 Good morning. So, let us continue our discussion. Last lecture we looked at different steps involved in a catalytic reaction. So, if you have a catalyst, typically the catalytic site will be present inside and a reactant will be there in the bulk. And this reactant has to diffuse or rather first overcome the external mass transfer resistance to reach at the external surface of the catalyst. And then inside the catalyst it has to diffuse through the pores and reach the catalytic sites. So, the steps involved I will just do a quick review external mass transfer, then internal diffusion, third chemical reaction on the surface sorry before that you have adsorption, then chemical reaction on the surface, desorption of the product or products. Then again the back diffusion, internal diffusion of the products and last step is external mass transfer back to the bulk of the products. So, these are the different steps and then we concentrated more or we focused our attention more on these 3 steps. Why? Because these are the 3 steps that take place when the reactant molecule is very close to the catalytic site. So, these 3 steps together will give us the rate equation for intrinsic or for the reaction to take place. And of course, after that like if we include the effect of external mass transfer and internal diffusion, then we get a overall rate equation for reaction that is taking place on the surface of the catalyst. So, in the last lecture we looked at these 3 steps and the equations associated with these 3 steps for the rate of the particular step. So, adsorption, chemical reaction and desorption. Today what we are going to look at is how to come up with the rate law for a particular reaction if one of these steps is the rate controlling step. So, what is rate controlling step? The concept of rate controlling step is very important if you have multiple steps occurring when a reaction takes place. So, the rate controlling step this is also called as rate determining step. So, R C S or R D S what is it? If you have many reactions or sorry many steps taking place for example, in our case we have adsorption, chemical reaction of the surface and desorption. So, these 3 steps are taking place in series. Now, the overall rate of the reaction of course, not considering diffusion and external mass transfer right now, but adsorption, desorption and reaction the overall rate will be determined by the slowest step intrinsically slowest step. So, it is very similar to what happens in the electrical circuit. For example, there are 3 resistances in series. So, you have a circuits the current that flows to the circuit will be dependent on these the magnitude of these resistances. If one of these resistances is very large that means, that resistance controls the overall flow of the current in the process and if I want to increase the current flow I will concentrate more on this which is the largest resistance than the other 2 right. So, for example, this is 50 this is 2 and this is 1 then I will try and reduce 50 that will make much more impact on the rate compared to the other 2. So, in short the rate controlling or rate determining step is a step which is intrinsically very slow. Now, who decides whether the step is slow or not there are the rate constants involved in that process. So, the rate constant for the reaction is very small in that case the chemical reaction the surface controls overall rate. If the rate constant for the adsorption is very small in that case the adsorption that controls overall rate or if the desorption rate intrinsically rate constant is very small in that case is the desorption that controls the overall rate. Now, how is this important because after all I want to get the rate equation which combines the effect of all these 3 steps. Now, why do I need to combine these steps because I want to get the rate equation in terms of the concentration or the partial pressure of the reactants in the bulk. Now, it is very important that you have the surface reaction taking place and A is getting adsorbed and A will have certain concentration in the surface in the form of say A s. Now, this concentration I cannot measure this concentration. So, if I want to design a reactor I know the concentration in the bulk right I know the concentration of A coming with a feed I know the concentration of A going out of the reactor I know the concentration inside a reactor, but that is in the bulk bulk in the sense in the liquid or gas mixture which surrounds the catalyst. But I do not know the concentration of the surface. So, that is the problem. So, I need to somehow get rid of all these concentrations which are unknown and come up with a rate equation that is in the form of the bulk concentration that is what we are going to do now. Rate law for catalytic reaction remember when I say catalytic reaction it is the solid catalyst reaction where adsorption takes place and in the chemical reaction and desorption. In a normal homogenous reaction that we have studied before in the first of part of reaction engineering there we just looked at chemical reaction there was no adsorption there was no desorption. Now, I want to combine the effect of these three steps and come up with the rate equation. So, first step adsorption we have looked at this adsorption A. So, let me let me say I have a reaction A reversible with B overall reaction. So, if I am setting outside a reactor sorry not reactor outside the catalyst then I am seeing conversion of A to B and this is the overall reaction. Here it is assumed that external mass transfer and internal diffusion are so fast that the concentration of fluid inside a pores in contact with the active sites is equal to the measurable bulk concentration. Actually this reaction will take place in different steps. So, adsorption A first A adsorbs on the catalytic site which is present on the surface then this is in equilibrium with A s. So, this is adsorption we already looked at it the rate of adsorption the rate of adsorption is equal to k A C A C s minus k A dash that is the that is the reversible step or the reverse step C s this is the rate of adsorption this is the rate of adsorption. Now, this step may be rate controlling or may not be rate controlling we will look at that separately later on right now I am just looking writing rate equations then the chemical reaction A s on the site gets converted to B s. This is my hypothesis remember this is my hypothesis I told you like they can be different ways I can write these steps and accordingly my rate equation will change later which one is this right rate equation that will decide based on the experiment experimental data we are going to we are going to look at it later. But right now I am just assuming this and seeing whether I will get a right kind of rate equation or not. So, the reaction that is taking place the as summarization reaction on the surface this site A s gets transformed to B s and this is a reversible reaction it has to be a reversible reaction because overall reaction is reversible. So, I am not quite sure whether. So, let me write it as reversible reaction itself and then we will see what happens. So, chemical reaction the rate of chemical reaction. So, let me write a rate of chemical reaction is R R see where I did R A then it will be negative sign. So, I have to consider the sign whether it is reactant or product and all that I am just writing the general equation right now. Fine this is K R C A s minus K R dash C B s simple. Then we will move to desorption third step B s is going to get converted to B plus s that is desorption and it is going to go to bulk right fine. So, now K desorption sorry rate of desorption is equal to K D C B s minus K D dash C B C s. So, we have three rate equations we have three rate equations I want to now combine these three rate equations of individual steps and get the rate the overall rate as a function of C A and C B very important. I do not want C A s here I do not want C B s here I do not want C S here in this because C B s C A s and C s these three concentrations I do not know the values of these three they are there they I cannot measure them they are there, but then I cannot go to the surface and actually measure this concentration. So, it is always better that I get a rate equation in terms of the bulk concentration that is the aim of this exercise. So, how do I get this now in order to get a rate equation for catalytic reaction I need to make assumption now that one of the steps is the rate controlling step. So, it can be either adsorption it can be chemical reaction or it can be desorption depending on the activation energy of the respective steps right now the way I have written all these reactions in the reaction form even if it is adsorption physical adsorption or it can be chemical adsorption of course, for the reactions chemical adsorption I am considering it as a reaction right. So, I am writing rate equations for these and one of these is the slowest depending on the activation energy or the rate constants of those steps right. So, let me make an assumption initially that the chemical reaction controls the overall rate what I mean by controls the overall rate this again one more way of saying that chemical reaction is the rate determining step or rate controlling step right. So, that is the slowest step that is the slowest step now if you remember when the chemical reaction controls I had written R R as K R into C A S minus K R dash into C B S right. Now in this equation C A S and C B S are unknowns. Now if the overall reaction rate is controlled by this I can make an assumption that R O is R R very very important because the rate of the reaction is very slow this is intrinsic rate is very slow the other two steps are very fast. So, the overall rate is governed by the reaction. So, it is like a relay race at I was telling about this that three people running in the relay race the slowest of them will govern the overall time required to complete the race. So, the first fellow does it in five minutes the second fellow does it in two minutes, but the third fellow takes normally does not happen, but if he takes say 50 minutes the total time is very close to 50 right. So, the rate of the third person is the rate of the overall process right that is how I am looking right. So, rate of overall process is the rate of chemical reaction. So, this is the rate now can I just say that this is my rate equation and stop here I cannot do that because this concentrations are not a bulk concentration I need to express these concentrations in terms of the bulk concentration. So, next step is again very important that I need to convert these we are going to list different steps how to derive the rate law for catalytic reaction I am just first explaining you how with the help of example how to get to a rate equation then we will systematically write different steps to be performed alright. Now, this is the rate equation C A S and C B S have to be expressed in terms of bulk concentration. Now, how do I do that now I take the help of the adsorption step and desorption step. Now, these two steps are instantaneous they are very fast what does it mean just look at those steps I have this adsorption step now I have this rate equation if this reaction is instantaneous if this this particular step is instantaneous then it is always at equilibrium I will repeat it is always at equilibrium what is what does it mean it means that A S is in equilibrium with A and S at any given time this reaction is so fast that the equilibrium is achieved what does it mean if the equilibrium is achieved that means the rate of adsorption once the equilibrium is achieved is 0 that means this is equal to this right. So, this is an equilibrium step because it is intrinsically very fast what about this this is not an equilibrium step because this is a slow step and the rate governs right. So, I am still going to write the same thing here whereas in this case I am going to equate this rate to 0 similarly for desorption for desorption this rate is going to be same as this rate because R D is very very fast that means this step is also this term is equal to this term the reaction or the desorption is the equilibrium step. So, the rate controlling step is the slowest step and that is going to govern the overall rate rest to R in equilibrium they are instantaneous they are in equilibrium. So, let me write those expressions by setting those rates of those individual steps which are intrinsically fast to be equal to 0. So, we have we have adsorption step I am going to set this rate equal to 0. So, what happens is R A D S is equal to 0 which is nothing but k A C A C S minus k A C A C S minus k A C A S is equal to 0 right what it means is that A S C A S divided by C A C S is equal to some constant capital K which is equilibrium constant right is nothing but in terms of small k A and k A dash it is k A divided by k A dash right what is this this is adsorption equilibrium constant. So, I got one equation which relates A S with A C A and C S. So, let me write let me remember this equation and get back to this later. Similarly, for desorption for desorption I have R D equal to 0 because rate of desorption is now 0 equilibrium is achieved it is very fast step which is nothing but k D C B S sorry small k D because these are the rate constants for desorption all right. So, this gives me C B S divided by C B into C S is equal to some capital K B right. So, this is adsorption equilibrium constant which is nothing but k D dash divided by small k D dash divided by small k D these are rate constants. So, that is similar to a normal reaction equilibrium or reversible reaction this is adsorption equilibrium constant for B. Similarly, I had adsorption constant before. So, let me remember this equation and make use of it later. So, I have got two expressions C B S for C B S one equation from desorption step and for C A S one equation from adsorption step this is adsorption constant. Now, I am going to make use of these two equations and express C A S in terms of bulk concentrations. So, let me write it down from equation one C A S is equal to C A into C S into capital K A from equation 2 C B S is equal to C B into C S into capital B right. So, these are the two equations I have got for C A S and C B S let me get back to the rate equation rate overall is equal to rate of reaction is equal to K R C A S minus K R dash C B S. Let me substitute for C A S and C B S which gives me K R into C capital K A C A S sorry C A into C S minus K R dash capital K B C B into C S. So, I have got a rate equation or the rate equation for the overall rate remember what I have done is I have just looked at the slowest step which is the rate controlling step I have retained its rate equation and whatever I do not know I am trying to substitute for it taking help of the equilibrium steps or the instantaneous steps which are in this particular case adsorption and desorption fine. Now, R O is equal to this do you know everything in on the right hand side I know C A I know C B do I know C S I do not know C S I do not know the concentration of the vacant sides empty sides on the catalytic surface ok. So, these concentrations are again needs replacement or rather I need to substitute for them such that my right hand side is the function of only the bulk concentration of course, there will be rate constants and adsorption constants which are constants anyway their parameters of the rate equation. But concentration should be the only bulk concentration measurable concentrations fine. So, let me do that how I get C S now I need to express C S in terms of the other concentration what I do for that now consider a catalytic surface there are many sides of course, this is a very simple presentation of the surface actually the surface is not in the form of a sheet or plate most of the time it is there inside a pores, but then I am just opening those pores and showing the surface as this and there are sides present now these sides this is a fixed concentration of these sides fixed number of these sides right. So, the total concentration of the sides C T which is a constant for the given catalyst this C T is a constant that is equal to at any given time C T is equal to there are some sides on which A is adsorbed there are some sides on which B is adsorbed what else there are some vacant sides right. So, we have three different types of sides present on the catalytic surface now this makes this side balance makes our job very easy now. Because in this expression C T is constant C S is something that I need to find out C A S and C B S they are also not known, but then I have got a expressions for them right what are those expressions for C A S and C B S these are the expressions these are the expressions for C A S and C B S all right. So, let me go ahead C A S is nothing but C A S is nothing but capital K A C A into C S plus capital K B C B into C S plus C A now this gives me the expression for C S in terms of C A and C B. If I take C S common then I can write C S is equal to C T divided by capital K A C A plus capital K B C B plus capital K B C B plus 1 what I have done here is the side balance very important step. So, remember what I have done so far I have looked at different steps I have decided that one step is rate controlling that governs the overall rate. So, I have written the rate equation for that step in that step there are some species concentrations unknown unknown the sense I cannot measure them there in the adsorbed form or whatever right. So, in order to replace those concentrations I have taken help of the equilibrium steps the steps which are not rate controlling like adsorption and desorption and from those equations equilibrium equations I have got the concentration of the adsorbed sides and I have replaced these concentrations in the form of the bulk concentrations right. So, let me substitute for C S now I have done side balance and got a concentration of empty sides or vacant sides in terms of the total concentration of the sides and the bulk concentrations. So, let me get back to my rate equation. So, my rate equation is this in this I do not know C S. So, I have I am going to replace it with the help of the expression that I have got just now it is C T divided by 1 plus K A C A plus capital K B C B this is the expression for the overall rate look at the right hand side is there something that I do not know see there are all constants K R K A right capital K A K R dash capital K B C T this is also constant and what is not constant or other the other concentrations of A and B they are in the form of or they are the bulk concentrations that is something I want. So, this is the final expression that I have got now of course you can club this and say that fine I have R O is equal to K of forward reactions K A sorry K F C A minus K of backward reaction C B divided by 1 plus K A C A plus capital K B C B what I have done is I have combined the C T K R and capital K A in K F and K R dash C T and capital K B in K B right. So, this is a form of equation now if I want to write it for A that is R A instead of this I have R A then this will be negative right because A is getting consumed R B this will be positive like say same as what we have done before in part 1. So, depending on the stoichiometric coefficient of the component I give the sign on to the right hand side. So, K B can be expressed in terms of overall equilibrium constant of the reaction in that case the number of unknown parameters will be reduced by 1. So, this is something which is equivalent to what we have done before like for example, a rate power law rate equation what is the power law rate equation R is equal to minus R sorry R is equal to K C A raise to N R is equal to K C A raise to N if it is R A then it is minus right. So, this equation that we will use so far rate constant and then the order of reaction instead of that I have got this particular rate equation. So, this is for a normal reaction all in a sense homogeneous reaction irreversible power law rate equation whereas, this is the rate equation that I have got for a solid catalyzed reaction which takes place on the surface as a result of adsorption chemical reaction and desorption, but do not forget I have got this rate equation by assuming that does chemical reaction is a rate controlling step clear. So, this is not a unique equation for any heterogeneous catalytic reaction it depends on the assumptions that I have made. So, assumption is reaction is rate controlling and other two steps are equilibrium steps, but what I am trying to tell you is these are the so this. So, what I have got is I have got something equivalent to what I used before here. Now, then I what I did later with this rate equation in the first part that is designing CSTR plug flow reactor in which I have used this equation. Now, what I am going to do now is like for a catalytic reactor instead of this rate equation I am going to use this rate equation that is the only difference. Of course, I need to consider external mass transfer then inter particle diffusion and other steps, but at this moment if suppose for example, like these two steps external mass transfer and inter fusion are not controlling or they are not there at all in that case this is my rate equation. So, to understand the difference between solid catalytic reaction and normal reaction homogeneous reaction where the rate equation is now in a very simple form whereas, in this particular step or in this particular case like you have a rate equation slightly complicated and again depends on the assumption that I am making. Now, the next obvious question is that whatever assumption I have made whether it is correct or not. So, who will tell me whether that assumption is correct for that I need to do experiments in laboratory and see whether this rate equation fits my data very well. I hope it is clear rate equation the validity of the rate equation that I have assumed is correct or not is decided only through experiments of course, there are theories to find out and all that, but the common practice is to do experiments in laboratory using the same type of reactors that I used the first part in differential reactors and all that or a batch reactor in laboratory or small c h t r and do experiments and see how the rate equation fits the data. If it fits well then my assumption is correct I go ahead and use that rate equation for my reactor design. If it does not fit the data well then what do you do you go back to your assumption change your assumption. Now, for example, in this particular case I have got this rate equation I have got this rate equation from by the assumption that reaction is controlling. Now, I make an assumption that the adsorption step controls. In that case the rate equation will change we will see what change will be see in the rate equation, but before that let us try and summarize and write an algorithm step by step procedure to get a rate equation when reaction takes place on the solid catalyst and is accompanied by adsorption and desorption. So, what do I do I have already told you about and let me write it. Step one rate equation for all the steps we need to assume some stoichiometry here or the scheme here like a plus s giving a s and so on. For all the steps that means adsorption reaction and desorption second assume one of the steps is rate determining step r overall is equal to r r d s. In the example it was a chemical reaction that we considered to be the rate determining step and this particular equation will have c a s c b s c s and all these concentrations are not known for that what I do is rest of the steps are in equilibrium so the rate for these steps is said to be equal to 0. What next what I get is all the known concentrations so the concentrations on the surface. So, from this particular assumption what I get is the concentration of the surface in terms of bulk concentrations terms of bulk concentrations in terms of bulk concentrations. What next write side balance why because I need to get expression for c s substitute for surface concentration and c s substitute for c a s c b s they can be many components which are going to adsorb and c s in r o is equal to r r d s sorry. So, this is substitute for c a s c b s in c s in this particular equation and then you have to just club the constants if possible and get a final rate equation expression. So, these are the steps involved in getting rate law for heterogeneous reactions or other reactions catalyzed by the solids in which adsorption and desorption are taking place. So, now I have shown you just one example very simple reaction a reversible with b that means isomerization reaction in that how to get a rate equation. But then there may be further complications that you may have a bimolecular reaction or you have a reacting with b giving c or c plus d what are how to get a rate equation for such cases. I have shown I have considered only chemical reaction to be controlling here, but there is a possibility that this rate equation does not feed the data well I can get back to my assumption for the same reaction a going to b I can say that fine now adsorption is controlling. Now, let us say what will happen if we have adsorption let us see what will happen if we have adsorption controlling over all that. I am not going to derive the rate equation again where so many things so many assumptions based on which every time we will get a rate equation different rate equation what you need to remember is the procedure you do not need to by heart this rate equation that you have got finally you need to understand how the rate equation is being derived. So, if adsorption controls the same procedure I will follow in that I will write rate equations for all the steps I am not going to write all of that, but then the next step is the rate determining step that is the adsorption. So, r overall is equal to r adsorption is equal to what is the adsorption a plus s giving a s. So, k a c a c s minus k a dash c a s this is my rate equation now I am not going to set it to be equilibrium because it is the slowest step. Now, this a s is unknown this c s is unknown I need to replace them how do I replace them by setting other two steps to be in equilibrium that is the chemical reaction and desorption. Now, chemical reaction is now equilibrium steps chemical reaction r r is equal to 0 is equal to k r c a s minus k r dash c b s minus k r dash c b s minus k r dash c b s minus I am going to set it to be equal to 0. So, c a s divided by c b s is equal to some equilibrium constant for the reaction capital K do not forget this is capital K is not small k this is equilibrium constant I have got a relation between c a s and c b s this is not enough this is one equation that I am going to use later in equilibrium again I have desorption to be very fast. So, r desorption is equal to 0 is equal to what is that desorption step b s giving b plus s. So, desorption is very fast same as what we have done before this is equal to 0. So, k desorption c b s minus k d c b c s right this is 0. So, c b s divided by c b into c s is equal to k desorption capital K and this is number 2. Now, these two equations together will give me c a s and c b s. Now, slightly different from what we have done this is not explicit relationship for c a s c a s because c b s appears here, but then I have an explanation for c b s I substitute it here and then I get c a s and c b s right in terms of the bulk concentrations and side concentrations and you know how to get rid of this this particular concentration that is by side balance. So, I am not going to continue this exercise further and get a final rate equation you can do it on your own right and will you get a same rate equation that we have got before? No, you are going to get different rate equation the form of the rate equation is going to be different from what we have got before by assuming the rate to be controlling that is before rate of reaction to be controlling before surface reaction and in this case it is adsorption that is control. Similarly, you may have desorption controlling the overall rate you will get a different rate equation all together. So, you got three rate equations three choices. So, your experimental data that you are going to generate in laboratory see where exactly this data falls and rather which equation that you have got fits the data very well right and this is not a end of it there are so many other assumptions that I can assumptions that I can make to get a different type of rate equation. So, that my data fits well in the rate equation I have I have just looked I have just looked at a possibility of single side adsorption. If you remember last lecture I told you that I can adsorb on two sides at a time. So, right now I have just taken rate rate for adsorption or the equation for adsorption as a plus s giving a s, but it can be a plus 2 s giving a s 2 right. So, in that case the situation will be quite different my rate equation will be quite different and I have choice of rate equation. So, like this you can make many assumptions and come up with different rate equations and see whether it fits the data very well or not. There are independent ways to know the mechanism, but we will not get into details that. So, there are techniques through which we know whether it is a single side adsorption or double side adsorption and so on. So, for a simple reaction of a being converted to b you have many possibilities of getting rate equation for a solid catalyzed reaction. Now, there is another possibility that you may have reaction in which two molecules are involved a reaction in which two molecules are involved where a is reacting with b a plus b giving c. This is my overall reaction overall reaction. Now, I am not looking at what happens on catalyzed surface, but if I am sitting outside a catalyst in the reactor then this is something I am seeing. Now, in this case again there are n number of or there are many possibilities and you can come up with different mechanisms. So, let me just write down the equations and look at the possibilities and I will derive the rate equation, but see there is a possibility that adsorption a plus s giving a s. Now, it is not only a being present there. So, you have b plus s giving b s reaction a s plus b s giving what c s, but one side is left out plus s. This is one possibility one can say that c is adsorbing on two sides. So, it becomes c s 2 instead of s I am just telling you many possibilities and every possibility is going to give you a different rate equation. So, this is reaction desorption c s giving c plus s. This is one possibility from this you can get a rate equation assuming that one of the steps is rate controlling. Now, how many steps are there earlier you had three steps. Now, in this case you have four steps 1 2 3 4. So, any one of them can control overall reaction. It can be adsorption of a, it can be adsorption of b, it can be adsorption sorry chemical reaction of the surface and it can be desorption of c. So, these possibilities will give you different rate equations. Now, I have written this particular mechanism here which is a Langmuir Hinshelwood type mechanism, where both the reactants are in adsorption state giving the final product in the adsorption state itself. But we have discussed this before as well that there is no or that there is it is not necessary that a s should react with b s. a s can be on the catalytic surface and will react with b which is sitting in the bulk. So, get back to this a s plus b s giving c s plus s instead of that I may write a s plus b giving c s. Instead of this I can write this in that case this becomes irrelevant. So, I have three steps a s a plus s giving a s, a s plus b giving c s and c s giving c plus s. There is no b adsorption surface, b remains in the bulk its adsorption is negligible. I will get different rate equation. So, this is called as Euler radial. We have discussed this before. So, this is one mechanism. It will give you another rate equation, it will give you another rate equation. So, there are many possibilities and you will get many rate equations and because of which you have another when you get rate equation there like we get so many rate equations you have to validate against the experimental data and in that case you see which rate equation is fits the data very well. So, in this case we have assumed that all sides are equal in terms of the activity in terms of the affinity towards the components. Now, there is another possibility that the sides on which a is adsorbed are different from the sides on which b is adsorbed. So, on the catalytic surface I have two different sides. So, one species or one component will get adsorbed on one type of sides and the other component will get adsorbed on the other types of types of sides and then the reaction will take place on the surface. Then the side balance will change instead of one concentration of empty sides C s I will have C s 1 and C s 2 right the two different types of sides. It is difficult for me to derive equations for every case, but I am trying to tell you many possibilities here which are likely to give rise to different rate equations and give the design engineer or reaction engineer or reactor engineer a choice of rate equation that fits the data very well. So, these are all called as semi empirical rate equations where I have taken help of theory to some extent, but I made some assumptions and I am going to do data fitting later. So, these are all semi empirical rate equations and most of the times they fit the data very well that is why they are quite popular as far as the kinetics of solid catalyzed reaction is concerned. Thank you. We will discuss how to use the experimental data to validate these rate equations in the next lecture by taking some examples. Thank you.