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0:01
- Rate law's gonna be the topic of this lesson, and we're gonna talk about rate constants, and we'll talk about reactant orders and the overall reaction order. But the most important part is we are going to determine the rate law of a reaction based on experimental data. And this is something students often struggle with a little bit initially, but we're gonna make it simple, I promise. My name is Chad and welcome to Chad's Prep, where my goal is to take the stress out of learning science.
0:25
Now, in addition to high school and college science prep, we also do DAT, MCAT and OAT prep as well. I'll leave some links in the description. Now this lesson's part of my new general chemistry playlist. I'm releasing several lessons a week throughout the school year. So if you wanna be notified every time I post one, subscribe to the channel and click the bell notification. All right, so let's dive in. We've got some vocab to talk about first. And so first, let's talk about what a rate law is.
0:48
And a rate law, when you think of like the laws of nature, like the law of gravity. Well, what laws ultimately allow us to do is make predictions. If I say, if I let go of this marker, what's going to happen? Well, the prediction is it's going to fall. (marker clunks) And that's based on the law of gravity. And so that's what laws ultimately let us do, is let us make predictions. That's what's gonna be true of a rate law as well. And so this is what a typical rate law looks like.
1:12
And it starts off with a rate constant, lowercase k, and it's important to note that it is a lowercase k, not a capital K, because in the next chapter, we're gonna see something else we use the letter K for, but a capital K as an abbreviation for, or a symbol for. So be careful, this is a lowercase k. It's a rate constant. It is simply a proportionality constant. We'll talk more about it in a little bit. And then typically, reactant concentrations are gonna show up.
1:38
So it's typically just reactants, not products, although that's not exclusive. That's just what's usual. And then you have these exponents right here, and these exponents are called the orders. And so x would be called the order with respect to reactant A, y would be called the order with respect to reactant B. And then there's also what's called the overall reaction order, which is the sum of all the individual reactant orders. And so in this case, the overall reaction order would be x plus y.
2:05
Now these orders are very commonly integers, and they're sometimes related to how many of that particular molecule is involved in what we're gonna call the slow step in a reaction, not necessarily in the overall reaction. And so it turns out sometimes these are gonna match up with the coefficient's interaction and sometimes they're not. So you can't just look at the coefficient's reaction, and be like, oh, those are the orders. It sometimes just happens to work out that way,
2:29
but it's not necessarily gonna be the case. So it turns out you're gonna have to determine these and get the rest of the rate law here from experimental data. You can't necessarily just tell by looking at a reaction what those are gonna be. All right, so again, these are very often integers, and zero, one, and two are the most common by far. However, in principle, if you're using some actual kinetics experimental data, these can come out to fractions or decimals and things of this sort.
2:54
However you're gonna find in your typical freshmen general chemistry course, most commonly are you gonna see zero, one, or two. However, if I make the statement that reactant orders are always integers, you should know that that is false. So however, almost exclusively, the ones you're gonna see in this chapter are going to be integers, which is why that makes that statement a tricky true/false evaluation. But they don't have to be integers, just most commonly that's what we're gonna see.
3:21
And again, they're most commonly zero, one, or two. So we're gonna take a look at a few reactions here. And so let's say I've got a reaction of A going to B, and I tell you that the rate equals k times A to the zero power. Now by telling you it's k times A to the zero power, that's telling you there that the reactant order with respect to A is zero. So however, with only one reactant here, so notice that's the only reactant showing up in the rate law,
3:53
then the overall reaction order is zero as well. And that's what I really wanna focus on here, is something that is zero order. And so in this case, also notice that anything to the zero power equals one, and anything times one, notice k times one, and again, it doesn't matter what concentration of A you plug in, anything you plug in there to the zero power equals one, and k times one is just equal to k. And so oftentimes, you'll just see this written
4:19
without any reactant concentration listed there at all, because it doesn't matter. And that is the hallmark of a zero order reactant is that its concentration doesn't affect the rate at all. You wanna put in one molar A, the rate equals k. You wanna put in 10 molar A, the rate equals k. And so the rate is not affected by a zero order reactant. All right, next, we're gonna take a look at a first order reactant.
4:46
So, and for this one, we're gonna take a look at a fictitious reaction, M going to N. I prefer to actually just do A going to B every time, but I really wanna distinguish this and I don't want this to get confusing. But M going to to N here. And so again, M's typically gonna be what shows up in the rate law. And so here, we're gonna have rate equals k times the concentration of M to some power here, and that power again is often integers, doesn't have to be,
5:11
but if I'm telling you it's a first order reactant, or reaction, in this case, then I'm telling you that that right there is going to be a one. And so we'll substitute a one in there. That's what makes it a first order reaction here with only one reactant. That one reactant is going to be first order. But notice anything to the one power is itself. So like three to the one power is three. Five to the one power is five. Well, the concentration of M to the one power is just the concentration of M.
5:38
And so you often see that completely omitted. And it's just implied that there's an exponent of one there. And so this is what typically a first order reaction looks like. And what you see here is that the rate is proportional to the concentration of M. If you double the concentration of M, it doubles the rate. Notice if I have one molar M, well, one times k is k. And then if I do a second trial of this, but I use twice as much, twice the concentration of M,
6:03
so two molar M, well, two times k is 2k, and 2k is double the value of one k. And so doubling the concentration of M doubled the rate of the reaction. And so this is the hallmark of a first order reactant, where again, in contrast to the zero where the concentration of that reactant doesn't affect the rate at all, so here with a first order, the rate of the reaction is proportional to that concentration. You double the concentration of M, the rate doubles.
6:27
You triple the concentration of M, the rate triples. You increase the concentration of M by a factor of 10, the rate's gonna go up by a factor of 10. And that is the hallmark you get on first order. And so when you're looking at experimental data, if you look at how... What we typically are gonna do is run two independent trials, where we, say, vary the concentration of M. We might double it or triple or quadruple it, or cut it in half, or we're just gonna have two different concentrations.
6:49
And then we're gonna see what happens with the rate. We'd actually measure the rate of the reaction and see what happens. And if the rate does exactly what that concentration does, that the hallmark of a first order reactant. So again, if you double that concentration of M and the rate doubles, first order. If you triple it and the rate triples, first order. All right, last but not least, we'll take a look at second order reactions.
7:13
So in this case, I'm gonna do X going to Y. And so X being the reactants, typically, what's gonna show up in the rate law.
7:22
And in this case, what makes it second order is it's gonna have an exponent of two.
7:29
And so in this case, what this ultimately means is that now the rate is not proportional to the concentration of X now. It's proportional to the concentration of X squared. And so this means that the reactant X has an even bigger impact than the reactant M here had on the rate. So now, if you double the concentration of X, well, doubling squared is four. And so if you double the concentration of X, it's actually gonna increase the rate by a factor of four.
7:56
And so whatever you do to the concentration, the same thing's gonna happen to the rate, but squared. And so if you triple the concentration of X, well, three squared is nine. That means the rate's gonna go up by a factor of nine. If you increase the concentration of X by a factor of 10, well, then the rate's gonna go up by a factor of 10 squared, or 100 in that case. And so, again, we're going back and looking at some experimental data. When we look at two different trials
8:20
and we say, oh, look at those two trials, the concentration of whatever tripled there,
8:25
and then the rate went up by a factor of nine. Well, tripled squared is nine, and that's how we determine if it's a second order reactant. Now, in principle, you can also have, say, a third order reactant. They're super rare, super uncommon, and you're not likely to see them, but in principle it would work the same way. If you had a third order reactant, well, then doubling its concentration would cause the rate to go up by double to the third power, and double to the third power,
8:50
two times two times two is eight times, eightfold increase in the rate in that case. Now, again, you're not likely to see that, but in principle, it works in a kind of the same sense of what we're doing here as well. Okay, so these are the three most common orders, but once again, fractions are possible, decimals are possible, but again, in this chapter, this is probably all you're likely to see. If you end up doing any kind of kinetics experiment in your lab, all bets are off though.
9:15
Just an FYI. All right, let's take a look at examining some experimental data. All right, so before we look at some experimental data here to determine a rate law, just wanna kind of give an analogy here. And so let's say that I've got a vehicle and I walk out and I turn the key on my vehicle, and this actually happened to me this week, and nothing happens. No engine turnover. Doesn't make a sound. And the question is, well, what could be the problem? And don't say out of gas.
9:43
If I was out of gas, the engine would still try to turn over, just wouldn't start. But here's nothing. Doesn't make a sound. So what could be the problem? Well, I could have a dead battery. That could definitely be the issue. Or other common option would be I have a dead starter. And those are probably the two most likely scenario. And so the question is, well, which one is the problem? And let's say I can't afford to take this to a mechanic and I need to fix it myself,
10:09
but I don't know how to diagnose which one it is. I don't have a battery tester. I don't know how to test a starter with a voltmeter or anything like this. So how do I figure this out? And how do I replace this at the most economical way? Well, what you do is you go to your local auto parts store and you buy a brand new starter for your vehicle and a brand new battery for your vehicle and you replace 'em both. And that's the most cost-effective way. Okay, that's not the most cost-effective way
10:31
because why replace both of them if only one of them is the problem? So again, you can't afford to take it to a mechanic, so you can't afford just to replace parts you don't need to replace. And so what you're gonna do here is you're actually still gonna again buy both of them brand new and you hold onto that receipt. And so what you're going to do is you're just gonna replace one at a time, because if you replace both and then your truck starts,
10:52
well, then you still don't know which one was the problem. But let's say you just replace the battery and the truck starts. Well, then you know the battery was the problem. But let's say you replace the battery and the truck doesn't start. Well, then you know the battery wasn't the problem. And in that case, you'd put the old battery back in and then you'd swap out the starter. And if replacing only the starter and the truck starts, well, then you know the starter was the problem.
11:16
And the idea is that you wanna only change out one of the variables here, battery or starter, at a time so you'd know what impact it actually has. Notice by switching them both out at the same time, if it fixes the problem, that's great, but I just don't know which one was really the one that fixed it, and I don't know which one to return back to the auto parts store to get my money back. So it's kind of the idea here. So what we're gonna do is we're gonna look at a reaction here
11:41
between NO and Cl2 to make NOCl.
11:45
And we're gonna look at three different trials we ran. So what we're gonna do is we're gonna do three independent trials. We've got three different test tubes or three different beakers or something like that. And we're gonna put in different concentrations of the reactants, NO and Cl2. And by differing those reactants, we're gonna see then what impact does that have on the rate? We'll measure the rate going on in each one of these beakers, test tubes, whatever.
12:07
And so in this case, what we wanna do then is compare two trials, and in the second trial, we just wanna change just one of the reactants. Only one. If we change them both, and the rate changes, I don't know which one to attribute the change to. Was it because we changed the NO? Was it because we changed the Cl2? I don't know. Was it both? Who knows? But if I change only one of them, and the rate changes, I know which one is responsible for that change in the rate, the only one that I changed.
12:33
And so in this case, like if you take a look at NO, if I wanna pick two trials, either the first trial, the second trial, and the third trial, two out of those three, and I want, out of the two initial concentrations, so that little zero there means at time zero. It just means initial. So initial NO concentration, initial Cl2 concentration. And if I want to pick two trials where only this is different, and notice the first two, that's not gonna work.
12:58
Now the second and the third looks like it'll work, but I gotta make sure that it's the only one changing, and that is totally the case. Now notice these aren't in any particular order. I could've compared the first to the third, but that would have been a problem. If I compare the first trial to the third trial, the NO changes, and that's good because I can evaluate whether that change has an effect. The problem is it's not the only change because the Cl2 concentration was changed as well.
13:20
So that would not be a comparison that's gonna benefit us at this point. But comparing trial two to trial three, that is gonna help us. The only change is the change in the NO concentration. And so in this case, as we go from trial two to trial three, we can see that the overall concentration here is going to double. And so if we take a look at what this rate law might look like here, we've got rate equals k. And again, we expect our reactants to show up here.
13:48
So NO, we don't know the order, and then Cl2. And again, we don't know the order. You can't just assume that these coefficients in the balanced reaction are the orders. It sometimes works out that way. It commonly doesn't. There's no way to know, although we'll see one caveat to that in a little while, later in this chapter. But the experimental data is ultimately what's gonna tell you those are. And so in this case, we can see that the overall concentration of NO doubled.
14:16
And so in this case, that NO concentration is doubling. Now the rate constant isn't changing. We'll find out that you can change this if you change the temperature or something like that, but it's not changing in this case, in principle. And we didn't change the Cl2 concentration in going from trial to trial three. So this is not changing. And so the only thing that's changing here is that NO concentration. And the rate's gonna have a corresponding change to the power.
14:41
But if that's a zero, well, then it's not gonna change at all 'cause anything to the zero power is one, and anything times one is itself. And so if it's zero order, then the rate's not gonna change. Well, if we compare trial two to trial three, the rate changed. Okay, well, then it's not zero order. Now in this case, if it's first order, and again, we're really just gonna consider zero, one, and two, the integers. So again, in principle, they could be fractions or things of that sort.
15:07
And in that case, like in the lab, we'd have a different way of kind of going about to solve what these orders are. But in kind of in the lecture, it's either zero or it's gonna be one or gonna be two, and maybe three. And we don't have to worry about some complex stuff involving log calculations and stuff like this, to figure it out. We're just gonna assume it's either zero, one or two to begin with, or possibly three.
15:27
And so in this case, if I know that the NO concentration doubled, well, the rate's either gonna go up by a factor of two to the zero, two to the one, two to the two, or really rare, two to the third power. And the question is really just which one did it do? Well, going from 0.36 to 1.44, it turns out 0.36 times four is 1.44. Or if you wanna do 1.44 divided by 0.36 equals four, probably the better way. But this is a fourfold increase. Not necessarily the easiest one to see in your head,
15:57
but notice 36 times two is 72, and 72 times two is 144, so 36 times four is 144, if you wanna look at it that way, but definitely use your calculator. But this increased by a factor of four. And so the question is what exponent lines up with that? So the concentration of NO again doubled, but the rate went up by a factor of four, well, two squared equals four, and that's how we know that it's second order. And so the key was you could have gone back and said, okay, well,
16:29
so what did the concentration do? Well, the concentration doubled so to the x power, according to the rate law, and then what did the rate do? Well, the rate went up by a factor of four. Then what must that exponent be? Two to what power equals four. Well, two to the two power, two squared equals four. And now we know that this order right here
16:49
is a two. Okay, so so far so good. Now, if we wanna find the order with respect to Cl2, again, if we can find two trials where Cl2 is the only concentration of either reactant that we change, that we allowed to be changed, then we got a comparison we can make. And so again, notice, two to three, that's not gonna work. And again, we said comparing the trial one to trial three, where they both change is not really gonna be the most helpful, but going from trial one to trial two,
17:17
so the Cl2 concentration, once again, doubles.
17:24
So just like the NO did in the comparison we made, and the NO concentration is not changing. And so, in this case, only the concentration of Cl2 differed between these two trials And so if the rate changes, and it does, I know that Cl2 is responsible. And so in this case, if we take a look at going from 0.18 to 0.36, what did that rate do? Well, it also doubled. And so in this case, the concentration of Cl2 doubled to some power. And what did the rate do? Well, the rate also doubled.
17:54
So the rate doubled, concentration doubled. What order is that consistent with? Two equals two to what power? Well, in this case, that's a one. Two to the one power equals two. And so we just figured out that the order with respect to Cl2 is a one, or technically, we wouldn't even have to write it in, essentially. Again, if there's no order written, it's implied that it's a one. So in this case, we've now figured out that the order with respect to NO is second order. We figured out the order
18:21
with respect to Cl2 is first order. And if I said, what's the overall reaction order, it would be a grand total of three.
18:30
Cool, now the last question you might get asked here, from experimental data, and you'd have to figure out the orders first, but the last question could be what is the value of the rate constant? And so now that we figured out the orders, we can rearrange this expression to solve for k.
18:50
And so if we do that here, we've got k equals the rate
18:55
all over concentration of NO squared times concentration of Cl2. That's the expression for the rate constant. And so to be able to figure out the rate constant, I need a situation where I know the rate and both concentrations. Well, great, I actually have three different situations where I know all three of those. I know the rate, and I know the concentrations of the reactants, and I can just pick any one of these trials and plug them in. Now, again, if you did this in the lab,
19:23
you might not get exactly the same k value trial to trial, but they should come out close. It is a constant, as long as we're not varying the temperature.
19:32
But here, when I make up the data and I make the data fit the equation, no matter which trial you use here, the k value's gonna come out exactly the same. And so, in your case, I would pick whichever trial has the easiest numbers to work with, and I'm just gonna use the first one. I like 0.1's.
19:48
Multiplying by 0.1 is dividing by 10, or dividing by 0.1 is multiplying by 10. It just makes the math easy without easy access to a calculator for me here. But you should be using a calculator. All right, so in this case, the rate is 0.18 in that first trial. And I'm gonna put the units on. I didn't put units here, but on the study guide, it's molarity per second
20:09
for the rates. And then the concentrations are all in molarity. And so the NO concentration is 0.1 molar.
20:18
And then the Cl2 concentration's also 0.1 molar.
20:23
And I'm gonna almost make a mistake here and forget to square this.
20:28
Cool, and we're just gonna keep track of the units here. And so if you look, we're dividing by 0.1 squared, which is same thing as dividing by 0.01, which is the same thing as multiplying by 100, and multiply by 100, we get this to 18. Dividing by another 0.1 would be multiplying by another 10, gets us to 180. And if you look at how the units work out, one of these molarities is going to cancel, but the other two aren't. And so we got a molarity squared on the bottom
20:54
and one over molarity squared is more commonly in these kind of cases gonna be written as m to the minus two. And then a per second is the same thing as seconds to the minus one. And so what you find out if you take the absolute value of adding up the exponent for molarity in seconds, or molarity in time, whatever time it is. It's commonly seconds. If you add those together and take the absolute value, it actually is just gonna correspond to the overall order, it turns out.
21:20
So it turns out the units on these rate constants can be different, depending on the overall order of the reaction. So however, again, negative two and a negative one is negative three, and the absolute value of that is three. And this just happened to correspond to third order overall. So there's a kind of a way of, if you see the units on your rate constant, of knowing the overall reaction order. And so you kind of see the way this works. If you have molarity seconds to the minus one,
21:47
well, molarity without any kind of exponent is just molarity to the one. That would correspond to zero order. If I just had seconds to the minus one, you're always gonna have time, like a per second or a per minute. You're always gonna have time to the minus one power. And that would correspond to first order. And if I had m to the minus one, s to the minus one, that's gonna correspond to second order. Or as in the case here, we had m to the minus two, s to the minus one, and that corresponds
22:13
to an overall reaction order of third order. And so the units on your rate constant can tell you what order, what's the overall order, of a reaction as well. And so sometimes you're gonna have to calculate a k value, and sometimes they're mean enough to give you just a difference in units. And either you can keep track of your units when you're doing the math, or you can realize, well, if I look at my units, this corresponds to zero order, this corresponds to the first order,
22:36
this corresponds to second order. So if you know your order, you can pick the right units. Or sometimes they'll give you a rate constant, and you can look at the units and they're expecting you, based on the units, to know what order it is as well.
22:49
Cool, so one last time here. We're gonna write this overall rate law now, now knowing the value of the rate constant. And so if we take a look here, we're gonna get
22:59
rate equals 180. We'll put the units on there as well. So molarity to the minus, two seconds to the minus one times the concentration of NO squared,
23:11
times the concentration of Cl2 to the one power, which is implied again. And that's our rate law. And we said at the beginning of this lesson, that the whole point of having a rate law is it allows us to make predictions. And so now, if I said, hey, if you started off with a concentration of NO of two molar and a concentration of Cl2 of three molar, how fast would the reaction go? Well, I mean, you could just put those into a beaker and then measure the rate and figure it out.
23:35
But now that you know the law, just like the law of gravity allows us to make predictions. Now that you know the law for this reaction, for the kinetics of it, you don't actually to run the reaction. You have the law. If I know that the initial concentration of NO is two molar and the initial concentration of Cl2 is three molar, I can plug those values in and calculate out a rate. It allows me to predict the rate of this reaction now without actually having to run it in the lab anymore.
24:00
Now, again, turns out we can't get these rate laws without having this kind of experimental data.
24:06
But once we have it, now I don't actually have to run that reaction anymore to know the rate of the reaction. I just plug it into that formula. Now, one thing to note, I said earlier that the coefficients in the balanced reaction don't necessarily have to correspond to the orders. In this one it does. And we'll find out that that's gonna be a special case, but again, don't just assume it's always gonna be that way. It will, sometimes it won't. You really have to determine them
24:31
from the experimental data most of the time. And again, we'll see one exception later in this chapter. Now if you found this lesson helpful and you wanna support the channel, consider naming your firstborn Chad. Not really. Please just hit that Thumbs up button. Lets YouTube algorithm know that other students should be seeing this lesson as well. If there was anything particularly helpful in this lesson or if anything is still unclear, just let me know in the comment section below.
24:55
And if you're looking for practice problems on kinetics and how to determine rate laws from experimental data, check out my general chemistry master course, includes video solutions, so if you're still struggling, you can have me work it out for you. The course includes over 1,200 practice questions. A free trial is available. I'll leave a link in the description. Happy studying.
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