 Good morning. We will quickly review what we have learned in the last lecture. I told you about what is catalyst? It reduces the activation energy. The two types of catalysts use in industry homogenous and heterogeneous. Heterogeneous catalysts are quite common because of their engineering benefits. They can be separated very easily, corrosion problems are less and that is the main reason like most of the reactions are catalyzed by the solid catalysts. Now, how to deal with solid catalysis? Because the rate expression can be different. There are many other aspects like mass transfer effects because we have multiple phases involved in the reactor and there is a possibility that you have a different kind of rate expression depending on how the reaction takes place on the catalyst surface. Now, today we are going to look at different steps involved in catalysis. So, when a reaction takes place, it is not as simple as like A is getting in contact with B in the presence of catalysts and reaction takes place because here in this case catalyst is or the catalytic active species is present in the solid phase and the most of the times the reactants and products are in the liquid or gas phases. So, if you have a reactor, say for example, the CSTR in which there is a reaction taking place in the presence of solid catalysts. So, it is like a slurry for example or it can be a fluidized bed reactor where the catalyst is in the fluidized form and the reactant goes from the bottom and fluidizes the catalyst can be a gas phase or liquid phase reaction or other gas phase or liquid phase reaction mixture or it can be a fixed bed where the catalyst is packed in the tube and then you have a flow taking place either bottom to top or it can be top to bottom. So, there are many possibilities. We are going to look at catalytic reactors in detail later and their designs, but then just to get an idea as to what reactant molecule has to do and the product molecule has to do as far as the reaction on catalyst surface is concerned. So, we have the reactants sitting in the bulk, whereas the catalyst is present in the solid form on which the active species are present. So, now the reactant has to go to the catalyst surface and the reaction has to take place there. Now, while deriving the rate equation or coming up with the estimation of the rate of reaction, we need to understand all these steps because most of the times these steps they take place in series and because of that like one of them if it controls the overall rate then we need to really concentrate on that phase. So, let us look at different steps involved in catalysis or a catalytic reaction to take place. So, I have a catalyst particle and I have a reactant sitting in the bulk. Now, there are so many particles like this. So, we just looking at one catalyst particle and A that is reactant is going to go and react on the catalyst surface. As I told you before like inside the catalyst you have pores. So, most of the time we are talking about the pores on which much of the surface area is there. So, A has to go to the catalyst surface first and then travel through the pores and get access to the catalytic sides. So, as I said in the lecture before or earlier lecture rather. So, you have this pores presence and through which the reactant is going to travel. So, what it needs to do first is it has to overcome the resistance offered by the environment near the external surface of the catalyst. This is the external surface of the catalyst. So, that resistance needs to be overcome and after that the molecule has to travel through the pores. So, there is a different type of resistance offered by the particle. So, there are two different resistances offered by this particular system for the reactant to go to the catalyst surface and react there and then subsequently adsorption will take place and then of course you have reaction and the same thing will happen in the case of product. So, what are the different steps? The steps are external mass transfer from the bulk to the catalyst surface from the bulk to the catalyst surface. Then internal diffusion that means the component or the reactant has to travel inside the particle through the pores the environment is different there. So, the resistance offered by a particle will depend on its own properties and of course the reactant diffusivity and all. After that suppose there is a catalytic site present on the internal surface then you have the actual reaction taking place there. But yesterday I told you or in the last lecture I told you that the component or the reactant will not directly react because the catalytic species or the active site will be present on the surface though the reactant has to go and adsorb on the surface first. So, if you have the surface on which the catalytic site is there A has to go and adsorb and form a complex which you may call it as A plus S giving AS. We are going to deal with this later in detail. So, this adsorption has to take place and all this is going to happen for the reactant first. It can be a reaction with multiple molecules it can be A plus B giving C or it can be A getting asomerized to B. So, it can be a single unimolecular reaction or it can be a bimolecular reaction. So, all this is going to happen for reactants right. Then once adsorption takes place the reaction will happen on the surface between the adsorption species or one of the reactant can be an adsorb state and the other reactant can be in the gas phase. So, actual chemical reaction will take place. So, once the reaction takes place the same thing will happen for the products. What is that? The product has to desorb from the surface because reaction is taking place on the surface product might be in the adsorb state. So, it has to desorb then diffuse through the particle and then it should come out from the film. So, again the external mass transfer for the product. So, this is for the product. So, these are the different steps that are taking place when a catalytic reaction on solid takes place. So, in order to understand the overall rate we need to understand each and every step what is it depend on because most of the times the rate is governed by one or multiple of these steps depending on what kind of conditions you are doing the reaction in. So, there are 6 or 7 important steps 7 important steps main chemical reaction is one of them and rest all are mass transfer diffusion inside a particle adsorption and desorption diffusion external diffusion again for the product. So, let us look at individual steps in detail of course to start with let us concentrate first on these steps that is adsorption reaction and desorption. We will talk about internal diffusion or sometimes it is also called as intraparticle diffusion and external mass transfer in detail later and how to incorporate the effect of these steps in overall rate. So, let us first concentrate on 3, 4 and 5 that is adsorption reaction or sometimes called as surface reaction, surface reaction and desorption. Why we want to study these together because these 3 steps will relate the overall rate of the reaction or actually the rate of the reaction with the concentration in the liquid adjacent to or liquid or gas adjacent to the solid surface. So, these 3 steps are going to give us the rate law. So, in a normal case I have only the reaction taking place whereas, in the case of heterogeneous catalysis along with the reaction we have adsorption and desorption both taking place simultaneously. So, the rate law that is going to come up or going to be derived out of this will depend on these 3 steps diffusion and intraparticle sorry intraparticle diffusion and external mass transfer will be dealt with as a separate exercise. So, let us talk about adsorption. Let us give a mathematical treatment to this process as you know adsorption can be of 2 types physical adsorption and chemical adsorption out of which as I said before chemical adsorption is quite relevant here. Most of the times the reactants they form a kind of a bond with active site and this bond or this particular complex is something that corresponds to a transition state or an intermediate which is at the peak of that energy curve. Now, this particular adsorption can be treated as a reaction. So, let us write I have a suppose I have a reaction that is taking place which is A goes to B say simple reaction isomerization reaction that is the overall reaction that is happening. So, this is A which is present in the reactant stream is getting converted to B. Now, what is the rate law for this reaction if the reaction takes place on the solid surface. So, in order to in order to derived rate law I need to understand different steps and we are looking at the first step that is adsorption. Now, you have active site S on which A is going to get adsorbed. So, let me write this as a reaction A plus S. So, this is the overall reaction and this is an individual adsorption step where A plus S gives A S. This also is a reversible process. So, that is why I have a reversible sign here. Now, I am writing this as a reaction. So, it may have its own rate adsorption as an individual step will have its own rate. Now, if the if the adsorption step is instantaneous at any given time this equilibrium is satisfied. That means, if the reaction is very fast the equilibrium is satisfied and what do we call this equilibrium as in the language of adsorption it is called as adsorption isotherm adsorption isotherm. So, can we come up with an expression for adsorption isotherm. So, let us write law of mass action for this reaction say K adsorption is equal to C A S divided by C A into C S. So, this is the law of mass action. Now, in this what do we have you have this as a concentration of adsorbed species A S. This as concentration of vacant sites what is S S is a site. So, C S is a concentration of vacant sites and this is a concentration of A in the bulk. Now, the units can be different this can be per unit weight of the catalyst or per unit surface area of the catalyst this would be per unit volume of the reactor it can be anything and accordingly the unit of K will get adjusted. But this is something that I do not know vacant site. So, can I express vacant site in terms of adsorbed species. So, when I do experiments in laboratory for adsorption what I would determine is the concentration of adsorbed sites and concentration of A in the bulk I would not determine this. So, let us try and express this in terms of other components. So, let us not worry about the reaction right now. So, let us talk about only adsorption that means there is only A present in the bulk apart from inert which is not getting adsorbed probably yes and let us try and write an expression for it. So, adsorption of A on the catalytic surface adsorption on surface and I am writing it as A plus S giving A S for which I have suppose I do a site balance that means on a catalytic surface I have different sites. So, the total concentration of sites C T is equal to C A S that means there are some sites on which A is adsorbed plus C S. So, there will be some sites on which no A is adsorbed or they are vacant sites. Please remember that here we are not considering reaction or any other species. So, there is only A present right and of course inert. So, A is getting adsorbed and there are some sites which are vacant right. So, we just trying to understand adsorption at the moment will try to connect it with reaction bit later. Now, we already written law of mass action for this reaction. What is that? It says K adsorption is equal to C A S divided by C A into C S. Now, if you calculate C S or other if you calculate C A S from this and substitute here what do you get? K adsorption into C A C S plus C S. Can we derive C S expression for C S from this? C S is equal to C T divided by 1 plus K adsorption C A that is it. So, this is the expression for C S if I substitute for this here then I get a relationship between C A S and C A. So, let us do that. So, K adsorption is equal to C A S divided by C A into C T divided by 1 plus K adsorption C A. So, this is nothing but a relationship between C A S and C A C A and rest all that is K adsorption and C T they are constants. So, in other words C A S is equal to K adsorption C A into C T divided by 1 plus K adsorption C A. So, this is nothing but the adsorption isotherm it relates C A S with C A. Now, how do we know that this adsorption isotherm is correct? This is all based on our assumption that A goes and adsorbs on a single side which follows a reversible reaction between A and S. If that assumption is correct then in that case this expression is. So, what do we do to understand or to know whether this expression is correct or not we do experiments in laboratory. This is only adsorption that I am talking about I am not talking about reaction here. So, just take A in some E naught if required and contact it with catalyst and look at how it behaves or what kind of adsorption relation I get. So, you generate data in laboratory and try and relate the data with this if your data fits very well in this particular equation then your assumption is correct and this adsorption isotherm is correct. But there is a possibility that A is not getting adsorbed on just single side. It is quite possible that A gets adsorbed on say two sides at a time. So, for example you have a surface on which you have two sides there and A is a molecule. So, suppose A is carbon monoxide. So, what is likely to happen is that C the atom from carbon monoxide will get adsorbed on one side and oxygen will get adsorbed on the other side. So, that different possible ways. So, what kind of expression or reaction I can write in this particular case because the two sides participating in adsorption. So, you have A plus 2 S giving A S 2. This is one possibility or if you want to write it in terms of actual component like for example carbon monoxide. I can write it as C O plus 2 S giving C S plus O S. So, if you start from here the expression that I am going to get later for adsorption isotherm would be quite different. It would not be as simple as what we have derived just now by considering A plus S giving A S. Now, it is A plus 2 S giving A S 2. So, you can follow the same steps and try and come up with the equation for adsorption isotherm. It will have a very different form. So, of course, these two are again different. So, if I start with this and try and eliminate the vacant side concentration in terms of C S at shock concentrations and the bulk concentration, you will get an expression which is quite different as I said from the earlier one. In this particular case, C S is same as O S. The concentration of C S is same as concentration of O S. So, I will give you the expression. You can try and derive it. So, for example, if you start with this particular case, the expression that you are going to get is of this particular nature. Say, C O S is equal to K A P C O raise to one half C T divided by 1 plus 2 into K A P C O one half. So, this is the expression. So, try and derive this equation based on this reaction as the adsorption mechanism. And one more assumption that you of course, one more thing that you need to remember while deriving this is that C C S is equal to C O S. And that is quite understandable because it comes from both O and S O and C they come from the same molecule. Now, let us look at this expression. So, I have given you the expression in the form of P C O. What is P C O? P C O is a partial pressure of carbon monoxide. And as you know, like if I am dealing with gas phase components, there is always a better way to express concentration that is nothing but partial pressure. So, concentration and partial pressure, they are interchangeable. So, P by RT is C or P C O by RT is equal to C C O. So, earlier we were dealing with concentrations now that we are talking about a gas phase component C O. I am expressing it in terms of P C O. So, do not get confused if you are seeing an expression in terms of partial pressure instead of concentrations. So, you can derive this try and see by following the same steps that what we have done by taking the side balance and all we are able to derive this or not. Now, why we are doing all this? Because now I have got a different rate expression sorry not rate expression it is an isotherm. And then again I am going to generate data and laboratory and see whether this fits well in the isotherm or not. And if it fits well then this is my isotherm equation and this is the adsorption mechanism that is the two sides participating. So, on atomic level the adsorption is taking place C is getting adsorbed on one side and O is getting adsorbed on the other side. So, because of this you have different types of isotherm equations that are possible and your job is to see which one is correct. So, you know the mechanism also and you get a right isotherm equation that you are going to use later in deriving the rate law. So, in this particular case if I have been talking about fitting isotherm in the data you can give a different form to this particular equation. So, this particular equation can be written in the form P CO raise to half divided by COS is equal to 1 divided by CT KA raise to half plus 2 P CO raise to half divided by CT. So, this is nothing but or this is same as this, but I have given a form which looks like a linear plot between this. So, this is y and this as x and if I have written this if I plot this then I am going to get this as the intercept and this as a slope. So, I am going to generate the data COS versus P CO in laboratory and put them in the form like this and if I get a linear relationship between these two then I am sure that this expression holds good. So, this is a way to know how adsorption takes place and what is the right kind of adsorption isotherm that is applicable in a particular case. So, it may vary from component to component it may vary from adsorbing to adsorbing. So, it is always better that we do this exercise before we go ahead and try and find out how the adsorption takes place and what is the adsorption isotherm equation applicable in that particular case. So, now after knowing how adsorption takes place let us talk about reaction, chemical reaction. We are going to revisit adsorption later when we derive the rate law I am just explaining you individual steps which are possible. Now, chemical reaction what is going to happen on the surface? So, you have now the adsorbed species that is AS which is going to get transformed to BS. Now, let us try and understand the chemical reaction that is taking place on the surface. What is the rate of this reaction? It depends on which concentrations and how to derive that rate in terms of the bulk concentrations. Now, if I want to write the rate expression for this reaction which is simple let us assume the reaction to be elementary and say that fine R AS is equal to minus K CAS minus say K R K R dash C BS because this is a reversible reaction. AS is going to get consumed that is why there is a negative sign here. So, this is the rate equation this is the rate equation if the reaction is instantaneous then that case these two are equal that is a that means at any given time the equilibrium is achieved. So, K E Q for reaction is equal to C BS divided by CAS. So, the ratio of adsorbed the ratio of concentration of adsorbed B and concentration of adsorbed A is constant because there is always instantaneous equilibrium if the reaction is very fast. If reaction is not fast or it is kinetically controlled then you have this expression this is how we give mathematical treatment to reaction. Now, again there are many possibilities like what I am saying here is that on a catalyst A is getting adsorbed. So, I have AS here and on the same side it is getting transformed to B that means this I will draw it on a separate paper. So, you have AS you have this situation it gets converted to BS on the same side on the same side. This is called as same side mechanism this is called as single side mechanism. Now, there is a possibility that another side may be involved in this it may be like this you have AS and there is a side here in order for AS to get converted to BS it needs help of this side how of course not you do not you should not go in that much detail, but let us assume that this side is required for conversion of AS to BS. So, it may you after reaction you may see something like this. So, here the equation is A. So, in this case the equation is AS in equilibrium with BS whereas here it is AS plus AS is in equilibrium with BS plus AS you may have three sides required. So, it all depends. So, not necessary that you always need one side for conversion. So, there are different mechanisms. So, that is how the reaction on solid surface is different from what happens in a normal homogenous medium because we have discrete catalytic sides available on the surface at several at different positions. So, there is a possibility that you may have this particular situation you may have two AS here. So, you may need one more side. So, there are many possibilities accordingly your equation will change. Now, in order for this reaction to take place I have two sides involved. So, if the rate equation for this particular reaction RAS is equal to minus KRCAS CS minus KR dash CBS CS. Now, you may say CS is common you can always club it with this no because CS will change with respect to time. It is in equilibrium if adsorption is fast it is in equilibrium with adsorption species. So, accordingly rate equation will change. So, there are different mechanisms possible. So, the first one is single side the second one is called as dual side it is a dual side mechanism. The scientist in who have come up with these are Langmuir and Hinshelwood. Of course, these are the concepts of Langmuir the Hinshelwood applied them to get these equations that is why it is called as Langmuir Hinshelwood. So, this is a Langmuir Hinshelwood single side mechanism or single side kinetics and this is Langmuir Hinshelwood dual side. Now, I am talking about just one isomerization reaction. Let us take an example of a bimolecular reaction or two different reactants reacting. In that case again there are several possibilities look at this you have a site you have a s and you have another site where another reactant is adsorption. So, I am talking about a multiple reactant case a plus b now b is a reactant and not a product. So, a plus b giving sorry c plus d or c whatever. So, a s. So, a will get adsorbed on this and b will get adsorbed on the other site b s and the reaction will take place between these two. So, a s plus b s giving possibly c s plus d s and then you can derive the rate equation. So, this is again a dual side mechanism. Now, there are again several possibilities if you want to further complicate it this one this site is of one nature say s 1 and this site is of another nature. So, on a catalytic surface you may have sites of different natures which are selective towards particular reactants many many possibilities and in that case again you can imagine things will change. So, all this exercise gives us a nice platform to come up with a right kind of rate equation for a particular step in heterogeneous catalysis. We want to as I have been saying all the time we want to put all these things together at a later stage to derive the single rate equation for a given reaction taking place on the solid surface. Right now we just looking at individual steps first understand them properly and then try and put them together later. Now, so this is a dual side mechanism there is another possibility that you have a s and it is going to react with b, but b is not getting adsorbed even if there are sites present here it is not going to get adsorbed b remains in the bulk and the reaction takes place between a s and b giving say c s plus b or whatever or it can be d s also some s will participate you know. What is more important here is that here the reaction is in between adsorbed species and the component present in the bulk and this is going to give you another rate equation. This is going to give you another rate equation which is not same as the one that we say it all depends on how we write the reactions. And finally, which equation is the so many expressions that we have got which equation is correct who will decide that it is the experimental data in the board. So, that experimental data which fits this equation well or rather in other words the equation that fits the experimental data well is the right equation that is a kind of mechanism for the reaction. So, this exercise helps us in knowing the mechanism of the reaction at the same time coming with the right kind of equation for the chemical reaction that is happening on the solid surface. And this mechanism is called as illy radial mechanism where the reaction is taking place between adsorbed component or adsorbed species and the bulk gas phase component. Now, the last step or the third step which is nothing but desorption now that is exactly similar but opposite of adsorption. So, it does not need much or rather adsorbed we need can give same treatment to desorption as what we have done for adsorption. So, again let us go back to a reaction overall reaction a giving b in that case suppose b is adsorbed or a is adsorbed and it forms b and b is in adsorbed state then desorption is b s which is formed after reaction gives b plus s again same thing this is opposite of adsorption. So, in order to know the adsorption isotherm we have to do the similar experiments that we did before for a and understand whether it is single site or the two sites participating whether it is on atomic level or molecular level. So, all these things are quite similar to what we have done for adsorption. Now, instead of a you have b here desorption is exactly opposite to adsorption. Now, that is for isotherm and for the rate of desorption again treat this as a reaction. So, rate of desorption is equal to k desorption c b s minus k d dash c b c s. So, this is rate of desorption. Now, in this rate equation you have these species b s and c these are the concentrations on the solid surface which is difficult to measure during the course of the reaction this is something that I can measure that is c b. So, the next step is how to deal with these concentrations these will appear even in the case of adsorption and reaction. So, all the rate expressions that we have seen so far are having these concentrations concentration of the adsorption species as well they are difficult to measure. So, we have to get rid of them or rather we have to substitute them in terms of the bulk concentration and that is the next exercise. So, let me summarize what we have learned so far. We have looked at three different steps adsorption reaction and desorption adsorption can take place on single side it can take place on two sides and depending on that will have a different isotherm equation. How do we verify the isotherm is correct or not? We generate data in laboratory for adsorption and see whether it falls in line with the expression that we have got and from that we identify the mechanism and of course, the rate expression or isotherm expression for the adsorption. Reaction it is again similar we have to write expression like what you do for reaction, but again it can be a single side it can be dual side or it can be iller ideal mechanism where the adsorption species can react with a molecule in the gas phase and accordingly we will have the rate equations and if at all the reaction is instantaneous we can write equilibrium relation also for the reaction and same treatment holds good for desorption as well where again like we can write expression for isotherm if the desorption is instantaneous or and of course, there is a way to verify whether the isotherm is correct or not and we can write the rate equation as well depending on what is the mechanism of desorption how many. So, are the vacant sides taking part or whatever. So, we just looked at B as giving B plus S again write expression for that. So, three steps we looked at them individually now we are going to put them together and see how the rate law can be derived for a particular reaction that is taking place on the catalyst surface. So, let us start with the same reaction and derive the rate law depending on depending on which one is the slowest step or in other words which one is the rate controlling step. So, we need to understand what is rate controlling step. So, let us spend some time to understand what is rate controlling is easy to understand it has it is quite logical that you have processes taking place in series. So, you have processes taking place in series then the overall rate is governed by the slowest of all. So, let us talk about rate controlling step. So, I have three steps that are taking place in series this is more like a relay race. So, there are three steps or three participants two of them are very fast and one is relatively slow. The time taken by first participant is say one second time taken by second participant is two second and time taken by third participant is fifty seconds. So, who is the rate controlling person this one right because he is the slowest of all I need to take more efforts on this. So that the overall time required can be reduced what the total time is about fifty three seconds. So, if I take efforts on these people I am not going to reduce the time much whereas, this fellow needs attention. So, what this fellow is doing is a rate or rather this is a rate controlling step as simple as that there is analogy between electrical circuit and chemical steps as well. So, you have three resistances in series. So, these resistances are nothing but steps and the total resistance is the addition of or the total of all these resistances. So, it is like this you have right. So, R 1, R 2, R 3 if R 2 is much higher than R 3 and R 1 I need to reduce this. So, that the current that is flowing in the circuit can be increased significantly right. So, that is nothing but rate controlling step. Now, let us come back to our reaction right. In reaction you have these three steps that are taking place adsorption chemical reaction adsorption chemical reaction and desorption. The overall rate will be governed by one of these sometimes it can be two or all three of them can govern, but that is quite rare. So, the one that controls overall rate that is called as a rate controlling step. So, we will use this concept later to come up with a rate law. So, we are going to derive a rate law for heterogeneous reaction taking place in the solid surface. Let us go back to our equation A giving B this is the overall equation overall reaction isomerization reaction right. Now, if I want to know the rate for this reaction I need to come up with a mechanism. So, I have adsorption taking place A plus S giving A S right. So, this then the reaction A S plus sorry A S giving B S and then the desorption that is B S giving B plus S. This is a single side mechanism for A going to B, but it is quite possible as said before that there are two sides involved in the rate determining steps. In that case the rate expression the derivation will change. So, these three steps they are going to take place simultaneously right and overall rate will be governed by the step which is the slowest of all intrinsically slowest. So, that is going to govern the overall rate and we are going to write the expression for the rate for individual steps the one which is slowest we are going to retain that expression and rest all we are going to treat them as equilibrium steps and come up with the rate law. So, we will continue this discussion in the next lecture. Thank you.