 Yeah, so the other, the design expressions what we have discussed, design equations for ideal reactors, okay. So for batch you have T by C A not equal to 0 to X A d X A by minus R A for P F V by F A not equal to 0 to X A d X A by minus R A M F equal to V by F A not equal to X A by or X A F by, this is also X A F by minus R A to be very precise I should write minus R A F because I have written here X A F because that is rate corresponding to only at the outlet, okay. Yeah, so these are the three equations which we have derived. We spent lot of time in understanding what is batch ideal, plug flow ideal, mixed flow ideal and all that. Finally you have very simple equations. The simple equations have come because we have very beautiful assumptions what is perfect mixing, what is ideal plug flow, what is ideal batch, because of those equations, assumptions only you could get very simple equation, right, good. So now if you go back to your, that information required for reactor design, this is input output kinetics and I have contacting then here we have batch continuous P F M F, okay that way this is physical chemical, right. So this is the design information required for the design, this all kinetics will give me minus R A, this will give me what? Reactor volume only but I think what volume you are talking, okay reaction is also not the one, that the entire thing will give you what kind of react you have to use for stand, contacting, okay. So you have to choose that whether you have to batch or continuous and out of that you have to choose now, you see so many times repeating also you are not able to get still the points, you know, yeah all this will give me the type of reactor which I have to choose. This suddenly will not give you the rate or someone was telling, Abdul was telling conversion. How can you get conversion the moment you take contacting, okay. So you need again for get conversion, you need what kind of reactor you are going to use, what is the rate expression and what is the input, okay. So now these are the, then output will be volume if conversion is known or conversion if volume is known, that is all, okay. So now by deriving these equations what are the quantities you have now with you? Very good, what kind of reactors we have discussed, so how do you choose we have discussed? So that means I have information on this, okay, yeah, then that is all, that is all. What is performance equation? Output equal to function of input, kinetics, contacting, yeah, okay, yeah. So what is the information till now we have covered for that using, yeah, for this equation if you want to calculate finally either volume or conversion, what is the information? You said contacting you know, that means you know how to choose a reactor, okay. So then you know again kinetics, we do not know, we have not discussed anything about how do you get this minus r A, okay, this minus r A, right. So then what about output, sorry, what about input? Input did we discuss? It is based on the output we can do, okay, yeah. So input we have never discussed, input we have never discussed, F A naught we have never discussed. Sir we can decide the input but we cannot decide the input, only by the rate we can decide the input. How do you get F A naught we have not discussed? That is the first thing we know before starting any plant. What Abdull, what has happened? Yeah, you have to produce some output for that you know what is the kind of reaction stoichiometry and then how much reactants you have to get, I was telling you so many thousands of times that F A naught we know, F A naught we know, F A naught we know, F A naught is input. In your language F A naught is what? So then why do you say that input is not known, we have not discussed. So many times we have discussed about input and then I was telling you that you know if you do not have time, appoint a MBA guy and then get all the, ask him to do the market survey and get what is the total capacity for this plant we have to design. So F A naught is the first thing you know, I think what I told is that you identify a product and find out which stoichiometric equation is giving you, from there you are calculating what is the amount you have to produce, amount you have to produce this product and now how much reactant you have to, I mean use, so that is what is F A naught we know in the beginning itself. So right now what you do not know is only minus R A, you know F A naught, right and you know, you do not know minus R A but this equation or this equation or this equation which you have, equation you have to use will come from contacting, right, whether you take batch reactor, whether you take continuous reactor or in continuous reactor again you take P F or you take M F, all these things we have discussed, so that is very clear for you know, okay, just to take this sample, Gopinath, when do you choose mixed flow reactor? Xothermic reactions, when do you choose mixed flow reactor? Xothermic reaction, Xothermic reaction, yeah, first of all liquid phase reactions, I mean high residence times will come, most of the time liquid phase reactions, okay, then even if it is gas phase sometimes if the temperature control is very crucial, then temperature, you know, when you have exothermic reactions, so under those conditions only you choose mixed flow reactions, okay, when do you choose batch reactor? Small scale production, when I ask thereby in, except thereby in everyone talks, okay, you mean, okay, so small productions like may be 10 tens, you know 50 tens, 30 tens you go for batch, but I think you know somewhere 80, 100 and all that we do not know which one is right, but definitely 1000 tens per day is continuous, okay, so that is why you have all the information, I think do not disappoint me because so much time I have now spent on contacting so that that will be permanently printed in your brain, but it might be printed, but you are not able to recall, you know that information recalling also is important, you know, what is that random access memory or AM, so that access is not there, you are printing somewhere, I think you know, computer also stores somewhere but to get that file out, your interest must be there, you should not forget so easily, okay, anyway, so all this we have spent lot of time in deciding which reactor you have to use, either this reactor, this reactor, this reactor, that comes from this, this is what, type of reactor and this is what, F A naught and now we have to find out what is minus R A, right, okay, so now let us talk about rate of reaction. I just taught it a little bit, rate of reaction, rate of reaction, mainly it is our key reactant minus R A we call and I may have a rate A going to R, general rate or it may be having any complicated rate, right, so how do I determine my rate and we are talking first about the homogeneous systems and then we can talk about heterogeneous, homogeneous. Homogeneous means either I have gas phase reaction or liquid phase reaction, everything is available in that phase, if there are two reactants, two liquids, thoroughly mixed, two gases, you do not have to do anything, they will anyway thoroughly mix, okay, so we are talking about only gas phase or liquid phase for homogeneous reaction. Now if I have homogeneous reaction between A and B giving me R L, how do I know what is the order of reaction, okay, R I may have A gas going to R gas plus S gas, I cannot have R and I mean L and G because that will become heterogeneous system, okay and before this we have also defined now what kind of rates you can express, like homogeneous means mostly it is liquid, volume of the liquid or, okay, volume of the liquid otherwise you have many possibilities for heterogeneous system, that will come a little bit later but this one, so if someone tells that, okay I have now A plus B going to, now onwards please write in the bracket what is the state of the reactant, when it is reacting, not at room temperature, when it is actually reacting, okay, you may have both A and B liquids at room temperature, so under actual reaction conditions both may be vapours, so that is again homogeneous but gas phase reaction, now you have to see whether the reaction time is very small or large, so depending on that, so because you have the reactants liquid phase does not mean that you choose only C S T R, we are talking about those conditions we are talking about whether gas phase or liquid phase at the time of reaction only, but the small things also you know sometimes you may not get there, I have now this is first order but how do you know it is elementary, when I just asked you how do you know it is elementary, Pooja you are telling something, how do you know it is, you cannot see where see the actual reaction whether it is taking place in single step or two steps, you do not know, that is very important, so you can never say anything about the rate equation or order of rate or what kind of rate equation you get unless you conduct experiment, so that is why chemical reaction engineering always associated with the experiments, so do not tell very proudly that no, no, no, I am I fund agai, so I only talk about mathematics and I cannot conduct any experiment, you have to conduct experiments, okay, so otherwise you will never get a rate expression, but if it is already someone did and then if you are copying, copying is very easy you can do many things, okay, so that is not the one, so always reaction engineering associated with experiments because the first thing I do not know is whether this reaction is elementary or non-elementary I do not know, then what kind of rate form I will get I do not know, right and I will ask you another question, I have 2 A L plus 2 B L giving me R L, what is the order of reaction? Can I take this one as elementary reaction? No, no, no, order greater than 3, why order greater than 3, I mean why are you angry with orders greater than 3 or why you do not like orders greater than 3? How do you know there are lot of intermediate steps? First of all why you are telling that orders cannot be more than 3? That you have to remember, in nature if I have only A decomposing easiest, right, so when I have that means 2 molecules come together 2 A's and again they will collide and then they decompose, okay some kind of activation energy all is required, so if you have A and B 2 different molecules coming they come together again collide, okay, so depending on the energetic collisions they will have more and more product coming but important thing is molecule collision, so when you go to third order that means 3 molecules coming together, so the probability will fall very drastically, 2 coming together is good, 1 is of course just decompose reactions are many, right, 2, then 3 probability is less, 4 probability is 0, almost may be 0.0001, so that is why you do not have 4 molecules together or 3 molecules also together and then coming and happily getting reacted, reacting is very very difficult, so that is why in nature people have not seen that orders beyond 3 and third order reactions also I told you, what is the order I gave you, 2 NO plus O 2 that is one thing NO third order that means 2 NO molecules are coming, 1 O 2 molecule is coming together they are able to collide and they are able to give you know the product, so that is why people observed that experimentally later proved that yes this is a third order reaction where 2 moles of this and 1 mole of this coming together and reacting, so that is why absolutely I told you in fact that is what I have been telling you many times, contacting is the easiest one to remember in chemical reaction area, it is the actual reactions which are very very difficult to remember or very very difficult to get the information on rate of reaction, this is simply homogeneous and the moment I go for heterogeneous you will have real help, it is not that easy for heterogeneous reactions to get, right, so simplest equations if you take even combustion, coal combustion, there are hundreds of equations, okay heterogeneous systems, right, so that is why the easiest one in fact is of course input is the easiest one you need not do at all ask MBA guy to get information, second easiest one is contacting because you have only 3 ideal systems, you know now at least you know under what conditions you choose what reactor and then the next one is to get the output is the kinetics and kinetics is the most difficult part, for example very innocent looking equation like this A plus B going to R, you do not know what is happening there, it can be very very complicated equation, it is not simple second order, right or it cannot be simple pseudo first order, by looking at that you can never tell that, so that means you have to go to experiments and experiments you have to prove that whether it is elementary with respect to A if it is first order with respect to B if it is first order then you may think that, okay one molecule of A B reacting with another molecule of B then I am getting product all that imagination comes after doing the experiment and then proving what is the order then your model will come or otherwise in the beginning itself imagine that A is colliding with B develop equations finally have some rate equation go to laboratory do the experiment and then check whether what you have written those equations exactly tallying with your experimental data that is another way I am not saying whether this is right or that is right so depending on your convenience you may start this way you may start this way either from experiment or from theory so that is why but what we know at this point of time is only one thing rate is a function of that is all rate of reaction is a function of concentration and temperature that is all I know now that functionality I have to find out so that is why for convenience sake I can also write this one as a function of C A and T which also can be written as some kind of function F 1 T and F 2 C A I have divided that function into two groups where the term which is depending on temperature and the term which is depending on concentration so by keeping the temperature constant that means at 100 degree centigrade exactly isothermal condition then I will vary various concentrations and all that and try to find out what is the concentration dependency on the rate dependency on concentration that is one thing so once I know that this is the okay for easy thing it is the first order then I know that term that functionality now I will vary the temperatures 100 120 140 160 then I will get now the temperature dependency or rate dependency on temperature concentration is already known so that is the two terms what we have to now learn and fortunately for us F 1 T is only K naught E power minus E by RT we have given this equation Arrhenius that is the only equation we have there are many finally it is distilled to that level right so how do you get this format like for example the that functionality you know that functionality T can also be called as K rate constant so that functionality is an exponential function how do you know that experimentally people have also found but from thermodynamics from collision theory from transition state theory people have different equations for rate constant okay so this if I say this one as rate constant K K now the functionality of K can be obtained from thermodynamics and also from kinetic theory of gases kinetic theory in this kinetic theory again we have collision theory and another one transition state theory yeah but finally all these things will give me that format and actually that format is a simplification of all these things except of thermodynamics almost rate you get so now let us quickly see how do we get that format because that is easy to start okay the functionality that concentration is much more difficult but here at least with the temperature we have only one term and that is easy to get and from thermodynamics first let me tell you how do you get that format this format means exponential term okay you know we are now trying to find out that K functionality using thermodynamic information and you know thermodynamic information temperature dependency constant from thermodynamics okay good if you have because you know simple equations we will take just to get the point A going to R maybe A gas B R going to I mean R also gas then I have here K 1 K 2 its equilibrium reaction right so you heard of front of equation front of yeah so the front of equation will give me yeah that equation in terms of differentiation capital K yeah this one here equal to delta H R by RT square okay that is the first equation I have written for kinetics right I mean these things are okay this equation 1 2 this is 3 yeah so this equation we can also write you know this K is nothing but K 1 and this is equilibrium constant capital K and I write small K like this okay and capital K like this there is no round there okay so this is capital K okay where capital K as K 1 by K 2 okay so now I can also write the same thing here this will be if I write substitute this is equation 4 if I put equation 4 in 3 then what I get is ln small K 1 by D T minus D of ln small K 2 by D T also equal to E 1 minus E 2 divided by RT square where you have this information where E 1 minus E 2 equal to what delta H R right yeah good so now I can take that similarities here for example this term is equivalent to this term this term is equivalent to that term okay so if I yeah so this is equivalent to D ln K 1 by D T equal to E 1 by RT square minus E 1 by RT because you know the answer you are telling but I think I also want mathematics already done it yeah so this may be a surprise test so if you simply write without writing anything in between that answer I will put zero because I think I cannot go to your mind and then see you know what you have done I want something on the paper right yeah so this is equal to minus E by or E 1 by RT so this is one of the forms so you know yeah ln K 1 ln K is minus E by RT instead of just writing E 1 and K 1 like that yeah so integration constant then I will write equal to K naught into exponential yeah that is what integration constant we should not forget yeah so numbers if I give this is 4 this is 5 6 7 8 so this is one conclusion from thermodynamics that the rate dependency k rate of reaction rate constant dependency with temperature maybe something like this but again it is not true unless you check with the experiments okay so now we can come and discuss with you know the kinetic theory kinetic theory particularly collision theory yeah in collision theory what is the assumption yeah so all the particles will come together they will be colliding but all the collisions are not energetic collisions they may not be resulting only some fraction of the collisions will give you the product that fraction is mathematically it can be proved as this term exponential minus E by RT by the way what are the units of this minus E by RT unit less okay so that exponential term will give me the fraction of collisions which are resulting the product so that is why that fraction this fraction will increase if I increase the temperature when you are increasing the temperature the energetic collisions will increase so that is why at high temperatures you will get high rate of reaction okay but from kinetic theory of gases we have an equation the equation is k is proportional to T to the power of half E power minus E by RT or if I remove the proportionality constant this may be some k0 dash T to the power of half E power minus E by RT that is the equation so equation 9 equation 10 okay so definitely this is different than the thermodynamic relation right okay good so now the next one next one because here I am not going into details I am just telling you we have kinetic theory I mean there is lot of information very very very old one I think 100 years back also they did that okay so finally there are equations and all that but our final information what we require is because we need an equation for k value which is depending on temperature so this is the information what you have to remember or if you want to remember I think actual derivations also you can remember right physical chemistry people would have done it already good so then yeah the next one is transition state theory what is the assumption in transition state theory anyone anyone who has come across this transition state theory transition state theory okay yeah that tells us if I have again you know a reaction A plus B going to R here our imagination is that first A and B again will collide that collisions must be there because they are moving they have to collide okay no gas will be at rest unless you go to okay because this A and B is always moving you will have the collisions now those collisions will not give you directly R right yeah then what it will give our assumption is that that is leading to some kind of intermediate then intermediate will have sufficient energy and then that will decompose to product this is another imagination you see you can have for the same experimental data thousands of models when you conduct for example A plus B going to R experimental data right and you measure what is the k values or what are the conversions or rates rates you can measure there so now you go to collision theory and now imagine that A plus B getting collided and then you will have rate of reaction some conversion or rate of reaction can be expressed now in terms of collision theory you will have a rate equation right so similarly now you have another model in your mind all models are in our mind why why they are in our mind yeah because you know exactly what is happening there it is there or not I do not know that is why now I am imagining probably this is what is happening like exactly I am looking at you I am thinking that I am doing a very good job but I think by looking at you but I think you may not understood anything till now what I have talked 14 class 4 okay so my concept you know in my mind what I am thinking is that you may be understanding but actually you feel that you are not understood anything how do I find out yes that is experiment correct no so similarly here also you have any number of models and now final test is doing experiment so once you have experiment if thereby and thus you cannot get same conditions you would not get you should not get different r a because that means you do not know how to do the experiment if you differently get at different places but conditions are same exactly same concentration same temperatures okay if you maintain all that anywhere in the world you do you should get the that is constant experimental data is the true truth but your model is not the true why this imagination of a plus b giving us a intermediate product then it is decomposing to the final product all that is a model your experimental data if it verifies there then you can say that okay my my transition theory is right but in this case if someone tries with okay thereby and tries with collision and swish method tries with transition state okay what do you do if both are correct in theory experiment they must be right you know so this is what is the current problem also because these models approximately tally all these they tally a little bit it is not that one theory is perfect for this another theory is perfect for this so that is the reason why even now we do not understand what is actual molecular you know collisions and then giving us the product that is why we are lucky as chemical engineers because we do not have to worry why at the end only I will see whether I get conversion or not how they are reacting okay yeah chemistry they have to you know break their mind saying that okay you know this molecule is not reacting that molecule is not reacting because that poor fellows they have to fight for that okay because our our objective is to produce the chemical somehow okay yeah so that is the reason why but still we should know the theory that is why we are telling that the transition state theory and one imagination as I told you here is a plus b giving me that is again a reversible step a b star that is the intermediate then this intermediate is it is not reversible intermediate is this is again an assumption so now I have here k 1 k 2 k 3 yeah okay all these constants right so intermediate is not permanently there because I think after sometime it will disappear but temporarily it will form and then go away so then another imagination for this is that I have energy and here we have reaction path reaction coordinates what we say and then reaction coordinate what we will say then you will have a little bit of dip here may be energy yeah okay this may be one situation and another situation you have again energy reaction path so it may be starting at low very small for some reactions you do not see that also much yeah like this so this one this one is e 1 this one is e 2 energy okay and here also I have this one e 1 this one e 2 did you see this diagram earlier and here we have the activated complex so it needs some energy to reach this peak and from there it is decomposing okay energy decreasing that activated complex energy is decreasing so then you will have in this case one type of reaction this case one of the another type of reaction okay it is e 1 minus e 2 where can you tell which is exothermic reaction which is endothermic reaction first one is exothermic why e 1 minus e 2 is yeah negative exactly so this is minus delta h r so which is exo and this one is minus okay delta h r no yeah plus delta h r which is endo this this diagram is also important because sometimes in the interviews and all that I may also ask in surprise test because surprise test is only for 5 minutes you have to clearly write draw the reaction coordinates and if you do not draw anything and then only draw those two lines I do not give any marks I am just telling you good so this is the one so this is another way of imagining the reaction rate but we are not going to that kind of so much details right but our functionality here is k is proportional to t to the power of sorry not to the power t into minus e by r t this is another equation so this is equation 11 and this also k is k not dash for example t e power minus e by r t this is yeah so now you see you have the real problem that is why I told you kinetics are really difficult for us okay so now you have three equations one from thermodynamics which gives me nice equation like this one from collision theory which gives t to the power of of some constant right and here another equation from transition state theory with you know k is proportional to t this is here k is proportional to t to the power of of this one here k is proportional to t to the power of 0 okay which one is right we do not know okay so people say that Arrhenius was very smart and then he told that it does not matter so the variation of this term because exponential term is much much higher when compared to the variation of this term k 0 into t to the power of of r that t so you take this entire thing as a constant which is independent of temperature so that means the variation in that constant will not be it is it is not much so that is why finally Arrhenius equation in this format it is accepted and that is what we use 99.99 times in all C R E books okay in all C R E books good so at least now you have one term very confidently known as that f 1 t function of you know temperature f 1 that term now we are taking only Arrhenius equation type that means k equal to k 0 into e power minus e by r t people call that k 0 as collision factor yeah so all kinds of thing we can yeah okay so finally that term f 1 of t will be k 0 e power minus e by r t right I mean actually we can also prove that you know Levenspiel does that I do not know in which volume he has done it yeah he does something like this k equal to k 0 t to the power of m yeah e to the power of m into e power minus e by r t yeah and differentiate with respect to you know take longs first longs of this so long k equal to long k 0 plus m long t plus correct minus e by r t now differentiate this with respect to with respect to temperature okay d oh sorry sorry thank you thank you who is telling that d l n k oh ok d l n k very nice so d l n k so then the equation what you get is I will leave it to you okay I will just leave it to you then finally this can be written as m r t plus e by r t square because I may ask this also in the separate test so that is why that mathematical details I have to see in your thing okay so I lost equations so this is 12 this is 13 this is 14 side business I have started here okay this is 15 this is 16 okay so now for many reactions what they found was m r t is almost negligible okay approximately yeah I do not say okay it is negligible when compared to I think I will write here okay negligible compared to compared to for this is very crucial m r t that m multiplied by r m is that exponent okay we think that m is only either it can be 0 0 means Arrhenius equation half means collision one means yeah it seems there are some models which m is 4 in some models some kinetic models kinetic theory of gases some more models also will give you not only half and one I think sometimes you have this information which model I do not know but I have that you know sometimes it can go to even 3 also 3 and 4 also okay but anyway we are happy to ignore this so that we have this kind of rate expression which is nothing but k rate constant equal to k not e power minus e by r t so this is the final term correct whole r t square whole r t square that is why you have to derive that and then you find out okay you derive that do not ask me now okay because if I ask I will tell the answer and you will give the answer because I also want people you know when they do not write also I am happy because correction is less so okay anyway good anyway it is very simple you can just differentiate and then take the l c m and all that you will get okay I am not really trying to say that you should not write anything I am happy if all of you get s real s not by copying and all that okay good okay so this is the one what we have in that first you know two parts we have the k part f 1 t and we have the other parties f f 2 c a now what kind of thing what kind of format you get for this f 1 f 2 c a okay if you are very lucky then this will be simple okay one one means if I have rate equation minus r a equal to simply k c a equal to 1 c a to the power of 0 so it is zero order reaction okay so zero that is why it is one we have told there is no fourth order in nature if they are elementary reactions that is very very important elementary reactions means they are behaving the way they are supposed to behave right but if they are not elementary then non elementary means that kind of mechanisms will come which one a plus b going to again some intermediate now it is straight forward that intermediate is going to only directly r so that way we will go to something else okay so like that you will have 4 5 steps the best example is h b r decompositions reaction to yeah h 2 plus b r 2 giving you 2 h b r it is not decompositions it is the actual reaction there how many steps and also there are some 4 5 types of mechanisms which are possible but experimental data is only 1 so you have to fit one of those things but that in a way I will tell you a little bit later so this is one or this can be c a okay I am telling about only that part because we told that minus r a is f 1 t into f 2 c a so now this one I know this is equal to k where k equal to this oh this equation number 14 oh 17 17 okay good so now this we are focusing that can be either f 1 f 2 c a can be c a or c a square c a cubed c a cube or in general it may be c a to the power of n where n can be 0 to 3 0 to 3 it can also be negative but negative orders we have not observed again in nature okay so that is why it can be but it need not be integer it can be even fraction fraction also so that is why I told you it is very very difficult to the kinetics part in reaction engineering if unfortunately if you want to design a reactor on your own from from scratch right so from scratch means you do not know how much you have to produce that is very easy to get then you do not know what kind of you know the kinetic stoichiometric equation but you do not know what is the rate of reaction minus r a what is the order and all that so you have to conduct the rate I mean you have to conduct the experiment to find out what is the rate and what form you will get you need not get all the time k into c a you may get any kind of you know crazy equation like you may have also c a by k plus c a are okay some other form k means again you may think rate constant ya so m plus c a that format so that is why absolutely you do not have any idea of you know what kind of thing you are you are going to get when you have this kind of you know in in kinetics that is why that is much more difficult for us to deal with when compared to contacting that is why I started contacting but you know I could have started either this or that does not matter but this time I started with contacting so that you will have 3 reactors and any one of them you can use to find out minus r a this one minus r a minus r a is f 1 t f 2 c a need not be always batch reactor it can be even in fact the c s t r is the best reactor to find out kinetics okay so now what we have to discuss is that yes if someone gives me I have a reaction stoichiometric equation okay a l plus b l giving me r l now I have to find out the rate minus r a this minus r a I have decomposed into these two f 1 t and f 2 c a but this fellow I know very well so this is nothing but minus e by r t but still I do not know c a how do I do that we have to do the experiment okay suppose I give you this a plus b going to r how do you conduct the experiment by the by what for you are conducting the experiment to find out the to find out the rate you should have that minus r a okay you should have that this is the problem with our all our labs you know without telling what is the exact aim you know without discussion much we give that you know to find only aim this much okay find kinetics but we do not know we do not explain to you why you have to find the kinetics and what are the problems in finding kinetics and all that and then we will also tell that okay go to lab you have a reactor you have a reactant a reactant what are the famous in the world reactants soda metaxa and 90% of the labs in the world conducts sodium hydroxide and ethyl acetate that is called estrification reaction yeah yeah safe means what is going to kill you means going to kill you or not yeah but not only that all that is apart the kinetics are known to us whether you are right or wrong you should know that is the reason why I think well behaved equation as we said all those things are right but apart from that we know the kinetics that is also one drawback okay because we know the kinetics you will somehow get the answer okay you do not really get the answer somehow analysis also is very but if you do it because you know k is this much at 30 degree centigrade somehow you will get it yeah but if I do not know all that and then if I am only talking about a totally new new stoichiometric equation how do you go about that everything is a problem I tell you I can give straight away mtech or ms or phd if I give you I think two components and then go to lab and then you conduct the experiment on your own we do not tell anything everything you have to do it you have to choose what is the size of the container you have to choose the temperature you have to choose the concentration you have to choose the analytical method I will tell you I think it is very very very difficult in fact we are not training you in that way that is why at least you know I ate amadas do not they may not listen to me when I say in our department but I told in amruta universities or for mtech students do not tell anything ask them to go on do some experiments let them define their own experiment I want to do only flow through pipes okay or that means he has to choose the pipe he has to use the flow rate he has to construct the you know the all that connections pump and all that pumping and all that then a distillation column means a small distillation column he has to do it then you learn a lot but unfortunately in no other education institutions we are doing that allowing you to do all that we tried in our 8 amadas experimentation so we gave fluid mechanics lab for btech students long time back I do not know still they are breaking so you know there was a column like for example pipes were there right fluid mechanics no or fluidized bed was there so what they did was they removed all the pipes they do not know how to put it back again so they waited 2 days 3 days 4 days nothing is coming then mechanic has to come and then put it back destroying is very easy creating entropy is very easy but decreasing entropy is very difficult right I do not know whether you understood what I said yeah creating entropy we are all entropy generators one small bomb here in this room that is all this will disappear and you will get the maximum disorder when you have maximum disorder what will be the entropy yeah very high yeah so but now order comes if I ask you again okay you have exploded that building now you construct in the same way same shape you bring it you know how much time it takes how much energy required that is what is the difficulty to go in that direction you know the decreasing in entropy increasing entropy is very easy that is what is happening throughout the world increasing entropy not only I think that is why the thermodynamics is a philosophical subject it is not destroying building or increasing disorder by moving or by sending cars and I know all vehicles and all that not only that it is philosophically also our brain also goes for more and more disorder that is why 100 years back the sincerity the kind of people what we have all those people when you think at least you know we would have not seen them but at least we have some information on that and how ethical they are how good they are how they were treating anyone when they come you know all over anywhere on this planet I tell you