 మృర్చ ఏ్స్ప్దార్పెట్స్ రీనేపిన్స్ రోటికర్చ్పెడారంద్పికోత్త్రి రీట్స్తాత్ప్స్రందాందావోనటోవ్రచోత్టకిక్రట్ tänరందివ్సార Okay. So what you want is a rate where 1 by minus r A versus you can calculate C A minus r A and 1 by minus r A. You can also convert that into conversion if you want but even without doing that also you can calculate. So now you also know that C A not equal to 1, not C A not total, C not equal to 1. So that means this is 1 minus C A equal to 1 minus C R. So now you can calculate what is C A C R, what is C A, what is C R you can calculate, you can assume C A equal to 0.99, okay C A equal to 0.99 it cannot be because C A not itself is 0.99. So like that you know 0.9, 0.8, 0.6, 0.7 and all that correspondingly you can calculate r and the rate is C A into C R. So you will get all the data. So once you get the data now you plot, this is 1 by minus r A versus C A also you can plot, X A also you can plot, whatever is convenient to you, right. So then you will also get here something like this. So you will have the C A C R the product is maximum at one point, I think it may come around 0.55 or something, 0.5, yeah you will get that maximum. So then you will have that maximum here corresponding C A, C A equal to 0.5, it may not be C A equal to exactly 0.5, where you get the maximum rate. C A is equal to C R at that point you are getting. C A equal to C R at that point because you have 0.99 and 0.0, maybe the effect is not much but actually that is not the one, okay. You can also calculate for example when do you get that maximum, how do you get? C A equal to C R, yeah, C A equal to C R, your total concentration is 1, where is C A by D R, rate is maximum we are telling, rate is a function, this is what I have been telling you, yeah, this is what you know the simple calculus we are not able to use and here in C R E at this level only those simple calculus, see most of the time area under the curves and D Y by D X are second area also we do not go most of the time, right, yeah. So D F minus R A by D C A equal to 0 if you do you can actually find out, yeah, okay. So then that is C A equal to 0.5, right, yeah. So then now the problem is I think 90 percent conversion, no, given, yeah. So if I write C A must be 0.1 here, okay, yeah. So that means somewhere here this is C A not this side, okay because it is increasing here more and then here less, no, this is 0 almost, okay. So now all the problems can be solved if you plot this and then try to find out what is the tau, volume is not asked, only tau is given. So if it is only ideal simple single mixture flow reactor, what is the area you take? This much is the area, this is for M F R, okay and for P F R 90 percent conversion, if this is 90 percent conversion, so this is 0.1, yeah, this entire thing, okay, yeah, all that. So like we have already discussed, if I need 2 then I have to have till here mixture flow and from here to here plug flow, it is not, yeah because that 0.1 is there, 0.1 is there that is the reason, yeah whatever, exactly calculate that, it is not, actually Abhishek it is not asked in 90 percent, Abhishek it is not asked in 90 percent, the product should have 90 percent R, 0.9 R, 0.9 R, okay. So you can calculate from this equation what is exactly C A from that, okay. So then you will get to know that small difference will be there, that is why I was telling it may not come exactly 0.5, if you remember correctly, differentiate, to differentiate and get it, yeah. See if it is straight forward reaction like this it is okay, but to check yourself that if it is correctly coming at 0.5 or what is that maximum, you have to differentiate. See you are asking me as if you know you are always interested in writing the exam, right. So your idea is that when I am asking if I do not differentiate I will get, I will lose marks, okay. So not only you, I am telling you know all Indians are brothers and sisters, not much difference, they are keeping quiet you are frankly asking. In the figure itself it is given the maximum rate is that C A is equal to C R. Which figure it is given? 11 spirit is given like that. That is only for you know, he has not given what are the initial concentrations. Problem is given that C A plus C R is equal to constant which is what. Okay. So naturally C A is equal to C R at points. Every time you do not have to deduct like that, I mean why do not you do that once and you know the differentiation and then find out what is the truth in that. Why are you getting 0.5? And I can pain you with another separate test saying that it is not at 0.5. I can also show that, okay. Yeah, so that is why I am trying to give you as much knowledge as possible and you are trying to convert that knowledge into marks. That is all, I mean that is also important for you but I think knowledge if you have marks automatically come. So that is the reason. Okay anyway. So like that you can get all the possibilities. I will just give the final answer. You just try to find out whether what you get right or wrong, okay. What is the first one P F R, right? Or you do not have that one, I think you would have torn it somewhere and then P F R. So what is the value you got? Yeah, 6.8 minutes. Or may be 6.79, okay, no problem. Okay, that is fine. Yeah, then what is the next one? It is M F R. Yeah, tau equal to 9.9, yeah. 9.9 minutes. Yeah, from here to the entire, from here because it is plotted as C A versus minus R A, 1 by minus R A. You can also plot X A versus minus R A also, 1 by minus R A. Okay, so the next one, next one is 2 setup. 2, that is M F R plus, yeah. So M F R plus P F R. So this will be actually tau equal to 2 plus 2.2. 4.2. Yeah, 4.2 minutes. And of course if it is only M F R, tau equal to 2 minutes. You should have a distillation column or some other mass transfer equipment where you can separate it out, okay. Autocatalytic reactions and recycle is a wonderful combination. The next one which I have not given you, which you are supposed to do is find out the optimal torque for same conversion. You found out? R. R as 4. Oh, you did it, okay. Did you do it graphically or analytically? Analytical also can be done. Everything analytical. Oh, everything analytical. Analytical also you can do this. You know, all this you do not have to plot. You can simply substitute this V by F A naught. I have been telling you many times this. This is R by R plus 1 X A F to X A F, yeah. This is D X A by this minus R A as K C A C R, okay. But only C A C R you have to convert into conversion. R you can also use, we have also this equation in terms of C A. Same equation, yeah, minus R A can also be written in terms of C A, right? This equation. So either, whatever you see, once you are expert, so then you can use your right hand or left hand. You know, I think in at least stories people say that Arjuna, one of the brothers of this Pandava, he used to, I think, throw the arrow, it seems with right hand, left hand with leg, head, anything. Because he is so much, so much expert. Whatever he, you know, way he wants, he can put it. So that is what you have to also do, okay. There is a Pravar bin, Telugu also, other langs also, it must be there. If you have hair, you can put whatever hairstyle you want. Correct? No. Yeah, so, but people like me, what they do? I think, you know, where is the hairstyle? Because there is no hair at all, first of all. So that is why when you have sufficient knowledge with you in your brain, then whatever way you can do it. You want to only prepare for examination, then you want to only know one method. If that one method is slightly wrong there somewhere, then gone. Because you do not have overall perspective, okay. I mean, many times I am telling, but I think, you know, the moment you cross the door, you will forget everything, okay. So this is what, very good. This is a nice problem and graphically if you want to do that recycle ratio, to find out optimal recycle ratio for optimal, for recycle reactor, then you have to now try to draw these lines where this area equal to this area and somewhere here you will get r, okay. Yeah, so, xA1, not r. So that xA1 equal to r by r plus 1 into xA, that is one method. Otherwise differentiate this, you know the condition, okay. You write this in terms of, actually you integrate first and that integrated expression can be differentiated as d, d of v by fA0 by dr equal to dr. This is what again calculus, you know, what is the function with respect to what you have to differentiate. This comes also in the next thing where you are talking about multiple reactions. In multiple reactions also you have to find out sometimes optimal conversions or optimal volumes, okay, multiple reactions. Like you have done this already, I think I gave you this in the first examination I give you a going to r going to s. No. Asked you to calculate 90 percent, I mean show that 60 percent conversion. No sir, no. What was the first problem? No, first problem is that steps. Derivation. Yeah, catalytic reaction or steps. Second one? No sir. Pressure. Pressure, total pressure was 10. That is also another nice problem. A going to r, r going to s. Find out dCA by dr, where you get the maximum. Yeah, I think you also have know, when I have this kind of A going to r, A going to r, r going to s, then if I am plotting concentrations, so CA0, CA0 is this and CA will decrease like this, r will increase like this and s will go like this, okay. Yeah, so this is the one I am talking. This also can be differentiated, you find out CA and then differentiate that with respect to time when I am plotting this versus time. So always graph should have this x axis, y axis, that is very important, okay. So then you can also differentiate and find out. Good. There are very nice simple problems where to test your knowledge, how to differentiate and what kind of problems, particularly depending on K1 value, K2 values, this is K1, this is K2. K1 values, K2 values. That is again, you know, if you have interest and if you are able to do all these problems with good mathematics background, then I think you can easily score in this subject. It is not that much complicated mathematical subject at this point of time. If you go to complicated things like, you know, the reactions are not simply a constant density or reactions are multiple and with variable volumes, then you will have more algebra only, that is all. But conceptually you have learnt. I tell you, again I am guaranteeing that conceptually you have learnt whatever is basic things that are required. The additional things are delta x where you just add here and there, okay. Good, okay. So this is what I think here we will complete this, this part now with recycle reactor. That means ideal reactors we started and ideal reactors we thought only it is batch, plug flow and mixture flow. In continuous system it is mixture flow and plug flow. You know, when you have to choose continuous, when you have to choose batch, I think all of you, every one of you should know. And then between the continuous two reactors, mixture flow and plug flow, when do you choose plug and when do you choose mix? Swami, when do you choose? Yeah, wherever the reactions are faster. Okay? Yeah, and definitely where you don't have that when you won't have good temperature control. You can never get good temperature control in plug flow reactor, okay. Even though it is efficient but as far as temperature control is concerned it is very, very difficult, okay. Good. So those two, I think mixture flow you know normally continuous is chosen for large capacities and within continuous again you have mixture flow whenever there is a temperature control which is very important otherwise you know it may explode or those conditions you always go for mixture flow, okay. Yeah, or otherwise if you want to use still plug flow and then try to have better temperature control recycle the reactor. It is not to increase conversion. This is the universal answer every student gives. To increase conversion put the recycle. No, you don't. Okay, you can't. Because it is now not the mixing the moment you are bringing back the recycle to the inlet stream when you are mixing that you are now trying to have control over mixing where there is no mixing an ideal plug flow. Absolutely there is no mixing and mixing is not good for reaction. Very simple concept is whenever you mix two concentrations then you will have less concentration. When you have less concentration rate of reaction will be less for again there is a condition greater than zero not one again. This is another myth many people have. Okay, you can calculate for 0.5 order and then show for mixed flow or plug flow which gives more conversion. Right, so that condition is very important if you have negative order reactions dilute as much as possible if concentration equal to zero you will get infinite rate. Infinite rate means what is the value? Zero because in all these design expressions that rate will be in the denominator please remember that. That is why we want to increase the rate as much as possible either through temperature or through high concentrations right or deliberately not mixing any streams that is plug flow whereas in mixed flow you are mixing two streams the fresh reactant comes and falls into mixture where already it is converted may be 90 percent 80 percent 70 percent the way you are getting out at concentration both are mixed mixing is instantaneous right but the conversion you will get under steady state conditions because that much average residence time where it is defined volume by volumetric flow rate as volume by volumetric flow rate as tau that much average residence time is required for the reaction to go on but why for that same average residence time why you are getting less conversion in mixed flow more conversion in plug flow I am trying to explain it other way I told you also already maximum yeah it is the bypass or indirectly residence time distribution because the moment you have mixed flow and mixing and coming out so some molecules without spending much time at all they are coming out conversion is almost zero in that right so whereas of course it may take also long time infinite time but there is no use of staying there infinite time it is already converted as product then again that acts as a dead space correct no I mean what we expect is the good the time for the reactants to react not for the product to stay there so that is the reason again if someone asks you okay I have the same residence time in both the reactors why mixed flow gives you more conversion less conversion and plug flow gives you more conversion the reason is residence time residence time of individual molecules average residence time is same okay average residence time is same but the individual molecules individual packets if I look some packets will come very quickly from the mixed flow whereas all the packets exactly must spend exactly same time in the plug flow that is why even if I break all that packets and then mix them then I will get exactly same conversion there is no dilution there whereas here I have dilution I have a packet which coming in one minute I have packet which is coming in 10 minutes when I mix these two that is what what you see at the outlet the combination of all these packets that is what what you see in the outlet okay so that is what is ideal reactors and you want to have recycle reactor you will have only whenever you need some control over the mixing and mixing can help you in controlling temperatures reaction rate reaction rate or mixing also can help you in getting maximum products particularly for multiple reactions when you have some combination of reactants A going to R, R going to S again A is going somewhere, R is going somewhere it is producing some other but only one of those products is R or S is your actual product that is required under those conditions also I mean now we know this equation that is all this is the equation and you can substitute that rate expression for A for a multiple reaction substitute here and then calculate conversion for a given volume from those conversions again you have the relationship between yield and conversion that is what we are going to do now so finally you will end up under some conditions recycle reactor is the best so tell you later which reactions are best for that but that is there it is too much mathematical it is not that easy to integrate so that is why beyond certain point the chemical reaction engineering becomes very very highly mathematical and as teacher if I start that level of mathematics I think by this time you should have switched off all your brains I do not know even now I think I am not but with minimum mathematics you can learn beautifully all the concepts afterwards it is mathematics whatever system you take it is only complications in mathematics and some techniques you have to use because you cannot integrate analytically most of the time so you have to go for some other technique or differentiation to find out the maximum minimum and all that so now I think we will go so the ideal reactors you know but semi batch reactors also will be helpful particularly in finding out the product distribution like you know if you have multiple reactions sometimes semi batch reactor will give you more yield how do you define yield yeah yield means you know the concentration of desired product by maybe concentration no no by undesired product if you take that becomes selectivity yeah so yield yield is at the input either CA not or CA converted that is what we are going to do now this is multiple reactions our idea here is if yield is clearly known to you so much you have to get you have to find out what is the volume of the reactor okay and yield and conversion also are related definitely so or otherwise if I do not know what is the yield then I have to take a particular reactor when you find out in that given volume what will be the maximum yield okay so these are the same problems given conversion finding out volume given volume finding out conversion but when do you know conversion or when do you know volume again when I told you I have to give the question and I have to answer myself when you have an existing reactor you find I am talking about simple reactions product concentration there are two problems I told you one is known conversion find out volume find out conversion but when do you come across this kind of two situations when you have an existing reactor when you already have existing reactor may be someone was using earlier and he has thrown out and you want to use it so taking that volume how do you maximize your how do you operate so that you will get maximum conversion otherwise totally you do not have a reactor you do not have a new plant where you do not have any equipment totally new plant so there everything starts with the reactor even though reaction engineering start in third year fourth year like that okay but the actual process always it starts with reactor that is what I think I also explain to you know in fact we have taken more time leisure space I have explained to you how does a chemical process start you would have cursed me why this fellow is taking so much time for that but that only gives you the overall picture of what is going on in chemical engineering otherwise if I come and start only reactors you do not know where you have to put the reactors okay and you do not know how to connect with reactors and distillation columns all that so that is the reason why atleast at this point of time I have to tell that it should be done actually in introduction to chemical engineering in the first semester BTEC first semester okay but I think many people may not tell that so that is why whenever I have a choice even if I go outside to give some talks I will first start only with that how does a chemical process start and I am sure 90% of the people do not know okay so that is why okay so the multiple reactions okay please take this I think I will give you some notes when a single stoichiometric equation and single rate equation are chosen to represent the progress of a reaction we have a single reaction when more than one stoichiometric equation is chosen to represent the absorbed changes what changes you are talking about yeah concentrations concentration changes okay absorbed changes otherwise if you are not able to follow right in the bracket concentration changes then more than one kinetic expression is needed to follow the changing composition of all the reaction components and we have multiple reactions actually these are the definitions which I asked you in the first 0th examination differentiate between multiple reaction and single reaction okay single reaction definitely is not simply a going to r right important thing there is you need only one equation and one stoichiometry to represent the changes it need not be only a going to r I may have a plus b plus c going to r plus s plus t plus till xyz how do I represent all these compositions with time if it is a batch reactor I need only one stoichiometric equation because now a plus b going to r a plus b plus c going to r plus s plus t plus u plus you know v plus all that is there but if I know one mole of a I know how many moles of again that side r is forming and all that so one equation is just enough but whereas if I go here I cannot use only one stoichiometric equation because this also can be written as a going to r also r going to s so if I want to find out the composition of a, r and s I need again minimum 2 also you can do it but I think not required by material balance you can find out the composition of the other one because the total mass anyway that is same right so that is the difference actually between multiple reaction and single reaction when you say you have a single reaction that means you have only one stoichiometric equation and one rate equation whereas multiple reaction means you have more you need more than one rate equation and of course stoichiometric equations also will be more that is the definition in fact that is what also I told there in that para even though without thinking you would have just copied whatever I said good so then the this can be classified as classification it is not that easy to easily classify but some simple things we write here classification of multiple reactions so we have series reactions a going to r r going to s parallel reactions are a going to r a going to s and we can also have for example a going to r side by side reactions b going to s both are there a and b both are there so a may go to 1 and b may go to another product so I mean any combination I have written here very simple thing a going to r r going to s so that algebra will be simple going to r and r again may give s plus t okay this is where you will have l in reaction engineering that is why you should be good in simple calculus because I think it would not go you would not go beyond dca by dt or you would not go beyond you know either differentiation or integration of first order one it is not more than that but I think unless you have that practice that mathematical practice only I am telling so I have given here single I mean simple complicated but our requirement is this is a step and the intermediate component is giving me another reaction then you have the series okay one after the other whereas here need not be one after the other a is separately giving r and also a some part of it is also giving s it is not first r and then s okay there are many combinations in fact in industry there are many many combinations like that r is going to s again a is going anyway to r and a also is going to s and again r may give me s see I think it is really very very complicated now your objective may be to get maximum r maximum s is fine it is not that difficult why anyway this fellow is going here so what we want is maximum here if you want to have this r maximum then you have to try to control either to temperature so that means you have to minimize this rate as much as possible theoretically 0 but it is not possible for 0 so that is why you have to see that what is the what are the parameters what you have you have temperature and you have concentration and also you have a reactor so that means how do I minimize my r as much as possible in a particular reactor when do you get low concentrations of some reactants in the reactors by definition C S T R gives you always low concentrations that maintain low concentrations P F R maintains high concentration this is also another reason why you have P F R giving you more conversions and M F R giving you less conversion that is what I told you if you have sufficient air then always you can try to put whatever hairstyle you want so that is why that thinking that brain thinking that is very important for us good so this is the one then now we also have series parallel this combination series parallel combination all of you must be knowing all this but still a going to r but now so like that any number also you can write respect to what it is parallel b it is parallel yeah and a going to r r again going to yes okay you can also write this in terms of the individual components like if I have b this is r okay then this is s how do I represent here I should have plus a and here you should have plus r that is parallel series also we can write how do I write series yeah plus b yeah r arrow plus b gives you yes so that is the series representation so that is how what we can easily imagine that and there are many complicated reactions also I will tell you some more b actions okay I think let me tell here itself so I also may have another reaction called simple thing we start first this is small v actually it is called fende fussier reaction is name of a person and we normally we call as you know van de wussier right because we know we were but in many european countries they call v as faw f fende fussier okay yeah fende fende I think there is a movie actor also know phantom a phantom action action phantom a phantom a slightly old actor now I think you know I think all action very clean move is only action I think nothing else okay good one I think phantom a that is what it is not van it is not car or it is not van van van they call as van phantom a yeah now I think you have to go go to google and see whether what is the correct name what are the movies he has done okay so this fende fussier reaction is a going to r r going to s and here again going to some product t normally we will represent these are the yeah these are the desired products that means I am interested in finding out yield of this what is the problem here and all these things you know when you are discussing we assume that we have elementary reactions because if you complicated things complicated things will come anyway through mathematics but to understand the basics the first you know the concepts we will go to the simple ones like this all these reactions are elementary reactions and also they are constant density system so that mathematics wise I can simplify because beyond certain point you know in some courses you would have felt that also after sometime if there are too many mathematics you lose hands up okay that means mind cannot you know perceive all that and then remember like for example I can tell you very simple demonstration this is the limitation of mind so how many check piece are there 1 now now you come and to find out 18 are there who said abhisheka abhisheka do see 18 and all that your guess but you do not know that means mind cannot pursue the easiest one is one beautiful so maximum one is 2 okay maybe another 3 okay 0 is easy because we like only 0s also that is why okay so the moment we have more then it is mind cannot pursue that is what is exactly same problem with mathematics after sometime you feel that everything is mathematics and you have forgotten about what you are talking you know the problem actual engineering problem so because you have to now find out what is the technique you have to use to solve that problem where you do not remember what is that problem so that is why you know that balance should be maintained by any teacher when you are talking about mathematics even transport phenomena same thing okay then transport phenomena also if you explain the physics and then write the equations you know most of the time you will get differential equation that is also one common thing across all transport phenomena processes what do you do because you first identify whether it is lumped parameter or distributed parameter lumped parameter is very happy okay why whole system is one system only because there is not much there is no change at all if it is distributed system then what you have to take is you have to take a small element write material balance and energy balance for that and integrate between entry and outlet that is why what we call boundary conditions this is the universal method for all problems okay but the happiness only comes not by solving them but by actually formulating the problem finding out which differential equation is the best and also which boundary conditions are you have to apply apply what boundary conditions should be applied afterwards it is a mathematical problem and mathematics also you have to learn I am not saying that they are required okay but if I write one differential equation and then send and try to solve that problem in 10 classes then how did you get that boundary conditions how did you get that original differential equation you will forget that is what I meant okay so that is why always you have to connect there so that is why we are all again taking very very simple problem just to give the concepts first afterwards of course mathematics once you are experts you yourself will like to solve actual values because finally the actual mathematical solution that gives me when I am talking about transport phenomena problems or temperature is changing or velocity is changing that is all know only 3 transfer mass momentum and mass momentum and heat okay so what do you do every time every time you solve this you know either concentration change or temperature change or velocity change and what do you do afterwards I mean course will be over by the time but actually what will be over therefore you are doing all that excellent and that is the ultimate I would like to find out what is the flux either that is coming out of your process or whether it is going into the process how do you find out flux if I give you temperature versus length we have over 4 fathers 4 years law right that is all know that is the flux what you get and you also have another grandfather and you have another grandfather new terms that is all I think you know overall picture of transport phenomena is that that is why I am taking more time in teaching because I am teaching entire chemical engineering that is what is the problem you know these are the that kind of overall picture of any subject you should know at this point of time atleast because at beta level I cannot tell because they would have not done transport phenomena yet they would have not done some other thing but at this point of time you put your finger everywhere right in your BTEC some amount of transport phenomena some amount of control some amount of heat transfer mass transfer everything you have done that is why I am able to tell you all that because what is the real perspective behind all these subjects I mean what are the important things what you have to learn from those subjects that overall picture atleast if you do not get after taking a course I can tell you whether serious problem with you or with others because that overall perspective has not come means what we require at this point of time is first that overall perspective what is the use of CRE in chemical engineering transport phenomena and chemical engineering fluid mechanics in chemical engineering or mass transfer in chemical engineering not to get marks not to draw macabre-thiele diagram and finally get 100 out of 100 you may get 100 out of 100 but what is the use if you are not able to tell how what is the basis for macabre-thiele diagram or puncture-savarite method so those are the things what you have to remember it is not the actual values it is not the actual mathematical equations no one expects you to remember all mathematical equations all derivations but that exposure of where is what and also when do you use what good so this is when do we say equation what is the problem here I have A going to R, R going to S then I have a parallel reaction here A is combining with another A to give me a side product which is not required for me now that means I have to minimize that so I have to minimize the problem is I have first order here because all elementary we said and I have second order here second order reactor will consume more and more A because C A square right so that is why even though it is second order reaction how do I now stop that reaction as much as possible that is why we have the best techniques again called catalyst if you are able to develop a catalyst or take a catalyst where this is inhibited and this is accelerated you do not have to do both atleast inhibit this, this will automatically accelerate if you are able to do both that is excellent that is why catalysis also is important if you are able to find catalyst it is excellent you do not have to do mathematics will be very simple again because as far as catalyst is concerned this is suppressed totally you do not have to do anything with that only you have to calculate using that series reaction right so actually this is one of the examples where recycle reactor will give you more under some conditions it is not all conditions okay so when I substitute this and this and then try to find out what are the expressions then you will have some conditions where because this is k1, k2, k3 you know k1, k2, k3 depending on the values of this k1, k2, k3 under some conditions under some recycle ratios you get the maximum for this r this r is not recycle ratio okay this r is the product another famous equation is denby equation you know denby know ya I think you have to remember all these people are in k this is also important okay so I have here a going to r r going to s then I also have this a going to t and this is going to some u ya so this is series and parallel combination if okay if this is k1, k2 k3 k4 and if k3 equal to k4 equal to 0 then I have only series reaction right and other hand if I have k2 and k4 equal to 0 then I have only parallel reaction all kinds of combinations you can get there but here sometimes either s is the desired product or sometimes you may have also r desired product r is more challenging correct no r is more challenging why this r goes to s and also r is going to u whereas here anyway r is finally going to s then I have to suppress this equation and this equation this and this okay but now if I have r as the desired product then I have to stop this reaction stop this reaction 3 reactions okay I have minimized stop means if you are able to get a catalyst where you can only produce only this r and all other things are suppressed excellent but life is not that easy life is easy only in movies and in novels all these things are not that easy problems the way we talk and that difficult thing will come to you only when you try to solve the problem so they are not that difficult problems again but difficult and all that is relative if you don't do any problem everything will be different that difficult if you have done some problems things will be easy so that is why please try to do problems I think we will stop here tomorrow we will discuss about the definitions of