 So, we have been looking at the approach of partial equilibrium approximation and as a template set of equations or reactions I am sorry we could let us let us look at about three forward and reverse reaction pairs leading up to a fourth reaction which we deliberately take as a term molecular reaction so that it is slow the idea being that if you now have these bimolecular reactions happening relatively faster and they are forward and reverse reaction pairs then we can assume that because they are happening quite fast they could stay in equilibrium with each other which means their respective forward and reverse reaction rates would match and therefore the ratio of the rate constants of the forward and reverse reactions in each case would turn out to be the equilibrium constant based on pressure right and so we can write these equilibrium equations for each of those reactions and therefore we now are rewriting these and numbering them is 1, 2, 3 so what is going on with this here is here we are saying we recognize we recognize a2 b2 and a2b as stable species and ab and ab as intermediates right we are interested in we are interested in d over dt of concentration of a2b that is the final product whose rate of production we are interested in right now looking at the fourth reaction where this is being produced we could write this as concentration of a times concentration of I am sorry equals kf4 times concentration of a times concentration of ab times concentration of m right but we need to know what these are the concentration of a and concentration of ab which are intermediates okay in terms of concentrations of the stable species in the system namely a2b2 and a2b if we were to suppose that we do not have all these things right there is all too much detail all I want to think about is a2 plus b2 gives you a2b or maybe a2 plus half b2 gives a2b right I would I would like to write the rate equation as dc a2b divided by dt equal to k times small k times concentration of a2 times concentration of b2 to the half but there the powers that we are looking at one and half should really be orders and not molecularities because that is like the global reaction and we are not sure that we can actually use these stoichiometric coefficients in the global reaction for the orders right that needs to be empirically found and that is what we are trying to do with the detail reaction and here the detail reactions tell us that this is actually in terms of the intermediates which we do not want to deal with we want to now express the concentrations of the intermediates in terms of the concentrations of these stable species when we say stable species we mean both reactants and products because it is possible that the concentrations of these intermediates depend on the concentrations of products stable products as well as concentrations of stable reactants. So we shouldn't distinguish between products and reactants of at the stable level we should simply say stable species right so how do we find these is how is why we write out these three equilibrium equations and you see what happens here the intermediates are a b and a b and we have three equations so we strictly speaking if assuming that we are we are okay with dealing with a2 b2 and a b a2 b right is it possible for us to write a b and a b in terms of the stable concentrations right that is what we want to do having said that we are interested only in these two we are not really interested in concentration of b because that is not part of the final answer so is it possible for us to eliminate concentration of b in these equations and look for only expressions for concentrations of a and a b in terms of a2 b2 and a2 b is the rigmarole that we have to go through yeah so eliminating eliminating CB from equations 1 and 2 you can do this so you can now write this with in terms of concentrations of a and b as if you are now looking at only a and b do not worry about a b let it be there and then we have a and did I make a mistake which I might have made a mistake let me just check this a b plus a 2 yeah sure yes I understand what you mean yes you are right okay so you are right so strictly speaking what I need to do is to factor in okay let us do that thanks plus K F3 being produced in the forward reaction C C a b C a2 minus K F key K b3 C a C a2 b C a yeah I was wondering about it myself because it is not so simple it is not strictly speaking so simple you could you could do this but what you would find is this is equal to this because we are assuming equilibrium alright therefore it gets cancelled right so it is always good to go through the details and yeah therefore we are we are stuck with only this right if we if this were not to be cancelled then we understand that this is unknown but this is to be treated as known and this is to be treated as unknown and this is unknown and still we would be requiring concentration of a b and a to be expressed in terms of a 2 b 2 and a a 2 b this is the procedure that we are looking for so if you now eliminate C b from equations 1 and 2 I am sorry here yes that is what I was searching for right so yeah so we now have this so now so you can rearrange these things do not worry about a b showing up which is an unknown right but you try to eliminate C b what you get here is then C a b equal to k p 1 k p 2 C a 2 C b 2 the whole of the half you can work out the mathematics and you will you should we should be able to find that this is the case then using 3 right so we have used 1 and 2 equations let us go to the third equation we should be able to find there is a C a sitting there and a b is something there so we can plug this expression there and so we can find C C a equal to I am just giving the final answer k p 3 k p 1 k p 2 to the half C a 2 to the 3 halves C b 2 to the half divided by C a 2 b therefore right if you now use if you now write this equation as 4 right that means we are primarily interested in this expression because those two terms got cancelled so what happens is in 4 in 4 d C a 2 b divided by dt is equal to you would have a k f 4 coming in and therefore start with a small k f 4 then you have everything else showing up k p 3 k p 2 k p 1 C a 2 squared C b 2 C m that is still there let me just write this in the same line divided by C a to b so that is the answer for you I just worked it out for you so you know you know you are not supposed you should now know how to do these things right so look at what has happened we now got an answer based on the stable species which is a little more surprising than what you would have expected for a global reaction all right which is first of all it is going a square of a 2 so you might argue let me write the global reaction is instead of a 2 plus half b 2 gives a 2 b how might write this as twice a 2 plus b 2 gives twice a 2 b all right but then that would have meant that I need to have a 2 here which I do not okay and then you would be able to say I can put a square here and a power 1 there but the other two are extra that would have that you would have never figured out right unless you went through the details and if you went through the details as it is without the partial equilibrium approximation it would be a lot more horrendous when compared to having these algebraic equations replace your ODE's for the rates of the intermediates right. So so you now have a Cm which is sort of like saying it depends on the container what is the total number of moles that is there in the entire container because as the pressure increases this now goes as the whole thing will now go as p2 ppp to the p to the 3 because this is now going to have a contribution to the pressure this is going to have a corresponding contribution to the pressure and finally all the four of them should gang up and say it is going to go as p p3 right so this acts in bringing that up number on the one hand second what is the a2b showing up in the denominator of something that we went through last class self inhibiting reaction right the more a2b that you produce the less the rate at which you are producing it is what this really means and that is because a2b is actually participating in the reverse reaction as well so more it is getting produced it is also getting depleted all right. So all these things kind of come up in these expressions based on the approximations that we do so let me also give you a simple example which is a little bit more realistic here so example just like how we had the what should I say the hydrogen bromide example for the steady state approximation here let us suppose that we have H2 plus O2 gives and takes with the K1 here K2 for the reverse 2 H OH on the other hand and to H2 plus OH gives K3 H2 O plus H okay so here we find that D over DT of CH2 O equal to K3 CH2 COH and I do not like COH sitting there in this I would like to actually have H2 is already there I would like to express COH in terms of H2 O2 and maybe H2 O that is permissible right that is what I want to do so I use the equilibrium approximation I use the equilibrium approximation and say Kp 2 I should Kp 1 let me use the notation Kp 1, 2 equal to COH2 divided by CH2 CO2 right all right that is equal to K1 divided by K2 all right. Now keep in mind when you write Kp strictly speaking we should be writing POH2 divided by pH2 POH2 okay it just turns out that the molecularity is the same for both the forward and the reverse reaction and therefore the total pressure gets cancelled out and you can write in terms of concentrations directly if the molecularities are not the same for both the sides then you will get into trouble so a more direct thing to do is to probably write Kc instead of Kp but if you are insisting on Kp then there is a conversion between Kp and Kc that we have gone through that is that you need to factor in yeah. So if you now use this then what happens is from here you try to evaluate what your COH is so COH then is Kp 1, 2 CH2 CO2 all to the half is what we have and therefore the CH2O is K3 let us now get rid of Kp 1, 2 use K1 and K2 because they are like given information in the reaction scheme and you can say this is K1 divided by K2 to the half CH2 to the half goes with CH2 there and so it is like CH2 to the 3 halves and CO2 to the half remains so you have a CO2 to the half right so that is the final answer what it means is if you were having this as the reaction scheme like a three-step reaction scheme for production of hydro water right then the partial equilibrium approximation tells you that this is partial equilibrium tells you that the rate of production of water actually depends to three halves the power of CH2 and one half to the power of O2 okay these are now the orders with respect to H2 and O2 respectively together if you now look at the total order right three half plus one half is two so you can easily see that this is like a second order reaction that we are looking at all right but the the division between the fuel and the oxidizer is not equal that is the that is a lesson that we learned from these things all right let us now do some chemical explosions based on what we have learned okay so this is the last topic that we would like to cover under chemical kinetics before we proceed on further with other things in combustion so so let us let us consider the following schematic reaction mechanism 1 x gives you r with reaction rate constant okay 1 2 r gives you alpha r with a rate constant K2 3 r gives you K3 P plus r 4 r goes to S and 5 r K5 goes to X now this is obviously looking a bit strange if it is not looking strange then you must be a genius right what we mean by a schematic reaction mechanism here is this reaction scheme is focusing only on the intermediates that means there are hidden stuff around that are stable on either side which we are not worried about okay why so because you can now clearly see that this is a chain in its initiation step right because this is the birth of an intermediate or is the intermediate that we are looking at or becoming alpha r for alpha greater than 1 means we are now branching more and more okay so this is like some intermediate giving rise to more intermediate right so this is chain branching of course alpha less than 1 would mean chain termination kind of reactions right but they do not they do not really show up like high up in the list okay they are kind of the end and r gives P plus r P is like a stable product like let us say we are interested in this finally okay but then what is what is happening you start with our you have something else and then you also have r so as far as intermediates is concerned it is neutral that means it is a chain propagation yeah so you have a chain chain propagation step in reality you would have a bunch of steps like this okay so each of these kinds of things could be like a bunch it is not just one okay r going to s is schematically representing chain termination at a surface whereas r going to x means chain termination at in the gas phase in the gas phase x is any molecule right so you could have started with any x giving rise to intermediates and r could go back to any x okay so in that sense this is not strictly like a reverse of that because x could be anything yeah but we are we are we are making a distinction on the way the intermediates are getting terminated either at the surface around anywhere or in the gas phase by colliding with other molecules around and disappearing and becoming some stable stable species so note that we make a distinction between termination at the surface at a surface it does at a surface and termination in the gas phase why why would you want to make the distinction how does it matter a termination means the intermediates disappear right how does it matter which way they disappear what are we looking looking at so whenever we were doing things in the past I asked you to keep an eye out on distinctions between unimolecular reactions to bimolecular reactions in a scheme or bimolecular to termolecular reactions in a scheme and so on and the reason was to look at the pressure dependence of the rates of these reactions okay or the differences in the pressure dependence of the rates of these reactions and so as you change the pressure from low pressure to intermediate pressure to high pressure the importance of these different reactions becomes different right. So effectively we were looking at what is it going to be in terms of pressure similarly looking at the termination either at a surface or in the gas phase is going to be dependent upon pressure because a surface termination is relatively independent of pressure a surface is a hard surface it is not going to it is not going to change its packing depending upon the pressure okay whereas a gas phase is going to actually get more and more packed give it with less and less mean free parts for the collision between the molecules as you increase the pressure. So what you would expect is as you increase the pressure the termination via the gas phase is going to be more and more predominant when compared to the termination via the surface okay. So this is this distinction is going to enable us to tell us how things are going to change with pressure right as the reason why we want to make this distinction so there is another way of distinguishing things as far as particularly intermediates is concerned okay. So the term molecular by molecular and so on is for any species it does not have to be intermediates but here we are specifically talking about termination for intermediates in two different parts and they are dependent pressure in different ways okay that is what we want to try to explore okay. So then let us look at how to write the reaction rates for these now I am going to write these things in such a way that we care only about concentrations of the intermediates we will not worry about concentrations of stable species okay they are not showing up in the schematic reaction reaction mechanism anyway okay there are still stable react species around which are which we are not simply bothered about because when we looked at a reaction mechanism before to characterize it as a chain initiation branching or propagation or termination we were primarily looking only at the intermediates okay we just disregard at the presence of these stable species you are always looking for our intermediates being created more of them created than what was being consumed or the same amount is created and we did not even distinguish between the intermediates like it could be OH on one side and H on the other side does not matter they are all very active radicals anyway right so they are going to kind of propagate the reactions if one of them got killed and the other one got produced right so this is the way we were looking at it and of obviously they were reacting with stable things like H2 or O2 and so on which we did not really worry about and these are all there so since they are they are hidden we cannot write real equations exactly depending upon their concentrations so we are going to write equations that are sort of functionally correct but not accurate right so that is what we will end up doing so let us say the first equation so the the rate equations for these these are one let us say we call this omega i for the chain initiation step all we are going to write is dcr over dt because we do not want to be worrying about the fact that it depends on the concentration of x okay we could write k1cx for example okay let me not worry about it okay I am just going to keep this as omega i right second omega let us now call this chain branching so we will call this omega b that is equal to dcr over dt as well for this reaction here this would be written as k2 strictly speaking I should write a-1 k2cr times other things which are hidden right so I since I do not know what those are I am simply going to write this as equal to fb times alpha-1cr alpha is a parameter here parameter fb is like it is going to take care of k2 dot dot dot dot is like concentrations of stable species okay which does not enter the picture so essentially what you are saying is let us let us deal with f rather than case because the f will couch things that are not really important what we what we want to explicitly keep a cr okay and so this is a production step so we will call this as omega p the third step this is dcp over dt this is equal to fpcr okay here the intermediates are neither getting produced nor consumed on the whole but so if you could say that this is this is basically fp is of the order of the same as k3 you do not but you could have something hidden right so we just write f instead of k and for let us call the surface termination reaction rate so we have a negative dcr over dt there and that is fscr we do not know what else is there other than or on the left hand side so we just write fs instead of k4 and similarly let us call this omega g for gas phase termination reaction rate this is again minus dcr over dt equal to fgcr all right okay. So we have explained what the fs are there and I also told you what the why this is a schematic here so the rate of production of the intermediates then the net rate of production of intermediates intermediates dcr over dt from all the reactions equal to omega i plus fv times alpha minus 1cr minus fscr minus fgcr the third reaction did not have a net rate of production of intermediates at all because it is a propagation reaction yeah therefore we have only four reactions that are contributing here two of them where things are produced if alpha is greater than 1 that is if alpha is less than 1 this would mean consumption anyway and obviously here the radicals are getting consumed yeah and keep noting that these things depend on concentration of cr because it is like chain branching or termination but omega i does not depend on concentration of cr because it is starting from things that are pre-existing that is they are not intermediates fine we now apply the steady state approximation right or if you want to be quite precise we should say quasi steady state approximation that means dcr over dt the net rate of production is approximately equal to 0 for bulk of the time when reactions are proceeding right. So this implies we now say that omega i plus fb alpha minus 1cr minus fscr minus fgcr is approximately equal to 0 from which we can try to find out the concentration of the intermediate so this implies cr is equal to omega i divided by fb alpha minus 1 minus fs plus fg okay. So the steady state approximation did this for us you go back and think about what we did what it means is we want to get algebraic equations that govern the concentrations of intermediates right so that we can evaluate the concentrations of the intermediates in terms of the concentrations of the stable species and then plug those expressions in the rate equation for the production of products that is what you are finally interested in right. So what we should then say then we are interested in dcp by dt that is a products it is kind of like this is kind of like life you know we should be interested in the in producing products but we keep going going talking about intermediates like in most of our lives that we just caught and get caught up in intermediates right so let us not lose sight of our goal our goal is actually dcp over dt right fp times cr of course in life money is one of the biggest intermediates that distracts us right yeah where I am losing you because let us admit that by swapping the denominator I do not like negative signs for concentrations right. So we say fs plus fg minus fb times alpha minus 1 happy you need to keep me happy as well do not get negative concentrations you know right your your your advisers are not going to like you maybe when you get them right so you want to you want to have a positive quantity there okay so ffp cr then is fp omega i divided by fs plus fg minus fb times alpha minus 1 okay I pointed out this for you earlier in the context of let us say the hpr reaction or example or I do not remember what but let me do this once again and this this becomes very intuitive for us what is happening in this expression for you is these are this is a this this division I am sorry this difference here is basically telling you the competition between termination and production okay which is being compared with omega i that is the initiation okay. In other words the rate of production of your final products depends on what is the interplay between the rate of production and rate of consumption of the intermediates for a given rate of initiation of the intermediates once you have intermediates initiated and left into the pool then the dynamics of how they are produced further by chain propagation chain branching steps versus chain termination steps okay that ratio is what determines your rate of production of products you see so it is sort of like telling us the entire physics of what is going on right just the expression looking at the expression fine if you did not worry about all that stuff and you now said this is the expression when you look at it what would you worry about again I am trying to kind of ingrain certain traits in you you now look at an expression and the and the moment you have a square root there you should be worried about whether the stuff inside the square root is going to become negative right. So similarly when you see a logarithm logarithmic expression you have to look at whether the argument is going to actually become negative or 0 and so on so these are all like things that got to be wired in your head similarly and though these are like the square roots and logarithms do not happen all the time but ratios do happen most many times right and the moment you see a ratio and you see the denominator you have to start looking at is this denominator going to become 0 right we are always afraid of infinity okay so we have to start looking at when would this denominator ever become 0 yeah so for alpha less than 1 the denominator is always positive right and what is meant by alpha less than 1 when you now have alpha less than 1 I first of all told you that this reaction should not belong in that place there it should be somewhere but below because it is beginning to actually annihilate itself the intermediates are actually consuming themselves right but that is not what we are talking about for the reactions being up there we are now talking about the alpha typically greater than 1 that is what chain branching is all about right. So when you now have alpha greater than 1 okay for alpha greater than 1 then you have a positive quantity here and a positive quantity there and it is possible that they become equal oh we have a problem right so f s plus fg equal to fb times alpha minus 1 implies denominator denominator goes to 0 and dcp over dt goes to infinity right this is what we are talking about as chemical explosion it is basically saying that once you have produced your intermediates okay at whatever finite non-zero rate so long as the intermediates are producing themselves more and more and consuming themselves at the surface or in the gas phase in such a way that the rate of production rate of the production matches the rate of the consumption right you are now going to be continuing to produce products like crazy the rate of production of products is just going to go on and on very high so that means you are going to produce a lot of products so the intermediates are actually playing a sustained role in a continuous production of products at a very infinite level infinite rate right so this is essentially the basis of chemical explosion and then what you are going to do is to say well let me begin to distinguish between fs and fg as we change the pressure okay so at low pressure we now expect that fs should predominate over fg and at high pressure we expect that fg should predominate over fs all right so this is what we are going to be doing as p increases as p increases fs decreases relatively right but fg is not too large right at for low p because you do not you do not have a vigorous gas phase termination at low pressure that is together fs plus fg decreases as p increases okay so you now have a limit you now reach a limit okay so we reach a limit that means we are looking for as you change the pressure how is this going to change relative to this such that at some limit you might equate equate right so we reach a limit where the radical generation just exceeds determination right so let us suppose that we start out with a very very low pressure all right and we now found that from there as you increase the p okay the termination actually comes down in this low pressure range because fg is not very active yet because the pressure does not increase a lot but fg is something very very small at this at this at this pressure so you now actually having a situation where the termination is coming down but the generation is going up and therefore you now reach a point where the generation just exceeds the termination and the moment when you say just that means this is just beginning to equal that so far this was higher this was lower this is now increased so it is getting to this and therefore you now have an explosion limit okay for starting so it is a termination starting from starting from some low p for a given t get you fix the other thermodynamic variable and you get this so this is this is what you would call as a first explosion limit okay and then as you now look at this dynamics we do not care which is higher which is lower all we are looking for is when things match all right so when this keeps on going as you change your pressure if this catches up again then you again have an equality which is bad right so as pressure increases further right fd increases fd increases okay which implies fs plus fd increases okay and here the termination catches up with the generation then this this this is what will lead to the second explosion limit all right we will stop here and pick up from here in the next class.