 So, let's do one of these problems, determining the rate orders. So it says, consider the reaction presented 2NO gas plus O2 gas goes to 2NO2 gas. Collection of the following kinetics data was shown. Okay. So, we got rate data, concentration data. The rate is actually in units of all over per second. Okay, so how do we figure out the rate of law? So, remember, you guys remember how to do the rate of law, write it out at all. So, we're going to have to remember that. The rate equals the rate constant times, well, the concentration, and now we've got to go back and look at what the reaction equation gives us. Okay. So, we see nitrogen monoxide here. Okay. So, that has to go into the rate law, and we're determining the rate order scheme. So, that's this superscript, that M, okay. So, the other reactant is oxygen, so we're going to have to say the molar concentration of oxygen to the end. Okay, so we're determining M, and that's what we're determining. Those are the rate orders. And then it also wants us to determine the rate order of the entire reaction itself. Okay. So, how do I do this? Well, I've got to do effectively a couple of problems for specifically. So, the first problems I have to do is find, well, where does find experiments or do these experiments, and then find data where the concentration of one of the reactants changes but the other reactant stays the same. So, for the first set, so for nitrogen monoxide, right, we see experiments two and experiments one where it changes, but oxygen's concentration doesn't change. You guys see that? Yes. You guys see that? Yes. Yes. Okay. Cool. So, those would be the ones we would want to use for N up here. Okay. So, let's go ahead and write out that part of the problem. Okay. So, let's just keep this up there to remind ourselves what we're looking for. So, I'm going to just say, and I always like to put the bigger number on top so we'll have a number at the end that's going to be bigger than one as opposed to one that's smaller than one. Decimals are harder to work with. Okay. So, in this case, 0.02 is bigger than 0.01. So, I'm going to put rate two over rate one like that. Everybody okay with that? So now, so the other thing about these problems is we have to be under the assumption that temperature isn't changing in between any of these experiments or the rate constant would change. Okay. So, if you remember the Uranius equation problem that we did before. So, K is going to be constant. So, anyways, so K, the concentration of N, L, M, so that's K, N, L, 1, or in this case, 2, sorry, 2, and then 0, 2, 2, N. Okay. So, those 2's are referring to the experiment, too. Is everybody okay with that? Yes. So, notice, I didn't put a 2 under being the K. Why? Because it's not going to do what? Change. Change, right? Okay, 1. So, this would be N, 0, what? 1. Very good. And 0, 2, 1. Okay. Is everybody okay with that? Yes. Okay, wonderful. So, all straight away, right, we can see that K is going to cancel out. Everybody's cool with that, I'm sure. Right? So, let's just cancel that out. Alrighty. The one thing we should realize is that the concentration of N, O cannot cancel out. Okay? Because there's a change. Does everybody see that? Yes. So, look at the concentration of O2 of them. Right? It's the same, it's the same. Okay? So, if this is the same and this is the same, and the reaction order doesn't change when you change the experiment either. Okay? So, that's going to be the same as what? Is everybody fine with that? Yes. So, what we can do effectively is cancel this with this, because those two numbers didn't change. Okay? So, what's the problem with that? Let me know right now. Should be writing this down. If you're not writing this down, I'm going to be looking at you twisted. Write it down. There's no reason not to. Okay. So, I'm going to just kind of reduce our equation to get all this business out of the way. So, we've got an equation that looks like this now. Concentration of N, O to the M. So, the second N, though, over O, 1 to the N. Everybody's cool with that one, right? Yes. That makes sense. If I have M here and M here, I can parenthesis and take it out. Okay? So, that's what I'm going to do. I'm going to put 0.02 divided by 0.01, and that's what muller, or some muller, whatever. So, those units are going to cancel. So, in this case, it's going to be, what is that? 2 divided by 1. So, it's 2, right? 2 to the, okay? So, at this point, we don't know what M is. Figure it out. Have any problems with what we've done so far? Yeah. Kind of wonderful. So, now I'm going to figure out these things here. So, we get this from the initial rating. So, lucky that's rate, rate. Okay? So, we're looking at experiment 2 and experiment 1. So, experiment 2, experiment 1. So, we'll just say this equals experiment 2, 0.20 muller per second divided by 0.05 muller per second. So, notice, in this case, there was no super script, so we don't have to worry about that part. Is everybody okay with that? Yes. That makes sense? Yes. That's wonderful. So, this, if you reduce that, is going to be what? 4. In your head? Good job. So, 4, that equals 2 to the m. Okay? So, we're going to set those two things equal to each other. If you can't do that in your head, I can show you how to, we'll do another recording of like how to do that mathematically. But this one you should be able to do in your head if you can't. Just remember, 2 to the what equals 4. That's what we're taking here. Okay? So, 2 to the 2 equals 4. Right? Is everybody okay with taking that for an m? Usually these problems are pretty straightforward where you can figure out what that number is. There's a lot of pretty easy, I don't know, power equations. So, in fact, what we found is that m here is going to equal 2. So, the reaction order for nitrogen monoxide is 2. Okay? So, I'm going to have to erase that to use that piece of board again. But I'll let you guys take a second to write it down. I think you guys think we're going to be looking for it. So, what do we have to look for here if experimental ones? Ones that, what does this have to do? Stay the same or change? If 1000 stays the same, that one changes, right? So, which experiments do that? 3 and 2, okay? So, we're just going to set it up like we did last time. Okay? So, this is essentially a rehash of that same problem. Okay? 1 and 3. Oh, sorry. Oh, yeah. Is that one? Sorry, yeah. Sorry, sorry, sorry. You guys got it. That was a test. So, 1 and 3. Okay? We would have figured it out once we got over there anyway. So, anyway, thanks for catching me. Okay? So, 3 and 1, right? Which one are we going to put on the top? Well, let's write out over here what we got. So, we're going to do rate equals k. And we know m is 2. If you want to put it there, you can. Concentration of O2 to the hand. Rate to concentration of O2 to the hand. So, now, which one is going to go on top? 3 or 1, right? That's the one that I would prefer to put on top because it's bigger. k, does that change when we do this reaction experimental conditions? No, so we don't have to worry about changing that. 3, 3. 1, 1, 1. So, k cancels straight away. No, didn't change. Yep, cancels. Okay? So, let's reduce this equation out here. Concentration of O2, 3, to the hand. O2, 1, to the hand. Like that. And remember when we have hand over hand, right? Like that we can. Any problems with what we've done so far? Okay, wonderful. So, O2, 3, 0.02, 0.02 molar. But 0.01, is that right, molar? I can't see. 2 and 1, right? So, to the hand. Cancel, cancel. So, what's that going to be? 2 to the hand, right? Any problems with what we've done over there? No. Okay, wonderful. So, let's do our rate 3 over rate 1. So, rate 3 is 0.10 molar per second, divided by 0.05 molar per second. Like that. So, cancel, cancel. Notice again, no superscripts there. So, what's 10 divided by 5? 2. Okay, so now we have to set 2 equal to 2 to the hand. Okay, very good. That's awesome. To the hand. Okay, like I said, usually you'll see that these ones are pretty easy to do in your head. If they're not, then you might have done something weird in your problem. Okay, so, like I saw over here, people getting excited saying 2 to the hand or whatever, that's 1. Okay, and you're absolutely right. So, the reaction order for oxygen in this particular reaction is 1. Okay, is everybody okay with that? Yes. So, let's just write n equals 1. Okay, so what have we found? We found the reaction orders for the individual reactants. Okay, so how do I erase all of this business here that we've been working out? Does everybody got that written down to where I can erase it? Because there's more to do with this problem, not much, but a little bit more. Because it wanted us to not only tell what the reaction order for the two reactants themselves were, but the reaction order for the reaction itself. Okay, so let's just write this down. So, reaction orders, that's going to be m, right? So, that's equals to 2, and for o2, that is n, so that equals 1. So, we can actually come back here and figure out, well, what is the actual rate law? Okay, so let's write in what m and m are. All we have to do, erase m and put a 2 there, and 1's, you don't have to show, okay? They're understood. Is everybody okay with that? So, that would be the actual rate law for this. Okay, so I think that was one of the things that asked us to figure out what the rate law is. So, we figured out the rate law, figured out the reaction order for mO, the reaction order for o2, and it wants us to figure out what the reaction order for the overall reaction is. So, overall, n equals 2 plus 1, okay? So, that's all you do is add the 2 m and n up. Okay, so that equals, so that's the overall reaction order. And we could probably even do one other thing, you know, let's do one other thing. Let's figure out what k is for this reaction, okay? So, we have now data from an experiment, and you can pick any experiment that you want. Let's just pick number 1 here, okay? So, we've got rate data, right? So, the initial rate there. We've got the concentration of nO2, and we've got the concentration of o2. So, if we rearrange that equation, we can figure out what k is. Does everybody follow what I'm saying? Yeah, you got that? Okay, so let's rearrange this equation. So, rearrange it on your own and see if you can do it the same way I do. So, k to rearrange, guys. k equals the rate of squared. And like I said, we're just going to pick one of these reactions. I just picked 1 because, well, that was just the first one up there. Okay, so the rate is 0.05 molar per 1 second. The concentration of nO is 0.01 molar squared like that. The concentration of o2 is 0.01 molar, okay? So, tan to the 1, 2, 3, 4, 4, okay? So, 1, 6, big, 1, 6, big, 1, 6, big. So, that's going to be 1, 6, big. Okay, let's figure out what our units are going to be, okay? So, this is going to be molar times molar times molar, right? And that's going to be a molar on top, okay? So, molar divided by molar q. So, we're going to have molar squared in the bottom, right? Is everybody okay with that? So, molar squared in the bottom. And then, also, seconds in the bottom, okay? So, it's going to be per second per molar squared. All of those various questions. Okay? Long problem.