 So, the last class we have been talking about something called global reactions and what we found about global reactions was that they do not really happen in reality they are happening at some sort of like a conceptual level in our minds and we are trying to represent what is happening in reality at the molecular level by these global reactions and therefore the kinetic parameters need to be obtained empirically for the considered species in those global reactions. We also found that global reactions need not be single step they could be multi-step as well in order to take into account some of the intermediates that are not necessarily highly unstable but those that we want to keep track of for some purposes but in reality what we are having is reactions actually happening at the molecular level and these are what we call as elementary reaction steps. So if you have to consider reactions happening at the elementary reaction, elementary reaction level then what we have to do is we have to look at what is called as detail chemistry. So in general what happens with these reactions is there are a typically a huge number of reactions that happen and involve a considerable number of species. So typical numbers we are looking at are tens of species or maybe up to 100 species and maybe about close to 100 or more okay or few hundreds of reactions that are happening. Now many times it is either difficult for us to actually know what these reactions are or second thing is it is difficult to actually get the kinetic parameters for each of those reactions okay. So these are the first two steps for us to actually get information from the chemical kinesis. The third problem is let us suppose that the chemical kinesis were able to actually give you everything that they could possibly give right and let us say that they say okay you are looking at this reaction mechanism and it involves at least about 1000 reactions these are all the reactions and these are all the kinetic constants for those the question is can we handle them. So many times it is not possible for us to handle so many reactions in the first place number one number two we may not need to handle all the reactions as well okay. So just because that is what is happening in reality does not mean that we have to engineer from an engineering point of view consider all of them right. So many times what is typically adopted is reduced mechanisms reduced mechanisms or most often adopted. So the difference between reduced mechanisms and global reactions or reduced mechanisms still involve a reduced set of elementary reaction steps at the molecular level right. Therefore we use molecularity in the law of mass action right that is we do not use orders for the reactions in the law of mass action we actually can go ahead and use molecularity for writing out the reaction rates for each of those elementary reaction steps in a reduced set right. So that there is no ambiguity there about what we need to do and other than obtaining the reactions themselves from an empirical manner or getting the element the kinetic constants empirically even for the molecular level reactions we are not necessarily looking at empiricism in order to get the orders it is it is basically based on what happens at the molecular level. So we just use molecularities rather than orders there but the key is what is this reduced set right how do you determine this reduced set now that is actually something that is very difficult we do not want to get into that in this in this course. So reduced reduced mechanisms are determined in a number of ways the easiest I am not going to say anything much more about this but the easiest for us to think about is look at the rate determining step the rate determining step is typically the slowest step that is happening. So if you now have a sequence of reactions that are happening right so I have now a example of a reduced set that is put up here right for something as innocuous or very simple as a hydrogen oxygen system and you would think that it is only hydrogen and oxygen there is nothing more right but look at how many things that there are going on over there and this is like a one typical example of a reduced set of reactions that are happening with just hydrogen and oxygen to get you water that is the ultimate goal okay. So we are looking at a global reaction as H2 plus half photo gives water or something like that but in reality you are having at least nine reactions that we need to consider for doing something sensible typically what is how it is being done first of all what we have to look for is what is initiating this so there must be like one reaction or at least one or two reactions that initiates the reaction that means you start with a stable reactant and M corresponds to any species that this can bombard with so it could be it could be itself it could be O2 to begin with or once this has got started this reaction would still be happening so it could be any of those OH or O or H or those or even walls of the vessel okay like the surface right and many times we have a dilemma for example M is like a gas phase species then its concentration is important in determining the rate of this reaction okay and we will go through these as we go along but essentially that is how you start the reactions that is you now have a one of the stable reactants begin to decompose into atomic level radicals or intermediates so this is H for example as an intermediate in the big global reaction that we are looking at H2 plus O2 gives H2 O in whatever proportions H does not figure in O does not figure in OH does not figure in HO2 does not figure in right so all these things are essentially then intermediates that we have to recognize as such so you now have a chain initiation step this is a chain reaction and in all these things you have to stop start talking about chain initiation chain branching chain branching chain propagating chain terminating and so on okay so it is actually a reaction chain that is happening so when you say chain branching what that means is it actually multiplies the number of intermediates you see you started out with one stable reactant and one intermediate but it results in two intermediates that means it branches out okay so the third one is also actually multiplying the number of intermediates the propagating steps are sustaining the number of intermediates that means you now start with H plus O2 plus M gives HO2 plus M so it is like you had H to start with O2 is stable MSA any third species which is staying there on the other side you started with one intermediate you are sticking with one intermediate okay so you look at HO2 plus H2 gives H2O plus OH H2 and H2O are stable but HO2 and OH are intermediates so you now propagating okay that means you are not really branching you are not terminating but you are just propagating the number of intermediates keeps preserved right so those are the propagating steps and finally you are also looking for some terminating steps and in some of these propagating steps is when you are actually beginning to produce the final product the stable product okay so if you are really interested in finding out the rate of production of the final stable product in what you could think of the global reaction in your mind you should keep these reactions if you throw away the elementary reaction steps that are producing the final product in your reduced mechanism then you are not producing what you want right so you will have to you will have to keep these and then you have to look at the rates at which these are happening and then what happens so you now look at H2O that is being produced it is depending on the concentration of HO2 and the concentration of H2 concentration of OH so you have to now look for those reactions which are producing OH and HO2 all right now you might find that there are there is only one reaction that is producing HO2 when compared to a bunch of reactions that are producing OH so obviously we might think that the root by which H2O is produced from HO2 is less important when compared to the root from which H2O is produced from OH okay that is one way of thinking about the other way of thinking about it is look at the reaction that is producing HO2 it is a term molecular reaction okay that means H and O2 have to bombard in the presence of M that means three bodies have to collide for you to be able to produce this this is a highly unstable intermediate does not happen very prevalently right so the reason for that is term molecular reactions are important mainly but high pressures so it is only at high pressures we are actually trying to put a lot of species together in a much more compact region because you are trying to pressurize right and that is when the probability of term molecular reactions actually increases otherwise it is pretty low therefore we could say at moderate pressures or low pressures let us not worry about this for example right so these are the ways by which you would think about trying to pick some reactions over the others among these one of the chief things that I am talking about is we now say k1, k2, k3 etc. Each of these are actually an Arrhenius expression that you can write which indicates the exponential dependence of the these on temperature so if you now have a temperature bath at which these reactions were to take place in reality for the flames it does not happen that way okay in reality for the flames the temperature continuously changes in space and for unsteady problems the way you want to pose unsteady problems it might change in time right so it is not like a nice constant temperature bath but if you want to do for kinetic purposes think about a particular temperature at which you want to compare the rates at which these things are happening which is what do you call simpler it is actually simpler to think about that way okay in reality it is possible that you have to have a change you know like last time I said only if CO is produced can it get oxidized and then it is going to compete with the oxygen that was used to produce it in the first place then in a flame you might find that depending upon the reaction rates of the two reactions that we are thinking about they could be spatially displaced you might have like as the reactants are coming in you first have to have one reaction happen and then feed and convectively feed into the next reaction zone so the reaction zones could be separate it is not necessarily at the same place and therefore the temperatures of these two reactions may be different and depending upon the temperatures at which these two reactions are taking place one of them could be faster than the other merely because the temperature was different right so comparing these things at the same temperature is not a very straightforward thing but if you think about it from simpler for the purpose of simplicity right let us say you did that other than having to sequence them because one has to be produced one has to produce H in order for it to be utilized okay so keeping the sequence in mind the next thing that you want to worry about is which one is actually the slowest okay so when many things are happening in parallel okay if you now want to say I want to look at the whole thing as like one black box I am putting in hydrogen and oxygen here I want to get water there all this stuff is happening tell me how fast this is going to happen then all I am going to be thinking about is which of those reactions is the slowest to happen that is the one that is going to determine the rate at which the whole thing is going to happen right because until that is complete I am not going to get out water okay so all other reactions could have already happened but it is waiting for this last reaction to happen you see so the rate determining step is the one that is actually the slowest with the slowest reaction rate so you could also pick these reduced mechanism reactions based on the reaction rates okay and a much more sophisticated way of doing these things is called rate ratio asymptotics that is something that quite a few people have done in the last many decades to try to reduce reduced mechanisms okay to actually compare rates and then look at asymptotic expansions based on those and so on and we do not want to go further into that because it is an area that can take pretty much all your time thinking about so in what we have done we have introduced a couple of ideas here first of all the right thing to do instead of adopting global reactions is to actually adopt reduced mechanisms because the chemical reaction rates that you would want to write the law of mass action would be less ambiguous but then what could be ambiguous is to how to reduce the reaction mechanism and there are people who work on this separately so if it is possible for you to actually look at the literature for your application let us say you are now looking at a partially premixed flame under some conditions or something like that so particular a situation there are people who work on this do experiments and deduce these things and so on and also do the analysis for this and try to try to recommend reduced mechanisms many times as we go along between the time I graduated or I went to grad school to now the last 15-20 years for example things have actually grown so previously when two reactions was considered to be too much okay now your chemist is going to say take these 50 reactions again just take it and then you really like can I reduce further or maybe you do not have to so you have computer power to handle those 50 reactions so just go ahead and try to use it and so on so it is a it is a it is an expanding area where things keep changing what is required keep changing with computer power and so on. So therefore the right thing to do is for you to actually try to keep as many elementary reaction steps as possible because you are closer to reality and the second thing that we have also come across is the nature in which these reactions happen right that is the initial they get initiated the branch off then propagate then terminate so that sequence right. So in this sequence what is essentially crucial there is the role of intermediates the presence the formation their presence and the role of the intermediates in each of these because when we are now trying to identify a reaction is branching or propagating we are actually looking at how many intermediates are showing up on either side you see. So we have now also started looking at intermediates as one of the important things in the reaction scheme and finally I would like to point out that we now have a reaction scheme that put together is the one that is now going to give you the reaction rate expression that depends on the reactant concentrations for the stable reactants the original reactants. If you are now basically looking at H2 and O2 giving H2O you finally want an expression for the rate of production of water which depends on the concentrations of H2 and O2 right but that is something that you want to deduce from this entire soup of reactions water is happening here it depends on this and this this and this so this depends on this so it depends on this and this depends on this and this keep going back ultimately to wherever you are showing up as H2 and O2 the reactants. So we want to find our job is finally to link up the rate of production of products to the rate of production of reactants through the rate of production of the net rate of production or depletion of the intermediates right and if you now are able to actually look at how my global activation energy for example is going to fare given the activation energies of all these you see because you have kinetic constants for these and finally we want to deduce a kinetic constant for like a global reaction from this let us suppose right. What is going to happen is if you have more and more chain branching and propagating reactions the global activation energy will come down that means its sensitivity to temperature actually diminishes. So there is a role for consideration of detail chemistry and the way you want to reduce the chemistry in how it influences your global activation energy or the net activation energy for the entire system net activation energy essentially tells you how the full system of border production from hydrogen and oxygen depends on temperature right how the non-linear dependence is right. So there is a significant impact that this has on the global performance of the system. What we will then do is come up with a couple of approximations which we can try out with these intermediates in trying to quickly put together this connection between rate of production of products and concentrations of the reactants okay. So we let us get into that game of getting like a global reaction rate based on the reactant concentrations through the intermediates right. So some easy ways of doing this is one steady-state approximation steady-state approximation right so in steady-state approximation what we are essentially saying is if you now are trying to plot your concentration of different species with respect to time right initially at t equal to 0 you are starting with a bunch of reactants at some concentrations let us say we have some reactant that we can follow and as the reaction proceeds the reactant concentration will decline right with time. So that is not a you do not get any prizes for this right and what happens to the concentration of the products it will increase right how does it increase it increases like that it should increase like this because ultimately at t tending to infinity it does some thought right. So well let us not worry about how is there any inflection here and so on but ultimately we are looking at a picture that will emerge like that now the story is about the intermediates it is not about it is not about what happens when we are born or when we die it is about like the life in between right so that is the intermediates life so all life is intermediate essentially so question is what do the intermediates do they are born and then they die right like all of us so the question is how do they behave they have a maximum right so it goes and comes down does it go up and then come down immediately or does it want to live for some time like us right so it does that right so the intermediates you know they produced and then they want to stay for some time as long as they can so it is not really like a peak at the maximum it is a plateau at the maximum okay so they produced and then they are like yeah I am going to be here for quite some time yeah sure let me let me have some fun that is the way it goes and then ultimately it dies okay. So what we are looking for is it possible for us to exploit the situation say let us not worry about the initial transient when the when the intermediates are just beginning to get initiated and branch off okay or let us not worry about when they are busy getting terminated okay can we look at the region when they were actually simply propagating when the concentration is going to be more or less constant and if that is the case then we suppose that that is around like a steady state things are not changing with time and so we approximate the concentration of an intermediate approximately equal to the rate of change of concentration of an intermediate with respect to time is approximately equal to 0 where a star is an intermediate this is the sub this is the essence of the steady state approximation okay. So what do you get out of this is what we have to see right so we use this we use this approximation to obtain global kinetic behavior what is meant by global kinetic behavior that is like linking rate of production of final products to react in concentration to react in concentration that is like the global kinetic behavior right. So let us suppose that we consider consider consider a first order reaction first order global reaction I will put this within codes first order first order global reaction a gives products we do not really worry about what the products are because the law of mass action says that the reaction rate does not depend on the concentration of the products it is only the reactants that we have to worry about not that we are really worrying significantly about the reactants it is just a so it is a template reaction right now we could examine the following detail mechanism involving an intermediate involving an intermediate that means whenever you want to propose a theory the quickest way to publish is to make your number of variables as little as possible right so just keep it one so you are looking at the role of intermediates just keep an intermediate with you do not get into the soup bowl of intermediates right. So this is due to Blinderman so let us suppose that we now have a plus a gives and takes a star plus a and we denote the forward reaction by the rate constant kf and the backward reaction by the rate constant kb and let us suppose that a star gives products a star gives products and let us call this second reaction rate constant is kf kf I am sorry kf this is actually a set of three reactions okay so a plus a gives a star plus a is one reaction a star plus a gives a plus a is another reaction okay each of them having kf and kb as the rate constants respectively and the third one is a star gives products with a rate constant of kf prime. And let us suppose that this is actually this set of reactions is fast and this one is the one that is slow that means the moment you have a it produces a star and it now starts getting in equilibrium with this right the more the more a star it produces a star combines with a to produce a back therefore there is an equilibrium that can be associated with this okay so and then these things are happening very fast so it is like you now finally have like a some pool of a star now what is the job of a star job a star's job is to actually produce products it is not a is job a is job is only to produce a star and get into equilibrium by that that is what the first two reactions do producing a star and getting into equilibrium okay the third reaction is where the a star is the one that is producing the products and that is low okay. So now we are beginning to think what would be the order now think about this what would be the order of the global reaction a giving products that is what is supposed to happen okay and all this is like details that are happening in reality okay so let us now think about how to deduce this okay so the rate controlling reaction is the slow reaction right so which is unimolecular which is unimolecular which is unimolecular a star giving products is a unimolecular reaction therefore we expect the mechanism to behave as a first order reaction globally right because it depends only on concentration of one of the species okay so what is meant by globally like for example if you now think about these things as gases which is what we have been doing all the time right the concentration of these is actually directly proportional to pressure right if you now had a bimolecular reaction then the law of mass action will have a p2 so the question is essentially are you going to have the reaction be reaction rate for the final global reaction go as linearly as p or square of p or somewhere in between right order for the global reaction does not have to be an integer right so what is it going to be so in fact the way to pose the problem a little bit more general is consider a global reaction n a gives products okay and we basically for the global reaction assign a value of n based on what is the order that is deduced from the detailed reactions the elementary reaction steps right so question is then n equal to what is it supposed to be first order or second order or somewhere in between or what or anything else that is exactly what you are thinking about with this being the detailed mechanism in this case okay so but you see the first reaction is bimolecular bimolecular which implies reaction rate goes as p square and that is important for us to think about because what is ultimately going to affect is as far as a reaction rate is concerned is the concentration of a raise to whatever the exponent is and the concentration of a shows up in the first reaction which is bimolecular okay even though the second reaction is the slowest and therefore the rate determining step it is actually in a star so its reaction rate is going to actually depend on the concentration of a star rather than a you look at the reaction that is actually depending on the concentration of a the first reaction tells you it is bimolecular right which means it has to go as square of pressure okay and recall or note okay note that term molecular reactions term molecule reactions have omega goes p cubed okay square of like extending from bimolecular if you have p square and term molecular reactions of course you are not looking at term molecular reactions here just mentioning this because what we have to consider is it is going to go as p or p squared p cube okay then what happens is if you now are working at a high pressure then p squared is going to be lot sensitive when compared to p okay or p cubed is going to be a lot more sensitive to pressure than p squared and then p but if you are working at low pressure so the pressures are low p squared will be lower than p and p cubed will be still lower than p you see that it becomes like anti sensitive whether p cubed or the term molecular reactions now become a lot more sluggish when compared to or they are sensitive in that sense that as you now decrease the pressure they suddenly become or very drastically become less important right and it is the p squared ones are less important when compared to the p reactions. So here at normal pressures normal pressures the forward and backward or reverse reactions okay reaction rates go as p squared and third reaction rate go as p but at very low pressures but at very low pressures then the first one actually becomes low the first two reactions become slow right the first two slow down and the third predominance right that is what we expect that means we cannot put a number to n for all pressures depending upon the mechanism which we have which in this case we have considered a particular mechanism we can see that the number for the order could be anywhere between one and two okay. So normally it could be like a second order reaction at low pressures it could actually behave more like a first order reaction okay so you now beginning to look at this possibility so how do you demonstrate this the answer is we now try to adopt the steady the steady state approximation so we may write dca over dt equal to – kf ca squared plus kb ca ca star so this is for the depletion of a this is for the production of a okay a is being depleted because of the first reaction that is being produced by the second reaction okay this negative sign with a coefficient one takes care of the fact that we have two molecules of a consumed versus one molecule of a being produced even in the first reaction so it is taking into account the fact that there is some partial production of a there alright and still there is like a net depletion that is the reason why you have a negative sign and so let us now say this is one and dca star developed by dt equals kf ca squared – now a star is actually being is participating in three reactions okay it is being produced in the first reaction and it is getting consumed in the second there is a reverse reaction there and the third right so kf ca ca squared is actually the production rate for a star but it is also getting depleted twice which is kb ca ca star – kf prime ca star let us call this two right our goal is to actually solve for ca and ca star that is what we supposed to do okay and this is as I said a system of two simultaneous ODEs and we have to supply initial conditions and of course I can gladly tell you that ca star initially is equal to 0 okay so I got it and then I will supply some number for ca ca and starting from there you want to now march in time and then see how these things grow and then try to fill up this graph essentially right that is what that is what you are supposed to do so what you could do is you know in the in this equation you could try to eliminate ca star let us say you are not you are not interested in ca at all ca star at all okay so you can eliminate your ca star from these two equations and then you can also try to get your dca star over dt which will involve a dca by dt when you try to eliminate ca star and then plug for ca star and dca star dca star by dt and you now get like a humongous looking non-linear second order ODE for dc for for ca that means you will have a d squared ca by dt squared term you could also potentially have because you are showing up these squares and so on you will also have some dca by dt the whole squared terms and then dca by dt and so on so you now have like one ODE which is like second order non-linear in dca by dt okay and then you can if you can you can solve that okay so trying to eliminate ca star and dca star by dt of from the above will result in in a second order ODE non-linear ODE for ca that is ca star equal to kf ca square plus dca over dt divided by kb ca that is something that you can get from 1 okay let us call this 3 and dca star divided by dt equal to kf divided by kb dca over dt plus 1 over kb 1 over ca squared dca over dt the whole squared plus 1 over ca d squared ca over dt squared and then you now try to put 3 and 4 back in 2 and then you will now get an equation that involves only ca and its derivatives okay but it is going to look so ugly that we do not want to consider at all instead what we want to do is steady state approximation approximation means that we now say dca star divided by dt is approximately equal to 0 that is exactly what we started out with to consider right so do this and then we will try to take it up from here and see what we get next class.