 Hello and welcome to the second module of chemical kinetics and transition state theory. In this module we are going to finish up a preview of the prerequisites. So, today we will cover how to think of elementary reactions, reaction mechanisms and finally end up with how to write rate law equations. So, again the resources are the same as that of module 1. We are following the chapter 1 of the Laidler's book. You can find essentially the same information in the NCIT book class 12 chapter 4. And I have also provided you here a link where you can find the information online as well. So, let us begin a brief summary of what we covered in the module 1. If you have a reaction of the type that is written here A A plus B B going to C C plus D D, the rate of this reaction we proved in the last module under the condition of constant volume is given below here. It is given by minus 1 over A D A over D T equal to minus 1 over B D B over D T equal to plus 1 over C D C over D T and so on. So, important to note we use a negative sign for reactants and a positive sign for products. So, now we move forward and look at how to calculate the rate constant and the reaction order. So, we start with the same reaction as before we look at A A plus B B going to C C plus D D. For this reaction more often than not one can experimentally observe that the rate that was defined in the previous slide can be written as K into concentration of A to the power of alpha into concentration of B to the power of beta where K alpha and beta are numbers. Our focus in this course will be on this parameter called K. So, K is what is called the rate constant. I emphasize again that this rate constant holds only when this empirical observation rate equal to K into A to the power of alpha into B to the power of beta holds. Another important thing to note which is often missed by many students is that this rate constant is independent of concentration of A and concentration of B. That is by definition K is simply a number, but it does depends on temperature. K does depends on temperature. So, we will be looking in this course specifically how to calculate this K as a function of temperature. So, there are reactions themselves can be divided into two parts a reaction that is elementary and a reaction that is complex. An elementary reaction is a one where the reactants combine together to give the products in one step without any reaction intermediate. So, essentially what is happening is that your reactants are combining together in directly as a form of a collision you can say and just giving the product there is nothing in between. So, for an elementary reaction one very important observation is that the rate is always given as K into A to the power of A into B to the power of B for a chemical reaction for an elementary reaction of the form. So, if a reaction is elementary this equation must hold always no exceptions. So, we often write the rate constant here above the arrow to signify that this equation will hold. So, then let us just take one example of a first order reaction. Again you must have seen this before. So, we are going to go through this slightly quickly just to revise what you already know. Reaction can be of many orders it can be 0 order first order or second order. First order specifically is of the form A going to some products B. The important thing is reactant is only one molecule. So, let us say I have this reaction and this is elementary. So, for this reaction two conditions must hold true as we have looked at. First is rate must equal minus 1 into dA over dt. So, note that the stoichiometric coefficient here is 1. So, I simply minus 1 here and the negative sign because A is a reactant and this will also be equal to dB over dt correct. If it is elementary another condition holds the second condition is that the rate must also equals k concentration of A. So, I will look at these two equations and I will just write down the solution we are recapping here. So, I am moving slightly faster the solution for this differential equation is given by this. So, I will leave this as a homework to verify if this solution is true or not. So, to do that you must feed in this solution into this differential equation the left hand side and the right hand side and verify. Well now we have gained some information after all these definitions if I make a plot of a first order reaction of any first order reaction it will always have this characteristic decay. So, an example of this is for example, radioactive decay if you take a uranium it is decaying and if I look at the population of uranium it will decay in this exponential form. So, moving forward how to write these kind of rate law equations we will take a few examples in today's module. So, the first thing that I want to look at is a reaction at equilibrium. A reaction at equilibrium we will look at an example of a reaction at equilibrium. A reaction at equilibrium essentially has a form. So, not only can you go from A to B with some rate constant kf you can also go from B to A with some rate constant kb. And I ask you that I tell you that this is elementary both sides and I want you to be able to write the rate law equation for this how do I handle that. The trick of this is to divide this equation in two steps. One we write the forward part only and two we write the backward part only with the corresponding rate constants. So, one is kf the other is kb ok. So, now we treat both of them independently to start out with and then we calculate the rate of this is minus dA over dt equal to dB over dt equal to kf into A. I emphasize this portion only has the reactant concentrations in it. For this first step A is the reactant. For the second step I will treat B as the reactant whatever is on the left side of the arrow. So, for this one I will write the corresponding rate as minus dB over dt equals dA over dt equals kb concentration of B. Now, if I want to calculate dA over dt how do I calculate that? The idea is we need to add the corresponding elements in the two equations. So, we look at the first equation. The first equation tells me dA over dt is minus kf A. So, we look at this term and note that I had a negative sign before dA over dt. So, I put that negative sign in the right hand side and I get minus kf A and I look at the second equation that also has a dA over dt. So, I take that and I add kB B. So, that is my first equation for dA over dt. I can write a corresponding equation for dB over dt. dB over dt again appears twice in both equations. And from the first equation the contribution is positive kf A and if you see that in the second equation I have a negative sign here. So, I have a minus kB. So, this is how you write rate law equation for complex equations. So, once you have elementary reactions what can you do with them? You can essentially construct mechanism of any reaction in terms of elementary reactions. So, in some sense elementary reactions act as atoms of chemical reactions. You give me a chemical reaction and I will be able to figure out some combination of elementary reactions that will give me that full reaction. So, a very well cited example is H 2 plus I 2 going to 2H i. This was long thought to be elementary, but evidence showed that for this reaction H 2 plus I 2 going to 2H i. The mechanism actually turns out to be that I 2 in equilibrium first dissociates into 2I dot. I dot reacts with H 2 to give H i plus I dot my apologies this should be H dot. H dot reacts with I 2 to give H i plus I dot back. So, this is an example of what is called a chain reaction. You generate these radicals and they will keep on propagating till the radical population goes away. So, you do not have to memorize this reaction or the reaction mechanism. It is simply an example and illustration of how reaction mechanisms are can be broken down into elementary reactions. I want to end today's module with a discussion of a very interesting case which was very deeply studied and hotly debated in 1920s which is the isomerization of cyclopropane. Cyclopropane is a three membered ring and as it turns out under appropriate conditions this can turn into propane. One fact that let me tell you this is unimolecular. There is no other molecule that this cyclopropane is interacting with to give propane. It is isomerizing on its own. So, that is experimental fact number one. Just for ease let me write this as A and let me write this as B. Well in 1910s and 1920s people were studying reactions of this form and this particular case of cyclopropane got a lot of attention. People were trying to understand its mechanism. People were trying to find it the elementary reactions behind it and well many people argued you have only one molecule. What else can possibly be the mechanism and so the most natural thing to write is that this reaction itself is elementary. Let us assume it is elementary. I do not know I am not claiming it is elementary but let us start by assuming that seems like a natural approximation after all. If I assume it is true, well we have already derived for a first order equation rate is equal to minus dA over dt equal to kA. Well let me just rearrange it in a form that was experimentally observed. So, this is my prediction. If I hold the hypothesis that A going to B is elementary then I predicted rate over concentration of A must be a constant, must be a number. I again highlight that k is independent of concentration of A. So, what people did is they were smart experimentalists. They plotted rate divided by concentration of A as a function of concentration of A. So, take a little pause and draw this graph on your own first on what you expect if this hypothesis is true. So, please pause the video and fill in this graph on your own. Hope you have an answer with you or graph with you. So, what I expect is basically a straight line. Well that is simple right because I have already shown that rate over A must be equal to k and k is a number at a given temperature. So, therefore, I expect it a straight line but low and behold what experimentalists observed they saw correctly this. So, at large concentration everything is good but the hypothesis is clearly breaking down at low concentrations and that led to a hot debate on what can be the mechanism. So, clearly my hypothesis is wrong. My hypothesis does not tests with the experimental results. So, what I will just briefly describe is what was put forward by Lindemann and Christensen in 1920s as a way to resolve this problem. And I am showing this example again do not memorize these things but this is an example to show how the knowledge of elementary reactions and the rate laws can help us understand what is going on an atomic level. So, this is the hypothesis that Lindemann and Christensen put forward. They said A and A A again is my cyclopropane. A actually two molecules of A first collide with each other they react with each other in a reversible fashion to give me an excited A star plus A. This is the hypothesis they are they are just playing around and they had put up two forward and backward rate constant. I am just following the language they had used. So, I am using k 1 and k minus 1 and A star which is an activated state gives rise to B. So, this is a more detailed mechanism now it is not as simple. Now we have a more complex reaction and what I want to discuss in this module is how do we write rate laws for this and can we perhaps resolve the contradiction with experiment using this hypothesis. So, let us get to how to write rate law equations. So, we will we have three components here A, B and A star. We will start with writing the rate law for B. Well it will become clear why it is the easiest to write A star is next level difficulty and A is complex. So, we will go from easy to hard. So, let us start with B. So, here what I have already shown you are the three reactions the forward reaction of the first step of the last type, the backward reaction and the third reaction A star going to be. So, to write rate law we will follow the same scheme as I told for equilibrium reactions. So, if I have to find D B over DT well the first thing to note in which of the equations above is B appearing. I note that B appears only and only here B here is a product. So, this will be equal to K2 A star yeah it is positive because it is product site and I use K2 into A star A star is the reactant. So, B is easy we move on to A star again I ask you where all do you see A star in all of these equations I find A star is here I find A star is here and I find A star is here. So, all I have to do is to take the components of each of them and add them together. So, I start I write D A star over DT D A star over DT for the first one will be K1 into A square. So, this one is positive because A star is a product here and I get A square because of the stoichiometric coefficient 2 behind A. Second one is this A star is a reactant here. So, I get a minus K minus 1 A star A. Finally, I use this one again A star is a reactant. So, we have discussed it for A star let us discuss for A and let us see where all A is coming. So, A comes many, many times and so, the challenge is how do you write a rate equation for this? Pause the video write the rate equation on your own this one is slightly more challenging it is a harder problem, but you have full information on how to solve it. Please pause the video now and I will show you the solution after you have solved it on your own. So, hopefully you have an equation with you if you are not able to solve it do not worry let us solve it together now. For this let us use the same steps as we have done for D and A star. Let us focus on this equation first and here is the question is A a reactant or a product? It is both what do I do? Well, let us not panic what you do is go back to the fundamentals how is rate defined? Rate is defined based on extent of the reaction it is a differential of extent of reaction with time. How is extent defined? It is defined on the consumption or production of a given element per molecule. So, what is the extent here? If 2 moles of A is consumed 1 mole of A will be produced. Therefore, the extent is 1 mole of A is consumed overall. So, what we have is a minus k 1 A square. So, the minus sign comes because A is consumed by 1 mole as the extent and this component is simply rate constant into reactants nothing more no more thinking on that and I have a coefficient 2 here. So, I write a 2 here irrelevant of whether A was in the product or not. So, I do the same thing on this side. Now here all correspondingly I will get a plus because for every mole of A that is consumed 2 moles of A is being produced. So, in effect 1 mole of A is being produced and I blindly write the rate constant into product of reactants ok. We will have more problems of this in the assignment where you will get more practice on how to solve these problems. So, how do we make progress beyond this? What we do is to make what is called the steady state hypothesis. What this hypothesis states? I am again moving a little fast. This hypothesis states that the intermediates concentration do not change with time and the intermediates are the one which do not appear in the overall reaction. So, in this case that is A star A is the reactant B is the product and the intermediate in between is A star. So, I have written back the equation corresponding to D A star over D t that I derived 2 slides before and what steady state hypothesis says is this is equal to 0. So, I will just use this equation and simplify a little bit. So, I just find out what is the concentration of the intermediate here and eliminate it out. So, this one you can verify is equal to. So, this is simple algebra you can look at it and convince yourself this is true. So, we get concentration of A star, but what we are after remember is the rate. Now, our overall reaction is A going to B. So, the rate is D minus D A over D t equal to D B over D t, but D B over D t we showed is equal to K 2 into A star remember we solved for both all 3 A B and A star and D B over D t you can look back comes out to be K 2 into A star. Well, I know now A star. So, I will feed that here and I get this. So, I have now an equation for the rate. So, remember in experimentalists were plotting rate divided by concentration of A. So, I have just divided that and I get this equation now and I want to make a plot of my prediction versus concentration of A. Well, you can just use a computer to plot this, but it is always fun to be able to plot this on your own at least the qualitatively what it will look like. So, we will think of this plot in two limits. One when A is very large in that limit I will assume is much much greater than K 2. So, that I will ignore K 2 in the denominator. So, in this limit I get something that is independent of A. So, at large A I get a flat curve which is equal to K 2 K 1 over K minus 1. Let us also look at the smaller limit when A is small such that K 1 K minus 1 into A is much much less than K 2. Then my rate over A will be equal to K 2 K 1 into A and in the denominator I will ignore the first term and I will get simply K 2. So, here I will get something that is proportional to A. So, you get a curve that looks like this similar to the one that you get in experiment. So, experiment by the way is not does not matches exactly with this observation and you need to do a bit more to get the experimental research exactly right, but that is a story of for another course. So, in summary what we have looked today it is what is an elementary reaction, how to combine these elementary reactions to give a reaction mechanism and how to write rate law equations. And if we want to solve using the rate law equations what is the net rate of a given complex reaction we use what is called a steady state hypothesis. So, we end here and we will move on to calculating rate constants as a function of temperature in the next module. Thank you.