 So, we continue with advanced reaction engineering. Now, we look at further considerations on energy balance. Now, the point of looking at this is the following. Let us for example, let us say we have a reaction which is instantaneous. What is meant by instantaneous? That reaction R R 1 and R 2 are very large. We can also say that R 1 and R 2 are such that it is in equilibrium. On other words, the reaction this is one way of looking at it. The other way of looking at the reaction is essentially at equilibrium. Which means what? Which means that the conversion at any is given by k by k plus 1, where k is given by the van tof's equation, which is d l n k by d t equal to delta h by R t square. This is one way by which we can understand that the reaction is instantaneous means that the conversion is k by k plus 1, where k is equilibrium constant, k is equilibrium constant, equilibrium constant and equilibrium constant is given by the van tof's equation t l n k by d t delta h by R t square. Now, what are the situations where you are likely to encounter such fast reactions? There could be many situations in which we can be looking at. First situation of course, the rate processes are very fast. I mean practical situations, let us say practical situations. So, rate processes for reaction rate processes are very rapid. This is one. The second instance, which would be of relevance to us is for example, when for example, let us say an endothermic reaction, endothermic reaction. It proceeds only at the rate at which heat is supplied. So, heat is an instance of a reaction, which simply moves depending upon the rate of heat supply. So, these are all situations in which we need to be able to use our basic equations with some care. So, let us look back at our basic equations, the energy balance and the material balance. So, let me write that means, we have a situation like this. We have a reactor. Maybe it is a catalytic reactor and it is cooled or heated by a fluid. This is a coolant or this is the reagents coming in T naught. Now, our energy balance, we have done written the energy balance before, which looks like this. If there is a going to be, this is the reaction. So, I am just writing it at r 1 minus of r 2 times minus of delta h 1 star. This is the, then this is 4 h by d multiplied by T c minus of T. So, this is heat generation shall I say, heat generation and this is heat addition. Now, if it is an exothermic reaction, of course, this is positive. Therefore, this must be negative. If it is an exothermic reaction, if it is an endothermic, this exo, exo means this is positive, this is negative. Therefore, then if it is endothermic, this is negative and then this must be positive. This is something that we all know. Endothermic means heat is to be supplied and it is supplied by this term. And this difference is what you see on the left hand side. Now, keeping this in mind, let us look at the material balance, which says d f, let us say d f a d v equal to r 2 minus of r 1 or f a 0 times d x d v equal to r 1 minus of r 2. So, you notice here, if I call this as reaction equation 1, call this equation 2, then we have v c p d t d v equal to which is r 1 minus of r 2. I will write here minus of sorry plus f a 0 times d x d v equal to r 1 minus of r 2. This is minus of delta h 1 star plus 4 h by d into t c minus of t and then you have this is equation 1 and this is equation 2. Now, what we are trying to point out here is that r 1 minus of r 2, if r 1 and r 2 are very large for example. So that r 1 minus of r 2, we do not know how whether it is large, small, we have no idea. So, in cases where r 1 and r 2, so large that we are not able to measure them, clearly you know this is not going to be easy to handle this equation. So, in such cases, we can replace this first term by this d x d v multiplied by minus of delta h 1 star plus 4 h by d t c minus of t. Now, the important thing here is that now it may be possible for you to measure this term left hand side d x, this term may be possible to be measured, which cause you can take samples of the components and therefore, you can measure the left hand side. You can even plot you may be able to plot this x versus v showing that some data may be available. Therefore, this term is something that you can experimentally determine, so that it is more convenient to replace this r 1 minus of r 2 in terms of what you can measure. And therefore, if you look at this equation, I call this as equation 3, there are a number of things you can do with this equation. Now, if I just sort of look at this in some detail, let me just write this once again just for the sake of explaining equal to f a 0 d x d v multiplied by minus of delta h 1 star plus 4 h by d t c minus of t. Now, this term d x this term we are able to measure experimentally. For example, if it is so happen that this reaction this reaction r 1 minus of r 2 is so rapid that reaction is at equilibrium all the time inside the equipment that then you can recognize that d x d v, which can be written as d x d t and d t and d x d t d t d v. What is d x d t? If it if our reaction x is given by k by k plus 1, then what is d x d t? d x d t is derivative of this, which is what 1 by k plus 1 minus of 1 by k plus 1 whole square d k d t. And what is d k d t? I will write here d k d t, d k d t is simply where are we where is the band of the equation. Just go back to band of the equation and then ask this question what is d k d t d k I will just write here d k this implies d k d t is what is 1 by k d k d t is delta h by r t square d k d t becomes k times delta h by r t square. So, what it means is that this simplifies as this is equal to 1 by k plus 1 whole square d k d t is k times delta h by r t square. Therefore, you can see here d x d v is simply this whole term 1 by k plus 1 times k delta h by r t square. So, you can substitute for d x d t the whole thing d x d v you can replace this in this way and then you can replace this as d t d v. Therefore, our equation I will call this as equation 4. So, I can replace equation 4 in this form let me just do this will not take too long. So, I have this energy balance equation d t d v equal to f a 0 your d x d v this d x d v term I am writing from here which is k which is k divided by k plus 1 whole square delta h by r t square d t d v. That is the first term that is this I have written this term I have multiplied by delta h. So, I will multiply by minus of delta h 1 star then plus 4 h by d t c minus of t. Now, notice here that this is d t d v and this is also d t d v therefore, we can take d t d v common you get v c p I will put a plus sign because this is a plus sign goes other side f a 0 delta h square I have taken these two terms. Then you have k by k plus 1 whole square I have taken this by 1 by r t square. This is the term here equal to 4 h by d t c minus of t. So, what is it that we have done what we have done is that for us for this case where the reactions are very fast and therefore, the whole thing is a equilibrium at every position in the equipment this equation represents what happens to the reaction. So, let us understand what this equation means. So, if you make a plot actually I should have this equation in front of us. So, I will write it here for our sake d t d v within brackets of v c p plus f a 0 delta h square k by k plus 1 whole square 1 by r t square equal to 4 h t c minus of t divided by d. Suppose I want to make a plot make a plot of let us say versus v how do I find out. Now, this differential equation I have to solve notice here that I can this I can write this I am putting it here. So, d t d v d t d v equal to what it I can take it to the other side. So, that essentially d t d v is a function of k is a function of temperature. So, it is essentially a function of temperature on the right hand side everything is a function of temperature and t c and of course, it also depends upon the tube size I will put tube size anything and then f a 0 and then volumetric flow. So, what are we saying what we are saying is that the temperature variation at different positions of the equipment depends upon temperature coolant temperature reactor diameter inlet flow rate and volumetric flow rate. So, basically if you can once temperature is known all the rest follows. So, on other words if you want to find out what is how t varies with position what you have to do you start with t 0. So, you start somewhere at t 0 and therefore, you can forward march because everything is known from the initial state. That means, from position here 0 position everything is known therefore, you can do a forward march and then calculate what is the temperature as a function of volume. And once you know temperature similarly, once you can plot x as a position of volume because once you know temperature you know van toff's you know this k k by k plus 1 already you know this therefore, moment you know temperature you know x therefore, you can plot x was volume. So, what we are trying to put across to you is that if you have a reaction where in the reaction rates are such that is very large reaction rates are very large. And therefore, it conforms to equilibrium at every position in the equipment then you are able to tell how t changes with volume and the equation that describes this variation is what we have derived. This is what we have done so far this is the equation that describes the variation of temperature with volume. And then once you know the variation of temperature with volume we said we can plot t versus v and therefore, you can plot x versus v therefore, the complete the whole system is fully specified. So, for the case of a instantaneous reaction we find that from the energy balance we are able to tell what happens. Now, second related situation which is of interest to us is suppose for example, let us look at let me write this once again just for the sake of explaining to you that v c p d t d v equal to r 1 minus of r 2 minus of delta h 1 star plus 4 h by d t c minus of t this is one form. The other form is f a 0 d x d v times minus delta h 1 star plus 4 h by d t c minus of t the reaction is a goes to b and b goes to a 1 and 2 this is the reaction. Now, there could be a second situation the situation there is that this term which I call this term 1 and I call this as term 2 term 2. Now, there are situations where term 2 is quite small the sense that we are not able to supply enough heat therefore, essentially the rate at which let us consider as an example that this is an endothermic reaction. What is an endothermic reaction you have to supply heat then only this reaction will move otherwise it will not move. So, this is an instance is an instance in which this essentially the reaction moves forward as you supply heat and if this reaction is reversible may not be instantaneous it is reversible. Therefore, to the extent you supply heat it will move forward therefore, even in such cases you are able to conveniently use this form if you find that is not convenient to use you can use this form and take this reaction on the questions of our interest. So, in all these cases if it is exothermic if it is exothermic then you will see it will be thing like this. So, this is the process industry there are so many instances of exothermic reversible reactions therefore, this is what is called as the hotspot it is of much interest to the designer because hotspot temperature of the catalyst is a specification on the catalyst you should not exceed it preferably and therefore, lot of design has to take into account that we designed. So, that the hotspot temperatures are not reached so this is something that I wanted to reemphasize what was said in the last lecture just to bring to your attention that this formulation is very useful from the point of view of dealing with very fast reactions where R 1 and R 2 may not be known to you. So, let us take one more example to illustrate how we can manage what is called as the energy balance. So, second exercise in which we want to look at once again it is a it is a reaction was lot of heat is involved in a lot. So, let me just describe the process and then we can look at the details what do we have we have we have we have chlorine coming in chlorines coming in this is chlorine and then products we have benzene plus chlorine equal to monochlorobenzene plus HCl and then monochlorobenzene plus chlorine equal to dichlorobenzene plus HCl. So, some data is given what is the data given the data given is that chlorine is it correct F c 0 by F b 0. So, F c 0 divided by F b 0 that is given as 1 by 1 by 1 by 1 by 1 by 1 by 1 by 1 by 1 by 0.4 this data is given. See if you look at problems like this how is energy balance important here is that you have a chlorination reaction. Now you know this benzene these are all compounds which get chlorinated lot of heat is released. And generally it is not very easy to handle such reactions in a in a plug flow vessel because there is not enough mixing. And therefore, the heat release is such that it is very difficult to manage the reaction in a stirred tank such as here this is a stirred tank. We can comfortably in an appropriate design put a cooler cooling coil and then remove the heat. On other words it is possible to manage highly exothermic and very very rapid reactions in a stirred tank by appropriate choice of the feed this is the benzene feed here. So, we can adjust the feed and the chlorine feed such that the heat releases within manageable limits. So, the first part of the question is what is the first part of the question is what is the find the temperature temperature which maximizes maximizes yield of monochloric benzene this is the first part of the question. And of course, once you have done that what is the what is the heat load heat load that the coil must handle that clear. So, there are two parts to this I mean gets chlorinated lot of heat is released. And what is the temperature which maximizes the yield of after many cases you are intermediate you are interested in the intermediate see what maximizes the yield of monochloric benzene having done that. So, what is the amount of heat that this coil this coil must handle. So, these are the two questions that we must address some data is given which says some more data is given which says this is reaction 1 this is reaction 2 which says that is k 1 tau equal to k 2 tau or k 1 by k 2 is 8 this data is given. Now, chlorinations are very common in the process industry and many of the refrigerants that we use have some chlorine and chlorine in a fluorocarbons. So, these are all reactions involving you know halogenation and this is very important halogenation. And then benzene chlorination is of course, very important because traditionally benzene chlorobenzene are involved as insecticides in several applications of course, there are lots of issues coming up these days. But that aside that it has served a long purpose of insecticides for a very long time it will continue to be so till we find alternates. So, we need to handle chlorination now benzene is one of them methane chlorination is another you know very important products that we use in our what is called as the refrigeration industry and so on. Now, that we have this question in front of us our question in front of us is what is the temperature? What is the temperature which will maximize monochlorobenzene? This is the question that we want to answer and of course, once you do that we will do the rest. So, let us do the material balance as usual our job is to write the material balance. Let us write the material balance benzene plus chlorine gives you monochlorobenzene plus HCl then monochlorobenzene plus chlorine gives you dichlorobenzene plus HCl. So, our our stoichiometry says FB equal to FB 0 FB 0 times 1 minus of X 1 this is reaction 1 reaction 2. Then comes F monochlorobenzene equal to FB 0 X 1 minus of X 2 plus FM 0 which we taken as 0. Then we have dichlorobenzene which is FB 0 times X 2 then of course, you want to find out amount of chlorine consumed Fc equal to Fc 0 minus FB 0 minus FB 0 minus FB 0 into X 1 plus X 2 plus X 3 shall we say is this clear what we have said. So, the amount of chlorine that is consumed is whatever is the chlorine that we put in minus whatever is consumed. So, I will call this 1 2 3 and 4 in this exercise what is said is that the chlorine is added at a rate. So, that chlorine is completely consumed chlorine is completely consumed and the fluid here is saturated in chlorine you understand fluid here is saturated in chlorine. So, what does it mean it means the following it means it means the if you look at let us say if you look at reaction 1 if you look at reaction 1 this is reaction 1. So, what is the rate constant says reaction therefore, if I say R say R benzene rate of formation rate of is I will put a minus sign here K times C B times C Cl what are we saying. And now since this C Cl 2 the C Cl 2 concentration in the liquid does not change we can I will call this K dash K dash time C I will say this K dash time C Cl 2 equal to K therefore, it becomes K C B. So, what is being said is the following that the chlorination reactions that are taking place in the liquid phase the it is second order reaction yes, but in view of the fact that the chlorine concentration in the liquid does not change we can assume it to be a pseudo first order reaction that is what is being said with this we can write the material balance for say let us say like for dichlorobenzene. So, material balance material balance for for D B. So, what is input what is output what is generation equal to accumulation and we are talking about C S T R at steady state at steady state. So, what is the input 0 what is the output F B 0 times X 2 what is the generation dichlorobenzene is generated in reaction 2 it must be equal to K 2 times M B times C L 2 concentration C L 2 concentration is constant therefore, it becomes a pseudo first order in M B. So, it becomes K 2 C M B times V equal to 0. So, we have got an equation for dichlorobenzene let us do for one more. So, that we can finish it off. So, if suppose I write the balance for benzene what is the balance for benzene F B 0 minus of F B 0 into 1 minus of X 1 minus of K 1 C benzene times V equal to 0. So, this becomes F B 0 X 1 equal to K 1 C B B is it all right. Now, what is C B is F B by small V volumetric flow which is F B 0 times 1 minus of X 1 divided let us say is V 0 V equal to V 0 volume changes are small this is equal to C B 0 times 1 minus of X 1. So, similarly C M B equal to F M B by V 0 what F B M B is equal to F M B by 0 times X 1 minus of X 2 divided by V 0 is equal to C B 0 times X 1 minus of X 2. So, why I have done this we have done this because we want C M B and we also want C B in terms of X 1 X 2 that is what we have got. So, we have got everything in the form in which we require therefore, we can now use it to go forward. So, we can substitute appropriately therefore, we have from benzene balance. Let us do from benzene balance what do we get from benzene balance we get F B 0 X 1 minus of K 1 C B 0 times 1 minus of X 1 equal to 0. Please I am not seeing anything new something that benzene balance. So, this is what I have written C B I have written as C B 0 times 1 minus of X 1. So, this we can write this as V 0 C B 0 X 1 equal to K 1 C B 0 1 minus of X 1. Now, you can collect coefficients and recognize on B is missing. So, with this can be written as X 1 equal to K 1 tau 1 minus of X 1 therefore, X 1 equal to K 1 tau divided by 1 plus K 1 tau. This is one relationship that we have got notice K 1 is given K 1 by K 2 is given all right. Once we do that similarly for dichlorobenzene. So, what is dichlorobenzene we got for dichlorobenzene F B 0 X 2 minus this is minus this plus K 2 times C M B is what C B 0 times X 1 minus of X 2 times X 2 times V equal to 0. So, we can collect coefficients and so on and therefore, we get V 0 C B 0 X 2 plus K 2 C B 0 X 1 minus of X 2 times V divided throughout by V 0 becomes tau. So, this becomes tau. So, this is because C B 0 can be cancelled off. So, we get X 2 equal to K 2 this is K 2 times tau X 1 minus of X 2 therefore, X 2 equal to K 2 tau divided by 1 plus K 2 tau multiplied by X 1. So, what we got we got our relationship both X 1 and X 2 we now know in terms of K 1 tau and K 2 tau. Now, if I ask you now the problem says the problem statement let us understand the problem statement once again. Problem statement is what is being said is the following what is being said is that in this chlorination reaction chlorine which is admitted into the C S T R which is maintained at an appropriate temperature all the chlorine gets consumed or in other words if chlorine input is F C 0 all the chlorine gets consumed. Therefore, if you look at if you look at the chlorine balance what are we saying we saying in chlorine balance all the chlorine gets consumed that means this is 0 which means what which means F C 0 divided by F B 0. Let us write it here F C 0 divided by F B 0 equal to X 1 plus X 2 plus X 3 and if the data says F B 0 by F C 0 equal to 1.4. Therefore, this is points 1 by 1.4 equal to X 1 plus X 2 X 3 is not there sorry plus X 2. So, this is 0.71 equal to X 1 what is X 1 K 1 tau divided by 1 plus K 1 tau what is X 2 K 2 tau divided by 1 plus K 2 tau multiplied by X 1 which is K 1 tau divided by 1 plus K 1 tau. Now, K 1 divided by K 2 is given as 8. Therefore, here this can be written as what K 1 by K 2 is therefore, K 1 equal to 8 K 2. So, I can put K 1 is equal to 8. Here I will get 8 K 2 tau divided by 1 plus 8 K 2 tau first term then K 2 tau divided by 1 plus K 2 tau multiplied by 8 K 2 tau divided by 1 plus 8 K 2 tau equal to 0.7. What are we done now? Now, we can solve this to find out what is K 2 tau correct. Now, let us not forget the context. The context is this generally it is possible to find out these ratios more easily than the absolute values that is the point. So, here what we have done these ratios have come out from the experiment. Therefore, you can find out by solving this what is the value of K 1 tau correct what is the value of K 1 tau by solving when you solve this we get 7 1 equal to K 1 tau by 1 plus K 1 tau plus K 2 tau K 2 tau by 1 plus K 2 tau multiplied by K 1 tau divided by 1 plus K 1 tau where K 1 by K 2 equal to 8. You can solve this it find out that K 2 tau turns out to be about 0.25 and therefore, K 1 tau turns out to be about 2.0 sorry it is 0.22 he says 0.22 and 1.76. So, with this we can find x 1 equal to 0.63 and x 2 equal to 0.18. Once you know x 1 x 2 you can find out f m b f benzene f dichlorobenzene f monochlorobenzene and so on all these can be found out because. So, why have you done this problem what is what is the moral of this whole story the moral of the story is that if you have a series reactions like this you can find out the compositions if you know the values of the rate constants that is what is being said. But we want to go a little bit further and find out what is the best temperature which optimizes the production of this what is it called as monochlorobenzene. So, what monochlorobenzene f m b equal to f m b equal to f b 0 times x 1 minus of x 2 or that is equal to f b 0 multiplied by K 1 tau divided by 1 plus K 1 tau minus K 1 tau divided by 1 plus K 1 tau K 2 tau divided by 1 plus K 2 tau. Now, we have to differentiate this that means this we have to maximize this means we need d by d t of x 1 minus of x 2 and we have to set this equal to 0 and find the condition under which this difference goes to a maximum from very fundamentals as well we should recognize that as benzene goes to monochlorobenzene monochlorobenzene goes to dichlorobenzene. So, therefore, this intermediate there would be conditions under which this intermediate would be maximized and our interest is always to find conditions. So, that our product of our interest is as high as possible that is our interest therefore, we like to see what is the temperature at which this x 1 minus of x 2 goes through a a maximum. Now, it is not very difficult to do this differentiation you know it is not messy, but there is some algebra involved etcetera. Therefore, I prefer this condition it is not difficult to show this that this condition becomes that this is the condition under which under which you will find that this goes to maximum. What are we saying that the maximum value of x 1 minus of x 2 are monochlorobenzene attained when you choose when you choose your temperature of operation you have to choose T temperature operation. So, that even see even in E 2 or the activation energy of the reactions 1 and 2 or well documented. What is not clear from here is what is the temperature which sort of when this equality becomes applicable. So, that we have to put the numbers for k 1 k 2 tau this everything and then that will give a value of k 2 and once value of k 2 is known suppose this this particular equation gives us k 2 and then from that you can find the temperature. Once you know k 2 then you can find temperature corresponding to that k 2 because E 1 and E 2 are the activation energies are known. So, what are we saying what we are saying is that if you have a reaction taking place in a stirred tank if it is a series reaction there is a choice of temperature at which this the desired product goes to the has a maximum value it takes the highest value. So, this temperature T tells you the conditions under which the intermediate which is monochlorobenzene goes through a maxima. The next question of course, is that how do you achieve this which means what is the question is that you have found out from this from our calculations and so on. And that you know you have to operate this at some temperature T at some residence time tau. Now, given this temperature T and residence time tau how do you ensure that you know all these are possible that we come by looking at the energy balance. So, this is the heat generation plus q minus of w of course, there is no work done. So, what we are saying is that if it is at steady state if it is at steady state then the heat taken up by the fluids this coming in and going out. So, this is the heat generation. So, you must supply so much of heat. So, heats heat to be supplied is simply what V naught C p T minus T naught minus sigma i equal to 1 to 2 r i times minus of delta h i star V. So, what we are saying then is that for a for a series reaction like ours once you have once you if you want to choose a temperature choose a residence time then you will have to adjust you will have to ensure that this amount of heat can be added or removed. So, that you can get the conditions of your choice. So, this is what been said keeping in mind the requirements of the process. So, just to sort of summarize what we are trying to say here is that there are several new considerations of energy balance that we must bear in mind. So, what we have try to do is that we have try to write the energy balance and then try to see carefully the right hand side of the energy balance to see how we can make better use of the available facilities to to reach our our requirements. So, we consider case one in which the reaction was instantaneous. We said if it is an instantaneous reaction then the composite the extent of reaction is governed by Euclidia. And therefore, you can use that condition to eliminate the reaction rate functions from the equation. And therefore, we get answers which we can handle and quite comfortably. The second case we consider was that case in which the reactions are reversible yes, but they are not instantaneous. In such situations what we said is that it could so happen that the heat additions is at a rate or heat removal is at a rate which is so small compared to the reaction rates. Therefore, the reaction just does not move it only moves to the extent that you are able to supply or remove heat. This is the second case that we considered. The third case we have considered here is that it is a disturbed tank in which you have a very very rapid reaction occurring where k 1 and k 2 are very large. But these are irreversible reactions in the case we have considered. So, how did we handle that? We handle that by recognizing that our material balance equation we can eliminate one of the rate functions rate constant in rate constants. So, that we can solve that to find out the values of residence time multiplied by rate constant k 1 tau or k 2 tau which is able to give us the final result of our interest. So, these are the three cases we considered by which we are able to better get better insights into the energy balance. So, we will consider one more exercise. This is a multiple reaction. What is this reaction? Let me just explain. You have a stirred tank. You have a stirred tank. There is this reaction m going to p 1. This is called initiation, the polymerization initiation. Now, p 1 reacts with m to give you p 2. This is the polymer. p n reacts with m to give you p n plus 1. This is also polymer. R m which is the rate of formation of this is k i m plus sigma i equal to 1 to n k p m p i. This is the rate at which monomer is consumed. And then the rate of production of polymer R p is given as k p p n m rate of formation. So, our job here is to set up equations that will tell us how the reactions will move forward. That means basically set up equations that will tell us what is p n and then what is x and so on. So, we want to find out based on our material balances what will be the amount of polymer that is formed at any instant of time and so on. So, let us write all the equations so that we can answer many questions not just. Now, the context once again let us not forget the context. The context is that when you look at polymerization because polymers are very long molecules. They are pretty viscous. And of course, we can handle viscous in plug flow devices through appropriate mechanisms. But it is easier to handle very viscous fluids when you have a stirrer. If you have a stirrer you can the stirring itself takes care of you know many of these are shear thinning in the sense when you start shearing it the viscosity comes down. They were much easier to conduct the reaction when it is under stirring. So, a stirred tank or a CSTR for a polymerization reaction is a convenient. And plus we can set up equations for p n and find out conditions under which p n goes to the highest value. So, when do we get the highest value for p n also we can find out through our equations. So, let us try to do one by one. Let me write a monomer balance what is the monomer balance you have input you have output plus you have monomer. Now, which can also be written as this form that is minus tau k i m plus sigma k p m p i equal to 0. Now, is this clear I have divided throughout by volume therefore, it becomes. So, it is it should be some b naught only when you divide throughout by volume it becomes tau. So, I will call this is equation 1 what is it saying that monomer input monomer output residence time k i is the initiation rate multiplied monomer concentration. And these are the all the polymerization reactions k p multiplied by m. So, it takes care of the i th monomer multiplied by m. So, let me simplify this where it becomes m naught just help me m minus. k i tau m first term is taken minus within brackets k p tau p 1 plus k p tau p 2 k p tau p n multiplied by m equal to 0. So, this is some equation 1 naught 2 whatever. So, we have write the monomer balance we write one of the polymer balance. So, that we can do all the simplification. So, let us write balance for p 1. So, you have p 1 p 1 0 is not there p 1 plus k i tau p 1 tau m minus of k p tau m p 1 equal to 0. I hope this is clear. So, this takes care of p 1 plus k i tau minus. So, we just sort of simplify and write say p 1 multiplied by 1 plus k p tau m equal to k i m tau. This is correct p 1 multiplied by 1 plus k p tau m into p 1. So, you have taken to the other side fine very good very good or p 1 is simply k i m tau divided by 1 plus k p m tau. It is also can be written as k i by k p within brackets of k p m tau divided by 1 plus k p tau m. I have done nothing very new here please. This is very elementary material balance only club terms containing p 1 together and then k i and then I have just divided and multiplied by k p nothing very fancy has been done. So, this is p 1. Now, the idea of putting it in this form we will very soon see why this useful. Let me write for polymer p 2. So, we want to refer p 2. So, what is p 2 plus k p tau m p 1 minus k p tau p 2 m equal to 0. Therefore, p 2 multiplied by 1 plus k p tau m equal to k p p 1 m tau. Therefore, p 2 equal to k p p 1 m tau p 1 minus k p tau m tau. So, divided by 1 plus k p tau m. I am not doing anything new here please. It is just elementary material balance for a stirred tank. So, that p 2 becomes p 1 k p m tau divided by 1 plus k p m tau. This is all right. This notice here p 1 and p 2 are equal to 0. So, this p 2 are related by similarly we can write you know for others. So, we write this. So, this is also we can substitute for p 1 from here. So, we can write this as k i by k p you can see here. So, this becomes k i by k p within brackets of k p m tau whole squared divided by 1 plus k p m tau whole squared. So, k i by k p k p m tau 1 plus k p m tau p 1 for p 2 it is squared. Therefore, you can write for p n. The object of sort of doing this form is only to show that. So, if you want to look at p n p n becomes k i by k p k p m tau to the power of n to the 1 plus k p m tau to the power of n. So, this is the idea of showing it in this form. Therefore, the n th polymer which is given by this equation. Now, let us just try to understand what I am saying. Try to understand this. What we are trying to say here is that in stirred tank polymer 1 polymer 2 polymer 3 polymer of n all of them depend on k i by k p multiplied by this term k p m tau divided by 1 plus k p m tau k p is polymerization reaction rate constant m is the concentration of monomer and tau is the residence time. So, it is in a sense that we can plot these functions we can plot these functions. So, p 1 versus tau p 1 similarly, we can plot for p 2 and p n and so on. And we can actually choose the residence time which is appropriate to our interest. So, essentially what we are trying to say is that we are able to now adjust the conditions of the stirred tank. So, that our product of interest is maximized. I think this is the point that is and stirred tank gives you the advantage that you can stir and maintain a good level of shear. So, that the processing can be continued number 1 2 it also ensures the good heat transfer. So, that whatever heat is to be added or removed becomes facilitated. So, two important features that are so crucial to success of a process stirred tank is able to achieve for you that is where our big interest comes. Now, the rest is fairly straight forward, but just to for the sake of completeness I will just write the final form it is very easily shown that the extent of reaction extent of reaction x is you can show this I will not show this minus of 1 minus of x times k i tau within brackets of 1 plus k p tau m naught into 1 minus of x equal to 0. On other words you can determine x that means x is determined is determined from this equation. So, this is the monomer balance this is the monomer balance monomer balance. So, what we are saying is that we are able to determine what is the what is the overall conversion we are able to determine what is the amount of different products which are formed. And therefore, you can now decide how best to make good use of your monomer in relation to the polymer that you are forming. So, just to cut this long story short what we have said is that this whole idea the whole idea of looking at energy balance from the point of view of trying to see it is new features. We have looked at 3 or 4 new features all are which tells you how the energy balance equation is able to appropriate it take care of the processes that we are interested in carrying out. Thank you.