit, whether it be minutes, seconds or hours for your k value, that's also an indication that it is a first-order reaction. Now, if we remember, we have done this in a previous lesson. If we're dealing with gases, and we know that PV=nrt, we could put our n and our V together and get P equals n over V times RT. This that I'm circling are the units of concentration. So there is a relationship between concentration and pressure. So all the same plots, linear expressions, could be obtained using pressure instead of concentration. So we could plot natural log of pressure versus time and also get a straight line if it's first order kinetics. Here is a depiction of this, natural log of pressure versus time for this generic reaction where I've got an A that's a gas. It would also give me a nice straight line. Well that's the end of our learning objective five, where we're looking at only first order kinetics. We're looking at that integrated rate law for it, we haven't yet done any calculations with it, but you might have to do some, get some information from graphs. And if you ever see a straight line plot, or natural log of pressure or concentration is plot along with the Y-axis and time along the X-axis. Whenever you see a straight line, you will automatically know that's a first-order reaction.
