We are given a mechanism in this problem and we're going to derive a rate law for this reaction. Is that a slow step which is a second step and a fast equilibrium before it. It's got symbols that we may not be familiar with, so let's get familiar with it. E stands for enzyme. So this is a substance that is a catalyst, a biological catalyst. S stands for substrate. That is your reactant in a sense that is going to be converted to P product. And then it has ES and the question there is what is ES? Well a substance that appears in an early step and then later gets consumed and therefore is cancelled out in the overall reaction is an intermediate. So ES is an intermediate of this reaction. If we look at the overall reaction here, we would be cancelling the intermediate, but we also cancel the enzyme, and that would give me the substrate being converted into product. So this reaction will happen in a biological system with the enzyme catalyzing the reaction. Now we were taught that the write a rate law for the overall reaction we can look at the slow step, which is the second step of the mechanism and use the coefficients of the reactant. Which shows a one coefficient, that will become the power. Goes to the first power here, for the reaction. But we never want to leave a rate law in terms of an intermediate, something that is getting produced and later consumed in the reaction. Whenever the second step which is the slow step has an intermediate in that rate law, we can use a fast equilibrium that precedes it to replace that intermediate with actual reactants. So let's see how it's done. Because the first step is an equilibrium we can write a rate law for the forward as well as the reverse reaction. So the rate law for the forward reaction would be the rate constant of that forward reaction times a reactants, which is e to the first power and s to the first power. As we look at that reaction and we know it's reversible, we can also write the rate law for the reverse reaction, thinking about it going in this direction instead. So the rate of the reverse reaction would be the rate constant on that reverse reaction times e times s. Now since this is an equilibrium we know that the reaction will have the forward reaction occurring at the same rate as the reverse reaction. So we can set this portion and this portion equal to each other because their rates are equal. Now the rate constants aren't equal, but the rates are equal. Now why are we doing this? We are doing this because we want to come up with something where I am replacing this ES with actual substances that are reactants. If I take this equation and divide both sides by K sub r, we get that e sub s is equal to everything we see on the left side. Well a Kf and a Kr are two constants. When you divide constants you're going to get a constant. So I'm just going to replace this with, I'll call it K1. For the rate constant of that first reaction would be [E][S]. Now, what I can do is replace ES, this, in this reaction. And that's going to give me rate is equal to K, and then let's put that K1. [E][S].. Well we can part combine our k's into another k. All right, so we have a k is equal to [E][S]. So now we have got a rate law in terms of enzyme and substrate. And it's okay to enzymes in rate laws, but it's not okay to have intermediates. So this would be the rate law, what we see written right here. Would be the right law of the overall reaction where a slow step has an intermediate in it and is preceded by a fast equilibrium.
