We're going to continue on our theme of first-order kinetics. In our last lesson we talked about the integrative rate law for first-order kinetics. Now we're gonna talk about the concept of half-life. And you're going to learn to employ the concept of half-life and see the relationship between a rate constant and the value for half-life, and be able to utilize those into calculations. So what is half-life? It's the amount of time, if you start with a certain amount of reactant, it's the amount of time it takes to drop that down to half of the amount that you started with. And we've got these symbols here to represent half-life. The half-life symbol is t one half. And here is the statement above in symbols. The amount of time it takes to get your concentration down to half the amount you started with. So let's derive the half-life formula. This is our integrative first-order rate law. And we're going to do some substitution. What are we gonna substitute? We're going to take what this represents here. So we're going to substitute in half life of A nought there. So this is going to give us this equation. Now of course the A noughts will cancel, and we have a ln 1/2 is equal to -kt, whereas we bring the minus to the other side, that actually gives me the ln 2 = kt. And then we will actually calculate ln 2, and that's .693, so this equation right is our half-life equation for first-order kinetics. So make sure, as you commit this to memory, that you realize that it only applies to first-order kinetics. And it's helpful, I mean you could maybe learn it from this stage if you could derive it kind of quickly. It's helpful to look at that derivation to help you remember this equation, now for my class, here at the University of Kentucky, I require them to memorize equations that are in boxed form there. So this is one that I recommend that you commit to memory. So this is only for, again, first-order kinetics, and it is independent of how much you start with. It doesn't matter, whether you start with a lot or a little. It's the amount of time to take however much you start with, and drop it down to one-half its original value. And that's what that statement says. Doesn't matter how much time it takes. So let's look at this representation here. In the top sphere we see 16 molecules, let's say that maybe each one of those represents a tenth of a mole, then we can have a representation of a certain amount. As one time interval, whatever the half life is, whatever that time interval is the amount of time it takes to drop it down to one-half its value is its half-life. And what we see with first-order kinetics, is that time interval is exactly the same to drop it down to its next half as much. And that time interval is exactly the same to drop it down to its next half as much, and that's unique for first-order kinetics. Course in there next time interval how much will we have remaining? Well, there would only be one blue dot in my circle. So let's think about this problem. And I'm gonna have you stop the video and ponder it. I have a 1st order reaction. I'm starting with an 8-molar solution. So there's its initial concentration. And the half-life is a minute. So I want you to think about how much is going to remain after 3 minutes. Did you say 1 molar? Well if you did, you'd be correct. What are we doing here? If it's 3 minutes we have 3 half-lives that have gone by. In the first half-life it's gonna go from 8 to 4. In my second half-life, it's gonna go from 4 to 2. In my third half-life it's gonna go 2 to 1. So we're going down into those as we go through those half-lives. If we look at our picture, here we see that we have said that these time intervals are in minutes basically. And we're starting here with 8, and going one minute down to 4, going the second minute down to 2, and then in the third minute, we're certainly going to go down to 1. So let's do a calculation here. We have a first order reaction, and they've given me the rate constant. And we know that we can obtain a half-life from that rate constant. At 0.693 over k which is 3.0 times 10 to the minus 3 seconds to the minus 1. Now remember that's our unit for first-order kinetics, always for the rate constant. So we can obtain our half-life from that formula. Now let's think about this. We want to know the time required for the reaction to be 75% complete. What happens after the first half life? First half-life it goes from 100% there to 50% there. And then it's going to go from 50% there, to 25% there. That's cutting it in half again. That would be 75% complete, so this is asking how much time is it going to take to go through two half lives. Well, if we knew how much time it took to go through one half life, we'd be set so what is one half life? When we do this calculation we find it's 231 seconds. So we'll take 231 seconds to go through that half life. Now we only really know it to two significant figures. So we could say it is 2.3 times 10 to the 2nd, seconds. And that is one half-life. Two half-lives would be twice that. So it would be 4.6 x 10 to the 2 seconds, or around 460 seconds. This problem is a little bit different, in that it doesn't have a nice multiple of half lives. It's 30% complete. Well if it's 30% complete, what we're gonna wanna do, if we can't figure it out by a certain number of half-lives, is we're going to go back to our first-order kinetics problem, our integrative first law. And this is really the first time we are seeing this equation being used. 30% complete is 70% still there. So what we know is, if we started with all of it, that would be 100, and that's not a concentration but this ratio would be the same. It works in percentages as well. If I started with 100%, if it's 30% complete, be very careful here, 70% is still there. So we've got the 70 over 100 or we have 0.7. And we figured out, this is the same rate constant. So we know, actually we don't. Does it matter? [LAUGH] 3.0 x 10 to the -3, seconds to the -1 here. And we can solve for the time it takes to do that. So let's see here. We will obtain ln 0.7. And that's a -0.357. We'll divide both sides by negative 3.0 x 10 to the -3. And that will give me a t value, a time of 119 seconds. Again we really only know that to two significant figures. So that'd be 1.2 times 10 to the 2 seconds. Should it be less than the half life? The half life was 230 seconds. Well, yes, because the half life is to get to 50% complete, and this is only 30% complete, so that number makes sense. Now I wanna go back to the previous problem for just a moment, because if you looked at this problem and you could not think, okay, 75% complete, that's two half lives, you could work this problem in a similar fashion as the one before, and if you started with 100 and at 75% complete, there's a 25 there. And then we have a -k, times t, and we could solve for t. And we would find that it would be the same value. So you didn't have to notice that it was two half-lives, but once you know the half life it's sometimes very convenient to look for that. Okay, so that finishes our learning objective. We've learned the relationship between half life and k. We can utilize that to go between the two, and then use it in a calculation.
