 welcome to the third module of Chemical Kinetics and Transition State Theory. So, in the first two modules we have covered the prerequisites a brief summary of the prerequisites. Today we will start with understanding how rate constant changes with temperature. And we will look at two analysis done by two giants of physical chemistry. The first by van Thoff he is considered to be the father of physical chemistry. He started chemical kinetics, he started chemical thermodynamics in any other fields and he got the first ever Nobel Prize in chemistry in 1901. So, we will look at what he analyzed how he thought of temperature dependence of rate constant and then we will look at what Arrhenius had to say on that Arrhenius was another giant. He was the Nobel Prize winner in 1903 by the way and chemical kinetics is essentially attributed to Arrhenius. So, let us start I will start by writing what is called the Arrhenius equation to get started and we will get into its more details. So, the very famous Arrhenius equation that you must have seen before looks like this. Sometimes people use KB instead of R. Here let me just define the constants T is temperature A is called pre-exponential factor for the lack of a better word, we chemists are not very creative. So, we just see it is before the exponents we call it pre-exponential factor. We have EA that is some very important term that is called activation energy and we will discuss this in some detail and R is the gas constant that you are well familiar with. So, today we will look at the origins of this equation. So, let us start by reading excerpts from what Arrhenius wrote in a very famous paper a translation of the paper written by Arrhenius can be found in the link above. So, let me read on what Arrhenius is saying. Arrhenius notes that for most reactions that were observed in that time each increase in temperature by 1 Kelvin or 1 degree C changes the reaction rate by 10 to 15 percent 1 Kelvin that is it you go from 25 degree C to 26 degree C and it has been noted that the reaction rate is changing by 10 percent. So, what he says is how do I understand this? What is going on the atomic level or molecular level that can explain this and that is what is the genius of Arrhenius? The first thing he noticed well one argument perhaps I can make is that at higher temperature the thermal speed increases Boltzmann has told us as much. So, maybe I can say that with increasing temperature molecules are simply moving faster and therefore reacting faster. But Arrhenius very beautifully argues against it Arrhenius says that cannot be true he says that it cannot be assumed that the increasing reaction velocity reaction velocity by the way is the same as reaction rate in 1800s it was simply called reaction velocity. The increasing reaction velocity comes from increasing frequency of collisions of the reacting molecules. According to kinetic theory of gases which we will study in this course the velocity of the gas molecules changes by only 1 sixth percent of its value. So, that cannot explain an increase of 10 percent we have to do something else. So, Arrhenius points out we have to assume something different and he says and I let me read verbally here the translation it must therefore be assumed to be consistent that the act other actual reading substance is not cane sugar he was looking at the inversion of cane sugar a specific example. Since the amount of sugar does not change with temperature, but is another hypothetical substance which is regenerated from cane sugar as soon as it is removed through the inversion. This hypothetical substance which we call active cane sugar. This is the origin of transition state this hypothetical substance this active cane sugar is what we understand as transition state and remember in those times there was no notion of a structure which we have not measured and it was simply the genius of Arrhenius when he hypothesized that such a substance must exist. Otherwise, how do I explain the chemical data we have taken so much data as a function of temperature and I cannot explain it any other way and however improbable it may seem this is the only hypothesis that it fits it was a very radical idea for a certain and it is exactly right and that is why Arrhenius is given the credit for getting this Arrhenius equation right and the idea of activated state correct. I also want to show you an excerpt from the work of Vanthoff this is slightly older than what Arrhenius had written, but I note as I had noted before Arrhenius equation was not written by Arrhenius originally it is a very interesting tree here for you. Vanthoff wrote that equation earlier. So, this equation that you see here that is there in Vanthoff's paper this equation actually Vanthoff wrote in 1884 before Arrhenius wrote it and do not get me wrong Arrhenius gave full credit for this to Vanthoff and Vanthoff gave full credit to Arrhenius for identifying this reaction and connecting it to the idea of an activated state. So, this equation is the same as Arrhenius equation as you can quickly observe. So, let us look at the what Vanthoff essentially argued what was the intuition of Arrhenius how Vanthoff was able to write this equation. So, this analysis you can find in the Laidler's book section 2.9 I am writing the name of the section here as well which is called the influence of temperature and reaction rates. The reason is if you have a different addition than mine minus addition 3 you can still find it in some other chapter number. Chapter number will change but not the content. So, let us look at what is Vanthoff's analysis. We will start with a one equation that Vanthoff had already derived as I have noted Vanthoff is the father of physical chemistry he just knew every single thing and he had derived several equations in chemical thermodynamics as well. So, one particular equation that he had already derived was the following. In this course we are not going to derive this equation this we assume to be true for this course. Here k equilibrium is the equilibrium rate constant and delta U naught is the change in internal. You do not need to get into to understand what is internal energy mean it is some form of energy that is sufficient for the purposes of this model. So, let us look at a specific example of a reaction and for simplicity let us assume that the stoichiometry is all 1 the argument will not change one bit even if the stoichiometry is not 1. So, for this k equilibrium how is k equilibrium defined it is defined to be product of product concentrations divided by product of reactant concentrations. We are also assuming here that this is elementary that is true what do we get what kind of let us write a few more equations and let us try to use the Vanthoff's equation to see where we go to and then you will realize the genius of Vanthoff. We will consider this reaction and we will write the rate law we will write the forward rate forward rate is simply k f into a into b and the backward rate is equal to k b into c into d. Well this equation is always true if it is elementary well this is also true at equilibrium then, but at equilibrium forward rate equal to backward rate. So, at equilibrium I get k f a b equal to k b c b you will soon note a few interesting relations note that k equilibrium is c d over a b, but from the above equation this is equal to k f over k b let us use the Vanthoff's equation now. So, what we have got is k equilibrium is k f over k b well ln of k equilibrium is then ln of k f minus ln of k b. Vanthoff's equation relates d of ln of k equilibrium over dt well this is then equal to d ln k f over dt minus d ln k b over dt and this is equal to delta u naught over RT square good. At this point Vanthoff basically looked at this equation and he said well this looks like too much of a coincidence to be true always true. So, he said well most likely this itself this is a rate constant remember not thermal equilibrium this itself is equal to some energy over RT square and this itself is equal to some energy over RT square such that e f minus e b is delta u naught ok. So, he just hypothesized it he is not proving anything but this equation suggested that to him ok. So, he in general wrote d ln of any k over d temperature must equal some energy over RT square. So, this you can simplify and show k is equal to e to the power of minus epsilon over RT into some constant which is the Arrhenius equation ok. So, that is the argument Arrhenius had put forward note this is a completely mathematical argument and e is just some energy he is not telling you what energy enters Arrhenius now. Arrhenius looks at this equation and I have already shown you the excerpt from Arrhenius. Arrhenius says well what can be this e a this energy that was there in Vanthoff's equation and he essentially argued that there is a hypothetical active state between reactant and product ok. So, he hypothesized that a transition state exists in between transition reactant and product and this e a must be the energy required to go from R to T s. So, what he is talking of although he did not drew this figure, but what you have famously seen is this kind of a energy profile and well what Arrhenius was pointing out although not in a diagrammatic way is that this is e a and this is your transition state and this we very well know to be true today. So, the question that we ask in this module in this course how do we calculate this rate constant k? How do we have an Arrhenius equation, but we do not know how to calculate a for example. So, is there a way for us to calculate this quantities from an atomistic picture? At the end of that a all we have is a dance of molecules happening and this dance is governed by a certain laws of physics, we know the laws of physics. Can we use these laws of physics to calculate this rate constant? So, that is going to be the focus of this course. So, in summary for this module we have looked at the analysis given by Vanthoff, his argument on how he got the rate constant to be something like a into e to the power of some energy over R T. And we also looked at how Arrhenius looked at that equation and presented a physical picture out of it specifically he hypothesized the existence of transition state which is critical in our understanding of rate loss. So, with that we end today. Thank you very much.