Out tutoring objective deals with reaction mechanisms. We are going to learn how to take reaction mechanisms and examine them to find what is called the rate determining step and from that predict the rate law of a reaction. So this is another way to obtain a rate law that we have not seen at this point. We start with reactants that turn into products but it doesn't happen usually in one fell swoop. In a one step process. but it doesn't happen usually in one fell swoop. In a one step process. It takes a series of steps. These steps are called elementary steps. So they are simple reactions, what is colliding with what to produce a intermediate? Those thing then react and run into each other collide to produce something else, on the way a reaction. So reaction mechanism is the sum of these elementary steps it is that whole sequences of steps that come together that take you from reactants to products. Lets look at this reaction. We have 2NO and O2 producing 2NO2. This is an observable fact. We you are monitoring that reaction you find that some N2O2 exists but just for a brief time. It is not a reactant, it is not a product. So what would that mean? Take cannot mean that 2 molecules of NO collide with one molecule of 02 and out pops some NO2. It means that on a molecular level that the two molecules are doing something else. They do not collide to produce it in a one step process. Here is what is proposed, and this would account for the observance of the N202. An NO collides with a NO that produces N2O2. Then the N202 can collide with an O2 to make the final product. These are the elementary steps. Those are your elementary steps of this mechanism. These are the elementary steps. Those are your elementary steps of this mechanism. They have to add up to give you the overall reaction. So these N2O2 will be able to cancel out when you add these reactions together. They are intermediates so they appear in the mechanism for a brief time, but they are not in the overall reaction. They get produced in an early elementary step and then consumed in a later step. So we see here, it is getting produced in this step and it is getting consumed in this step. so we see they would cancel out as you add those together and give you the overall reaction. Your elementary steps must add up to give you an overall reactant that you are looking for over wise it is certainly not a good mechanism. You use experimental evidence to help you support the steps that you choose for a mechanism. You are not going to be given a reaction and asked to determine the mechanism. But you ought to be able look at a mechanism and see if it is a realistic or plausible mechanism for a reaction. [Here is] some terminology that goes along with this concept. An elementary step is uni-molecular if it only has one reactant. Its bi-molecular if it has two molecules. Two molecules have to collide in order to produce a bi-molecular That is by far the most common elementary step. A Tri-molecular, which is very rare, would be three molecules colliding into single space in order for a reaction to take place. It is rare because it is difficult for three molecules to actually collide if they are moving about randomly, into a single space and time. Here is a question for you to consider and answer. We have that mechanism what are the elementary steps in this mechanism. Well if you choose bi-molecular, the you are correct. Now when you look at, elementary steps and you want to come up with the rate laws elementary steps, you actually can and I will put a star by that. You actually can use the coefficients to determine the orders of those elementary steps. Now you could not do this for an over all reaction, but you can for a elementary step. If this were an elementary step reaction, but you can for a elementary step. If this were an elementary step you would see a one coefficient for the reactant A we can raise it to the first power here. And you know we do not write the first power down. A plus B going to products we have the reactants A and B both raised to the first power. If and elementary step looks like this, A has to collide with an A to produce products then that would be the same as 2 A, but we usually do not write it that way simply because we are showing the collision of what collides with what when we write our elementary steps. But that would give us a second power for A. Now remember, and this is important to note, because students want to go back to and overall reaction and use the coefficients. This does not apply for an overall reaction. How do you determine for an overall reaction? Experimentally, you have to determine it experimentally. Once you determine it experimentally, we are going to see that our rate law needs to support that experimental information. we are going to see that our rate law needs to support that experimental information. One of the steps in our mechanism is going to be determined is called a rate determining step, and it is the slowest step. It is what is going to determine how fast the overall reaction is. It can only be as fast as the slowest step. This elementary step, in which we determine to be the rate determining step is whatever rate it goes at, whatever speed that slowest step proceeds at is the speed or the rate of the overall reaction. Now to give you a good feel for this we are going to look at a specific example. So you are going to see me constructing these little booklets and from that process be able to determine, your rate determining step, and see how that actually determines the over rate of the reaction. determine, your rate determining step, and see how that actually determines the over rate of the reaction. We are going to simulate a mechanism and talk about a rate determining step through the simulation of this mechanism. It is going to have three steps our final products is going to be our little booklet. So we are going to staple this booklet together. The Booklet is made up of an orange sheet and a blue sheet. There are going to be three steps to this mechanism. One person is going to collate the sheets together. One person, that is going to be me, is going to fold them into three parts and hand it off to the next person, who is going to staple the booklet, to give us this final product. So what we just demonstrated was a process that represents a chemical reaction taking place. was a process that represents a chemical reaction taking place. It is representing a three step mechanism. We have the collating of the pages, the folding of the pages, and we have the stapling of the pages. If we are going to look at the over rate of the reaction we do not look at from the beginning to the end and say 'How Long did that take? We look at it coming off the assembly line and say 'How many can we produce in a certain amount of time?' and monitor that. So that is what you first need to learn about rates. Its not start to finish, but coming off the assembly line. So we produced six booklets Its not start to finish, but coming off the assembly line. So we produced six booklets in 45 seconds so if we were to find the rate we do 6 divided by 45, and that would be booklets per second. Now we want to think about what is the rate determining step for that process. We have the process of collating the putting two pieces of paper together. And if we speed up that process it would not come off the assembly line any faster We also had the stapling end of the process which if he could just staple just lightning fast and staple as fast as he could it still would not come off the assembly line any faster. But my step, which is to meticulously fold it into a tri-fold, to put them together and toggle them, and fold it, that was a slow step. Now if I could figure out a way to do my step faster that would speed up the overall process. So we call this the rate determining step. Now that you have an understanding of the rate determining step and how it drives the reaction lets examine another mechanism. We see the overall reaction above. We see that someone went can experimentally determined this to be the rate law for the overall reaction. Here is the proposed mechanism it is a three step mechanism. We have got some thing here to observe. First thing we want to figure out in which step would be the rate determining step. First thing we want to figure out in which step would be the rate determining step. Well they way that you would figure this out, is to come up here and write the rate laws for each of these elementary steps. For this elementary step, rate is equal too, and we are going to call the rate constant K1 because it is just step 1, times H2O2 raised to the first power. And I minus [I-] raised to the first power. For this reaction, rate is equal to K, raised to the first power. And I minus [I-] raised to the first power. For this reaction, rate is equal to K, and I am going to give it a K2, for it is a different rate constant. times OH minus [OH-] times H plus [H+], both raised to the first power. times OH minus [OH-] times H plus [H+], both raised to the first power. This one is rate is equal to, and I will name it K3, HOI times H plus [H+] times I minus [I-] We have to look and say which one of these rate laws match the over all rate law? What that would be the first one here. It matches that and that will tell me, that the first step is the rate determining step. If none of those matched that overall rate law something else is going on we have not discussed or you have a bad mechanism. We see underneath here, I have written that this one would be the slow step because it was the rate determining step. Its rate is exactly equal to the rate of the reaction. That is the slowest step, and as fast as the reaction can proceed. Its rate is exactly equal to the rate of the reaction. That is the slowest step, and as fast as the reaction can proceed. What you need to understand, that it is not the sum of these rate. We do not add them up to determine the rate of the reaction it is how fast it is coming off the assembly line. It is going to come off the assembly line as fast it is how fast it is coming off the assembly line. It is going to come off the assembly line as fast as we can do this reaction here. It is going to come off the assembly line as fast as we can do this reaction here. We see down at the bottom of the page, the mechanism that was on the previous page, that we determined the slow step for. I want you to look at the mechanism and determine which substances are the intermediates. Remember, the intermediates are the substances that disappear in the overall reaction. They would get canceled out. that disappear in the overall reaction. They would get canceled out. The get produced in an early step and re-consumed in a later step. OK, if you chose number 4 then you are absolutely correct. Those are our intermediates. So we have this substance, being produced and then consumed, so it is not in the overall reaction. We have OH minus [OH-] being produced and consumed. and then consumed, so it is not in the overall reaction. We have OH minus [OH-] being produced and consumed. Everything else is part of the overall reaction. OK so this is end of out learning objective 10 where we were determining a mechanism OK so this is end of out learning objective 10 where we were determining a mechanism looking at mechanism, and seeing how we can predict rate law from those mechanism and learning about the slow step, and the rate determining step.