 So, now we had this template notation for a reaction earlier but now let us suppose that we consider that we have forward and reverse reactions alright that means we are saying that we have one set of reactions that are going the forward direction another set of reactions that are going the backward direction now we could have actually written this as two sets of reactions alright all of them going in the forward direction this is essentially that means like for example if you now went from k equals 1 to m you now write these as two sets and from k equals m plus 1 to 2 m you will now have the other set of reactions okay. So, if you now write it this way then this m is half as before that is what it amounts to if it is possible that in the previous set some of them were not going through reverse reactions then those reverse reactions will have new I k single prime as well as new I k double prime equal to 0 and in the previous set alright so let us suppose that in the previous set we had a full set of which some of them were actually reverse of the other but some of them were not okay the new I k single prime and new I k double prime for them both will be 0 for a hypothetical reaction there is a reverse which does not really happen at all. So, it is possible for you to write this in terms of what we did before also that is what I am saying. So, we can write omega I k equal to new I k double prime minus new I k single prime omega k keep in mind we have this yesterday and we are using the symbol omega right and this is for a particular species omega I k I is for the ith species and k is for the kth reaction right. So, when you are able to actually write it for a particular species you might be able to go from a molar reaction rate to a mass base reaction rate okay. So, it is from moles per second per meter cubed to kilograms per second per meter cube by multiplying by the molecular weight alright. So, we could do that and that is what will actually be used in the mass balance because the mass balance is looking for number of the amount of mass that is produced to consume rather than number of moles that are produced to consume we will go through a molar conservation equation also okay at that time we will try to use number of moles but when you try to use mass we will change from Greek to just English English alphabet. So, from omega will become w small w so those are things that we will talk about but this is something that we did yesterday yeah so here here omega k is now this is for the kth reaction out of 1 to m you see therefore there is like a forward reaction rate and a backward reaction rate right. So, there is like something that is going forward and then some of it is getting consumed therefore it has to actually be subtracted out yeah. So, then we can say this is kfk p i equals 1 to n c i nu i k single prime minus kbk p i equals 1 to n c i to the nu i k double prime right by the way it is customary to actually indicate the specific reaction rate of the rate constant on top of the arrow for a given reaction to indicate that that is the kinetic data that we are looking for okay. So, pretty much k contains the kinetic data your A and E or B M and E what we saw yesterday all the stuff is contained in this so to say that you have this means that you are supplying the kinetic data for this particular reaction rate specific reaction rate constant. Here what we see that this is like c i to the nu i k single prime means we are looking at the forward reaction having the nu i k single prime the species as reactants with the nu i k single prime as the coefficients you say c i k c i to the nu i k double prime we are now looking at the reactants over here as a reactants for the backward reaction that is reason why you are using nu i k double prime right then at equilibrium at equilibrium omega k equal to 0 that means the rate at which the forward reaction takes place should be equal to the rate at which the backward reaction takes place okay then you have a equilibrium it is a dynamic equilibrium okay that means it does not mean that the reactions are not taking place the reactions are taking place you have a forward reaction going on you have a backward reaction going on but the rate at which the forward reaction is going on is the same as the rate at which the backward reaction is going on okay so it is a dynamic equilibrium reactants are being produced and consumed all the time in the situation. So we then say k fk p i equals 1 to n c i to the nu i k single prime equal to k b k p i equals 1 to n c i to the nu i k double prime right therefore k fk divided by k b k equal to p i equals 1 to n c i to the nu i k double prime divided by p i equals 1 to n c i to the nu i k single prime which is nothing but p i equals 1 to n c i to the nu i k double prime minus nu i k single prime. Now what does this look like we have seen this a little bit before maybe about couple of classes back except we did it for like let us say hydrogen oxygen reaction specifically or something like that anybody what is it k p capital Kp so this is actually it is not Kp here because this is actually based on concentrations whereas we wrote this in terms of partial pressures earlier okay so when you have partial pressures we had like partial pressure of a product raised to its stoichiometric coefficient times partial product pressure of another product raised to its stoichiometric coefficient if you have two products divided by partial pressure of reactant to its stoichiometric coefficient times partial pressure another reactant if you had no one more react to the its stoichiometric coefficient and so on. So we had this before that is what we are going through but in terms of partial pressures we call it Kp which is exact which is basically the capital Kp which is the equilibrium constant based on pressure or partial pressures here we are writing in terms of concentrations therefore this actually is k ck which is the equilibrium equilibrium constant based on concentration okay fine since ci is equal to pi divided by RT how do you get this number of moles per unit volume is essentially molar concentration number of moles of species I divided by the volume is the molar concentration of species I so if the ideal gas equation of state is valid for each species therefore ni divided by capital V is equal to pi divided by RT right therefore we can write ni divided by capital V is ci we can relate the partial pressure to the concentration and further we did not like partial pressures even last time we only get the total pressure therefore we use the mole fraction which is been defined now properly for us divided by RT in fact this is this is Ru we are talking in terms of number of moles molar concentration and so on this is Ru for ideal gases right therefore k ck equal to pi just plug this in there and then say i equals 1 to n xi to the nu ik double prime minus nu ik single prime p over Ru t to the this p over Ru t is going to be constant it should be like a common factor for each of those xi's right so xi goes through this the fate of what ci went through but then when you now have the pi over RT that is common for all of them then it gets summed over all the species right so you now have i equals 1 to n nu ik double prime minus nu ak single prime so in effect this is nothing but k pk which we were familiar with to begin with times p over RT to the power sigma i equals 1 to n nu ik double prime minus nu ik single prime trying to squeeze there okay. So from here you can see that the kinetics is related to equilibrium and this is something that you are not unfamiliar with you must have gone through this in your high school for regular reactions and so on but here we are trying to put all of them together in the algebraic notation that we are coming up with for a general set of forward and reverse reactions yeah and that is a relationship between equilibrium constant based on concentration and equilibrium constant based on so this is equilibrium constant based on partial pressures okay before we proceed from here the next thing that I would like to discuss here is order of a reaction versus molecularity okay molecularity of a reaction order of order of reaction versus molecularity of reaction so many times we come across we come across these things like we have a first order reaction or a second order reaction okay and sometimes you also hear this thing like we have a unimolecular reaction or a bimolecular reaction or a pteromolecular reaction and so on okay. So what are we talking about any feel for this we have to try to sort of anglicized this right so that means we have to say what we are what we are hearing in English in typical sentence it is not very difficult to think about for example what is what is a bimolecular reaction that means we are looking at a reaction where you have two molecules that are colliding with each other to react the pins as far as the reactons are concerned we want to have two molecules of different species one molecule each one molecule each of two different species have to collide with each other can we say that or could it be like two molecules of the same species have to collide with each other yes no obviously if you want to have a species react molecules are to collide that is something that we started talking about is as an axiom so even if one particular species just disintegrates why would it disintegrate it would disintegrate because molecules of itself stop colliding with each other right and then you could think well maybe I need two molecules to collide instead of just one molecule design disintegrating by itself without any collision huh would that be by molecular okay so let us have this thing in our minds okay of course now if you can extend this confusion to let us say termolecular which is pretty easy you can now say I want to have three molecules of different three different species each three molecules each of three different species in a in a termolecular reaction or a trimolecular reaction or I could have like two from one and one from the other and so on okay so we have that situation so what would you do for example let us suppose that you had a reactant a and I say that I want to have to a gives products how do I write the reaction rate where I now say that the reaction rate for this reaction is equal to k times concentration of a squared or can I say well let me divide it by half I mean to divide the whole thing by two and then say a gives half of products yeah and therefore the reaction rate should be equal to k times concentration of a raise to the power 1 there is a big difference you cannot have a rate either depend linearly on the concentration of a or a square of the concentration they are vastly different so which one would you do and then let us suppose so you have something like to H2 plus O2 gives products that is not worry about the products because the the law of mass action is only depending upon the reactant concentrations right so to H2 plus O2 gives products versus H2 plus half O2 gives products how would you write your reaction rates would you write this as k times concentration of H2 to the power 2 times concentration of O2 or would you write it as k times concentration of H2 times concentration of O2 to the half the two expressions are vastly different right so what is the reality which is the right answer could you just recklessly divide the reaction by whatever number you want like I just say divide by two all over and then I now come up with a new reaction rate constant expression I do not know which is right then until now we did not care we could we could divide and say okay half a mole here one mole there does not matter but now we have an implication these things are now going to show up as exponents of the concentrations of the reactants in the reaction rate expression and we cannot be fiddling around with this so what is going on right the answer is if you now look at a reaction where you are talking about like let us say H2 plus half O2 gives H2 O you know that it is not really happening that way because you cannot have half a molecule of oxygen react with one molecule of hydrogen okay it is not like oxygen is half heartedly reacting with hydrogen right so we have always been saying it should be half a mole and half a mole contains a large number of molecules so you now look at actually the molecules what are they doing obviously you need to have them react one on one you cannot have half a molecule do with another one or another half right so if you now have anything that has like a half you know that it is not a molecular level reaction that we are talking about it is it is what is called as a global reaction okay and for a global reaction you could now write the law of mass action globally and there you could have the exponents to the concentrations raise to whatever value it does not have to be integers you get the point so let us try to write this we have something called global kinetics right so consider the stoichiometric reaction does not have to be a stoichiometric reaction but let us consider this reaction CH4 plus 202 gives CO2 plus H2O right okay this does not happen as it is okay this does not happen as it is although thermodynamically feasible thermodynamically feasible okay does not happen in reality yes you are going back to this can you say what you are asking let us find that did I make a mistake should we have this no no no no let us just keep it this way no I think I made a mistake you are right you do not have the P here that is what you mean yeah thank you because what she is saying is this pressure should actually go along with this for you to have a PI small PI yeah okay fine you go back to what we were talking about fine so what I am saying is this this reaction actually does not happen the way it is written here that means it is not it is not like CH4 a molecule of CH4 reacts with two molecules of oxygen that is not what is going on in reality when you now get methane and oxygen together okay what else is going on what happens in reality in reality unfortunately for you you now have a huge set of reactions that are that kick in the moment you get this to go okay so the first thing that happens is like you now have methane break down into methyl like CH3 CH2 CH those kinds of radicals and also release some H radicals okay and then you now have a pool of these radicals right oxygen could split it to O and O can actually get together with H to form OH lots of such things lots of such things such things are happening so similarly similarly H2 plus half O2 gives H2 O does not does not happen as it is and we saw this we cannot we said that we cannot have half an oxygen molecule get together with hydrogen molecule we are okay with half a mole of oxygen getting reacting with hydrogen on the whole okay but essentially when you say on the whole then these reactions are now representing what is called as global reactions that are representative of a huge set of reactions that kick in at the molecular level so those reactions that are happening at the molecular level or what is called as elementary reaction steps in a reaction scheme so this is the terminology that is typically used so you have a reaction scheme that means you have scheme of lots of reactions going on and each of those reactions if you talk about in reality would be like a elementary reaction step that is happening at the molecular level right but all of it together could be represented by a global reaction if you now have this kind of a global reaction then so let us now go back to this example right then if you want to now write the let us say the production rate of CO2 concentration right so DC CO2 divided by DT that is like saying DC I over DT where I is for CO2 okay you can now write this as K CH4 power P C subscript O2 power Q we would not write we would not write for this reaction P equal to 1 and Q equal to 2 no okay because P and Q are to be found empirically many times in many applications you will find that these values are more like P is of the order of like 0.15 0.15 and Q is like about 1.65 so you see that first of all they are not integers number 1 number 2 it is got nothing to do with the coefficients of these reactants that are appearing in the global reaction because a global reaction is like a representation of what is happening in reality through a huge number of elementary reaction steps in a reaction scheme and to give you an idea of what the what is how huge it is for the methane oxidation you could be thinking about like at least about 100 reactions 100 reactions minimum okay and the number of species that it produces because it is now splitting into CH3 CH2 CH H O OH all those kinds of things it could be as many as about 40 species okay we are now looking at fairly large number of reactions and not so hope mostly not so large number of species but relatively large for this right again here you will be thinking about like a few tens of reactions that are really happening and a let us say handful of species at least okay 5 to 10 species that are involved at the molecular level you can go back and write the law of mass action as it is looking at a reaction and figuring out the coefficients there and then plugging it in here only for the molecular level reactions okay. So if you now do that if you now assure that a reaction is actually a molecular level reaction and not a global reaction then you look at the reaction you can write the law of mass action the rate of production of a particular species or the reaction rate as k times concentration of the reactants raised to the respective coefficients that are seen in the reaction those will typically be integers because we have to have integral number of molecules of each reactant react right and then we can say you have a molecularity of the reaction is essentially the number of molecules of reactants that are participating there. So there if you now add up those exponents that are happening at the molecular level that gives you the molecularity of the reaction okay. So if you had a term molecular reaction where you now saying 2h plus O gives you H2O let us say that is like that is one molecular level reaction that is happening then you can say it is a term molecular reaction that because that is because you need to have two hydrogen atoms collide with one oxygen atom to form a water molecule but you could also now distinguish and say the reactions molecularity is 2 in hydrogen atom and 1 in oxygen that means you should now be able to say what is the molecularity in each of the reactants that is the next level of refinement and trying to identify the molecularity right. It is also possible for you to say this for what is called as a order of reaction and order of reaction is typically like what we are talking about for a global reaction. So in a global reaction P plus Q will now give you the order of the reaction which are empirical constants right and then you can further say the reaction is of pth order relative to methane and Qth order in oxygen you see you can say that that is all common practice to identify what is the reactions order in each of those species each of those reactants for a global reaction okay. Of course for a elemental reaction step that is happening at the molecular level the molecularity and the order of the same but the order is not the same as the molecularity when you are looking at a global reaction and it has got nothing to do with the stoichiometric coefficients that are obtained for mass balance essentially when you are trying to do a balancing the reaction you are essentially doing a mass balance okay you want to balance the same number of nuclei and each of those nuclei have some mass right that is what you are trying to do good. So n plus n we now say n equals P plus Q is the order of reaction and need not be integer you need not be an integer okay so P and Q need not be integers can they be negative what does that mean if you now throw in a reactant the reaction rate actually comes down and why would it why would why would you have the reaction in the first place yeah it is possible under some circumstances and I think we will proceed further and I will try to explain that to you because we are trying to sort of globalize a series of reactions that are happening and under different conditions you can now have different emphasis for the role of each of those reactants sometimes it is also possible for the reaction rate or the rate of production of your products or rate of depletion of your reactants whatever to be depending globally speaking depending on product concentrations also so that looks like a violation of the law of mass action okay and many times under those circumstances the exponent for the concentration of product dependence on the reaction rate will be negative which is what is called as a self inhibiting reaction okay that is more and more the product is formed the less and less the reaction is going to happen because as the concentration of the product increases the reaction gets inhibited because it is like a self inhibition mechanism okay we will also look at some examples for that okay as we go along but these are basically looking at the global reaction at the molecular level we do not have any problem we can just go here and do whatever we did with the law of mass action the previous class right fine can we do we have to have a global reaction that is only one step can we have like multi-step global reaction yes or no no why not while we are having fun might as well have more fun this is not fun at all right so global kinetics need not be single step that is we could write the same reaction as above in two steps okay CH4 plus 202 okay and then let us say that this has a rate constant K1 you CO plus 2H2O plus half O2 and then you have CO plus half O2 gives K2 with the rate constant K2 it gives you CO2 right keep in mind I have a gap here so I am going to fill it in with something right question is why would you do this why would we understand why we want to have a global reaction right that means we do not want to go through the details we do not want to go through what is happening in at the molecular level we do not want to burden ourselves with those 100 reactions and 40 species and all those things we can I get away with chemistry no I cannot that is combustion well you can it is possible if you now instead of thinking about global kinetics we now have something called an infinite rate chemistry there is something that I will talk about in the context of diffusion flames as we go along we try to try to sidestep chemistry completely okay but short of sidestep in chemistry the best we can do is to globalize right means you just have one single step that represents all the details unfortunately the K will still be represented as an Arrhenius reaction and the A and the E that we have with the B M and E that we have as well as the P and Q all have to be obtained empirically because it does not happen in reality right we do not do quantum mechanical calculations to obtain those not that we do that for all the other molecular level reactions but it can be done it is tedious but it can be done okay but we do not do it for global reactions so that means we have to get numbers from literature for global reactions which are empirically obtained but while you are there why would you want to do like a two step global reaction and of course when you do two step you could do three step you could do five step not hundred step and so on right why would you do that why would you want to now split the global reaction which was convenient nice reaction and then have a splitting headache over it right any ideas in this case we are not just interested in this this reaction going on we would like to see if we can track how much CO is being produced so if you are particularly interested in a intermediate right I would like to now write my global reactions in such a way as to include that species showing up as a product in one reaction and a reactant in the reaction possibly the two rates are not going to be the same right so you are now going to have a CO that is being formed as an intermediate and I like to know how much it is I like to see if it is actually going to be completed completely consumed as much as it is getting produced which means I now start looking at what is my K1 what is my K2 right so I can then write here so we will try to fill the gap here DC H2O for example over DT equals you could write this for CO as well maybe maybe maybe yeah okay let us just do this K1 CCH4 to the P1 CO2 to the Q1 and here we say DC CO2 divided by DT equals K2 CCO P2 to the P2 CO2 to the Q2 right and then we will have to empirically find out what is your K1 and K2 which in turn means we have to now know what is our A1 and E1 or B1 M1 and E1 A21 M2 or B1 M B2 M2 and T E2 for these these are the kinetic constants and then you have these P1 and Q1 here P2 and Q2 here which all have to be empirically supplied but provided we can actually try to get those empirical relationships what this tells you is if you were to have this situation it is pretty much like the rate of production of water and rate of production of carbon dioxide are actually the same but here we will now be able to distinguish that the rates of production of water and rate of production of carbon dioxide are different depending upon the individual rates at which these things happen and look at something very very interesting that is going on you see that in this reaction the first reaction carbon monoxide has to be produced before the second reaction can happen so the second reaction rate will depend on the concentration of carbon monoxide which is produced in the first reaction if you the first reaction did not happen the second reaction cannot happen right and when the second reaction happens it competes for the same oxygen as the first reaction they are happening simultaneously right so you have methane getting oxidized carbon monoxide also is getting oxidized with the same oxygen in the pool so it is as if like methane was trying to get oxidized it produced some carbon monoxide the carbon monoxide now begins to actually eat up the oxygen that is trying to oxidize the methane and then reduces the rate of that reaction because it is consuming some oxygen but if it reduces the rate of that reaction it cannot get oxidized so it reduced so you see what is going on there is like a competition between these two but it has to be sequential it is sort of like mother and daughter competing with each other okay one of them gives rise to the other but then they compete right so it is pretty interesting dynamics here and it will depend on what your K1 is what your K2 is so think about K1 being much lower when compared to K2 see what happens then you have like it takes forever to get this going but the moment this this happens it will immediately form the next reaction the moment the next reaction immediately tries to happen it tries to rob O2 and then slows that down that means this also has to slow down because CCO has to come down you see so there is a nice dynamics that you can think about in these things which will finally fetch you a certain carbon monoxide that may be remaining that means you are trying to track carbon monoxide and that means you have to bring that in as part of your global reaction set now you are now beginning to talk about global reaction set rather than just one global reaction as a reason why we are thinking about a two-step global reaction there are much more complicated systems that are there so typically in chemistry or that you want to encounter like hydrocarbon oxidation methane is like the simplest hydrocarbon okay one carbon atom then you get into step into things like ethane ethene and ethane and all those things right and then you have acetylene sorry not acetylene and propane and propene and butane butane and all those things so keep on going and then you get into heavier hydrocarbons heptane and so on then you now go to do cocaine and all those things that's like you now begin to get into things like petrol diesel those kinds of benzene is like somewhere in between there and so on so you keep on growing and growing this number of the elemental reaction steps becomes longer and longer right but we are still talking about a carbon hydrogen oxygen system if you want to now use air to oxidize and then you are interested in NOx that means you have to now factor in oxides of nitrogen being formed okay then you are now throwing in one more atom type into the pool and the set of reactions that you have to consider there at the molecular level becomes lot more you suddenly are looking at about 500 reactions happening okay. So I used to be working with ammonium percolate reacting with polybutadiene right so the polybutadiene is like a polymer that is based on like a butadiene as a monomer and so on you are looking at about four carbon atoms in a monomer but you now have a change it is about having a molecular weight of about 22,000 okay which has to break down and then react with ammonium percolate and then there you have ammonia and percolaric acid that are formed and then you are have a chlorine that is coming up as a new atom type and apparently ammonium percolate can actually react by itself it is like a monopropellant and it is estimated that there are like about 1,000 reactions that are actually happening in a very thin layer of about 5 to 10 microns thick which we cannot even see for most part because the temperature gradients are so steep right. And typically it is approximated by five global reactions so it is like the typical number you are looking at okay so from 1,000 we try to reduce it down to something like five so that we can handle it okay or there are better ways of handling it in a more realistic manner we will wait and watch Monday.