 Welcome back to the lecture which is devoted to connecting equilibrium constant with molecular partition function. In the previous lecture, we started with an expression for the Gibbs function or Gibbs free energy and we express delta G naught in terms of delta E naught and the molecular partition function as presented here in this expression. And then we were discussing that since delta G naught is equal to minus RT log K, we can equate the two. Once we equate the two, the new equation that I have is minus RT log K is equal to delta E naught minus RT log pi J Q J m naught by N A Avogadro constant and raise to the power stoichiometric number. This equation can be rewritten as log K is equal to minus delta E naught by RT plus log pi J Q J m by N A raise to the power stoichiometric number if I can further extend this as I can write log exponential minus delta E naught by RT plus log this product Q J m naught by N A raise to the power stoichiometric number if you carefully examine what I have done that I have expressed minus delta E naught by RT as log exponential minus delta E naught by RT. Why I did was that I have a logarithmic function on the left hand side and this is log A plus log B which is equal to log A into B and then I can get rid of the log on both sides. Keep this equation in mind and from this equation I can write K is equal to product of Q J m standard state divided by N A stoichiometric number into exponential minus delta E naught by RT. This I am getting from the previous one log K is equal to log exponential this into this log A into B and then you can get rid of the logs and you come up with this expression that equilibrium constant is equal to product of molar partition function under standard condition for species J divided by Avogadro constant raise to the power stoichiometric number into exponential minus delta E naught by RT. Once again reemphasizing that this stoichiometric number is positive for products and negative for reactants going back to your discussion in classical thermodynamics. Equilibrium constant is a dimensionless quantity and here if you now examine each and every term you will see that the dimension the units they cancel out and equilibrium constant turns out to be a dimensionless quantity. In classical thermodynamics you use to express equilibrium constant as the ratio of activities of the products and reactants and here if you see this N A which is included within the bracketed term whatever is the stoichiometric number this makes the entire thing dimensionless quantity. Therefore, in order to evaluate the value of equilibrium constant by using statistical thermodynamic principles what you need to know you need to know the molar partition function. You need to know the stoichiometry of the reaction you need to know 0 point energy difference and gas constant temperature Avogadro constant are anyway the quantities which are required. What we will do now is that this equation which we have developed is the same equation now written in a typed form that k is equal to pi j q j m naught pi N A raise to the power stoichiometric number into exponential minus delta R E naught by R. Now we can think of applying this developed equation to the variety of reactions. The first reaction that I will like to discuss with you is dissociation reaction x 2 gas forming 2 x items. Let us see we apply to this and in future I will avoid writing g because we are anyway discussing the situation in the gaseous state alright. Now our system is x 2 going to 2 x. So, we have to use this expression. So, k will be equal to this nu j is positive for products and negative for reactant for product it is plus 2. So, what I will write q of x m naught by N A this number stoichiometric number is 2 and then I will have in the denominator for the reactant x 2 that is q of x 2 m naught by N A this is going to be for the reactant into exponential minus delta E naught by R D. Now if you do the mathematical rearrangement of this you get q x m naught square divided by q x naught for 2 into 1 over N A because 1 over N A will come from excess N A in the numerator part and exponential minus delta R E naught by R D. We have come up to this expression that is easy so far. Now when you talk about the dissociation reactions like this then delta R E naught is twice the internal energy of x at absolute 0 minus internal energy of x 2 at absolute 0 and what is actually this is x 2 going to 2 x this is dissociation. So, instead of writing delta E naught I can as well write dissociation energy at absolute 0 of x x bond. So, that means, I need to have information on dissociation energy of x x bond alright. So, going back to my expression it was q x m naught square divided by q x 2 m naught into 1 over N A was excess into exponential minus into exponential minus d 0 by R D where from I get this expression from this one q x m square over q x 2 m naught 1 over N A exponential minus delta R E naught this delta R E naught we have said that this is equal to d naught and we have this expression. Now you carefully examine the numerator is for x is an atom the denominator is for x 2 x 2 is a molecule an atom can have translational degree of freedom and electronic degree of freedom there is no rotation and vibration in atom. For molecule there is translational degree of freedom there is rotational degree of freedom there is vibrational degree of freedom there is electronic degree of freedom. That means, for for x I will have q translational of x and I will have g x which is the degeneracy of the ground state electronic states I require this information for molecule x 2 I will require q translational for x 2 I will require q rotational for x 2 because being a diatomic molecule it can undergo rotation I will require q vibrational and I will also require the degeneracy of the ground state I repeat in the expression for k you notice that the numerator is q x square where x is an atom an atom can have translational degree of freedom and you also require knowledge of the degeneracy of the ground state why degeneracy of the ground state because electronic states are assumed to be or are far separated. So, therefore, their contribution is usually coming from the ground state term x 2 being a molecule diatomic you will have translational degree of freedom you will have rotational degree of freedom you will have vibrational degree of freedom you will have the electronic term that is the ground state electronic term degeneracy of the ground state. Now, with this information what I have I go back to this k is equal to q x m not I am just not I am not writing because we assume it is standard state this square over q x 2 m into 1 by n a into exponential minus d 0 by r d this is where we are now q x m is equal to V m by lambda x q fine and it will also have a contribution from degeneracy of the ground state overall partition function is multiplication of the partition functions q x 2 m is V m here I will put standard state left hand side you assume as standard everything is actually in the standard state over lambda x 2 q into I need q r into vibrational into degeneracy of x 2 what I have done is I have included translational contribution rotational contribution vibrational contribution electronic contribution what we will do is we will substitute these both q's into this expression and let us see what it turns out to be. So, what we will do is let us put k is equal to q x m square that means V m not square by lambda x it is a square. So, raise to the power 6 into g x square now divided by V m not lambda x 2 into q r x 2 into q v x 2 into g x 2 and there is exponential minus d 0 by r t molar volumes are same p v m is equal to you need to use p v is equal to n r t or I will say p not standard state into V m v by n is equal to V m is equal to r t that means according to this V m not is equal to r t by p p not rather this and this will cancel. So, what we have remember that there is one more n a term and we should include that n a somewhere here into n a I will just put here. So, what I have now k is equal to V instead of V m not I will put r t by p not into I will put g x square into lambda x 2 of course, this has to be q divided by lambda x raise to the power 6 q r x 2 q v x 2 g x 2 into n a into exponential minus delta e not by r t. Next is use r is equal to k times Avogadro constant. So, that means this r by n a I can just use k r by n a use k that is the k this temperature is there and g x square use this g x square into lambda x cube divided by p not p not is there then you have g x 2 is there q r x 2 q v x 2 lambda x 6 and into exponential minus d 0 by r t. Now try to understand that how to arrive at this expression from this expression the shape of this expression of the form of this expression is similar to that you used in chemical thermodynamics because there also what was it equilibrium constant was equal to the ratio of the activities of the products and reactants weighted by their stoichiometric number. Here also this actually is ratio and how it is ratio is since your reaction is x 2 going to 2 x. So, it is q x square divided by q x 2 and since each q is divided by n a you have 1 by n a in the denominator then exponential minus d 0 by r t. Now after that you have to decide that the numerator and denominator or for a given species what type of contributions can enter if you are dealing with only atoms then you will have translational degree of freedom electronic degree of freedom. If you are dealing with the molecules then the molecules since we are dealing with the gases you have translational degree of freedom you have rotational degree of freedom you have vibrational degree of freedom and you have electronic degree of freedom. And we already showed that for x 2 going to 2 x delta E naught is simply the dissociation energy and this dissociation energy you can obtain from the 0 point vibrational energy. We will take up an example later on to show that how dissociation energy at 0 Kelvin can be calculated from the vibrational wave number data. But for the time being remember that depending upon whether your reactant or products are atoms or molecules you need to take their partition functions accordingly. Remember that the number of terms appearing here or in any of the expression is going to depend upon your type of reaction. Here we have taken a very simple dissociation reaction there can be reactions in which you have atoms also you have diatomic molecules also you have triatomic molecules also and you have multiatomic molecules also and there can be more than 1 is to 1 type means there can be reactions of a moles of a plus b moles of b is equal to c moles of c plus d moles of d. That means you will have to account for the partition function of each one a b c and d each one of them that is a b c or d can either be simply an atom or it can be a diatomic molecule it can be a triatomic molecule it can be multiatomic molecule. The examples include for monatomic you have just ideal gases helium neon argon etcetera diatomic you will have carbon monoxide nitrogen cl 2 I 2 etcetera triatomic you can have water and similar other molecules carbon dioxide is another diatomic molecule then you can have multiatomic molecules. So, the situations can become more and more complex as the number of atoms involved in a molecule increase therefore, you will have to very carefully decide how many terms and which terms are to be include in the denominator or are to be included in the denominator. We will discuss this in more details when we take specific numerical problems and then I will explain to you that when you are using a particular formula when you are using a particular expression whether it is allowed under those conditions of temperature or not, all those we will discuss in the next lecture. Thank you very much.