 we get started with plug flow recycle reactors. Today, I look at some illustrative examples just to say how this system will behave. So, our system we have said earlier looks like this you have a plug flow vessel here it may contain a catalyst or it may have a inert for through which you have reactance of flowing. And there is a recycle that recycle ratio is defined as F F 4 divided by F F 3 this is our recycle ratio. An example that of importance to us is use of auto catalytic reactions in in mix flow because mix flow is a form in which this context becomes. So, what I have got here is there a design equation for C S T R where the reaction is occurring where the rate function is given by C A C B and the reaction is A it is auto catalytic. So, A giving B, but it is auto catalytic therefore, it depends on both A and B. Now, point that I had wanted to get across to you is that when you look at this equation you find that even if C B 0 is 0 the size of the equipment is finite. The reasons it is not obvious from here why is it that even though you do not put C B 0 in the feed that the reactor gives you a finite result because you would expect that there should be C B 0 in the feed. Now, we explain this so just explain this once again just to put it in the context to understand this what we said is that let us look at the unsteady state. When you look at the unsteady state what we have material balance for A material balance for B and this term which represents the reaction rate function when you put C A I as 0 and C B I as finite what we said is that as long as C A I because C A is coming in therefore, the reaction rate is 0 at 0 time, but as time proceeds since C A is coming in this function becomes finite. As a result what happens is that at steady state there is the effect of C B which is there when you started up the equipment. So, what you are trying to say is that as long as material is there when you start the equipment C S T R will always give you a finite result. So, that is on other words what we are saying is the C S T R is an instance of infinite amount of recycle. Therefore, the material that you have started at some time in the past history always lingers there however, small it may be that is the point that we must remember. Now, suppose we look at the other example which is a P F R for example, there is no recycle where you have material coming in material going out it is a P F R which means the recycle ratio is 0 and it has initially let us say certain amount of C A I certain amount of C B I as an example. Let us see what happens. So, I have just written the unsteady state equation for a P F R where del f del v is volume reaction rate and then this is the accumulation term I have written for both component A and component B. Now, if you look at the what is happening to C A at 0 0 means at the inlet at any time it is it means what is coming in a C A 0 similarly what is coming in a C B 0 and at any position inside at 0 time I have taken C A I and C B I on other words what we are saying is that we have a plug flow vessel which initially contains C A I C B I to which you start with C A 0 and C B 0 what happens to this process. What I am trying to say is that if C B 0 is 0 if C B 0 is 0 what will happen is that the reaction will proceed as long as there is some C B I, but this C B I will get knocked out after some time and moment C B I gets knocked out there is no further C B inside the process and therefore, this rate function term will always be 0 and therefore, autocatalytic reaction in a P F R will not take off because there is no C B whatever C B was there has been knocked out and as soon as the C B is knocked out the reaction stops. So, in a C S T R autocatalytic reaction always goes forward while in the case of a P F R autocatalytic reaction does not move because there is no product. With this let us quickly look at one or two small examples to understand how well how the numbers look like see after all finally, it is numbers that matter this small exercise what I have got here is that you have given data there is some data which is given which says there is a P F R in which this reaction A goes to B is taking place and experimentally you find out that it gives you a conversion of 0.66. It is an experiment you have done and you find that the conversion is 0.66. Now, you put a recycle pump and put it back that means the same reactor of volume V this is also same volume V now it is there is a recycle the recycle ratio is given as 4. What is the likely extent of reaction or likely conversion you will expect now of course, we know that when you recycle your conversion suffers because the fact that you know it is it is a measure of mixing. So, how do you find out to find out which we recognize something that we said a little earlier in the earlier classes that if you have a plug flow device the relationship between k tau P and x 3 by x 3 I mean x 3 is this at the end of the reactor. So, this we have derived already and similarly, if it is a recycle reactor then this is the kind of relationship between residence time rate constant and recycle ratio this also we have done earlier. So, what are we saying now the same reactor you are using with the recycle of 4 which means what tau P equal to tau r same reactor you are using as a recycle tau P equal to tau r. So, what it means that in this in this exercise in this exercise x a is given as 0.66 which means if you put x a x x a means x 3 is 0.66 you can find out k tau P which if you plug in here you will be able to find out the conversion you understand that means essentially the data for a PFR is given x 3 is given. So, you know k tau P since the same reactor is used k tau r is specified therefore, you can find out the conversion. So, I have done this calculation and the answers you get is not 0.66, but it is lower is what is to be expected. So, what is the message that we want to take from here is that when you recycle you will find that you lose some amount of conversion because the recycle actually is form of mixing and mixing reduces the extent of reaction. This is another example which we have taken here this I want to spend some time on this. So, what I have got here you have a reactor it is operating at 570 c most of this data is taken from an industry and it goes to a condenser the reaction is a plus b going to c and then it goes to a condenser where the condensable gas c is a condensable gas each condensers. So, this is condensate now you are putting a recycle after the condensate you are putting a recycle and then you heat it up because the reactor is 570 the condensate this condensate is operating at 45 c the data is given therefore, you have to reheat it and put it back into the equipment. Now, this is fairly common kind of device you will see in the process industry that there is a reactor and the reaction does not go to completion you try to recover whatever is useful to you and whatever is not useful to you put it back into the process something that we will do anyway correct. So, data given is that the rate constant is given by looking at the form of the rate function rate constant it is a second order reaction. So, it is k times C A C B is a rate function. So, rate function is k times C A C B and this is 1 kilo mole per hour this I made a mistake is 1 kilo mole per hour please make that correction not 1, but 1 kilo mole per hour vapor pressure of c is given as 0.2 atmospheres and moles of a and b are equal. So, 1 kilo mole per hour sorry 1 kilo mole per hour reactor is at 1 atmosphere recycle ratio is 4 this is always given because of this condensation what is the effect you anticipate in the process. Suppose this condenser was not there or if the condenser was there it is an irreversible reaction I am not saying it is irreversible it is irreversible reaction what is the effect that you would anticipate yes product will go back, but this is not a reversible reaction the reaction is not reversible. See the rate function says k times C A C B therefore, even if there is C present here it does not affect the reaction at least this is how it is formulated. So, the way it is formulated it appears that it does not affect I mean is that right I mean would that make sense to you what effect do we anticipate to the to the extent that C is occupying part of the gas phase the reaction what this is now let us to understand this just to understand this I have done a small calculation I want you to appreciate this what I have done is I have written the stoichiometry this is the stoichiometry A and B are going in there all gases A is a gas B is a gas and C is a condensable gas it is a condensable gas is that clear. Now, what is the total moles I just added them all up this is the total moles R plus 1 F A 0 this is our convention our convention is that inside the recycle loop we take R plus 1 F A 0 as our basis R plus 1 X R plus 1 and the product is R plus 1 F A 0 X. So, what I have done is at the incipient condensation what is meant by incipient condensation when the gas phase is saturated with C. So, at the incipient condensation since the pressure is 1 atmospheres the mole fraction of component C in the gas phase must be given by this relationship amount of C divided by the total moles S and O. So, if you solve this you find that excess is 0.33 what are we saying at incipient condensation the gas phase the excess is 0.33 what does it mean you know what we are saying is till X becomes 0.6 0.33 there is no condensate till X becomes 0.33 there is no condensate your condenser has no it does not perform any function because till the gas is saturated only after the gas is saturated it will start condensing. So, this effect that X equal to 0.33 and therefore, to that that gas phase after saturation only it starts condensing will affect the gas flow that goes back to the reactor. How do we take that into account to take that into account what I have done is before condensation what is the stoichiometry after condensation what is the stoichiometry before condensation is what everything is in the gas phase everything is in the gas phase. And after after it started to condense what will happen the gas phase will have the excess see the amount of amount of see in the gas phase is given by this relationship excess of that will condense is that clear in the gas phase the so much material will be there in excess of that will be condensed. So, therefore, what we will find is that the moles of F t that you will see going through the equipment is given by this is this correct what I am saying F t r plus 1 F a 0 2 minus 2 x plus x s this is what goes through the equipment is it alright. Now, notice what we have been trying to do what we have been trying to do in this exercise is trying to set up a procedure for doing reactor design calculations for the situation where there is a condenser when there is a condensate is taking place. And what we want to draw your attention is that in case of situations like this we must first calculate what is the conditions under which the condensation begins. So, that we can take appropriate decision regarding how to do the calculation this is the exercise that we are trying to do. So, let us recognize let us recognize that before condensation before condensation the total molar flow at the reactor exit is given by r plus 1 F a 0 2 minus of x this for x less than x s. And we show that x s equal to 0.33 we have already shown that now after condensation we find that x s F t is r plus 1 F a 0 2 minus of 2 x plus x s this is after condensation where x s is 0.33 moment we recognize this all the rest becomes fairly straight forward. Now, what is volumetric flow v by v 0 after x x greater than x s because our interest is x equal to 0.5 and x s being equal to 0.33 our interest is x greater than x s. So, we know what is v by v not from our gas law we know it is F t by F t 0 t by t 0 p 0 by p z by z 0 if it is an ideal gas we know that this is 1 this is 1 this is 1. Therefore, F t becomes F a 0 r plus 1 F a 0 2 minus of 2 x plus x s and F t 0 we already given that it is a molar flow of a and b where a is and therefore, it is 2 F a 0. Now therefore, v by v 0 is simply v by v 0 is F t by F t 0 we can put this as equal to what is already mention the r plus 1 2 minus 2 x plus x by 2. What is C a by definition F a by v therefore, we can substitute for F a which is F r plus 1 F a 0 1 minus of x and what is v we already calculate v 0 times this whole thing. What is v 0 v by v 0 is this that is therefore, we have substitute for v as v 0 times this whole thing that is why we get this fairly simplified expression. So, what we have tried to do here is that after all any rate expression requires concentration in terms of conversion and therefore, we have derived concentration in terms of conversion. We are now therefore, in a position to substitute our rate function in terms of conversion and we can integrate and find the reactor volume. So, what is the reactor volume the reactor volume if you want to calculate for a recycle reactor for a recycle reactor we said d F a d v is r a r a is k c a c b and we know what is c a what is c b just now. So, we can substitute for all that and therefore, we get an expression for the residence time in a recycle reactor in terms of all these variables. We can simplify this and show that residence time is simply r plus 1 by k c a 0 integral x 1 to x 2 of all these expressions. Notice here that x 1 see our our recycle reactor our recycle reactor looks like this looks like this y this is 0 this is 1 this is 2 and this is 3 and we shown already that x 1 x 1 is r by r plus 1 x 2 same as r by r plus 1 x 2. We know we have shown all that therefore, we can do this integration I have put in all the numbers v 0 is what in this particular case 2 kilo moles per hour and so on. I have calculated the molar flow rate as so many cubic meters per hour and concentration at the reactor inlet c a 0 p by r t that comes out to be 3.6 10 minus 3 gram moles per liter. So, we know c a 0 and k value is given as 10 liters per gram mole per second I will write here k equal to 10 liter per gram mole per second. So, k is also given we can calculate all these numbers and that is how I got this number as 23 seconds when k is equal to 10 liter per gram mole second. So, what have we done what we have done is that for a recycle reactor we have put set up the rate expression and then substituted for c a and c b and then replace this f a as r plus 1 f a 0 1 minus of x and that is how we got the residence time by this expression we have done this integration and found out the residence time to be 23 seconds where x 1 the lower limit is taken as r by r plus 1 x 3 same as r by r plus 1 x 2 where r value is given as r equal to r equal to 4 r equal to 4 is given in the problem statement. Now, that we have done the residence time what is the reactor volume reactor volume is simply given as reactor volume is given as volumetric flow multiplied by the residence time what is the volumetric flow is 138 cubic meters per hour I divided by 3600 to get in seconds cubic meters I have converted to liters and then so many seconds. So, you get 900 liters per for the processing of 1 kilo mole per hour. So, what we are trying to say here is that it just quickly recall what we have tried to do what have we tried to do what you have tried to do is we have a going in at 1 kilo moles per hour b going in at 1 kilo moles per hour into a reactor where the second order reaction r a equal to k c a c b is taking place at 570 c and where rate constant is 10 liters per gram mole per second where recycle ratio is r and we want x a equal to 0.5. Since this component a a plus b equal to c I will write here a plus b equal to c and since c is a condensable gas you will find that the this exit here moment after condensation you will find that this gas always saturated with c. So, that effect is all that we have to take into account. So, that we do the calculation as per procedure this is what we have tried to do and for the case of k equal to 10 liters per gram mole per second we find for 1 kilo mole per hour of flow we find that you require a volume of 900 liters. So, in the recycle plug flow what we have tried to demonstrate is that the basis that you have to choose is r plus 1 f a 0 number 1. Second point that we tried to say is that because of recycle the there is mixing inside the equipment therefore, you lose on conversion and therefore, you can use the recycle reactor device as a way of understanding what is the extent of mixing that takes place in the equipment. Third thing we said is that if it is not a catalytic reaction the recycle is required otherwise the reaction does not go forward. But the fact that you know there is recycle and therefore, the reaction goes forward if it is infinite recycle like that of a stirred tank it might appear that you have not put any recycle in the feed, but actually because of the fact that there is so much of recycle internally you do not have to put it continuously. So, these are the points that we have made in our presentation. So, now with this we will go on to our next topic which you want to do is multiple reactions. Now, we have been writing our stoichiometry for single reactions. So, we want to write our stoichiometry I just take an example where there are two reactions we want to write our stoichiometry for the case of two reactions taking place. Now, there is lot of material written here, but let me just run through this quickly a reacts with b to give c and d coefficients a b c d given again a reacts with c to form another product e. So, there are two reactions taking place. So, taking a as the basis I have written here a a plus b by a b c by a c d by a and similarly f by g c equal to e by e. So, when you write this stoichiometry which means what what happens to all the species a b c d e and i and so on. So, these are the input values these are the input values all these are input values and these are the values at any instant where the conversions are whatever they are. So, what I have done is that since component a is reacting in reaction 1 and component a is reacting in reaction 2. So, I have taken let x 1 and x 2 be the fraction of a that reacts in 1 and 2. So, that is why say that the amount of a that remains unreacted is amount of a that you put in multiplied by 1 minus of x 1 minus of x 2 this how we postulate what is happening. Once you do this what happens to b b is reacting by this reaction therefore, I have written what is the b unreacted is given by this. Similarly, the c form c is form from reaction 1 and reaction 2. So, c is form from reaction 1 c by a f a 0 x 1 c is consumed in reaction 2 there I put a minus sign d is formed in reaction 1. So, d is d by x x 1 e is formed in reaction 2 therefore, f a 0 x 2. So, when you add up all this this gives as f t 0 this gives as f t 0 this is f t and when you add up all this you get f t 0 here. And then there are 2 terms as you can see very clearly f a 0 x 1 which is multiplied by the stoichiometric coefficients involved in reaction 1 f a 0 x 2 which is multiplied by the stoichiometric coefficients involved in reaction 2 where delta a 1 and delta a 2 is only what is written is essentially stoichiometry that is written with respect to component a. So, the number of moles of that you will see at any position in the equipment is what you start with multiplied by plus there are 2 terms 1 due to reaction 1 1 due to reaction 2 moment you have done that we recognize that we are able to now write what is the concentrations of species a b c d in the equipment. Why is it required it is required because our rate functions are always given in terms of concentrations therefore, we need to express concentrations in terms of variables that we can measure. So, c a c a is f a by v we said is v by this is gas law this is something that we know gas law. So, f t by f t 0 t by t 0 z by z 0 p 0 by p therefore, c a is what f a 0 multiplied by whatever is the amount of a that is remaining the amount of a remaining is given by f a 0 1 minus x 1 x 2. Therefore, and then v 0 is what v is v 0 times all this terms which takes into account the effect of temperature pressure and all that. So, this ratio essentially tells you how c a depends on x 1 and x 2 that is the most once you can do for component a you can do for any other component. So, we are able to express all the compositions a b c d in terms of 2 unknowns x 1 and x 2 what is x 1 and what is x 2 to be able to find out x 1 and x 2 we go back and say how do you find out x 1 and x 2 you have to make 2 measurements any 2 measurements if you make either on a and b that is enough to find out what is x 1 and x 2. Now, let us just look at this in some detail now suppose we have a reaction like this a goes to b and b goes to a. So, we write it as reaction 1 and reaction 2. So, and we write our design equation for a plug flow in this form d by d v of f a is the rate of formation of component a and component a is formed due to 2 reactions we say which is reaction 1 plus and the reaction 2 where terms r 1 and r 2 are the intensive rates of the 2 reactions. What is meant by intensive rates by intensive rates we mean that if you want to find out the rate of formation of a component you multiply the intensive rate of that reaction by the appropriate stoichiometric coefficient. So, if there are 2 reactions 1 and 2 you multiply by appropriate stoichiometric coefficient to find the actual rate at which that particular component is formed. Therefore, d f a d v becomes r 2 minus of r 1 and it makes stands to reason because component a is formed from reaction 2 and it is consumed in reaction 1 therefore, you get this. So, what we are now saying is that when there is a reaction like this a goes to b and b goes to a and we express this in the form of 2 algebraic equations b minus a equal to 0 and a minus b equal to 0. Why do you say this is b minus a equal to 0 or convention is products are positive coefficients and reactants have a negative stoichiometric coefficient. Therefore, we write the first reaction as b minus a equal to 0 and second reaction as a minus b equal to 0. When we express this in this matrix form we find that the determinant of this matrix is 0 showing that the rank of this matrix is 1. So, what we are now saying is that if you have a set of algebraic equations as we have written down just now you can find out number of those algebraic equations which are independent or how many of these react of these equations are independent that will determine the number of reactions which are independent that is what is being said. Moment we write this reaction a goes to b b goes to a in this form we find that rank of the coefficient matrix is 1 which means that there may be 2 rate processes, but there is only 1 independent reactions. What is meant by independent reaction which means that you need to make only one measurement which is required to characterize the system given the initial state. If you know the initial state or the feed state you need to make only one measurement which that is what mean by there is rank is 1 or number of independent reactions is 1. So, what we are now want to do is that use this idea for a multiple reaction network like something like this. That means you have a network which says alpha 1 1 a 1 alpha 1 2 a 2 after alpha 1 equal to 0 like that there are p reactions. Notice here the first subscript is reaction the second subscript is species. So, when we say alpha 1 1 first reaction first species alpha 1 2 is first reaction second species. This is the nomenclature of writing a network of reactions. Now, when we express this in the form of a what is what we call as this stoichiometric matrix which is alpha 1 1 1 2 alpha 1 n alpha p 1 to alpha p n. So, this matrix is the stoichiometric matrix this matrix has determines what happens to this reactions. Therefore, the rank of this matrix will tell you how many reactions are independent. The rank will tell you how many of these reactions. So, if the rank is r what does it mean that in this network if you want to understand the composition of the system at any instant of time after beginning then you only have to make r measurements. So, if your r measurements the entire n component system can be fully specified this is always meant by the rank of this matrix. Having said this. So, what is it that we can learn from this what we can learn from this is that if there is a component j. If you have a component j therefore, a j minus of a j 0 represents a change in this component j. How can this change come about this change comes about because there are r independent reactions. And there is a contribution to this component j change because of each of these independent reactions is that clear what we say delta 1 j is the change in a j because of reaction 1 delta 2 j is change due to reaction 2. Similarly, delta r j is change due to reaction r. Now, we recognize that this term delta 1 j by alpha 1 j what is this mean delta 1 j by alpha 1 j the change in component j in reaction due to reaction 1 alpha 1 j is the stoichiometric coefficient of component j in reaction 1. This is change due to reaction 1 this is the stoichiometric coefficient. Similarly, delta 1 n divided by alpha 1 n what is it it is change in component n due to reaction 1. And alpha 1 n is the stoichiometric coefficient of component n due to reaction 1. So, this ratio by definition is constant because this belongs to reaction 1 therefore, this is equal to beta this is clear change due to reaction 1 component j divided by this stoichiometric coefficient of component j in reaction 1 that is a constant that we call as beta beta 1. Similarly, for reaction 2 similarly, we reaction r. So, what is beta 1 beta 2 and beta r these are change essentially this is the what are the units of this moles change is moles. So, with this relationship we recognize that this this whole this whole change a j minus a j 0 can now be written like this. We have written this what we have written is that this delta 1 j we can write it as delta 1 j is beta 1 times alpha 1 j. So, delta 1 j is beta 1 times alpha 1 j delta 2 j is beta 2 times alpha 2 j similarly, delta r j is beta r times alpha r j. So, we are now able to express the change in component j in terms of what we call as the molar extent of reaction multiplied by the appropriate stoichiometric coefficients. Now, it is customary for all of us to always talk about reactions with respect to a reference we always find it convenient because we are used to saying that it is the conversion is 50 percent, 40 percent, 80 percent and so on. So, if you have a reference is much easier to comprehend what is going on therefore, if you have a reference component a 1 0 with respect to which you want to understand the change in a j then you can divide throughout by a 1 0. So, that now this term beta 1 by now this refers to change with respect to a reference species. There is an advantage in keeping a reference species because with respect to the reference frequently or numbers can be made to go to 1 because it is if you choose the reference appropriately otherwise it will go on to 1. Therefore, by looking at a reference component a j minus of a j 0 we are able to write in this form x 1 alpha 1 j x 2 alpha 2 j up to x r alpha r j. So, what is it that we have done by an appropriate definition we are able to express the change in a j a j minus a j 0 in terms with respect to a 1 0 in terms of x 1 alpha 1 j x 2 alpha 2 j x r alpha r j. There are r independent values of x of x 1 x 1 to x r. So, what does it say here that in a network we have to choose a set say after all we have a network. So, the rank is r which means what you have to choose from here a set or some reactions say may be 1 2 3 1 2 4 whatever. So, that particular set that you have chosen must have a rank r which should not be different. So, chose the set you choose must be a complete set that is important. So, if you chosen the right kind of set. So, that the rank is r then you find that this the kind of relationship that we have set up works nicely for you change in a j is given by x 1 to x r multiplied by the stoichiometric coefficients. See in this very important question very important question the number of react rate processes are p rate processes. If you recall I mention this little earlier how many rate processes are here there are 2 rate processes there is only 1 independent reaction. Similarly, in a network you can have p rate processes, but rank k be r and r is less than p delta 1 j by alpha 1 j what is meant by delta 1 j by alpha 1 j. Let us just look at that once I just what is delta 1 j it is change in component j due to reaction 1 change in component j due to reaction 1. You are dividing that by alpha 1 j this stoichiometric coefficient of component j in reaction 1. So, this ratio is something unique to the reaction which is reaction 1. So, therefore, it must be equal for every component that participates in that reaction that is why I have set delta 1 k divided by alpha 1 k. Every component that participates in the reaction this ratio is fixed number of the number of reactions may be p, but the number of independent reactions are r and you need to take only r reactions to understand the system because you do not need p measurements you need only r measurements r is less than p. On other words what we are saying is that when there are 2 reactions see examples are here when there are 2 reactions we need only one measurement that is sufficient to characterize this system. The reason is the independent reactions are only 1 r is the number of total number of reactions that is the now there is a procedure for finding out when you do matrix elimination you do have done in your school I am sure you have to do a matrix elimination and the elimination procedure will tell you which are the reactions they are independent because that will tell you by looking at that itself will tell you react 1 2 3 are independent 2 4 5 are independent. The elimination procedure that may have been taught to you in otherwise we can talk about it, but it is not a very difficult thing. So, it is this point clear what we are saying when you have a network of reactions there are r independent reactions and then these are independent reactions x 1 to x r and therefore, you need only r measurements and therefore, if you want to find out change in any component with respect to a reference this becomes what we have written here x 1 to x r multiplied by the appropriate stoichiometric coefficients. Now, advantage of this technique is that given species j we are able to write the species j in terms of the x s x 1 to x r therefore, it makes the network easier to deal with otherwise it is a little bit complicated. So, let us illustrate this by an example we illustrate this by this example we have chlorination of benzene what do we have we have chlorination of benzene. So, you have benzene reacts with chlorine to give monochlorobenzene it gives you HCl benzene react monochlorobenzene reacts with chlorine to give dichlorobenzene dichloro with give you trichlorobenzene is it. So, you have component c is component 1 b is component 2 m b is component 3 and so on all the components are listed. Now, the question that we have to address is how do you write this stoichiometric table for this multiple reaction we want to write this stoichiometric table and hence express concentration of all the relevant species in terms of conversion which we can measure how do we do this. So, what is done here is that you have written the material balance a j minus a j 0 by a 1 0 equal to x 1 just recall our equation here just a j minus a j 0 is x 1 alpha 1 j x 2 alpha 2 j up to x r alpha r. How many independent reactions are here we can say it is 3 because we are familiar with chlorination we are say it is 3, but it may happen that you may have to determine the rank to be sure in situations which you are not very clear you may have to determine the rank in this particular case we know the answer therefore, you have not done. See the procedure for determination of rank there is a Gauss elimination method is there you have to make all the diagonals 1 and below the diagonals 0 and you have to count the number of diagonals which are 1. So, it will be number of components is 6 there are 6 components there are 4 reactions therefore, there are 4 1 2 3 4 and then 6 reactions. So, it will be 3 rows and 6 columns 3 rows and 6 columns 3 rows and 6 columns. Is there now we go forward. So, what we are saying is that we can write our stoichiometry this form that means a 2 minus a 2 0 I have taken a 2 minus a 2 0 as an example this is given as x 1 alpha 1 2 x 2 alpha 2 2 x 3 alpha 3 3 is this clear what we are saying. So, what is the value of alpha 1 2 there in value of alpha in reaction 1 component 2 reaction 1 component 2 what is the value of x alpha is minus 1 what is alpha 2 2 reaction 2 component 2 reaction 2 component 2 0. Now, alpha 3 3 x 3 what is x 3 it is not mentioned here x 3. So, what is the value of component 2 in reaction 3 0. So, therefore, it is minus of x 1. So, similarly you do for component 5 you can do for all the components I have just taken only 3 components because it is convenient. So, reaction 2 you have a 5 minus a 5 0 divided by a 1 0 what is it x 1 alpha 1 5 x 2 alpha 2 5 x 2 alpha 3 5 what is x alpha 1 5 0 alpha 2 5 0 alpha 3 5 what is plus 1 plus 1. Now, if you want to do this for a 3 minus a 0 a 3 is what that is. So, it is x 1 alpha 1 3 x 2 alpha 2 x alpha 3 3. So, what are the values x 1 minus of x 2 on other words based on our stoichiometry we are able to express these species in terms of our unknowns x 1 and x 2 this is the important point. So, what it says is that if you take what is our reference chlorine is our reference chlorine is our reference suppose you take benzene as reference no harm is done no harm is done. So, what we have done here a 2 minus a 2 0 is x 1 alpha 1 2 and x 2 alpha 2 2 x 3 alpha 3 2 is that clear. So, what we are able to now do is that express a 2 a 5 and a 3 in terms of x 1 to x x 3 multiplying by the appropriate stoichiometric coefficients. Now, let us illustrate one more problem. So, what we have tried to do here is that if you have a multiple reaction you must be able to express your results using the standard procedure that is the most important. Now, this is q 2 where we want to set up the stoichiometry and then see whether we can set up the equations that are required to describe the process at work. How do we do this when you get a monster like this you should first find the number of independent reactions how do we find the number of independent reaction we said you have to make 0s along the diagonal by row manipulation. And if that is the case the rank is the number of independent reaction you can see below the diagonal. So, I have set up this matrix for this which looks like this ammonia oxygen nitrous oxide that water and NO 2 N nitrogen. So, how did I get this I got this by putting the appropriate stoichiometric coefficients here I will put minus 4 means it will be minus 4 there itself. Now, let us just look at these reactions. So, reaction 1 gives you NO reaction 2 produces nitrogen reaction 3 produces nitrite reaction 4 produces what is it produce reaction 4 ammonia 6 NO gives 5 H 2 is balanced is it 10. Now, please recognize that there are some reactions which throughout nitrogen may be into the atmosphere this reaction is also that this reaction also postulated through into the atmosphere. So, this 4 NH 3 plus 3 O 2 there is nitrogen gas produced similar to what we talked about anamox and so on at earlier stage. Now, what I want to do is I want to see after all if you want to do a design or if you want to do a kinetic measurement you need to be able to relate what your collection to the equipment which you are done the experiment with. So, you have to generalize your result so that people can use it. So, keeping that in mind what we have done is we written this stoichiometry for a 1 a 5 and a 6 all the sets of equations for example, what is it that I have written based on based on this the rate at which a 1 changes with respect to volume is given by minus 4 r 1 minus 4 r 2 minus 4 r 4 is it correct can you please tell me d by d v of a 1 is it correct can you just check and tell me a 1 d by d v of a 2 a 5 a 5 and a 6 tell me if it is ok. How do we check this in our material balance the left hand side is about what happens to our measurement the right hand side is what happens to the reaction. So, the left hand side is d by d v of a 1 therefore, we can replace that from our stoichiometric table our right hand side are the rate processes. So, if I ask you which are the rate processes in which a 1 is involved a 1 is what according to a 1 is ammonia. So, where are the ammonia is involved in reaction 1 reaction 2 reaction 3. So, reaction 4 reaction 4. So, you can say reaction 1 is minus 4 reaction 2 is minus 4 a reaction 4 is also minus 4. So, this stoichiometric coefficient of ammonia is minus 4 the reason is minus 4 is because the reactant. So, similarly we have written for other species I have chosen a 1 5 and 6 because I get some neat results because our the results are very neat in the sense that it gives simple answers that is why I have taken it like that. So, I have done this here please see if it is a 1 a 5 and a 6 gives us a 5 gives you to x 3 a 6 gives you x 2 and a 1 using these 2 you can eliminate x 2 you can get x 1. So, what we are able to do using this stoichiometry is that we are able to express we are able to express our left hand side in terms of x 1 x 2 and x 3 a 1 depends upon x 1 x 2 a 5 depends upon x 3 and a 6 depends upon x 2. So, the left hand side you are able to replace in terms of x 1 and x 2 and x 3 what is right hand side these are rate functions r 1 r 2 r 3 r 4 r 5 r 6 we know that the rate functions depend upon concentrations. We know how to express concentration in terms of conversion in this case in terms of x 1 to x r. So, the right hand side we know how to express in terms of x 1 to x r the left hand side also we know how to express in terms of x 1 to x r. Therefore, you get a number of differential equations in x 1 to x r on the right hand side is a very non-linear function of x 1 to x r, but at v equal to 0 x 1 equal to 0 x r equal to 0. Therefore, it is an initial value problem where the initial state is specified. Therefore, you can forward integrate and get your results. So, when the final result looks something like this I have done the manipulation what it finds is that d x 1 by d v looks like this d x 2 by d v looks like this d x 3 by d v looks like this where the right hand side is a function of x 1 x 2 x 3 and temperature. So, these functions how do you find out these functions these functions come from an understanding of the rate process r 1 to r 6 for which data will be available to you. This should come from an independent set of measurements that some people would have done. So, that the left hand side x 1 to x r is related to these functionalities which is known to you. Therefore, you can integrate forward and then determine what happens to x 1 as a function of volume x 2 as a function of volume x 3 as a function of volume. Essentially what returns to you from this integration are these results and this is what we want in a design or simulation or in a control problem this is what we want. We want to tell what happens to x 1 x 2 and x r as we change our system I will stop