 We were looking at this model reactions this is basically three reactions a plus a the forward reaction having a rate constant kf giving a star plus a and the third reaction is a star which is of course it has a backward reaction that is a second reaction which is a reverse of the first one the third reaction is a star giving products with a rate constant of kf prime the second one having rate constant kb so we could write these rate equations for the production rates of net production rates of species a and a star like like this and of course what you can do now is okay we have two equations in ca and ca star dca over dt and dca star over dt so let us try to see if we can try to eliminate anything to do with a star therefore we now write from the first equation ca star in terms of ca and dca over dt and then you plug that back in the equation for dca star over dt wherever ca star appears and then you get a complicated looking expression that involves again only dca over dt and ca wherever it shows up so one possibility for us is to now plug back 3 and 4 into 1 okay and so wherever you have a ca star you can now plug this over here and then what happens is you can get a very ugly looking equation in dca over dt in terms of ca right so plug 1 and 2 to get a an equation relating dca over dt to ca this is going to be a non-linear second order ODE right say very highly non-linear second order ODE so you are going to have trouble solving that okay so not easy to solve so the alternate approach the alternate approach is the steady state approximation this is a simplifying approach so this is a simplifying approach there that is we now say dca star over dt is approximately equal to 0 the moment you say that so we are now essentially trying to adopt the steady state approximation and say dca star over dt you should you should probably say this is quasi steady state this is quasi steady it is not really steady but it is kind of like freeze frame a particular situation that is intermediate where the concentration of the intermediate is apparently not changing okay changes sort of slowly there in this intermediate region so it is sort of like a quasi steady state approximation rather than just steady state alone so if you now try to say that this is what we want to apply call this equation 5 and so what this means is a star is produced quickly quickly in the first reversible reaction and consumed slowly right so the moment you have a hitting each a molecules hitting each other you produce a star alright but its consumption with a the backward reaction as well as its decomposition to form the products is relatively slow in that case then you now are producing a lot of a star but for a while it is there with a certain concentration that does not change a lot therefore we can now say on a quasi steady basis this is approximately 0 then so using 5 and 2 if you now apply 5 to 2 go back here right and say DC a star over DT is approximately equal to 0 we get ca star equal to kf ca square divided by kb ca plus kf prime not very difficult to figure that just to pull out C a star wherever it up here set this equal to 0 on the left hand side so you can collect terms together and do this now let us call this 6 and substitute 6 and 1 the first equation to get to get let us say minus DC a over DT equal to omega a the reaction rate based on a that is equal to kf ca square divided by kb divided by kf prime ca plus 1 okay so we got something we got we got an answer there for DC a over DT so what we have done basically is we now have an expression for C a star from 6 which we plug in here and then you can rearrange because you now have a C a squared there and then you have a you can plug in over here and then you should be able to get your DC a over DT right so what you are getting basically is kf ca square divided by kb divided by kf prime ca plus 1 okay so what is this what is this where does this take us okay what is our original aim we wanted to find out if this reaction set of three reactions whether behave like a first order reaction or a second order reaction or whatever whatever okay as far as like a global reaction is concerned so the global reaction is a giving products but a star is like an intermediate right that is showing up in a three-step reaction scheme and how do you now look at how the global behavior of the rate of depletion of a is going to depend on the concentration of a that is essentially the global question and the global question is was it going to be a first order or a second order and now you have an expression you see it is not so obvious we are not we are now having some some soup of an expression there with C a squared showing up in the numerator and ca showing up in the denominator okay if you did not have the plus 1 right you would have cancelled ca top and bottom and you will have the rate depend only on ca alone not ca squared right but if you did not have this term and you had only one in the denominator then you can see that dca over dt is going to depend on ca squared so in the first case you would have deduced that it is going to be a first order reaction in the second case you would have deduced that it is going to be a second order reaction. So depending upon whether it is going to be first order or second order what you will have to look for is how does this compare with unity there is a reason why we have rearranged things like that it is always good to compare things with one okay to say whether it is going to be much larger than one on much smaller than one yeah so if it is going to be much larger than one we ignore one if it is going to be much larger than one then we I am sorry I said the same thing if it is going to be much larger than one we ignore one if it is going to be much smaller than one you ignore it right so that is exactly what we are going to do so at originally we said that this is going to be the fast and this is going to be slow so with this in mind we can now look at when would you when would you have which condition as far as this term comparison in comparison with unity so at high pressure at high pressure kb ca over kf prime kf prime is much greater than one okay. So what you are saying is kf prime is a unimolecular reaction okay so it is going to go more as p whereas this is the reverse reaction the competition is now between look at look at basically what is going on the kf ca squared is coming from producing a star with this forward reaction the first forward reaction the kb ca is coming from the reverse reaction okay where a star is getting consumed and the kf prime is actually coming from this one the third reaction that is also consuming. So essentially in order for us to understand whether this is going to be larger than one or smaller than one much larger than one and much smaller than one we are essentially looking at how the competition between the two reactions that are consuming a star okay far in relation to the one that is producing a star so there are what there is one reaction that is producing a star there are two reactions that are consuming a star so between the two reactions that is a competition on which one is doing a better job of consuming than the other okay how is that competition faring when compared to the production rate that is what this entire expression is all about so if you want to now look at how this competition in the consumption is happening you have to understand that this reaction the backward reaction that is consuming a star is a second is a is a biomolecular reaction which is going to go as p square yeah whereas this third reaction that is consuming a star is a unimolecular reaction which is going to go as p as for its reaction rate therefore at high pressure you would expect that the backward reaction which is a second reaction is going to predominate over the third reaction that is unimolecular and therefore we now expect that the top the numerator in this case is going to be larger than the denominator okay therefore we expect that this should be much larger than 1 and therefore if you now plug that there then omega a then approximately becomes you just go back and get rid of the one okay in favor of this expression and so you now have a k kf kf prime divided by kb all right and then the CA gets cancelled with one of the one of the two CAs in the numerator to get only one CA this will behave like a first order all right at low pressure right when the pressure is low then the square of the pressure will be lower than the pressure raised to 1 okay that is what low that is what mean that is what has been by low right so when you now say at low pressure you have the other situation which is kb CA over kf prime is much less than 1 then therefore omega a so what you do is you drop the complete expression there in comparison with 1 and therefore you now simply get this is this is equal to kf CA squared right so this is this is first this is second order so with just applying the steady-state approximation we should now be able to understand how the pressure behavior is going to happen when you are looking at two competing consumption reactions of the intermediates of different molecularities which will behave differently at different pressures this is essentially what you are trying to do that is not like a summary of what is going on good then let us now try to see how this works in the case of the HP or example the HP or example now let us look at the global reaction the global reaction H2 plus Br2 is k twice HP or right if this were a molecular level reaction if this were a molecular level reaction then what would you say for the rate that the reaction rate you would say the reaction rate would be DC HP or divided by DT is equal to we still have to have the new double prime minus new single prime for HP or new double prime is to new single prime is 0 for HP or so you will have twice yeah k CH2 times C Br2 that means they will be the concentrations will be raised to powers one each right if this were a molecular level reaction so if this were a we would write we would write DC HP or divided by DT equal to twice k CH2 C Br2 but this is wrong another why is it wrong it is because it does not happen at the molecular level not because of what we wrote right this does not happen this is not what is observed this is not what is observed that means if you were to do an experiment where you had hydrogen and bromine and you now allow the reaction to happen you now start sampling a hydrogen bromide okay at different times you do not find that the rate of change of hydrogen bromide concentration to go linearly as concentration of hydrogen or concentration of bromine as you now do this experiment with varying concentrations of hydrogen or bromine this is how we would deduce this you actually find that the rate of rate of production of rate of change of concentration of bromine is not going linearly as concentrations of hydrogen and bromine so in reality yeah so experiments indicate the following results DC HP are divided by DT is some constant C1 CH2 C Br2 to the half right well you know now that that is not the whole story because we now have this bar but as it is we are now finding that the rate of the rate of production of hydrogen bromide is dependent on the concentration of bromine to the half rather than linearly right but that is not the whole story hold your breath this is getting lot more interesting so we now have CH Br divided by another constant C2 let us use capital C's here for these constants because we are using small C's for the concentration so this is just in case you are getting confused we will now have this little serif font so this is C1 and this is C2 C Br2 what does that mean this is a pretty wacky expression there okay now if you were to be so intelligent has to immediately start thinking about what we did here right the moment you see a 1 plus something you are like going to say I know what to do okay I am going to look at when will this be actually much larger than 1 okay and then throw away the one and then I am going to you know rearrange things and what is going to happen so you are not going to say well this you know this constant times BC Br2 is going to go up there and then it is going to become 1 ½ it is not going to become like 1 that was bad so the original prediction was bad to bad still okay but that is not the most interesting thing the most interesting thing here is you know how a CH Br in the denominator is that okay what will you allow what would you what would you want to keep in your final expressions for the production of the production rate of the product the answer is like what an experimentalist would do which is you have to allow for the concentrations of stable species that is participating in the reaction and the stable species would be either reactants or products as well and here we now find that the concentration of the product is showing up as influencing the rate of production of the product itself and how it is showing up in the denominator right what does that mean it means that if you now have a increase in the concentration of HBr the rate of production of HBr is going to decrease you start out with no HBr you started started out with only hydrogen and bromine and you started producing HBr so the moment you start producing HBr you are now going to actually impede the reaction because the rate of production of HBr is going to come down because the concentrations of HBr is increasing that is what is called a self inhibiting reaction okay so this is an example of what is called as a self inhibiting that is the more the product is produced well if you are not scientists refer like advertising managers or something we would probably say the more the product is produced the less the product is produced yes that is not what we should say the more the product is produced the less the rate at which it is produced okay so we are now looking at how the rate of its production is affected by how much of its produced so the more the product is produced the less its rate of production becomes right now our problem is not just trying to reconcile a self inhibiting reaction our problem is more basic if I were to say that the actual experimental data were to show that the rate of production of HP are dependent on the concentration of bromine power half instead of one or one or a half so the other possibility that I did not talk about this or if this entire thing would actually much less than one then you still have the concentration of the rate of production of bromine to be depending upon the concentration of bromine to the half which is not the case is what would the global reaction says right so you have a half or three halves depending upon this is great much greater than one or much less than one just like the overall dependence if it is comparable to one you have to keep this as it is and we cannot resolve exactly between half and three halves it is somewhere in between right for the concentration of bromine depending influencing the rate on top of it we have to explain the presence of CHPR so how would you do this right so this is possible the above result the above result the above result can be explained can be explained by the following five step mechanism right and we say five step mechanism we are now talking about things are happening at the molecular level so far this was a global reaction that was not happening at the molecular level clearly because if you if it were this would be the rate react rate equation but that is not what it what is found in the experiments so something else is happening at the molecular level let us look at the five step mechanism so if you now say there was some compound x prime some species x prime that bromine collided with to form 2 Br that is an intermediate right with a rate constant K1 and let us call this reaction one this is a chain initiation step and the intermediate Br now attacks this table reactant H2 to produce HBr plus another H which is a which has a rate constant K2 and let us call this reaction to this is a chain chain propagating propagation step because we start started with a intermediate reacted with a stable reactant produced a stable product but produced another intermediate so like the net gain and loss of intermediates is the same okay so it is just propagating the chain and let us say we have a third reaction where H is reacting with Br2 to give you HBr plus Br this is this is the Br counterpart of H the previous reaction so this is let us say reaction number 3 this is also a chain propagation step and then we have H plus HBr that gives you H2 plus Br right look at what is going on right this is starting with a intermediate it is reacting with a stable product and producing a stable reactant and producing another intermediate so it is intermediate neutral if you just count intermediates blindly this is still like a chain propagation step alright but what is but it is actually consuming a stable product and producing a stable reactant and simply that is simply because this is actually the reverse of reaction to so you have reversible reactions that are happening okay in trying to get into equilibrium under the given conditions right and what is the flip side of that some of the stable products that you produce might actually get consumed right so in this case we found that a plus a there is a stable reactants produces a star plus a right we are actually not producing any more reactants than we consumed stable reactants we have been steps produced intermediates that is alright okay that is like a chain initiation step and the reverse reaction a plus a star gives to a that is again producing stable reactants it is kind of like saying yeah I started with reactants and then did a lot of things and then got the reactants back okay so the story obviously is not over there you now have a star giving products now let us suppose that this was actually a reverse reaction then you would be consuming products to give you intermediates and that is when you will start looking at self inhibiting reactions so when you are having the stable products being involved in reverse reactions right then they get consumed and their concentrations affect your global reaction rate right so this is a for alright but this is a this is a chain propagating step alright but this is this is the one that is reverse of this is an inhibition reaction reverse of reverse of 2 okay and finally we now have x prime plus Br plus Br gives course the previous reaction had a rate constant k4 and this is a react rate constant k5 Br2 plus x prime okay now what is going on here we are we are talking we are starting with an intermediate two intermediates as a matter of fact and then producing a stable species the stable species so we are actually killing the intermediates but this way of killing intermediates is to actually produce back the stable reactant yeah so this is a chain terminating step and it is a reverse of one so you are having two reverse reactions in this out of out of five say effectively this is having a reverse here this is having a reverse here this is the one that is not having a reverse reaction yeah and there are two reactions that are producing there is one that is consuming HP or alright and finally something that you have to keenly note you now have a term molecular reaction here so the moment you have a term molecular reaction as opposed to anything else that is bimolecular okay the first thing that you have to think about immediately is effective pressure so anytime you see a unimolecular reaction versus a bimolecular reaction or a bimolecular reaction versus a term molecular reaction whenever you are making these kinds of comparisons the first thing that hits and hits your head is pressure what is the effect of pressure okay so that is one of the things that we will have to be looking at just like how we did previously for this example so the answer the answer is I am not going to work out the details that is going to be a homework for you so here is the answer on how so what you do you now take each of those reactions write the rate equation for whatever is its chief products okay in terms of its reactants concentration this is a molecular level thing so law of mass action directly applies right and we can show we can show using steady state approximation that capital C1 that we had before right in the in the global reaction rate expression is twice k2 times k1 divided by k5 to the half and C2 equal to k3 divided by k4 so this is a homework for you so whenever I say we can show basically mean you can show okay so just go ahead and show yourself don't don't don't let's not worry about this so you understand right that means if you now write the rate equations for each of these and then pick out those rates that are based on intermediates that is dcbr over dt or dch over dt typically these are the two intermediates that are happening here okay and then set them approximately equal to 0 you now get two equations and you have to try to eliminate ch and Cbr in this in this expression and you have to pray to all the gods that Cx prime would vanish as well all right and then you will get something that looks like this after massaging all the remaining equations and so on okay don't don't over massage your because whenever you know the answer you try to actually try to fit it in but I have done this when I was a student okay so it is not terribly bad you can you can show this yeah so what else can we do so this is an example of how the steady-state approximation looks like so now let's look at another approach where we try to simplify which is called the partial equilibrium partial equilibrium approximation yeah partial equilibrium approximation now this is a bit different from what we have done for the steady-state approximation but effectively leads to the same same idea the idea here is we want to try to get algebraic expressions for concentrations of intermediates as a function of concentrations of stable species does not have to be reactants okay so this is our goal we do not want to deal with a ODE for the concentration of intermediates we do not want to deal with a expression there for DC A star over DT and then do a time integration of that and try to find out how the A star changes in time and so on okay the previously we said no change in time okay approximately 0 for the rate and we just go go with a algebraic expression is there another way by which we can obtain algebraic expressions for the concentrations of intermediates as a function of concentrations of these stable species reactants or products right the answer is we will now try another approach the remember these are based on some physics it is not it is not like we are trying to do this just for the sake of getting these algebraic expressions okay the previous case we said the intermediates are quickly produced and then they are there as a pool for some time when they are getting interchanged through the chain propagation steps right and takes a while for them to happen during this time they they their concentrations approximately constant okay so that was a physics there there is another thing that we can try to exploit which is the occurrence of these reverse reactions to forward reactions right so typically when you now have two reactions that are forward and reverse pair and let us suppose that they are fast okay like the first six first example that we had no a plus a gives a stop plus a forward and backwards and that is fast and the second reaction a star giving products was the one that was slow so what happens there you now start with a a bombards with itself like one molecule of a bombards with another molecule of a and quickly produces a star you now have an equilibrium that is established with a star and a like a soup now a star is around and it is it is an equilibrium with a right and it just slowly gets consumed during the third reaction for for it to for it to produce the final stable products right so during this time is it possible for us to exploit the equilibrium between a star and a between the forward and the backward reactions so if it is now so effectively what we are looking for is if you now have fast reactions that are forward and reverse is it possible to actually apply equilibrium equation rather than rate equation the rate equation is the one which has a DC a star over DT equals a function of CA and CA star okay that is an ODE and we do not like the ODE we want an algebraic expression but the equilibrium equation for that assuming that it is actually in equilibrium right would be you now have a KP or a KC that is equal to concentrations of reactants of the forward reaction divided by concentrations of the the reactants of the reverse reaction raised to their respective stoichiometric coefficients right that is an algebraic relationship we do not have a we do not have a differential equation there right so this is essentially what you are trying to do so here where forward and it is a fast forward and reverse is not fast forward this is fast forward and reverse okay reactions occur so that they could be considered as in equilibrium actually we are saying something about their rates it is not just that they are fast and therefore those those reactions are in equilibrium essentially when you say something is in equilibrium we are saying that the forward reaction rate is equal to the backward reaction rate reverse reaction rate okay so we are basically saying let us not worry about how fast it is all fast okay so they are all equal right that is exactly what you are what you are saying here so and this leads to leads to algebraic relations relations for intermediate concentration in terms of stable species okay using the equilibrium constant so now you have the equilibrium constant we know is a ratio of the forward to the backward reaction rates rate constants okay so many times we find these kinds of ratios showing up you see so here as well we find these kinds of ratios that are showing up k1 over k5 for example is actually ratio forward to backward right so that that is exactly the equilibrium constant so we will simply be dealing with equilibrium constant instead of the ratio of the forward and backward reactions as if the two were in equilibrium so let us now look at an example so example or since a example let us say let us look at consider template reactions template reactions let us say a plus b2 gives let us say this is forward and backward so we call this a b plus b this is kf1 this is kb1 all right and let us call this one and b plus a2 gives and takes gives and takes a b plus a and we call this kf2 kb2 let us call this reaction 2 and a b plus a2 gives and takes a2b plus a right and let us have rate constants kf3 and kb3 we will call this reaction 3 and finally let us have a plus a b plus m gives a2b plus m that is reaction 4 that means the last reaction is not a reverse reaction so it forward it does not have a reverse reaction so strictly speaking the way you should look at it this this is actually two reactions two reactions two reactions and one reaction so totally it is actually seven reactions it sort of like having five reactions but if you were to write it like this it would have been only three reactions this this and this going together is forward and reverse this and this going together is forward and reverse and the last one right so it is just a different way of writing this and so let us call rate constant for the fourth one is kf4 we do not have a kb4 yeah so all we are interested now it depends on how you are I mean if you are like a chemistry kind of person like I want to see reality what is this a and b okay they are not really happening in reality you might want to throw hydrogen for a and oxygen for b okay and finally you might get you might be looking for H2o right and then you can say a b is OH and so on so essentially what is going on is now you have a atomic species colliding with a molecular species to give rise to two atomic species so this is like a chain branching right and this is again starting with one intermediate and this is this is supposed to be a stable species you are now getting two intermediates so chain branching and this is a stable species here this is an intermediate this is the final product and that is an intermediate so that is a chain propagation it start with an intermediate reacting with a stable reactant producing a stable product and leaving another intermediate and m is any third body okay and you are finally getting a stable product so this is like a chain termination step so you can see that sequence of events that happen that starts with the chain initiation branching propagation termination and so on now what we are interested in is to find out what is the net rate of production of the final product and how it depends on the concentrations of the stable reactants and the presence of the intermediates is spoiling the show we want try to get rid of them yeah so what we want to do then is we say here we say K F 1 C A C B 2 equal to K KB 1 KB 1 C A B C B right that is for the first reaction okay or since you have the same molecularity on either side okay this will go as p2 this will go as p2 so concentrations can be written in terms of most fractions and most fractions can have pressures the total pressure so the pressure dependence is going to be the same on both sides so does not matter where you use a equilibrium constant based on pressure or equilibrium constant based on concentration I am just going to go back and use concentration equilibrium constants based on pressures okay so so this simply means that we now have C A B C B divided by C A C B 2 equal to capital P 1 right capital K so capital K subscript P 1 capital K represents equilibrium constant based on pressure and we have subscript 1 to denote this is actually the equilibrium for the first reaction similarly we can write we can write for the second and third we can write C A B C A divided by C B C A 2 equal to K P 2 and the third one C A 2 B C A equal to sorry divided by C A B C A 2 equal to K P 3 so what was the purpose now do we still understand we want to try to get rid of C B and C A in favor of and C A B as well we want to get rid of C B C A B and C A in favor of C B 2 C A 2 and C A 2 B all right so this is something that we will do next class.