 We will get started with multiple reactions. We have already shown the following result just quickly state the result. There if you have a multiple reaction like alpha 1 1 a 1 plus alpha 1 2 a 2, alpha 1 n a n equal to 0, alpha 2 1 a 1, alpha 2 2 a 2, alpha 2 a n equal to 0, alpha p 1 p 2 a 2, alpha p n a n. So, there are p reactions, p rate processes and components and are independent reactions. When we said independent reactions what we meant is that this system requires r quantities to be specified to be able to characterize the system as it changes in the course of the reaction. Now, what we want to do now is take one example and then understand what we have done so far. So, to write this stoichiometry we said we can write a j minus of a j 0 can be given as some x 1 alpha 1 j x 2 alpha 2 j up to x r alpha r j. Sometimes we normalize it with respect to a reference in which case each of these x's have no dimensions. If you do not normalize each of them will have the same units as h a depending upon how we do the exercise. Based on this we are able to write the design equations for various equipments including batch CSTR and so on. So, now we will take an example to see how to make use of this result in our designs. So, for that what we have is this is a multiple reaction a goes to b, b goes to c and c goes to a. So, I have written 1 and minus 1 showing the forward and backward reactions and we are trying to do this in a stirred vessel or a CSTR and some numbers are given k 1, k 1 is simply k 1 by k sorry k minus 1 similarly k 2 and k 3. All reactions are instantaneous all reactions are instantaneous and if a enters at this rate what is the composition here what is f a, f b and f c this is what we want to calculate. And we want to use this procedure plus we also want to use procedure which we think is common sense and see how best all these procedures come together. Now, to do this what we normally do is the following. Suppose these reactions were not reversible the a goes to b, b goes to c, c goes to a assuming they are not reversible that means there are only 3 reaction sets which is written in this form b minus a equal to 0, c minus a b equal to 0, a minus c equal to 0. Assuming that they are not reversible then the matrix the what is called as the stoichiometric matrix looks like this. Now, if you take only the reverse reaction then the stoichiometric matrix looks like this. Is this clear what we are saying we are only a goes to b, b goes to c, c goes to a assuming that is not reversible then this is the reaction this is the stoichiometric matrix. If only the reverse reaction is occurring that means b goes to a, c goes to b, a goes to c then this is the stoichiometric matrix. Now, we can actually do the determinant and find out for ourselves in a very simple determinant we can do right now find out the rank of this matrix please find the rank of this matrix minus 1 multiplied by this please find the rank just have to see what is the determinant. If you find the determinant of this matrix you will tell you it is 0 you can just expand it and find out for yourself the determinant is 0. So, which means that this matrix is singular showing that you know all the three reactions are not independent only if you take any second order matrix for example any two we can see it is independent. So, showing that there are two reactions are independent is this clear how to find out. Now, you can combine forward reaction and backward reaction and write a much bigger matrix and also try to see what is the rank of this matrix it is it will take much longer time, but the fact is that is very obvious that the rank is still 2 it is not going to change the rank correct. So, the rank of this matrix whether it is single reaction or the reverse reactions and so on the rank is 2 therefore, to be able to understand what happens to this reaction set in a process you only have to take two of these three reactions any two you can take and write your stoichiometric. So, what I want to do now is do the same thing in two or three different ways just to illustrate how things play out. So, I have taken first independent set A goes to B B goes to C this is our independent set example first example. So, that what is F A what is F B what is F C assuming that X 1 is reacting here I have written F A is F A 0 times 1 minus of X 1 if X 2 is reacting here. So, I have written the difference in F A 0 X 1 is formed F A 0 X 2 is reacting and therefore, F B is so much similarly, F C whatever is formed due to this reaction. On other words what I am saying is that if this is X 1 and if this is X 2 then this is the stoichiometric table which tells us how much is the material flowing at any point in the equipment. Now, all these reactions the instantaneous means what we can say that each of those rate processes are in equilibrium. So, this kind of relationship should hold which means what A and B are in equilibrium B and C are in equilibrium. Therefore, C B by C A is K 1 C C by C B is K 2 and therefore, we can calculate what is I mean I have just written down what is C C and what is C B all that we can just look at here. So, F B is F A 0 times X 1 minus X 2 F A is this F C is this and F B is this. So, this tells you 2 equations in X 1 and X 2 involving K 1 and K 2. So, you can find K X 2 from 2 this equation you can find X 2 straight you can see straight away it is K 2 K 2 X 1 divided by 1 plus K 2. And you can take it forward and from equation 1 if you look at this equation and then we can resolve this and say that X 1 is this X 1 is given by this relationship X 2 we have already shown X 2 is equal to K 2 X 1 divided by 1 plus K 2. So, this comes from the equilibrium relationship. So, since K 1 is given as 1 K 2 is given as 2 you can find out X 1 and X 2. So, what are we done we have there are 6 rate processes even then we have taken only 2 reactions we have taken arbitrarily A going to B B going to C. And then we set up the and then we got the results and once you know X 1 and X 2 you will find calculate F A F B and F C. Now, we can do the same thing in a slightly different way. So, that means, what I am saying is that let us say as an example in this in this exercise in this exercise in this exercise I said F A 0 is 10 that is why we set up all this where F A 0 is the basis. Now, suppose I say no it is F C it is actually material coming in F C is 10 moles per liter F A and F B are 0 what changes how do you formulate the same problem what I have done is let us say C goes to A A goes to B. So, the F C is 1 minus of X 1 F A is X 1 minus of X 2 F B is F C 0 times X 2 is the logic clear. Instead of the previous case where I said A is 10 mole per liter mole per second now I am saying same system F C is 10 mole per second. So, stichometry does not change but we write the stichometry in this form how much is C F C 0 multiplied by 1 minus X 1 why is it 1 minus of X 1 I have taken this as X 1 I have taken this as X 2. So, you can write how much is A is this is formed here and consumed here therefore, I put X 1 minus of X 2 and B is formed in reaction 2 therefore, plus X 2. So, once again you can go through this is solve but when you try to solve this you find the way it plays out is not identical to what we have done because now. So, I have solved this here just look at I have just taken this C B by C A is K 1 and then C C by C B is K 2. So, you have these two equations X 2 by 1 minus X 2 equal to K 1 and then 1 minus X 1 by X 2 it comes from here only both. So, you have K 1 equal to X 2 by 1 X 1 minus X 2 and K 2 equal to 1 minus X 1 by X 2 two equations you can solve. So, when you solve this what you get here is this X 1 turns out to be this X 2 turns out to be this. So, it is fairly elementary you know this is equation 1 this is equation 2 X 2 and X 1 you can solve this which I have done. So, you got here X 2 is this X 1 is this once you know X 1 and X 2 your stoichiometry is already set out. So, you can calculate this is the stoichiometry what is C going to be A going to be. So, you can calculate what is F A F B and F C is this clear procedure is the same in one case F A 0 is coming in another case F C 0 is coming in that is the only difference. Now, when you do when you substitute values of F A and F B and F C values of X 1 and X 2 the what you get for F A F B and F C is these 2.5 2.5 and 5 adds up to 10. So, it is sort of satisfies the material balance also. Now, what we have done in these two exercises is that by looking at the form of the of the chemical reactions we have taken some things as our reference. First case we took A as our reference, second case we took C as our reference. Suppose you are dealing with the biological reaction where a thousand reactions are occurring very difficult even identify which is the reference species we should take it will not be easy to detect. So, you need something which is able to handle thousands of reactions how do you do this. So, let us do the same problem in a slightly different way. So, what is the problem we want to solve we want to solve for the case let us I have just taken this example C is coming in as 10 moles per second. So, I want to solve this problem I want to find out what is the value. So, F A F B and F C we already solve this problem we know the answers, but we want to repeat this using the general approach that we have given in our class. What is the general approach what we said was that if there are large number of reactions occurring let us say a j minus of a j 0 is x 1 alpha 1 j plus x 2 alpha 2 j up to x r alpha r j where r is the number of independent reactions. So, x 1 is the extent of reaction in first independent reaction x 2 second independent. So, what I have taken for this network a going to b going to c let us for the moment choose 2 reactions I have just taken b going to c c going to a as an example you can choose any. So, there are 2 reactions. So, this b going to c this is x 1 the extent of reaction is x 1 c going to a extent of reaction is x 2 correct. So, when you write a j minus of a j 0 as x 1 alpha 2 j now I want to write the stoichiometry for component a component b and component c. Let us try to do that that means a j minus of a j 0 is f a minus of f a 0 what is x 1 alpha 1 j here for component a what is alpha 1 j alpha 1 j is 0 what is alpha 2 j plus 1 plus 1. So, f a minus f a 0 equal to 0 plus x 2 now f b minus f b 0 f b is in reaction 1 f b has a negative sign minus 1 reaction 1 b occurs in reaction 1 it is minus. So, I put minus x 1 x 2 is it does not occur in the second reaction. So, I should put 0 now go to the next one f c f c f c minus f c 0 equal to x 1 alpha 1 j c is plus 1 x 1 and then x 2 alpha 2 j c is minus 1. So, it becomes x 1 minus of x 2. So, what we have done now we have taken a general case b going to c c going to is we have just selected 2 of the reaction that occurring what I am trying to say is that in your when you are dealing with large number of reaction this is what you will do you just select some reactions which you think is an independent set based on your matrix analysis having done that we have written the stoichiometry. Now, the problem specifies that the reactions are in equilibrium. Therefore, c b by c a must be f b by f a is k 1 what I have done is please you point is well taken what is given to us is a to b is k 1 b to c is k 2 it is given I can I can calculate the rest I have not done that. So, what is given is that equilibrium constant a to b is k 1 that means c b by c a is given as k 1 and b to c is given as k 2 which is c c by c b is given as k 2 that is why I did not make any change there because that is what is given anyway. So, it this does not alter the result anyway. So, what is given is c b by c a is k 1 c c by c b is k 2 this is given to us the now you have to write this in terms of our stoichiometry what is our stoichiometry it says minus of x 1 f b by f a is minus of x 1 by x 2 is k 1 is it all right minus of x 1 by x 1 by x 2 is k 1 and then we have f c by this is f c by f b correct. So, f c is what f c 0 plus x 1 minus of x 2 and then divided by f b is minus of x 1. So, we have 2 equations 3 and 4 which to solve for x 1 and x 2 can you help me please solve for x 1 and x 2 for the case where minus x 1 by x 2 is k 1 and then this is equal to k 2. We can solve for x 1 and x 2 now please solve and tell me the result I will write the result just you tell me whether this is I am just write the result and then you can check and then the other one also you have to do. So, this is the result that I get x 1 is minus of k 1 x 2 1 plus k 1 plus and x 2 is this do you all get this x 1 also is correct x 2 x 2 involves f c 0 anyway do you all get this x 1 x 2 it is important because it is something important to be said that is why I am saying please remember x 2 x 1 comes from k 1 x 2 this comes from here minus not x 2. So, this is the result I get this is correct this is correct x 2 is what this is what I get x 2 this is correct. Now, we know x 1 and x 2 correct x 1 is known x 2 is known. So, this result is correct. So, please find out what is x 1 k 1 is given as 1 k 2 is given as 2. So, we can find out x 1 and x 2 what is x 1 what is x 2 x 1 is where are we minus 2.5 x 2 2.5. So, what is f a b and c f a f b f c this is the result I get f b f c f c is 5 we had solve the same problem little earlier if you recall and then we got our results. So, I am just showing those results see we have f a 2.5 f b 2.5 f c 5 correct this is the same result that that we have got last time when we assumed we did the same problem slightly differently, but the results are the same. So, what I am trying to put across to you is that frequently we find it convenient to choose a basis and it is generally as long as this problem is small when the problem is very large number of reactions are very large you will find that this technique of choosing a basis is not very convenient. Therefore, we will have to go with this general procedure where all these are not important you just simply select a set of our reactions which are independent. How do you do this we have done the matrix analysis we can set select the our independent reactions from our matrix analysis. Once you have done that you can set up this stoichiometry and then proceed with this some of these excess may be negative nothing to worry because the choices are like that some of them may be negative some of them may positive it does not matter as finally, all the numbers that you get would all be perfect just like here f a f b f c you know although our 1 x was negative, but f a f b f c were consistent there is no problem. So, what I try to do through this exercise is that the general procedure that we have set out may be a little inconvenient for small problems because the basis is not obvious, but the advantage is really in very large systems particularly if you are dealing with biology by the reactions are very very many very useful what is meant by instantaneous and overall yield you know this is something that comes in extremely useful when you are doing what is called as process development you see frequently what happens is that you know you have to set up a process and then you are looking for a desired product and. So, many side reactions do occur and therefore there is an undesired reaction and now when there is a desired product there is an undesired product then clearly you need a way by which you can maximize whatever you have in your mind whatever you want to maximize. So, this whole procedure is trying to set out an experimental procedure by which we can maximize our objective whatever that objective may be. So, do this what has been said that a goes to desired product a goes to undesired product therefore, when we conduct this reaction in a P of r where therefore, I mean we can write d f d by d v is r d the rate a similar d f a by d v is r a that means rated which f d goes to r d and f d to r a. So, this ratio gives us what is called as r d by r a that means rate of a formation of desired product to the rate of consumption of the raw material is the ratio of the reaction rates r d and r a on other words the advantage of this way of looking at the whole problem is that the right hand side r d to r a is a state function. The right hand side is a state function therefore, I mean this can be determined in various ways because the state function and therefore, it is denoted as minus of phi to indicate that r d and r a have opposite signs this is consume this is formed just to keep it positive. So, we have this phi minus phi where phi is a positive number. So, phi has a meaning of instantaneous yield because it is occurring as ratio of two reaction rates that is why it has a meaning of instantaneous yield. We can integrate this and represent it like this or what I am doing now is simply integrate multiply this side d f d you multiply phi by f a and integrate both sides. When we do that what we get here is d f d which is f d minus f d 0 when we integrate this d f a because of this relationship f a is f a 0 1 minus of x the right hand side becomes f a 0 of integral phi d x a where phi is the instantaneous yield. Frequently our interest is what is the overall yield by definition is how much is the product formed divided by how much is the raw material consumed that is how we look at overall yield correct. So, product formed to the raw material consumed. So, overall yield if I call it as capital phi it is raw material formed sorry product formed raw material consumed. So, that is can be represented as this f d minus f d 0 what is this term it is coming from here. So, it is f a 0 times phi d x a and this term is f a 0 which is given as x a f. On other words what we are saying is that the overall yield that you will get from a process is an integral of phi d x a where phi is a state function. And phi d x a integral 0 to x a divided by the overall yield that gives you the overall yield. This representation will see shortly it has several advantages. So, what we are saying here overall yield in a batch in a in a plug flow kind of device is integral of phi d x a divided by x a f. Now, we can do the same experiment in a in a CSTR. So, you have input output generation equal to 0 input output generation equal to 0. So, we can take these two ratios r d by r a equal to f d 0 is this correctly written please check have written it correctly f d minus f d 0 equal to minus of r d f a 0 minus f a equal to minus of r a is it correctly written yes is it fine with all of us. Now, what meaning can we attach to this r d by r a we said it is instantaneous yield. What meaning can we attach to the right hand side overall yield what are we saying in a stirred tank instantaneous yield is equal to overall yield in a stirred tank single stirred tank instantaneous yield is same as overall yield while in a in a p f r instantaneous yield overall yield comes as an integral of phi d x a divided by x f. On other words overall yield shows you an average in the equipment in a p f r while here it shows up directly this is something that we all know. Now, let us look at at this system carefully phi what is phi phi is a ratio of r d to r a correct it is instantaneous yield suppose I want to measure what will I do I will run a CSTR experiment I can run a CSTR experiment. So, that I measure r d to r a simply by this ratio because I can measure f d n f a and therefore, I know the right hand side therefore, I can make a plot of phi versus x. So, if I want phi versus x data what would I do I would run a CSTR experiment is this clear to all of us if I want to get a data on phi to x what will I do because CSTR directly gives me phi and this gives you x also and phi. So, CSTR gives you good data on selective instantaneous yield versus conversion. So, you are able to plot phi from your CSTR data x versus phi correct is this clear. Now, what is this data minus a 1 by r a versus x can we get the data from the same experiment the same experiment there are 2 x c you are doing an experiment. So, you are able to measure f a 0 you are able to measure f a you are able to measure r a therefore, your 1 by r a versus 1 x is also known to you because it also comes from the experiment. So, a CSTR experiment gives you 2 valuable results 1 result is that it gives you minus a 1 by r a versus x second result it gives you is that how phi is related to x is that clear. So, both the data comes out of your CSTR experiment on other words what we are saying now is that through your CSTR experiment you can generate data on the system you are trying to investigate without any knowledge of the chemical kinetics because the chemical kinetics in many cases are so complicated that the effort involved in getting the that function is not easy. But this may not be so difficult because this is easier experiment to do. So, you can generate phi versus x data and minus a 1 by r a versus x data. Now, you have all the data in your hand various kinds of questions associated with reactor design reactor optimization etcetera can all be answered now because you have the data and you have of course, experimental points therefore, most of your numbers will come out of graphical integration kind of answers, but in any case it is supported by experiments. So, with this suppose I now ask you suppose we have a combination of reactors which means what you have a stirred two stirred tanks put together. So, let us see what is the overall yield that you will get when two stirred tanks are together what we are trying to say is that we already have this data phi versus x this data is already with us. Now, we want to see how we can make use of this kind of data by a fundamental understanding that we have about stirred tanks. Now, what I have got here I have written for tank 1 input output generation equal to accumulation this is for desired product input output plus r d I have written on the right hand side. Similarly, for component a input output equal to minus r a 1. So, we can take a ratio you get the result which you already know what is that f d 0 minus f d is given by this result showing that the overall yield is same as instantaneous yield something that we already know I am stating it once again for tank 1 our result is that phi 1 equal to small phi 1 showing the overall yield in tank 1 is same as instantaneous yield that we already said we just stating it once again. Suppose, let us go to tank 2 what does tank 2 cell as input output plus generation I have taken it to the other side input output I have to outside. So, this is the component a that ratio if you take what does it tell us it tells us that f d 1 f d 2 f a 1 f a 2 r r d 2 by r a 2 what is the meaning of r d 2 by r a 2. Because, it is the stirred tank it is a stirred tank it operates at the exit condition therefore, r d 2 by r a 2 is phi 2 it is phi 2. Therefore, f d 1 what I have done f d 1 I have replaced it from here f d 1 we have got already here f d 1 comes from here. So, I have just replaced it from the previous page. So, what you get here is that f d 1 in terms of phi 1 x 1 f d 0 stays f d 2 equal to phi 2 x 2 f a 0 x 2 minus of x 1 is this clear to all of us how this phi 2 f a 0 x 2 minus of x 1 comes. So, what it means is that now f d 2 minus of f d 0 now comes in terms of phi 1 x 1 phi 2 times x 2 minus of x 1 you understand. So, what we have now said is that the product that we are making in the second tank f d 2 is the product that comes out of the second tank f d 2 is the product second tank that depends on f a 0 phi 1 x 1 phi 2 f a 0 x 2 minus of x 1. So, it is now expressed in the convenient form in a very convenient form which is what overall yield equal to phi 1 x 1 plus phi 2 times x 2 minus of x 1 that means if you have this function here overall this is the data that you have got from our experiments correct. So, in the first the overall yield that you will get in a 2 tank sequence is what phi 1 x 1 that is this triangle this rectangle plus phi 2 times a phi 2 is this point phi 2 times x 2 minus of x 1 this rectangle you understand see this very nice procedure which is developed by process then b in 1940s. What is being said is that if you generate this curve of phi versus x then you can use the same procedure that we have used for you know staging and multi you know reactors I mean size of the equipment that we have done for a long time the same procedure applies here also. That means if you want to find out the overall yield of a 2 tank sequence. So, first tank gives you this area second tank gives you this area is that clear. So, same procedure applies this is what is the interesting point of this procedure that has been set out this is clear this is for a 2 tank sequence. Now, let us look at one more example what is this example you have a CSTR followed by a PFR. So, what is the overall yield now based on whatever procedure we have set up straight away you should tell me the overall yield will be the first will be this area second will be this area that means phi 1 x 1 plus integral phi d x a by x 2 same procedure what we have said earlier here we had a 2 tank sequence this is a 2 tank sequence. So, we said that it will be this rectangle this rectangle and this rectangle. That means second rectangle is constructed on this on this first rectangle is constructed on this point, but if you have a CSTR followed by a PFR it is what CSTR is phi 1 x 1 this rectangle you can see this rectangle and then second one is integral this integral the integral of this area here phi d x a divided by x 2 which is the final conversion. So, this is exactly what we have learnt in the previous I mean reactor design approaches now if I ask you what is the overall yield for the case of this you will simply say this is the answer. So, various combinations for which we want overall yield we can straight away get because we know this data. So, this is what then we pointed out that when we have most of this results came out when they were trying to develop explosives during the war where kinetics etcetera are very difficult to do number one and under the urgency it may not have been worthwhile. So, that what we wanted was a design which would actually produce the product of your interest. So, in most cases you will find that kinetic data comes after a very long time, but process data that is required for design this is sufficient for you to give you the information that is required. So, with this background I want to quickly look at this exercise. So, we have an exercise here where this is also taken from del B only what it says is the following this is read out what is the problem statement is that experimental data on a reaction system has been found to look like this, but the instantaneous yield is related to conversion by this functionality. Now, how does it come we have said just now it comes out of an experiment CSTR experiments are relatively easy to do even if it is an explosive reaction it is quite easy to control temperature and make measurements this is the great advantage of CSTR. You can measure even very difficult reactions you can manage to measure, because you are able to maintain temperature quite well through an appropriate cooling. So, 5 verses x data is given to you now what is asked is if this reaction is to be terminated when 5 is 0.5 that means this is when 5 becomes 0.5 you want to terminate this reaction why do you want to do this, because beyond that it does not give you any benefits in terms of the product of your interest. So, what is the overall yield in a batch reactor is this question clear if the reaction is to be terminated when 5 instantaneous yield 5 is 0.5 what is the overall yield to be expected in a batch reactor is the question clear to all of us what is the overall yield to be expected in a batch reactor. We said that just now if you have a PFR our overall yield is integral 5 d x a divided by final conversion if it is a batch reactor also it means the same thing 5 d x a divided by x a final conversion. So, overall yield in a batch reactor is simply integral 5 d x a divided by final conversion this is all right what we are saying. So, this function that is given to you this function that is given to you it says what is the overall yield in the batch reactor how would we approach this 5 is given as 0.5. So, what is the expression for overall yield what is the expression for overall yield we say it is integral 5 d x a divided by x f correct. So, how do we do this what what will you get what is what is the value of x point how do you find out we put 5 equal to 0.5 in this that tells you that we have to stop at x equal to point whatever that number is 0.45. So, once you put x equal to 0.45 that tells you 5 becomes 0.5. So, you know the value of x. So, how do you find overall yield. So, we find overall yield by our same expression that we have derived overall yield is 1 by x 0 to x 5 x d x this is all right what we are saying. This comes from our understanding that 5 d x integrate 0 to the final value divided by the final is the overall yield for a p f r or for a batch reactor is this all right. So, I this result correct what I have got 0.71 can you please tell me let us go to the next. Suppose, we want to do the same thing now second exercise is if this reaction is carried out in a two tank in series there are two tanks in series 1 and 2 series what conversion in the effluent from tank 1 would lead to highest overall yield. What are we saying let us let us just recognize this we have this is our this is our 5 verses x curve it says it is a two tank sequence. We want to terminate this somewhere in between. So, that define what what is it is a what conversion in the effluent from tank 1 would lead to highest overall yield what would be our strategy see basically overall yield is determined by the area. Tell me what choice of the intermediate point will give us the highest area what choice of x 1 would give us the highest area let me let us try to answer this now. So, this is our function. So, this is how the curves looks like what choice we want to make a choice somewhere here. So, that the area the overall yield is simply area correct, but if it is since is the stirred tank does it say stirred tank it says two tanks two tanks both the stirred tanks both the stirred tanks correct what shall we do we want to choose x 1. So, what I have done here is that we do not know what is that. So, what I say all right we know that phi is given by I should write x 2 here phi x 2 I forgot to write. So, phi x 2 equal to phi 1 x 1 phi 2 times x 2 minus x 1 this is the statement of the overall yield. So, we want to maximize what it says is that we want to have choice of x 1. So, that the left hand side is maximize x 2 is fixed. So, suppose we differentiate d by this overall by x 1 which do not know what is x 1 and set it equal to 0 then you can find out the value of x 1 at which this goes to a maxima is this clear what we are saying. So, x 2 is a constant. So, it therefore d phi d x 1 this is the right hand side d phi d x 1 actually x 2 can stay here this does not create any harm. So, let us differentiate the right hand side with respect to x 1 how many terms you get x 1 d phi 1 d x 1 second term is x 1 and then when you differentiate this you get minus phi 2. So, you have to set it equal to 0 what is d phi d x 1 x function is phi function is given here phi function is given. So, you know d phi d x 1 is that clear. So, d phi d x 1 is known therefore, you should be able to tell me now please tell me the result I get a value of 0.286 please tell me whether this is correct. I have done this I have done this d phi d x 1 equal to 0 which is equal to x 1 d phi 1 and x 1 plus phi 1 minus phi 2 I have put in the value for d phi 1 d x 1 all the numbers etcetera I have done this here this is the result I get please tell me whether it is correct. This is result correct it is please this differentiation is all that you have to do see what I have done yes or no this is the differentiation is all right. So, this comes from here x 1 x 1 d phi d x 1 that function is known. So, I have put all these numbers here 0.286 is correct. Now, that means the highest value for phi we get when you choose x 1 as 0.286. So, what is that value? So, question now this what is that overall yield x 1 is 0.286 x 2 is given as what is x 2 0.445 whatever the number this what I have got. So, I get totally 0.669 please see if it is correct phi 1 x 1 x 1 is 0.286 phi 1 function is given and then this phi 2 times x 2 minus x 1 which is given here is it 0.669 is what I get. So, what is so complicated about this calculation? So, what are we saying that if you have a two tank sequence and the best choice is to maximize the overall yield function with respect to that intermediate point that is the point that is being said. The next result next point which you want to answer is frequently in our problem is not just maximizing the instantaneous and overall yield. You also may have some constraints on what is the extent of reaction that you have to reach. I mean after all we cannot throw a reactance correct we have to make use of them. So, if this exercise is about if the effluent in the plant is to correspond to 50 percent conversion which means we must make use of 50 percent of the material. What is the highest overall yield which can be obtained? It is question clear. So far we are concerned only about maximizing the overall yield. Now, what is being said is that no there is another aspect which is important that we want to make at least 50 percent conversion is required. So, what is the maximum highest yield that you can get? If you want to achieve 50 percent yield what is the highest yield you can get? Let me just state this in a slightly different way. What is being said is the following. You want to choose a reactor we do not know what reactor to use. You want to choose a reactor such that the overall yield conversion should be 50 percent 50 percent you should and then the what is the highest overall yield we can get. That means you are going to stop at 0.5 this is fixed our conversion is 0.5. Now, we have to choose the reactor system. So, that the area under the curve is the highest is that point clear. We have to stop the system at x equal to 0.5, but now we have to choose what reactor can we choose see let us do a small experiment. Let us say I do one here. So, what is the area? This is the area correct yes or no? Now, let us do one more experiment here let me. So, very clearly if you choose at the maxima at the maxima and then here a p of r is probably be the x equal to 0.5 and this maxima some where it occurs we will have to find out this is x 1. So, this will give us the is this clear what we are saying. So, this is. So, this is a p of r and this is a CSTR. So, in this case this is phi and this is x. So, what we are try to say in this exercise is that when we have a reactor process development problem frequently this procedure would be extremely value. Because you can construct a start tank and therefore, construct this curve phi versus x even if it is a very complicated reaction. Once you have this curve in your hand now we can play around with how to put your equipment. So, that it will give you what you are looking for. So, there might be instances where this curve will go like this and your force to go up to x equal to 0.9. I mean because of pollution control issues and so on, but this is something that we accept nowadays you know you cannot allow our reagents to go into the ground. So, we accept all this. So, all the issues related with the reactor design answers can be found out of this. So, this is the advantage of this procedure which was developed of course, long back very useful particularly when you are working in a development laboratory. I will stop there.