 to the next lecture on statistical thermodynamics. We have been talking about equilibrium constant and by now we have developed some relations between equilibrium constant and molecular partition function. And based on those derived equations, we have also solved some numerical problems. Numerical problems starting from simple dissociation of a diatomic gas to a little complex systems in which the reactants and products are combination of linear or non-linear rotors. Please remember that whenever you are dealing with such systems, it is very important to look at the stoichiometry of the reaction and make appropriate representations there in the equation. Equilibrium constant is connected to change in standard molar reaction Gibbs free energy change. And delta G naught when we write equal to minus RT log k, remember that delta G naught is also equal to delta H naught minus T delta S naught. So, therefore, there are two contributions. One is enthalpy change and the other is entropy change. And you know from your concepts of chemical thermodynamics that both the enthalpy change and entropy change make a difference in the overall sign and value of delta G naught. The overall spontaneity is decided by the sign and magnitude of delta H naught and T delta S naught. But those are the concepts that we discussed in classical in chemical thermodynamics. But when we are talking about the discussion in statistical thermodynamics, today we will discuss that the same principles apply here too. We cannot just go by delta G naught. Both delta H naught and T delta S naught make a contribution to delta G naught. So, therefore, in today's discussion, we are going to talk about contributions to equilibrium constant. In order to elaborate upon that, let us consider a gas phase reaction R in equilibrium with P. R stands for reactants, P stands for products. And let us try to understand the physical basis of equilibrium constant. Now, take a look at this figure. This array of R which I mean by R by reactants, this is an array of energy levels for reactants. And this is an array for energy levels for products. And also note that there is a difference in their 0 point energy levels. The example that I am covering here includes the 0 point energy level of product higher than the 0 point energy level of reactants. So, that this delta E naught is a positive number. Now, when you have this mixed array of energy levels, some corresponding to reactants, some corresponding to products. At equilibrium, all these energy levels are accessible, but these are accessible to different extents. If you consider the system comprising of energy levels as shown here, you can see the spacings in energy levels for reactants represented by blue and the spacing of energy levels for product represented in violet. This population, the relative population of the states corresponding to reactants or corresponding to products will depend upon temperature. And equilibrium composition of the system reflects the overall Boltzmann distribution of populations. So, at a given temperature, suppose there are n number of molecules. At a given temperature, these n number of molecules will distribute in this way. And this distribution is given by a single Boltzmann distribution. Now, the relative population of reactants and products is going to depend upon temperature as well as this separation delta E naught. Obviously, as seen in this figure, if delta E naught increases, the population of R becomes dominant. Because if delta E naught increases, then the population in product is going to be less. And therefore, R the reactant becomes dominant under equilibrium. What we are discussing over here? We are discussing the set of energy levels corresponding to reactants corresponding to products. And then we have created or we have shown the difference in 0 point energies for the reactants and products. If the product is higher than the reactant, then the reaction is going to be endothermic. You can similarly show for exothermic reactions and so on. Obviously, the separation between the 0 point energy levels is going to decide whether R is going to be dominant at equilibrium or P is going to be dominant at equilibrium along with the temperature factor. Now, you consider different set of energy levels. Notice that R has a big difference in energy levels. See from 0 to 1, this is a big difference than 2 etcetera. The product, the density of states is high. These energy levels are very close to each other and then again becoming far separated. In this kind of arrangement, it is easy to observe that the occupation of the energy levels corresponding to product is more than the occupation of molecules in the reactants. That means, under these conditions, the population of product dominates in the mixture even though P is lying above R. Just it is dominating because you see the density of states is higher and if you count the overall population that will be more than the population of molecules in the energy states corresponding to reactants. I take you back to the previous case. Here the energy levels were similar, but you see in this kind of arrangement, the population of the states for reactants is dominating at equilibrium and here the population of states at equilibrium is dominating for products at equilibrium. Now, if you note the comments written over here, what the comments say? It is important to take into account the densities of the states of the molecules. Here it is dense. Even though P might lie well above R. Here you can see P is lying well above R. That means, delta E naught is large and positive both large and positive. Then P might have so many states that its population dominates in the mixture that is what we were discussing. Now, look at the next comment. In classical thermodynamic terms, we have to take entropies into account as well as enthalpies when considering equilibrium. That is what I was talking to you that delta G naught which is equal to minus R T log K. I can write this also equal to delta H naught minus T delta S naught at a given temperature and that is what the comment is being made over here that in classical thermodynamic terms, we have to take into account the enthalpies as well as entropy changes. What is taking into account the enthalpies? This one and what is taking into account the entropies? This one. So, therefore, the discussion is similar, but here in one diagram in one discussion, we can relate the changes in standard molar Gibbs free energy to the enthalpy change and entropy change. Now, let us take the example which is shown on this slide and let us take a look at the comments. The first comment is the populations of the states are given by the Boltzmann distribution and are independent of whether any given state happens to belong to R or to P. What it means is when you have this combination of R and P, the population of these states combined states is going to be given by Boltzmann distribution. It does not matter whether a given state belongs to reactant or belongs to product, the population is decided by the Boltzmann distribution. Therefore, we can consider the single Boltzmann distribution spreading without distinction over the two sets of states. Now, in these two states, states corresponding to R and states corresponding to P, if you see that their density of states or their separation energy separations are similar, these spacings are similar and just a look at this diagram suggests that under the conditions of temperature, given conditions of temperature, if the distribution is like this, just by looking at you can predict that R is going to dominate at equilibrium or R will dominate in the equilibrium mixture. And then if I replace the P by a certain product P representing very high large number of densities of states, in that case P can dominate in the mixture even though it is lying above R as discussed earlier. So, why I am talking about all these things, you know and I am bringing in Boltzmann distribution also into our discussion. That means, by using Boltzmann distribution, I can now quantitatively come towards the same expression that we have just discussed in couple of lectures before. And as pointed out, you consider this total as single Boltzmann distribution without distinction over two states, let us do that. Now, we have the reactants, we have products, we have their energy levels for the reactants and for the product. We can calculate the number of molecules in the product and number of reactant molecules under equilibrium conditions. Once we have the numbers, this number is directly proportional to concentration. How does the number gets converted into concentration? Suppose if I have x gram of product or reactor, then I from x gram I can calculate divide by the molecular weight, I can calculate the number of moles. One mole is equal to Avogadro constant number of molecules. Therefore, from the given number of moles you can calculate how many molecules will be there. So, therefore, if I can know this ratio n P by n R, then I can get equilibrium constant. I am going to show that this ratio is equal to Q P divided by Q R into exponential minus delta R E naught by R T. Let me take you back. The reaction that we are talking here is R in equilibrium with the P. And if I were to use this expression K is equal to pi j Q j m naught by n A raised to the power stoichiometric number into exponential minus delta E naught by R T. This is the expression for equilibrium constant. So, for this what I will write K is equal to partition function of product of course, standard state divided by n A whole divided by partition function for reactants divided by n A into exponential minus delta E naught by R T. Rearranging this I have K is equal to Q P over Q R into exponential minus delta R E naught by R T. If you compare these two look at this. This is the same expression as this. So, that means, now my next step should be to find out an expression for number of molecules product molecules and number of reactant molecules. If I can find that expression then I can derive this expression that n P by n R is equal to Q P by Q R into exponential minus delta R E naught by R T alright. So, this is the same expression which you get by using the result that we have obtained a few lectures ago. Y n P by n R let me just reiterate again Y n P by n R you can take any example n P. If you just look at this distribution this talks about there are more number of molecules in the reactants and this distribution this talks about there are less number of molecules in the product. So, therefore, if I can know this number of molecules in the reactant and number of molecules in the product I can the ratio of that can give me a value of equilibrium constant. How do I get now these numbers? Now you recall that what can give you the number of molecules Boltzmann distribution and what that Boltzmann distribution is if you remember your discussion we had earlier n i upon n is equal to exponential minus beta E i upon Q this basically is population n i upon n is population of ith state. That means, from this expression I have n i is equal to n by Q into exponential minus beta E i where i is any ith state. Now how do we get the total number? Total number of molecules will be obtained by summation of this you can take some of this n i that is summation n i i n i this is n and if I apply this to reactants only reactants not to the product only reactants then I will get n r do not get confused between this n and this n r this is the total number of molecules and this is the number of molecules which are in the reactant side going back. That means, I am talking about the number of molecules which are here in the reactant side then this will be equal to summation I will only invoke reactants exponential minus beta E i let me put i reactant. This expression is very important because this is what I need to use to obtain an expression for equilibrium constant over here. Summation n i is equal to n corresponding to reactants and then this summation I will only apply to the reactants. Now very carefully listen when I apply this to reactants my 0 point energy is this this is the 0 point energy of the reactants and then you have distribution and there are certain number of molecules and then if I write the same thing for the products then your product 0 point energy starts from here which is well above the reactants and then remaining energy levels are corresponding to this 0 point energy levels. We will use this information when we derive next equation but what we have got is we have now understood how to calculate the number of reactant molecules. Once we know how to calculate the number of reactant molecules then we will know how to calculate the number of products molecules. Once we have N p once we have N r then take the ratio of both and we will have the expression for equilibrium constant. The important point to note here is that our discussion has restricted to occupation of different energy levels and we have said that this occupation or occupancy rather depends upon temperature and also depends upon delta E naught that is what is the difference between the 0 point energy levels. We have obtained an expression for N r similarly we will obtain an expression for N p and as I said when I talk about products the 0 point energy of product lies above the 0 point energy of reactant and this difference we are going to consider in the derivation of the overall partition function. But that we will do in the next lecture. Thank you very much.