We've now observed this beautiful result that we call the law of mass action, which describes chemical reaction equilibrium. And our task next is to build a model that would explain this experimental observation. Remember, the law of mass action says for a general chemical reaction, as is illustrated here, this particular ratio of the pressures of the products divided by the pressures of the reactants, each raised to their own coefficient turns out to be constant, regardless of the initial conditions that we begin with for that particular chemical reaction. But that's a strange result. So we want to know why is that true? And as usual when we ask why it's true, we're going to try to develop a model that might account for it. Let's think then carefully about this chemical reaction that we studied in the previous result. Or in the previous lecture. Here we have the decomposition of N2O4 to form NO2. And we noted that this ratio of the pressures was, in fact, constant. But let's sort of go back and look again at what this looks like in terms of time. And imagine then, what we are plotting here are the pressures of either the products or the reactants. What this suggests to us, is if we start off with an amount of the reactant which is here, and it declines in time, eventually it appears that the reaction stops. And similarly the reaction rise, the products rise rapidly, but then they stop as well. So why would this reaction eventually stop once we have achieved the appropriate pressures corresponding to the equilibrium. We did a lot of work early this semester asking the same question about why would say, evaporation stop once we have achieved the equilibrium vapor pressure of the corresponding substance in the vapor phase. Why does it stop? And the answer to that question was it doesn't stop, of course. Evaporation always occurs and condensation always occurs. And what causes the properties to change is not that those processes stop, but rather that they come into equilibrium with each other with balancing of rates. That suggests that the answer to our question of how we achieve reaction equilibrium is also the same in that we achieve dynamic equilibrium. And dynamic equilibrium means that the reverse process achieves an equal rate to the forward process. In other words, once we have started to create the product NO2, NO2 itself can begin to create N2O4. And in fact, we need to draw an arrow in both directions representing the fact that both of these processes are occurring. And once equilibrium is achieved, they are occurring at the same rate. How does that give rise to the equilibrium constant then? Well fortunately, we have studied what controls the rates of these reactions. So, if the forward rate is given here by the rate law for the forward reaction, assuming it occurs in a single step, then you'll remember a rate law is a rate constant, multiplied by the concentration of the reactant raised to its stoichiometric power. Or in this case, it's simple first order reaction. The reverse reaction rate is the, is a rate constant multiplied by the amount, or the concentration of the product raised to its stoichiometric coefficient. So if I analyze these two, and keep in mind, the rate constant for the forward direction and the rate constant of the reverse direction don't have to be same by any means. I should set the forward rate, the rate going forward, equal to the rate in the reverse direction. And the rate in the forward direction is the rate constant in the forward direction, multiplied by the concentration of the reactant. And the rate in the reverse direction should be the reverse rate constant multiplied by the product raised to its stoichiometric coefficient squared. That gives rise to this equation. Well, we're well on our way at this point. Because if I now move, this concentration to the other side of the equation and move this right constant to the other side of the equation, I wind up with this simple expression. That, the ratio of the two rate constants here kf over kr is equal to the concentration of the product squared, divided by the concentration of the reactant. That looks very, very familiar. If all I do now was multiply in the numerator, rt over v squared, I'm sorry rt squared. The concentration of the gas times rt is the pressure of the gas. And the denominator will take the N2O4 concentration, multiplied by rt. And let's see over all then, in order for that to work out, there's an rt I've added to the right side of the equation, so I need to add that to the left side of the equation. This equation is just like the equation above it, except the right side of this equation now is the pressure of the NO2 squared, divided by the pressure of the N2O4. The left side of the equation apparently is therefore equal to the equilibrium constant for this reaction. What that tells me then is, that this term is of course constant. For a constant temperature we know that these rate constants are constant. So in fact, we've now established that this particular ratio of the pressures, each raised to their stoichiometrical coefficient, must be constant. And that the condition for that being true is simply, that the forward rate and the reverse rate need to be equal to each other. In other words, the equilibrium constant that we observed experimentally, is a consequence of the dynamic equilibrium between the re, the forward reaction rate and the reverse reaction rate, exactly as we might've expected. We can generalize this with a great deal of work, but the purpose of this single illustration is to show that if I use the rate laws that we derived in the previous concept study, and set the forward and reverse reaction rates equal to each other, I immediately get the equilibrium constant predicted by the law of mass action. Now we have an understanding of why the product is in the numerator, the reactant is in the denominator. And why each of the product and the reactant are raised to their own stoichiometric coefficients. There's an interesting point here, which kind of suggests that somehow or another if I just varied the temperature, I might simply vary the equilibrium constant. But we also know that the rate constants here also depend upon the temperature, and therefore the equilibrium constant itself is temperature-dependent. We'd now like to explore how it is temperature-dependent. How does K, depend upon the temperature. Well, in order to figure that out, let's take advantage of the fact that we have done this dynamic equilibrium. We know, from the work that we have done over here, that the equilibrium constant is equal to the forward reaction rate constant, divided by the reverse reaction rate constant. Remember as well, let's write this on the other page, that we know what the temperature dependence of a rate constant is. It is equal to a pre-exponential factor, A, that depends upon a variety of things. Exponential of minus activation energy divided by RT. If I now consider that for the forward reaction, it has its own set of constants. So Af and f for the activation energy. And likewise the reverse rate constant, has its own pre-exponential factor. E to the minus activation energy for the reverse direction, times RT. Let's take the ratio between those two. So what we're going to analyze here, since we know from our previous work that the rate, the equilibrium constant is the ratio of the rate constants, then we wind up with just this simple result by taking the ratio of the two sides. We can now simplify this equation quite a lot. First, we'll of course set the left side of this equation equal to K. And since there's an exponential in there, it makes a lot of sense for us to take the logarithm of this. Doing so might mean that we would like to partition the exponential terms away from the pre-exponential terms. If I take the logarithm, the exponentials are going to go away, and I wind up with this equation. The logarithm of K minus the logarithm of this pre-exponential ratio is equal to the logarithm of this combined ex, exponentials. Putting the arguments of the two exponents together, I wind up with the result that we see right here, that the logarithm of K is varying with the temperature according to this one over T relationship here. That's an interesting relationship. Let's go back and look at the relationship between the activation energy in the forward direction, and in the reverse reaction. Remember, we drew this energy profile when we were discussing the kinetics as related to the activation energy. Let's draw a particular curve, say for the endothermic reaction, which would look something like this. What is the activation energy in the forward direction? Well, it's this number, right? Here is the activation energy in the forward direction. The activation energy in the reverse direction is this amount. If I take the difference between the activation energy in the forward direction, and the activation energy in the reverse direction, notice that that difference is actually equal to the difference in energy between, the reactant here, and the product here. In another words we'll call that the delta E of the overall reaction. If I take it from that perspective, then I can replace these last two terms here with delta E, for the overall reaction. When I put that into our equation here, and simply take this equation for two different temperatures, I can replace the this difference in energies here with delta E. I can take it to two different temperatures, in which case this term is going to cancel out, and I just have the difference between the inverses of the temperatures here. That actually gives us then the temperature dependence of the equilibrium constant. Later on when we've done some more thermodynamics, we'll actually rewrite this equation in a somewhat simpler form. Although it's the same equation, the logarithm of K is the change in the energy times minus 1, divided by RT plus a constant. And it turns out the constant is delta S over R. This is an interesting result though, because remember, chemical reactions can have either an increase in their energy as is shown in this particular case or they may show a decrease in the energy. Let's draw the complementary profile for the exothermic reaction. For the exothermic reaction we know that the reactants have a higher energy than the products. The activation energy in the forward direction is this one. The activation energy in the reverse direction is this one. Oops, reverse. The change in energy between reactants and products is still, the difference between the activation energy in the forward direction and the activation energy in the reverse direction. Except in this case, delta H is a negative number, whereas above delta H is a positive number. That means in this equation here, the way in which the equilibrium constant varies with the temperature depends upon what the sign of delta H is. Notice, if I were to plot the logarithm of K versus 1 over the temperature, I should get a straight line. But the slope of that straight line is going to depend upon whether delta H is a positive number or a negative number. Let's take a look at that. Here are two different chemical reactions written side by side, in which we compare the temperature dependence of the equilibrium constant for an endothermic reaction on the left and an exothermic reaction on the right. Remember, our equation tells us that the logarithm of the equilibrium constant is equal to the delta H for the reaction over R times a minus 1, plus we'll just take as a constant for the time being. That means if delta H is a negative number, sorry I missed a T here, then the slope here is going to be a positive number log of nK versus 1 over T is a positive number for the exothermic reaction. Here's the positive slope, so for this particular case it must be true that delta H is a negative number. Here, we see, a negative slope, and that tells us from the equation that it must be true that delta H is a positive number. So, the reaction on the left, hydrogen plus iodine making 2HI, is an endothermic reaction, in which the equilibrium constant is increasing with increasing temperature. And the reaction on the right, nitrogen and hydrogen to form ammonia is a exothermic reaction, in which the equilibrium constant decreases with increasing temperature. We're going to study variations in the equilibrium in a variety of different ways, after we've learned how to do a few chemical calculations using these equilibrium constants in the next lecture.
