In our previous lesson, we looked at rate laws, and we looked at rate constants and we talked about orders. We mentioned that you have to experimentally determine those rate laws. You can't just look at a reaction and know what they are. So in this lesson, we are going to have the learning objective of trying to see how you take experimental data and obtain a rate law. In other words, obtain the orders of the reactants by using that data. And the best way to demonstrate this is with an example, so we are going to do a couple of work examples as we go through this process. Here's our first reaction. We are reacting NO and CL2 to produce NOCL. We know that the rate law will take on the form of a rate constant, which we use a lower case k, times NO raised to some power and we'll call it x. I don't know the power. And CL2 raises some power y, I don't know the power. My goal is to determine those two powers. So for x, what we wanna do, in other words, is the order of NO. We need to choose experiments that hold chlorine constant. Now what this does is it takes chlorine's effect out as we run these experiments, and I want to number them one, two and three. As we run these experiments, by holding chlorine constant, we can see how did the changing of the concentration of NO affect the rate and from that be able to determine the power. So, weren't we looking at experiments one and two for this. Take a ratio of the rates for those two experiments and the ratio of the concentration of the NO for those two experiments. Okay? So, that's what we're going to be doing. Now, I generally like to take large divided by small because it keeps me from having to work with fractions. And that's how you use essentially four for the ratio of the rates. Now, I'm looking at the concentrations up here again in these two values. I take the ratio of rates. Now make sure you do it in the same order as your ratio rates were. We're doing a ratio concentration here. I had done experiment two over experiment one. Make sure you do experiment two over experiment one for this portion, as well. And that gives me two. Now, what do these numbers tell me? I am doubling the concentration, there's the two. I am doubling the concentration of NO. And I'm seeing a quadrupling in the rate. So, we say two to what power, that's my x, will give me 4. And certainly, x is going to be two. If I raise it to the second power, it fulfills that, and that tells me now that the order of NO is second order, and I can put a two there for that. Now, we're ready to look at y, the order of chlorine. Again, we want to chose experiments that hold the other substance constant. So we wanna hold NO constant. So look at that data and decide which experiments do that. Well, I see and I wanna just erase these, get them out of my way here. We see that NO is being held constant in experiment two and three. So once again, for these two experiments, we're going to do a ratio of rates and a ratio of concentration for experiments two and three. Again I'll choose the large number divided by the smaller number. In this case, it's essentially two. And in that same order, that was experiment three over experiment two. So, make sure you do concentration experiment three over experiment two. Always be larger over smaller, it is in this example. Okay. So now, we see that this is telling me that if I double the concentration, I will double my rate. And so, we say two to what power gives me two. Well, y is certainly equal to one. So now we know that it is first order in chlorine. And we could write the finished rate law to look like this. Rate is equal to k NO squared Cl2, and we don't write the first power. What's the order overall? Well, it's third order overall. Sometimes you'll be asked then, what's the value of the rate constant k, and to do this, you choose an experiment up here. You plug in what you know and solve for k. If you get a real good k, you do it for every experiment you run and do an average. We're not gonna calculate that on this example, but we will learn to calculate k in the next example. Okay, so here's our second example. This ones got some different features. One of them is this little thing sitting up here over the arrow. That is a catalyst. And a catalyst is something you add to the reaction to speed up the reaction. It does not get consumed. So, it's not written as a reactant, but it could. Its concentration could affect the rate. So we see that in our experimental data, we not only have the reactants OCl minus and I minus, we also have included in here some data about OH-'s concentration and how it affects the rate. So the general form of the rate law will be rate = K and the reactants, OCl minus raised to some power, the reactant I minus raised to some power, we'll call it y, and the catalyst, OH minus raised to some power. Now if it doesn't have any affect on the rate, it'll end up with a zeroth order. But that's looking a little messy already. So, I'm gonna erase that out and rewrite OH minus raised to the power of z. So we can determine that. So part A is asking me to determine the order. In other words, X, Y, and Z for this. So we're gonna follow a similar procedure, and it's a little bit more complicated. So for x, x was the power for OCL minus. Y was for I minus. And Z was for OH minus. I'm gonna just write those up there. For X, what we need to do is we need to hold the other two substances constant. So we need to hold I minus and OH minus, both constant. I'm gonna write in black through this thick pen here. Hold I minus and OH minus constant. So look at that data and see which two experiments you would like to choose. It looks like we could choose experiments, let's number them. What would that be? Experiments three and experiment one. Experiment, I'll write it one and three. Hold the i minus the OH minus constant, and then we'll do a similar work. We're gonna do ratio of rates and a ratio of concentrations. Ratio rates of 4.8 times ten to the minus four, that was experiment one, divided by 2.4 times ten to the minus four. That is experiment two. And that gives me a value of two ratio of concentrations. We have for experiment one, 0.0080. For experiment three, I said two three. For experiment three, it's 0.0040. And this is also equal to 2. So a doubling of concentration yields a doubling the rate. Two raised to what power gives me two? Well, x is equal to one, and now we know it's first order in OCL minus. Okay, now let's do it for Y. For y, we need to hold the other substance as constant, so we need to hold OCl minus constant and OH minus constant. And let's figure out what experiments do that. I see that experiments two and three hold OH minus and OCL minus constant. So I'm going to be looking at experiments two and three. We'll do the same process, ratio of rates and ratio of concentration. I'm going to do five times ten minus four over 2.4 times ten minus four, and I'll get essentially a two. It's not exactly a two, but its experimental data. It's close to a two. It's close to the doubling of the rate. And then, I'm going to do a ratio of the concentrations. Now, I did experiment two over experiment three. Make sure I do that same order here. We have 0.0080 and 0.0040, and so again, I have a doubling. So doubling the concentration raised to some power will give me a doubling of the rate. And so, that tells me that y is equal to one, as well. So now I know it is first order in I minus. Now, we're ready to do z. But do we need a whole constant? Well, z is OH minus, so we have to hold OCl and I minus constant. Which experiments do that? Or, you might be able to find a couple of them that do that. I'm not sure, but looks like to me it could be this one, experiments three and four, okay? Experiments three and four meet that criteria. So let's do a ratio of rates. And a ratio of concentrations. I'm gonna do large over small, 4.6 times ten to the minus four divided by 2.4 times ten to the minus four. And that's close to two. Now, this was experiment four over experiment three. So let's make sure we do the same order for that. Experiment four is 1.0, experiment three is 2.0 and we end up with 1/2 or 0.5, which ever way you wanna think about it, one or the other. Okay. So now, we have to say to ourselves 1/2 raised to what power gives me two? Well, that would be negative one. So here's a case where we have a negative power. We know that 1/2 when raised to the negative power, we're taking the reciprocal. So that would give me a value of two. So now, we know up here that it is minus first, minus one for z. So, our generic form of this rate law so far takes a form. I mean, it's no longer generic, it is actually rate law. K OCL minus I minus and OH minus to the negative one. Now what I want you to do is determine, we go back here to the previous question, part b says, write the rate law, which we just did, and determine the value of the rate constant, k. Now to determine the value of the rate constant, k, you can take one of these experiments, plug in the values and obtain a value for k. I want you to try to work on that and come up with an answer to this question. Now, if you pick number three, maybe you only picked it because you see right here that that matches the rate law that I did before. But this would be the same thing, wouldn't it? A negative one would be to put that in the denominator. Those are equivalent at that point. You probably came up with a value for 30. Those are equivalent there. The only difference between these two answers comes in the form of the unit. So let's think about the unit for just a moment. We know that rate was equal to k OCl. Let's look at the units of these things. Rate is in units of molarity per second. K, we don't know, we're trying to determine its units. OCL minus is molarity. I minus is molarity. And OH minus is molarity to the minus 1, or 1 over molarity. Now, if we take all of these units to the other side, we'll have k is equal to. I'm gonna take each one of these over, so I'm gonna divide by that M. So I have m over s divided by that M. I'm dividing by this M, and I'm multiplying by that M. So the molarities are canceling, and I am left with seconds in the denominator. Now, what's the overall order of this reaction? Well, it would be one, the order of OCL. One the order of I minus, and minus one, the order of OH minus. For an overall order, it's first order overall. And what I want to tell you that you probably need to put to memory but we'll mention it many times. Every time you have a first order overall reaction, that will be the unit of k every time. If it's first order overall, the unit of the rate constant is always one over seconds. Okay. So, we have gone through our learning objective. We have taken experimental data, and we've used that experimental data to come up with a rate law for the reaction. Now, we also took it one step further. We determined the value of the rate constant, but know that this rate constant only applies to the experimental conditions that that reaction was run under. So whatever its temperature was, that is the rate law, its rate constant value.