[Music] [Music] we were looking at this U 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 a rate constant KB so we could write these rate equations for um the production rates of net production rates of um species a and a star like like this and of course what you can do now is okay we have um uh two equations in uh CA and CA star uh DCA over DT and DCA star over DT so let's try to see if we can try to eliminate anything to do with a star uh therefore we now write from the first equation c a star in terms of CA and DCA over DT and um then you plug that back in uh the equation for DC a star over DT wherever c a 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 three and four into uh one okay and uh so wherever you have a CA star you can now plug this over here and uh then what happens is you can get a very ugly looking equation in uh DCA over DT in terms of CA right so plug one and two to get a an equation relating DCA over DT to ca this is going to be a nonlinear second order OD right it's a very highly nonlinear second order OD so you're going to have trouble solving that okay so not easy to solve so uh the the alternate approach the alternate approach is the steady state approximations this is a simplifying approach so this is a simplifying approach there that is we now say DC a star over D DT is approximately equal to zero the moment you say that um so we are now essentially trying to adopt the steady state approximation and say DC a star uh over DT you should you should probably say this is quasy steady state this is quasy steady it's not really steady but it's kind of like freeze frame a particular situation that's intermediate where the uh concentration of the uh intermediate is apparently not changing okay it changes sort of slowly there in this intermediate region so it's it's sort of like a quasi steady state approximation rather than just steady state alone so if you now try to uh uh say that this is what we want to apply call this equation five and uh so so what this means is uh a star is uh is produced quickly quickly uh in the first reversible reaction and uh consumed uh slowly right so the moment you have a hitting each a molecules hitting each other you produce a star all right 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 you now are producing a lot of a star but for a while it is uh then with with a with a certain concentration that doesn't change a lot therefore we can now say on a quasi steady basis this is approximately zero then um so so using uh using five and two if you now apply five to uh two go back here right and say DC a star over DT is approximately equal to zero U we get c a star equal to KF CA squ divided by KB CA plus KF Prime not very difficult to figure that just to pull out CS star um wherever wherever it appears set this equal to zero on the left hand side so you can collect terms together and um do this now let's call this six and uh substitute 6 and one the first equation to get to get um let's say minus DCA / DT equal to Omega a the reaction rate based on um a that's uh equal to KF c² divided by KB divid by KF Prime CA + 1 okay so we got something we got we got an answer there for DCA over DT so what we've done basically is we now have an expression for CA star from six which we plug in here uh and then you can rearrange because you now have a c Square there and then you have a um uh um you can plug in over here and then you should be able to get your DCA over DT right so what you're getting basically is KFC is c² divid KB uh divid by KF Prime CA + 1 okay so what is this what is this uh where does this take us okay what is our original aim we wanted to find out if this reaction set of three reactions will it behave like a first order reaction or a second order reaction or what what whatever okay uh 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 uh 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's essentially the global question and the global question is with it going to be a first order uh or a second order and now you have an expression you see it's not so obvious we are not we are now having some some some soup of an expression there with c squ showing up in the numerator and CA showing up in the denominator okay if you didn't have the plus one right you could have cancel CA top and bottom and you will have the rate depend only on CA alone not C squ right but if you didn't 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 squ so in the first case you would have deduced that it's it's going to be a first order reaction the second case you would have deduced that it's going to be a second order reaction so depending upon whether it's going to be first order or second order what you will have to look for is how does this compare with unity that's the reason why we have rearranged things like that it's always good to compare things with one okay to say whether it is going to be much larger than one or much smaller than one yeah so if it's going to be much larger than one we ignore one if it's going to be much um larger than one then we I'm sorry I said the same thing if it is going to be much larger than one we ignore one if it's going to be much smaller than one you ignore it right so that's ex exactly what you're going to do so at uh originally we said that this is this is going to be the fast and this is going to be slow um so with this in in mind we can now uh look at when would you when would you have which condition as far as uh this term comparison in comparison with unity so at high pressure at high pressure um KB CA over KF Prime KF Prime is much greater than one okay so what you're saying is KF Prime is a unimolecular reaction okay so it's going to go more s p whereas this is the the reverse reaction the competition is now between look at look at basically what's going on um the kfca squ is coming from producing a star with this forward reaction the first forward reaction the KB um CA is coming from the reverse reaction okay where a star is getting consumed and 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 or much smaller than one we are essentially looking at how the competition between the two reactions that are consuming a star okay fair in relation to the one that is producing a star so there there there's one reaction that is producing a star there are two reactions that are consuming a star so between the two reactions that are consuming G there's a competition on which one is doing a better job of consuming than the other okay how is that competition faing when compared to the production rate that's what this 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 bimolecular reaction which is going to go as P squ yeah whereas the 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 the second reaction is going to predominate over the the the third reaction that is UN molecular 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 one and uh therefore if you now plug that there then Omega a then approximately becomes you just go back and uh get rid of the one okay in favor of uh this this expression and so you now have a uh K KF um KF Prime divided by KB all right and then the ca gets canceled with uh one of the one of the two CA in the 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 uh pressure raised to one okay that's what that's what mean that's what is meant by low right so when you now say at low pressure you you have the other situation which is KB CA over KF Prime is much less than one then therefore um Omega a so what you do is you drop the complete uh expression there in comparison with one and uh therefore you now simply get this is this is equal to uh KFC squ right so this is this is first this is second order so with with just applying the steady state approximation we should now be able to understand how the pressure behavior is going to happen when you're looking at two competing consumption reactions of the intermediates of different molecularities which will behave differently at different pressures this is essentially what you're trying to do that's that's like a summary of what's going on good then let's now try to see how this works in the case of uh the hbr example the hbr example now let's look at the uh Global reaction the global reaction H2 plus B2 gives K twice hbr right if this were a molecular level reaction If This Were a molecular level reaction then what would you say for the rate the the the reaction rate you would say the the the reaction rate would be um DC hbrr divid by DT is equal to we still have to have the new dou Prime minus new single Prime 4 hbr new double Prime is two new single Prime is zero for hbr so you'll have twice yeah uh k c h 2 * C B2 that means they will be the concentrations will be raised to Powers 1 each right if this were a molecular level reaction so if this were a we would write we would write dcbr ID DT equal to twice K CH H2 cbr2 but this is wrong why is it wrong it's because it doesn't happen at the molecular level not because of what we wrote right this doesn't 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 a hydrogen and bromine and you now allow the reaction to happen you now started sampling hydrogen bromide okay at different times you don't 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 you 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 R so in reality yeah so experiments indicate the following result DC HP divided by DT is some constant C1 C H2 cbr2 to the half right well you know now that that's not the whole story because we now have this bar but as it is we are now finding that the 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's not the whole story hold your breath this is getting a lot more interesting so we now have C hbr divided by another constant C2 well um let's use Capital C's here for these constants because we are using uh small c's for the concentration so this is um so just in case you're getting confused we will now have this little um Ser font so this is C1 and uh this is C2 cbr2 [Applause] 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 one plus something you're like going to say ah I know what to do okay I'm going to look at when will this be actually much larger than one okay and then throw away the one and then I'm going to uh you know rearrange things and what's going to happen so you're not going to say well this you know this this constant times bcb2 is going to go up there and then it's going to become one and a half it's only going to become like one that was bad so your original prediction was bad to bad still okay but that's not the more most interesting thing the most interesting thing here is you now have a chbr in the denominator is that okay what would you allow what what would you what would you want to keep in your final expressions for the production of the the production rate of the product the answer is like what what an experimentalist would do which is you have to allow for the concentrations of stable species that are participating in the reaction and the stable species would be either reactants all 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's 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 started out with no hbr you started started out with only hydrogen and bromin and you started producing hbr so the moment you start producing hbr you're now going to actually impede the reaction because the rate of production of hbr is going to come down because the concentration of H hbr is increasing that's what is called a self inhibiting reaction okay so this is an example of what's called as a self inhibiting right that is the more the product is produced well if if you are not scientist we were like uh advertising managers or something we would probably say the more the product is produced the less the product is produced that's not that's not what we should say the more the product is produced the less the rate at which it is produced okay so we we are now looking at how the rate of its production is affected by how much of it produced so the more the product is produced the the less its rate of production becomes right now our problem is not just trying to reconcile a self inting 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 HPR depended on the concentration of bromine power half instead of one or one and a half so the other possibility that I didn't talk about is what if this entire thing were actually much less than one then you still have the concentration of uh the rate of production of bromine to be uh to be depending upon the concentration of bromine to the half which is not the case of what what 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 that's like the overall dependence if it is comparable to one you have to keep this as it is and we can't resolve exactly between half and three halves it's somewhere in between right for the concent concentration of bromine depending uh influencing the rate and on top of it we have to explain the presence of chbr so how would you do this right so this is possible the the above result the above result the above result result can be explained can be explained by the following five step mechanism right and we see fstep 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 Rea rate rate equation but that is not what what is found in the experiments so something else is happening at the molecular level let's look at the five St mechanism so if you now say there was some um compound X Prime some species X Prime that bromine collided with to form um 2BR R that's an intermediate right um with a rate constant K1 and let's call this reaction One um this is a say chain initiation step and uh the intermediate BR now attacks the stable reactant H2 to produce hbr plus another intermediate H which is a which has a rate constant K2 and let's call this reaction two this is a chain um chain propagating propagation step because we stop started with a intermediate reacted with a stable reactant produced a stable product but produced another intermediate so like the net um gain and loss of intermediates is the same okay so it's just propagating the chain and uh let's say we have a third reaction where H is reacting with br2 to give you uh hbr plus BR this is this is the V counterpart of H in the previous reaction so this is let's say reaction number three this is also a chain propagation step and uh then we have h plus hbr that gives you h2+ br right look at what's 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 all right but what is but but but it actually consuming a stable product and producing a stable reactant that's simply that's simply because this is actually the reverse of reaction two so you have reversible reactions that are happening okay in trying to get into equilibrium under the given conditions right and what what is the um what is the uh uh 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 those are the stable reactants produces a stop plus a right we are actually not producing any more reactants than we consumed stable reactants we have instead produced uh intermediates that's all right okay that's like a chain initiation step and the reverse reaction a plus a star gives 2 a that is again producing stable reactants it's kind of like saying yeah I started with reactants and then did a lot of things and then got the reactants back okay 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's when you will start looking at self-inhibiting reactions so when you're having the stable products being involved in um reverse reactions right then they get consumed and their concentrations affect your Global reaction rate right so this is a for all right but this is a um this is a chain propagating step all right but this is this is the one that is reverse of uh this is a inhibition reaction um reverse of reverse of two okay and uh finally we now have X prime plus Dr plus p r gives of course the previous reaction had a rate constant K4 and this is a Rea rate constant K5 B 2+ X Prime Okay now what's going on here we are um 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's a reverse of uh reverse of one so you're having two reverse reactions in this out of out of five so effectively this is having a reverse here this is having a reverse here this is the one that's not having a reverse reaction yeah and there are two two reactions that are producing hbr there is one that is consuming assuming hbr all right and finally something that you have to keenly note you now have a ter molecular reaction here so the moment you have a ter 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 termolecular reaction whenever you're making these kinds of comparisons the first thing that hits hits your head is pressure what's the effect of pressure okay so that's one of the things that we will have to be looking at just like how we did uh previously for this example so the point the answer the answer is I'm not going to work out the details that's going to be a home work for you so here is the answer on how so what you do you you now take each of those reactions write the rate equ rate equation for whatever is its Chief Products okay in terms of its reactants concentration this is a molecular level thing so law of M action directly applies right and we can show we can show using study State approximation that Capital C1 that we had before right in the um in the uh Global reaction read expression is tce K2 * K1 divided by K5 to the half and uh C2 equal to K3 / 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 U let's not worry about this so so you understand right that means if you now write out the rate equations for each of these and then pick out those rates that are based on intermediates that is dcbr over DT or DC h H over DT typically these are the two intermediates that are happening here okay and then set them approximately equal to zero you now get two equations and you have to try to eliminate c h 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 don't over massage you because whenever you know the answer you try to actually try to fit it in but but I I have done this when I was a student okay so it's not it's not terribly bad uh you can you can show this yeah so what else can we do so this is this is an example of how the stady state approximation uh looks like so now let's look at another approach where we try to simplify which is is called the um partial equilibrium partial equilibrium approximation yeah partial equilibrium approximation now this is a bit different from what 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 doesn't have to be reactants okay so this is our goal we don't want to deal with a a a a OD for the concentration of intermediates we don't 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 zero 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 the stable species reactants or products right the answer is we will now try another approach remember these are based on some Physics it's it's 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 and during this time they they their concentrations approximately constant okay so that was a physics there there's 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 first example that we had no a plus a gives a Star Plus a forward and backwards and that's 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 and now a star is around and it's it's it's an equilibrium with a right and it just slowly gets consumed during uh 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 it if it is now so effectively what we're looking for is if you now have fast reactions that are forward and reverse is it possible to actually apply equilibrium equation rather than raid equation The Raid equation is the one which has a DC a star/ DT equals a function of CA and C A star okay that's an OD and we don't like the OD we want an algebraic expression but the equilibrium equation for that assuming that is actually in equilibrium right would be you now have a KP or a KC that's equal to concentrations of reactants of uh the um forward reaction divided by concentrations of the the reactant of the reverse reaction raised to their respective stochiometric coefficients right that's an algebraic relationship we don't have a we don't have a differential equation there right so this is essentially what you're trying to do so here where forward and let say fast forward and reverse there is not fast for forward this is fast forward and reverse okay uh reactions um occur so that they could be considered as an equilibrium actually we are saying something about their rates it's not just that they are fast and therefore uh 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 the reverse reaction rate okay so we we basically saying let's not worry about how fast it is it's all fast okay so they're all equal right that's exactly what you're what you're saying here so and uh 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 uh we find these kinds of ratios that are showing up K1 over K5 for example is actually ratio of forward to backward right so that that's exactly the re the 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's now look at an example so example or shouldn't say example uh let's say Let us look at consider temp template reactions template reactions let's say a + B2 gives um let's say this is forward and backward so we call this a ab+ b uh this is uh kf1 this is kb1 all right and let us call this uh 1 and uh b + A2 gives and takes gives and takes AB plus a and we call this kf2 kb2 let's call this reaction two and uh AB plus a A2 gives and takes A2 B plus a right and let's have rate constants kf3 and kb3 we'll call this reaction three and finally let's have a + a plus M gives A2B plus M that's reaction four that means the last reaction is not a um a u reverse reaction see it U forward it doesn't 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's actually seven reactions it's 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's is just a different way of writing this and uh so let's call the rate constant for the uh fourth one is kf4 we don't have a kb4 yeah so while we are interested now it depends on how you are I mean if you if you are like a chemistry kind of person like I want to see reality what is this A and B okay they're not really uh happening in reality you might want to throw hydrogen for a and oxygen for B okay and um finally you might get you might be looking for H2O right and uh and then you can say a B is O and so on uh so essentially what's going on is now you have a atomic species uh colliding with a molecular species to give rise to two Atomic species so this is like a chain uh branching right and this is again starting with one intermediate and this is this is supposed to be a stable species you're now getting two intermediates so chain branching and U this this is the stable species here this is an intermediate this is the final product and that's an intermediate so that's a chain propagation you started 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're finally getting a stable product so there 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 to try to get rid of them yeah so what we want to do then is we say here we say um K kf1 CA C uh CB2 equal to k kb1 kb1 c a b CB right that's for the first reaction okay or since you have the same molecularity on either side okay this will go as P squ this will go as P squ so concentrations can be written in terms of mole fractions and mole fractions can have pressures uh the total pressure so the pressure dependence is going to be the same on both sides so doesn't matter whether you use a equilibrium constant based on pressure or equilibrium constant based on concentration I'm just going to go back and use Concentra equilibrium constants based on pressures okay so so this this simply means that we now have c a CB divided by CA CB2 equal to Capital P1 right capital K sorry capital K subscrip P1 capital K represents equilibrium constant based on pressure and we have subscript one 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 um c a divided by CB C A2 equal to KP B2 and the third one C A2B c a equal to sorry divided by um c a b c A2 equal to kp3 right so what was the purpose now do we still understand we want to try to get rid of CB and CA in favor of and cab as well we want to get rid of CB cab and CA in favor of CB2 C A2 and C A2B all right so this is something that we will do next class [Music] [Music]