 We are looking at advanced reaction engineering, practice problems in gas solid reactions. The specific applications you are looking at is Spongia technology. Of course the context is very simple steel is a very large commodity around the world and blast furnace is the technology which is commonly engaged and we all know that blast furnace is a very large capacity kind of technology and investments are very large, footprints are very large, environmental impacts are also significant and so on. And of course it requires movement of materials, availability of raw materials and so on. Now Spongia is technology which is quite suitable for very small scale therefore is able to use ores of local origin and so on. So it is something that is gaining ground in different parts of the world particularly in India. So we look at this very briefly you have an ore which is iron ore as you can see. It is reacted with a gas hydrogen to form iron and water. So this is the reaction the equilibrium constant at 25 C at 25 C is 0.1 showing that you know it is an endothermic reaction. So you need high temperatures to reach reasonable velocities, reasonably favorable equilibrium constants and so on. So but if you look at a reaction a particle I mean generally particles are of various sizes you know it is never spherical but you know from the point of view of doing analysis. We assume that it is a spherical particle and this particle will react there is an external diffusion as you can see there is a resistance in the external layer as the gas moves through ok. So there is an external diffusion resistance, there is an internal diffusion resistance that is the resistance to the flow of gas movement of gas through the product layer. As the reaction proceeds the product layer keeps on becoming bigger and bigger therefore this resistance becomes larger and larger as the reaction proceeds plus the reaction takes place at the interface here let us say and there is a resistance due to reaction. I mean this is a very simplistic model of what might be happening what happens is much more involved than what I am saying therefore there are three types of resistances that are important for a reaction of this nature. We have at an earlier stage looked at problems like this and we have formulated this a gas plus b b solid giving products a reaction of this form in this case for example our in our case you have a gas so this is gas. So this a b b solid let us say b b solid so giving you products the two products are given here. So a gas b b solids giving you product we are looking at and then idealized form of this is what we are looking at and we have at an earlier stage if you recall in the course we have shown that the time required for this reaction to undergo a certain extent of reaction let us say if x b is the extent of reaction which is r c by r then given x b you can find out the time that is required for this extent of reaction. So this we have already formulated and derived at an earlier stage we will not do it again. What is important to recognize here that there is a reaction time there is a external diffusion time and there is a internal diffusion time all three of them together determine the actual time that is taken for the reaction to occur. Now we also recognize that all these resistances all these we have shown this earlier in our classes and then we are just using those results at the moment. So you have the total reaction time which is some of the reaction time due to individual resistances where the reaction time for time for complete consumption if it is a reaction control film diffusion control or product layer diffusion control these are also been derived in class. Just to contextualize what we are saying rho b is the rho b is the particle density rho b is particle density particle density you have r is particle size then this c a g is concentration gas in contact with solid with solid. Now the formulation the way the formulation has been done is that the gas concentrations have been assumed to be so large that it does not change during the course of reaction that is an example that we have taken but we will try to relax all this as we go along. Now k s is reaction velocity k g is mass transfer coefficient and then d is diffusion coefficient or internal diffusion coefficient. So what we are saying is that what we are saying here is that the time required for consumption of the particle is determined by all three factors the reaction external diffusion resistance and internal diffusion resistance. So we can do a small calculation just to get a feel for the kind of numbers we will get in the case of what we call as sponge ion. So I have got some numbers here let me just put down those numbers you have k s equal to 1.93 10 risks for 5 exponential of minus 12000 by t this is in centimeters per second then diffusion coefficient d in product layer that means hydrogen diffusion coefficient hydrogen diffusion coefficient this is for hydrogen. This data is given as 0.2 percent of the 0.3 centimeter square per second this is given film diffusion coefficient film diffusion equal to 10 centimeters per second these are all data given to us. So recall that the time required depends upon tau r tau f and tau d and tau r tau f tau d depends upon the concentration of gas which we will calculate shortly and of course k s k g and diffusion coefficient. So basically this data is given to us we have to do the calculation to find out what is the actual time that is required that is the point. So this is k g. So let us do this calculations and then find out what is the actual time that is required for this reaction. Let me just do a small calculation for you what is k p k p we know this l n at say 2 we will say at k p at 298 and we were our reactions are taking place our reactions are taking place at 600 c we want k p k p at 600 c. So we want this we know is delta h by r this this is Vantos equation. So we know all this. So you can put your numbers delta h is given as plus 7 kilo cal per mole this is given. So you put all these numbers. So k p at to sorry at 600 c sorry 600 c comes out to be 228. So please put the numbers and you will see for yourself this is 228 showing that you know at this temperature the reaction is quite favorable. Now our reaction is what Fe 2 O 3 plus 3 H 2 equal to twice Fe 2 O 3 plus 3 H 2 equal to twice Fe plus 3 H 2 O. This is gas this is also gas. So k p can be put as p H 2 O to the power of 3 divided by p H 2 to the power of 3. This is something that we know from our basic physical chemistry. So this is nothing new now let us go through the whole exercise one by one. Now k p equal to so if you look at this reaction if this reaction is a gas plus b solid giving you c solid plus d gas. So I can this there is a 3 here which you can divide. So it becomes 1 by 3 and then this is 2 by 3 and there is a 3 here by 3. So we can look at our stoichiometry F A becomes F A 0 times 1 minus of x we know this this is gas and then F D equal to F D 0 which is given it is not given. So we take it as 0 F A 0 x A I will put it x A. Therefore k p we write as C A 0 x A we can put an R T here divided by C A 0 times 1 minus of x A to an R T here to the power of 3 that is equal to x A to the power of 3 divided by 1 minus of x A to the power of 3. This is equal to k p. So we can solve this k p is given as 228. So we have this is equal to this 228 is given solve. How do you solve this? Take the cube root of 228 let me just see we have the calculator alright I do not have it here right now. So we can take cube root of 228 and then find x A to be solve and x A comes out to be 0.86. It is valid the elementary take cube root of 228 straight away you will get the answers alright. So since it is equilibrium I am putting a star here I am putting a star here putting a star here showing that this is the equilibrium conversion this is the equilibrium conversion possible this is implies equilibrium conversion. Now clearly in practice we do not reach this we can only approach this. Let us go further you know x A star now we have gone through all that exercise and so on. So let us not spend too much time we have shown see the problem that we are trying to solve is you have 2 mm particle it says 2 mm particle and if there is a rotary kiln which is rotating solids are coming in solids coming down. So you want here 95 percent equilibrium conversion this is the problem statement 600 C is the temperature of the kiln now the hold up here epsilon r hold up hold up of solids is 0.1 that is also given. Now we have gone through all the exercise we have set up the equations before. So there is no need to do the exercise again. So we know that this is given as l n we know this. So I am just writing the final form l n of x A star by minus of x A we know this equal to k g epsilon r 3 times divided by r tau g. So this is a result that we have gotten much earlier. So I am just putting that result right now. So this is something that we know. Now all the numbers are given here. So you can put all the numbers and do not find out tau g. So when you putting numbers I am just putting numbers what are the numbers here x A star is given as 0.86 and x A is given as 0.95 of x A star. So you know x A star you know x A k g is given and then epsilon r is 0.1 size of the particle is 2 millimeter that is also given putting all the numbers I find that tau g comes out to be 0.2 seconds. So essentially what we are saying is that the residence time gas residence times are quite small. So gas residence times are quite small. Now what we want is solid residence time. So how do you relate gas residence times to solid residence time? Solid residence time residence time is what by definition it is volume of equipment divided by the solids flow. In this particular we mean from our stoichiometry we know it is tau g V g by V s volume of the equipment is gas residence time minus the gas flow divided by V s. Now that is equal to tau g and then sorry and V g what is V g? V g is F A 0 by C A 0 and then C B 0 by C B 0 is that clear? So what is given F A 0 is given as equal to F B 0 given this data is given. So that if you put all these numbers now putting numbers F A 0 is. So you get this F A 0 equal to F A cancels of C A 0 and C B 0 put all the numbers it turns out that tau s equal to 285 seconds. How do you get this? Tau g is given we have got in the previous case C A 0 C B 0 all the data is given. So what are we trying to say that if you have a rotary kiln context once again let us not forget context is that you have a rotary kiln particles are coming in hydrogen is fed like this hydrogen and this is the ion ore and then what you get here is F E and this moisture escapes is it ok. So we are able to tell what is the kind of size of equipment that is required for your application. We have we have only done we have not yet finished what is called what is the suppose we do the same process in the laboratory that part we have still not done let us do it quickly. So that we can find out what is the time required for. So we said total time is equal to tau r into 1 minus of r c by r plus tau f 1 minus of r c cube r c cube by r cube then we said tau d equal to 1 minus twice r c square by r square plus twice r c square by r square plus twice r c cube by r cube. Now it is all this we know tau r equal to rho b r divided by b k s C A g rho f equal to rho b r divided by thrice b cube k g C A g. Now this particular case what is given is that the size of the particle is given as what 5 millimeter size. So particle size r is given as 5 mm and then k b all this k s and all that is given putting numbers putting numbers putting numbers we get tau r equal to tau r turns out to be about 10700 seconds putting numbers. Similarly putting numbers here putting numbers we get tau f turns out to be 71 seconds and similarly tau d I will write here tau d tau d equal to rho b r square let me write here rho b r square divided by 6 b d C A g putting numbers it comes out to be how much 5 9 5 2 seconds. Now C A g I did calculate C A g is p by r t that turns out to be p by r t p is 1 atmosphere r is 0.082 and t 600 600 273 is about 873. So that turns out to be 0.014 10 minus 3 mole per per milliliter. So basically what I am trying to put across to use that we can now calculate since you know tau r tau f and tau d we can now calculate what is t. So what is the question that is asked is the question that is 95 percent of equilibrium conversion. So we have been asked to find x A which is equal to 0.95 times x A star we have got x A star already. Let us just do that calculate the important calculation let us just do that our reaction is F e 2 O 3 plus 3 hydrogen equal to twice F e plus 3 H 2 O. Now F A 0 x A by F B 0 equal to thrice x B. So x B equal to 2 by 3 x A that is 2 by 3 of x A what is x A? So x B equal to we have got x we have got x A as 0.86 times x A star 0.86 x A star. So we can put all these things x A star is we already derived that x A star is 0.95 x A star sorry and x A star is 0.86. So we can put all these things 0.95 multiplied by 0.86. So that comes x B comes out to be 0.55. Now 1 minus of x B by definition is r c cube by r cubed. So x B is known this implies r c by r equal to 0.7 double 6. Now once you know x r c by r now we can calculate once you know r c by r we go back now our expressions are that is all here. So r c by r is as you can r c by r for our application is I will just write here just for sake of simplicity r c by r becomes 0.7 double 6 once you put r c by r values here you should be able to find what is the time for company for a for the extent of reaction desired. So what in a sense we are saying is that the time required the time required for our case. So the time required if you put all these numbers the time required comes out to be something like this let me write down t equal to 10700 multiplied by 1 minus of 0.7 double 6 plus 71 1 minus of 0.7 double 6 whole cube is 0.45 plus 5952 1 minus 3 times 0.7 double 6 whole squared plus 2 times 0.7 double 6 whole cube. So when you do all this it comes out to be 3333 seconds. So close to approximately 1 r. So what we are saying is that see in this in this exercise what we have tried to do is the following that we have we have looked at a problem which is of commercial interest which is of of considerable interest because small scale industries are important and then it is where we make what we sponge iron see we are making sponge iron. So what we have tried to calculate here is that what is the time that might be required what is the time that might be required to get a certain extent of reaction. If you do an experiment in the in the batch equipment but what we also pointed out is that when we do this experiment or do this in a rotary kiln when it is in a rotary kiln what we find is that because of the rotation of the kiln the external surface gets sheared off the product layer gets sheared off and to an extent and more importantly a shearing off may not be all that good but more importantly because of the fact that lot of gases are coming out and then it is quite porous and as a result those resistances may not be significant. I mean in our that is the kind of assumption we have made in this calculation I mean in reality these mean may not be correct these resistances will have to be taken into account and so on. So what we have tried to do here is that if we can manufacture sponge iron using a technology like this in a rotary kiln where it gives you continuous productions and so on and then you have to of course ensure that the temperatures are managed and then heat losses are minimized and so on. So this example illustrates how we can use the procedures that we have developed during the course of last several months and how we can apply them to understand a fairly important commercial process. Now we will go on to another application we are now going to look at what we call as a practice looking at advanced reaction engineering once again. Now this is the application we are looking at is practice problems practice problem in residence time we call it RTD. You have to take an example of RTD let us see how we can make use of this. Now the example in front of us is the following it says I mean this problem reads like this there is an arbitrary vessel to which this is this is the reactor. Now in this reactor you have you are getting a conversion of x a equal to 0.88 it is given. So reaction is a goes to b this is the reaction it says the residence time this residence time here is 10 minutes that is also given. Now the rate constant this rate constant k is given as 0.3 per minute the rate constant is 0.3 per minute. So it is modeled as a tanks in series tanks in series 2 it is modeled as dispersion model. Third it is modeled as dispersion model as a recycle reactor model. What is being said is when you model an arbitrary vessel as assuming that it is can be described by a tanks in series your parameter is n number of tanks. So the question that is in front of us is that there is an arbitrary vessel which gives you a conversion of 0.88 for a residence time of 10 minutes rate constant 0.3. Now if you model it as a tanks in series model how many tanks will be required this is the question. Now if you model it as a dispersion model what would be the Peclet number that would be able to describe this experimental data. Now third thing if you describe it as a recycle reactor what will be the recycle ratio that will be able to capture this experimental result. On other words you know what in essence we are trying to do here is that in an arbitrary equipment we may have designed it as a plug flow reactor and so on. But the fact is that it behaves in a certain way we do not know what the behavior is. We have models which is able to account for deviations from ideal behavior. This is what we have been doing when we talk about non-ideal reactors how does it deviate from ideal behavior for which we have these models. So let us do one by one let us do this one by one let us see how what kind of answers we can think of. Notice let us look at the simplest and then go on to the more complicated. First we will see how our system can be described by tanks in series model. So what is a tanks in series model let us just understand what is tanks in series model. What is tanks in series model tanks in series model essentially says that our arbitrary vessel can be represented as a number of stirred tanks number of stirred tanks. And therefore, whatever we get here at the end of these tanks here in this particular case this x a is 0.88. And the total residence time tau is how many minutes 10 minutes is 10 minutes and this rate constant is given as 0.3 per minute. So this reaction a goes to b is taking place in this equipment. Of course, it is broken up as three stirred tanks or four stirred tanks, but the fact remains is that it is a single equipment which is being model this way. Now we know if this is c a 0 if this is c a and let us say this as n tanks in between assuming that there are n tanks. Our result says that c a we know this from our experiment c a by c a 0 equal to 1 by 1 plus k tau by n to the power of n. This result we know this result that we know from our understanding of stirred tanks. Now what is c a by c a 0 it is 1 minus of 0.88 you know this 1 by 1 plus k is 0.3 tau is 10 it is given what is not known is this n solve for n. So when you solve you find n becomes 3.3. So I mean you might say how can you have 3.3 number of tanks I think that is not the issue the point is you have an arbitrary vessel. This arbitrary vessels performs in a certain way now how do you try to describe this arbitrary vessel in terms of models that you have. Therefore, whenever you looking at these models you should recognize that this n is only a parameter it is a model parameter it could take any value depending upon the situation. Therefore, n equal to 3.3 only means that here is an instance of an equipment which performs similar to number of tanks between 3 and 4 that is what is to be understood. Now let us take the next example which is instead of tanks in series model we have dispersion model what is dispersion model what we said is that suppose there is a fluid flowing through that is say there is a packing here. Now what we said is that fluids a plug flow means what every fluid moves together they move together they move together move together that means a fluid elements which is at this level and a fluid element at this level they move through without recognizing the existence of each other that is an instance of plug flow. But, when there is dispersion what happens is some fluid elements move faster than the other because of dispersion they are moving faster than the other. Therefore, the kind of time of residence that fluid elements spend inside the equipment is different from what we might think and the basis of the plug flow model in which all the fluid elements have the same amount of residence time. So, this effect we had taken into account and then we have derived in our equations a certain equation that tells us tells us that the constant suppose this reaction a goes to b and then the rate constant is k first order rate constant k we have shown this I am just writing the final result there is no point in deriving it here by 2 divided by 1 plus q square exponential of p q by 2 1 minus q whole square exponential of minus p q by 2. Where q equal to the time called a number equal to 1 plus 4 d a by p. Now, we have we have done this in an earlier class and if you have not done this we do not recall this you do not just look up some books on actual dispersion models it is all derived fairly simple things. Now, if you want to find out conversion at the exit of this equipment essentially what you need to know is that d a equal to what is d a d a is equal to k times tau now in this particular case k is given as point 3 per minute we know this and then your residence time is given as 10 minutes. So, in this for this equation the left hand side is 1 minus of point 8 8 equal to the right hand side the unknown quantity on the right hand side is Peclet number there is nothing else only Peclet number is unknown. Now, it might be more complicated to calculate left hand side from the right hand side is easier the reverse is more complicated. So, you have to do a trial and error you have to do one trial to find the answer. Now, when you solve. So, when you put your numbers and solve you will find so putting numbers you can put numbers here is no great problem in putting these numbers. So, Q putting numbers find Peclet number corresponding to the left hand side LHS equal to point 1 2. So, Peclet number corresponding to about 10 or so you will find roughly LHS equal to RHS I mean you have to or you can read out these numbers from charts which are available and so on. So, what have we done we what we have done is that we have recognized that you can describe the performance of this arbitrary vessel this arbitrary vessel conversion here it is point 8 8 it is given point and then rate constant is given residence time is given. So, what is the parameter Peclet number that will describe this data the answer is that if you put Peclet number around 10 you will be able to say that the left hand side is roughly equal to right hand side. So, what are we trying to say what we are trying to say is that for this particular instance for this particular instance the number of tanks equal to 3.3 corresponds to a Peclet number of 10 is what we are trying to say in this particular case. Now, there is another way by which we can describe this data there is another way by which we can describe this data there is another way by which you can describe this data. Now, if you look at a recycle this is the recycle reactor now we said if you recall at an earlier class we have said that if you have a recycle reactor this is the reactor. Now, depending upon this recycle ratio recycle ratio the design equation looks like this r by r plus 1 I will call this 0 1 2 and 3 r by r plus 1 x 3 to x 3 d x by minus of r a. So, when we went through this exercise at an earlier stage we said that r equal to 0 means plug flow PFR r equal to infinity means CSTR. So, we said at that stage that basically recycle ratio is a way of describing mixing within the equipment. So, no mixing is r equal to 0 very large amount of mixing is r equal to infinity and therefore, we can describe various kinds of reaction equipment between no mixing to infinity amount of mixing or PFR to CSTR by adjusting the recycle ratio. We said the same thing when we looked at number of tanks in series we said when there is one tank two tank and infinite number of tanks we said when there is a infinite number of tanks we are talking about a plug flow reactor. That means n very large means plug flow reactor n very small means we are having CSTR ok n equal to 1 is a CSTR n equal to infinity is PFR. So, on other words r equal to 0 that means PFR is equivalent to n equal to infinity you understand. So, this is what is the kind. So, just like n number of tanks is a way of quantitating the level of mixing just like Peclet number is a measure of mixing within a reactor recycle ratio is also a measure of mixing ok. So, it is a first order reaction first order reaction we can integrate this whole thing and the integrated form is available to us the integrated form we will just write down the integrated form. So, the integrated form for a recycle reaction. So, the integrated form is k times tau divided by r plus 1 equal to l n within brackets 1 plus 1 plus 1 plus r C A by C A naught divided by C A by C A naught multiplied by r plus 1. Now, of course, we can put it in various forms and so on. But the important thing here is that C A by C A naught is given as 0.88 this is known ok. Now, k is given as 0.3 per minute and this is tau is given as 10 per minute. So, 10 minutes are 10 minutes. So, putting all these numbers we can find out value of r. So, solve putting numbers numbers solve and then solve and then we if I find that it is around 0.9. When you put the circle point this the it is sort of it satisfies. So, what are we saying let us just summarize this is an important point what we are saying is the following. We have looked at an arbitrary vessel we have looked at an arbitrary vessel. So, we have an arbitrary vessel here coming in and going out x A is 0.88 tau is 10 minutes k is 0.3 it is all given we find Peclet number is about 10 number of tanks is 3.3 recycle ratio is about 0.9. So, what does it mean what it means is that the we are able to see this suppose you make a plot of r versus x r versus n r versus p p. So, I mean we can plot this. So, we can sort of get an idea as we as r becomes larger and larger it is CSTR. So, we should expect curves like this when n becomes as x that means when r increases it is going towards when n is equal to 0 n increases. So, we can plot on other words what we are trying to say is that as r increases it is going towards like this n increases it is like this and then Peclet number what do we expect Peclet number becomes larger and larger it is sort of it is higher the Peclet number you will have higher is the it is approach towards a PFR correct. So, let us understand this one by one when r becomes large when r becomes let me draw this once again I am not satisfied with this let me just do this once again. We have a we have a stirred tank x is 0.88 then k is 0.3 per minute tau is 10 minutes our results are n equal to 3.3 Peclet number is approximately 10 recycle ratio is 0.9. So, all these numbers describes this machine level of mixing in this equipment. Now, we want to understand this in a slightly better way what we expect. So, suppose let us say we increase for the same volume if we increase the number of tanks we will expect that conversion would approach plug flow that means the suppose you have similarly Peclet number if you draw you will find that it will approach plug flow this is what we expect. Now, but if you increase r if you increase r we will find that you will do this. So, this is recycle reactor this is dispersion model this is tanks and series model. So, what we are saying is that these are all numbers which can be used to describe a certain level of conversion. Therefore, if r conversion is let us say is 0.88 see this is 0.88 we can see here these are the numbers we will get this is for dispersion model this is for tanks and series model this is for recycle reactor model. So, by from this you can tell what is the parameter values for different models cut this long story short what we are trying to put across in this question this question essentially says given the data determine this parameter values which you have done. Then it says sort of try and understand how these numbers relate to each other. So, this relationship comes out of this it tells us how conversion changes with the parameter values. So, it gives you qualitatively tells you how these parameters affect extent of reaction. So, this this approach to understanding real vessels is very useful when you have an operating process and if you quickly get some data and you can get from the data what is the kind of mixing that is appropriate to your process. So, it is very useful to get a feel for what is the real life problems you want to go on to the next exercise interesting exercise, but something that not very difficult to do. So, this exercise we want to do now is advanced reaction engineering this is what is called reaction network analysis. So, we want to take an example in this reaction network analysis to illustrate how our techniques become. So, useful in some in many instances here is an example of A going to B. Going to C going to D. So, it is a reaction network some data is given A goes to B goes to C goes to D A going to B this equilibrium constant is given as 1 B goes to C B goes to C where are we is given K 2 is given as 2 B goes to C goes to D K 3 is given as 3 delta H 1 delta H 2 delta H 3 all these are given delta H 1 is 10 20 and 30 all in units of kilo cal per mole. So, this is the reaction network and this reaction is taking place in a stirred tank the stirred tank feed is only pure A is coming going out and then it is a coil to which which is added. Now, the question that is ahead of us is how much heat is to be added or removed. So, no that is not the question question is this is operating at 25 C operating as 25 C all these reactions are instantaneous all these reactions are instantaneous all reactions are instantaneous that is 1 and we are maintaining the temperature at 25. What does it mean to be able to maintain temperature at 25 when there is endothermic reactions are taking place it implies that we are putting in heat through our heating system. So, here is an instance of continuous input of heat into a reaction system to ensure that these reactions are in the at equilibrium. So, the question in front of us is what is the extent of reaction. So, let me ask this question to you once again what is the extent of reaction that we should anticipate. So, what is the extent of reaction we should anticipate when we maintain a temperature of 25 where the reactions A B C D are instantaneous which means that at 25 C they all exist in their equilibrium is that clear. So, we have set up this stoichiometry long back. So, there is nothing new for all of you. So, but just 2 for the sake of completion I can say if I say this is x 1 this is x 2 this is x 3 and then our stoichiometry says that f A equal to f A 0 times 1 minus of x 1 correct f B equal to f A 0 times x 1 minus of x 2 correct f C equal to f A 0 times x 2 minus of x 3 and f D equal to f A 0 times x 3. So, no what are we saying what we are saying is that is the multiple reaction there are only 3 independent reactions is obvious I have not proven to you 3 independent reactions if x 1 x 2 x 3 are the extent of these reactions this is the stoichiometry. And it means what it means that since they are all in equilibrium you know that C B by C A is k 1 C C by C B is k 2 C D by C C is k 3 yes or no this is the definition of equilibrium we know this. Now, since you know these 3 now we can using this relationship since k 1 k 2 k 3 are given since k 1 k 2 are given we should be able to find out x 1 x 2 x 3 there are 3 equations 3 unknowns we should be able to find out. Let us do that. So, that we know what the numbers are. So, help me now what is C B C B by definition C B equal to f B divided by V what is f B which is f A 0 times x 1 minus of x 2 what is V V equal to V 0 which is C A 0 times x 1 minus of x 2. Similarly, C A which is f A divided by V which is f A divided by V 0 which is C A 0 times 1 minus of x 1. Similarly, C C which is f C divided by V which is f A 0 times what is f C notice here what is f C f C is x 2 minus of x 3. So, we can put it here which is x 2 minus of x 3 divided by V 0 that is equal to C A 0 times x 2 minus of x 3 and then C D is what f D divided by V which is which is f A 0 divided by V 0 into x 3 which is C A 0 times x 3 is that clear. Now, what it means now is that now we know from our problem statement they are all in equilibrium. Therefore, C B by C A is k 1 therefore, from each of these equations we can determine and solve let us do that quickly. So, what are we having C B C B by C A is k 1 k 1 is 1 equal to k 1 C B is what C A 0 times x 1 minus of x 2 C A is C A 0 times 1 minus of x 1 equal to 1. So, our x 1 minus of x 2 equal to 1 minus of x 1 that is equal to what you get x 2 equal to or x 1 we will say x 1 equal to sorry 2 x 1 equal to correct 2 x 1 is equal to 1 minus of x 2 or x 1 equal to 1 minus of x 2 by 2. Now, we can substitute this in the next equation then second now we have C C by C B equal to k 2 equal to 2 C C is x 2 minus of x 3 divided by x 1 minus of x 2 that is equal to 2. Similarly, C D C D is what C D by C C equal to k 3 equal to 3. So, that gives us C D is what x 3 divided by C C is x 2 minus of x 3 equal to 3. So, let us 3 has x 3 equal to 3 x 2 minus of 3 x 3 therefore, 4 x 3 equal to 3 x 2 so x 3 equal to 0.75 x 2 0.75 x 2 now we can we can use this result. So, we have x 2 minus of x 3 divided by x 1 minus of x 2 equal to 2 and then we have in the sense we have not do this now because it is very elementary. So, solving we get x 1 equal to 0.9 x 2 equal to 0.8 x 3 equal to 0.6 this is something that comes out of our. So, we will stop here and little bit more to be done we will have to take it up and we meet next time. Thank you very much.