We are now ready to look at the third learning objective in our kinetic's lesson. And in this learning objective we're going to be examining what's called a rate law. We're going to identify the necessary components and parts of the rate law and we're going to, if given a rate law ,be able to determine how the change in concentration of reactants will affect the rate of the reaction. So here's the definition of a rate law. It expresses the relationship of the rate of the reaction to a rate constant and the concentration of reactants. Okay, this is different than a rate expression, which we use changes in concentration and changes in time. This has nothing to do with time in there. It has a rate constant and it has only the concentrations in there. So here is a reaction, a very generic reaction, of A and B as the reactants and C and D as the products. The Rate Law for this reaction would look like this. It has a rate constant k, so that's what we use always for the rate constant is a lowercase k. And it's a concentration of reactants A and B raised to, and see the word some, raised to some powers. It is not the same powers as the coefficients. My coefficient here was A, the power is x. I'm using a different representation here. X and y can only be determined experimentally. You cannot look at a reaction and know what those powers are. You have to perform some experiments to determine them. Now, we're gonna learn how to determine those powers In our next learning objective in our next unit, our next series of lessons. But for now we're going to examine some and see how we can use that rate law to make a judgment about the reaction. Now here's a reaction, A plus B going to C plus D, I have one coefficient and somebody went and did some experiments. And they noticed that as the reaction takes place, and they did it and monitored it many times, they saw that at certain concentrations of A, the reaction took place in various rates and they found a nice straight line. This straight line tells me it's a linear relationship and tells me it's a direct proportionality. So we could say that the rate of reaction is directly proportional to A. Or we could replace that proportionality symbol with an equal sign and say that rate is equal to, and we put the constant in there of K equal to K times the concentration of A. So it's technically raised to the first power there, but when it's raised to the first power we don't write it down. So as we examine this, what does it tell me? Well k is called the rate constant, I've already told you that. Let's talk about the units of k. The units of k change depending upon what rate law we're dealing with. And it's obtained by solving the equation for k, so if we took rate divided by the concentration and put the units in for rate, which is always molarity per second. And the units for concentration, which is, in this case, well, it's molarity. Then that would give me, with the molarities canceling, the units for k. You can also obtain a value for k. So if you actually knew a value for rate, how fast it was going, and a value for a, you could obtain a value for the rate constant. Now every time the reaction takes place, and this is key, at a constant temperature, the same value of k will be obtained. k is a constant. But know that this constant is only a constant for a reaction taking place at the same temperature. If you go and change the temperature, it will change the value of this rate constant. We'll see how and why In lessons to follow. Let's look at a very specific reaction. Here is a reaction NO plus hydrogen producing nitrogen and water. Now, somebody went into the experiments they did the monitoring and they found that the rate law was this, rate is equal to and there's rate constant k NO raised to the second power and H2 raised to the first power. Note again, this is not the same as the coefficients. This one happens to be the same, but this one is a one so be aware of that. Some language that goes along with rate laws is the reaction is said to be second order in NO. The order is the power. So second order in NO. It is said to be first order in H2. So that's just the language we use. And it is third order overall, and to get that you take the two plus the one and you get a three, third order overall. There can be zero orders if the concentration of a substance has no effect on the rate, then we'd say it's zeroth order. Orders can be fractions. Orders can be all sorts of different things. It could even be negative numbers. I didn't put that in here. It could be negative. Just be aware of that. Okay, so let's look at this rate law. Let us say we go in here and do the reaction twice. And in those 2 reactions we don't change the NO concentration, but we double the H2 concentration. Think about this reaction, if we kept this part the same, but the double the H2 concentration on those two reactions. How would that affect the rate? Would you say that the rate wouldn't change if we doubled hydrogen concentration? Would you say the rate would double, The rate would be cut in half, or the rate would quadruple? If you said it would double, you’d be correct. One way to do that is to plug some simple numbers in. Let’s do the reaction once, and let’s do a one and a 1one. And let’s do it a second time. We don’t change NO because it’s remaining constant, but we double this number. That is going to give me, for this one, just k, for this one, 2k. Let's ignore that little guy there. So we see that it is doubling the rate, okay? Let's try it another way. Now we're going to do the experiment, and we are going to do it twice, we're gonna keep hydrogen constant but we're gonna double the NO. Now if we do that, would the rate stay the same? Would the rate double? Would the rate be cut in half or would the rate quadruple? If you chose quadruple you'd be correct. Again, let's use some simple numbers. For experiment one, we're going to use ones in both place, and that would give me k. For experiment two, we're going to put a doubling of the NO, but not change the H2 and that's going to give me 4k. So, we see the quadrupling effect. Now, when we go to our next learning objective,we are going to be able to use this information. They're going to do experiments, we'll see the experimental values. Andthey'll say, okay we're doing this experiment. We're doubling the concentration of one substance and we see that the rate quadruples. That tells me it's a power two. So we'll examine that more closely. But for this learning objective we simply learned what a rate law looks like, we learned about words like order, and we learned that if you have a rate law you can predict how a change of concentration will affect the rate.
