In the previous lecture, we'd begun our study of reaction of rates by defining what we mean by a reaction rate, and then trying to relate those reaction rates to the concentrations of reactants. You'll recall we had data like the following, in which we would measure, for example, say, the concentration of a particular reactant as a function of time. And we could observe that the typical sort of curve look something like this, as the reactant disappears as a function of time. We then define the rate of the reaction at any particular time as being the slope of the graph at that particular time. Which meant that the rate, if we write it in calculus terms, is the derivative of the concentration as a function of time, or it could be approximated by the amount by which the concentration is changing as a function of time. We then noticed or observed that the rate of the reaction could be related to the concentration of the reactants. And at that the particular case of the C60O3 composition reaction, we found that the rate was actually proportional to the concentration of the materials. We also noted that experimentally as observed, that the rates of the reaction are commonly functions of the concentrations of two different reactants. So if we have a reaction taking place which is the reaction of A with B, then the rate depends upon the concentration of A, and depends upon the concentration of B. And we need to find these integers n and m because those are going to be the functional dependencies of the reaction rate on the concentrations. So, what do we need to do? We need to find a way, if we're going to understand the relationship of rate to concentration, to find these integers n and m. One way we can do this is with a method that is referred to as the method of initial rates. Which says that we could actually look at the concentrations as a function of time right at the beginning of a reaction. So imagine we've got some reaction of A reacting with B to go on and become products of some sort or another. And we can set up an initial system in which we have, say, an initial concentration of A and some initial concentration of B, and we watch the concentration of A disappear as a function of time. The initial rate is simply the slope of this curve, right at the outset. Which could be measured. We could measure the A as a function of time, and then figure out what the slope is back here at zero. And that would then give us a means by which we would measure the initial rate as a function of the initial concentration of each of the reactants for this particular prototypical reaction A plus B. Let's imagine that we've done that experimentally here, and here's what a set of experimental data might wind up looking like. For example, perhaps we start with an initial concentration of reactant A, which is 0.1 molar, and an initial concentration of reactant B, which is 0.1 molar. And let's imagine hypothetically then that what we have observed is the experimental reaction rate. Turns out to be one mole per liter per second the rate of change of the concentration as a function of time. Then what we could do is change the concentration of A while not changing the concentration of B. And when we do that we might observe hypothetically. That reaction the reaction rate changes from one to two moles per liter per second when that change takes place. Then we could revert back to the original concentration of A, but this time double the initial concentration of B, and again measure. How much does this slope change when we have changed the initial concentrations. And this is just sort of a prototypical reaction, but what we've noticed is in this set of hypothetical data, the reaction rate has gone from one to four by doubling the concentration of B. These data should be sufficient now to tell us what the values of n and m are in the rate law expression. Let's see how that would work out. Here's how we could proceed, alright. In the case of the rate for experiment one, the first trial that we run, we know that that value is equal to 1.0. This is moles per liter per second are the appropriate units, and let me get my paper straightened out here. That needs to be equal to the proportionality constant k, what we call the rate constant, times the initial concentration of A raised to the N times the initial concentration of B raised to the M. And then we can compare that to the rate in experiment two again now, the rate in experiment two is observed to be two moles per liter per second. Same rate constant, different concentration of A now, same concentration of B. Remember the value here, let's actually plug these values in, is 0.1 here and 0.1 here, and the second experiment it is 0.2 and 0.1. Let's compare these two. If I take the ratio between these two equations, I can see that I double the rate by doubling the concentration of A and holding the concentration of B constant. In essence I can can-, cancel out the concentration of B and the rate constant, and the doubling of the reaction rate is due entirely to the doubling of the concentration of A. That tells me immediately that N is actually equal to one, because the, the amount by which I've increased the concentration of A originally, doubling it, has gone straight over and into doubling the reaction rate. Let's contrast that to the rate for experiment three. Now the rate is 4.0 moles per liter per second. That's K multiplied by the initial concentration of A in this particular case, raised to the nth, which is of course now we know one. But that value is 0.1 Times the concentration of B initially raised to the m, and that value is now 0.2. Let's compare now the first experiment and the last experiment, and we see that when we hold the concentration of A fixed and double the concentration of B, the rate goes up by four, not two. That means in taking the ratio of the concentration of B, I must have squared it, to take that factor of two, and turn it into a factor of four. So immediately I can tell that M is equal to four. From this, for this particular reaction, we can tell then, that our rate law, is simply the rate constant multiplied by the concentration of A raised to the first power, times the concentration of B raised to the second power. And we have successfully determined now a reaction rate law. In general what we'll discover is that when we are working through, reaction rate laws, we'll find a variety of different things of this type. In some cases, what we may discover is that reactions are what we will call first order. In this case, we've referred to the reaction as first order because the exponent here on the reaction reactant concentration is just one. That could happen, for example, as we saw in the case of the C60O3D composition. But of course it could also happen that we wind up with a second order reaction. Now we call it second order, sorry we call it second order, because the exponent now is two instead of one. Now in the case of, for example, A and B, it might be possible that the reactant, reaction rate in fact probably will depend on both A and B. We have a different type of second order reaction now that we see here. Second order. And it's second order because we in fact have not one but two reactants here. Each one of which is raised to the first power. Overall, then, the sum of the powers, one plus one is two, and we'll refer to that as a second order reaction. These are some of the more common rate laws which are observed experimentally. Let's spend a minute looking at those rate laws and thinking about what happens. Let's go back and talk now about the first order reaction. Remember the equation for a first order reaction is this one here from the previous slide. If in fact we spend a little bit of time doing some Calculus, if you're familiar with Calculus. We can actually figure out what the function, A, must be for that reaction in order to satisfy that particular equation. I could, for example, rewrite this. Let's write the equation again. Minus, oops, I've got minus on both sides of the equation, let's move it to one side. We'll have, if we move the A to the other side of the equation, one over the concentration of ada, is equal to minus kdt. And that equation can in fact be integrated to give us the logarithm of the concentration of A, minus the logarithm of the concentration of A0, is equal to minus kt. Or in the alternative we can rewrite this equation as, the concentration of A is equal to the initial concentration of A multiplied by e to the minus kt. This is all true for a first order reaction. And it's interesting to look at this particular equation because, we could actually plot this equation. This is just an exponential function. If we draw the equation for an exponential function, the concentration of A is a function of time. When T is equal zero, we have the concentration of A0. And we get a normal kind of exponential decay curve, which asymptotically approaches the the, the, the t axis. That's exactly the kind of thing that we observed in the case of the C60O3d composition. We refer to this particular form of the equation as the integrated rate law, as you can see in this table here. These two equations are in fact completely equivalent to each other. I can get one from the other, they're just different ways of writing the rate law, and either one of them might be useful. As we saw in a previous slide, it's actually possible to have a second order reaction as well. This is a first order reaction. If we go to the second order reaction, the rate law depends upon the concentration of the reactant squared, instead of the concentration of the reactant to the first power. We can similarly integrate that rate law as we did in the example given just a minute ago by writing again. If DADT, the change in the concentration of the reactant A is given by minus K, times the concentration of the reactant squared. I could rewrite this as one over A squared DA is equal to minus kt. And now when I integrate this, I wind up with one over the concentration of A, minus one over the concentration of A at t equal to zero, is equal to kt. Or the alternative, I can rewrite this now as one over A is equal to one over A0 plus KT. What that suggests is if I were to plot one over the concentration of A, as a function of time, I'd actually get a straight line. For a second order reaction, one over the concentration of A as a function of time is simply a straight line. And you can tell actually that one over A increases linearly with time. So, that's an example of an integrated rate law for a second order equation, and you can see we've included that in our table as well. We could actually keep this up in a couple of other ways, but one of the things we'll actually considered is in fact, what these equations predict for what is called the half life of a reaction. The half life is how long does it take before the material, half of the material has disappeared, Let's consider the case of the first order reaction. First, where we have A is equal to A0, E to the minus KT. Let's let the time at which half of the material has gone away be the half life. We going to call that how long does it take for that to happen? At that point, the concentration of A is equal to half the concentration we started off with. We can plug this back into the above equation and say, the concentration of A is equal to one half of the initial value, which is equal to the initial value times E to the minus KT. We can cancel both sides here, maybe take the logarithm of both sides. We have the logarithm of one half is equal to the logarithm of e to the minus kt is minus kt. If I put the negative sign into the logrithm, I've got a logrithm of two is equal to kt. I should note this t one half right? Because this only is true at that time where we fit the half life. So, the half life for a first order reaction is simply equal to the logarithm of two, divided by k. Notice that this is independent of the initial concentration of A. It does not matter how much A I start off with. The amount of time that it takes for half of that to go away is a constant that depends only upon the rate constant of the reaction. And this logarithm two is just a proportionality here. That's a fascinating result. Half lives of first order reactions are independent of the amounts of material which are present. We can get lots and lots of different kinds of experimentally observed reaction rates using the methods that we've talked about in this particular lecture. Here are some examples of them listed for a variety of different reactions. And there are some interesting results here when we study these. Because one of the things we notice is that for some of the reactants, if we look at say this one in particular. There are two nitric oxides and one oxygen, and the rate law corresponding to that, we'll notice is second order in nitric oxide and first order in oxygen. That makes it look a little bit like the stoichiometric coefficient on the nitric oxide which is two. Matches up to the exponent on the nitric oxide. And the stoichiometric coefficient on the oxygen, which is one, matches up to the exponent on the oxygen here, which is also one. That might seem like a general result, but we can immediately dash our hopes about that by looking at the second illustration. Here we have two nitric oxides reacting with two hydrogens. Again, what we observed is that the stoichiometric coefficient on the nitric oxide lines up, but the stoichiometric coefficient on the hydrogen does not, and in fact that's generally the case. Look again, icl reacting with h2, we don't get the expected exponents back over here from stoichiometric coefficient. Sometimes we do, sometimes we don't. Let's look at the last one down here. This is a particularly interesting reaction in which ozone reacts with chlorine, chlorine atoms, and notice that the rate law does seem to line up. Where each of the reactants appears to the first order. But some of these can even be more complicated. Look at this particular reaction, here, hydrogen and bromine reacting together. It turns out, in this case, we wind up with a rate law in which the coefficient, I'm sorry, the exponent on the bromine is not even an integer. It comes out to be a one half. So rate laws can actually be quite complicated. Part of our task is going to be to understand why is it that in some cases the rate laws seem to line up with the stoichiometrical coefficients, and in others they don't. And in general, what is it that determines what the values of n and m are? Those powers, those exponents, those orders on the concentrations of the reactants. And that's what we're going to to take up in the next two lectures.