 Good morning. So, in the last lecture, we looked at different steps involved in catalysis and based on these steps, we wrote down the equations and synthesized the rate equation, the overall rate equation depending on which step is rate controlling. So, it can be a chemical reaction which controls the rate or it can be the adsorption or it can be the desorption. So, there are three distinct steps which are nothing but adsorption, chemical reaction, surface reaction rather and the third one is desorption. And we have written down an algorithm as to which one of these steps like controls overall rate and based on that how to derive the rate equation. Now, the adsorption step can be controlling, it can be a reaction that can control overall rate or it can be the desorption that can control overall rate. And the rate equation that we get is of the form say r is equal to some rate constants into concentrations, the bulk concentrations, it can be K dash C B for a reaction A in equilibrium with B. And in the denominator, we are going to get an expression something like this. This is one particular rate equation depending on what assumption I make probably the rate of reaction on the surface to be controlling. So, this is one rate equation. I am just showing you the form the rate equation. It can be slightly different depending on which step controls the overall rate. So, if you just closely observe this particular rate equation what you see is in the numerator you have something similar to what you would have otherwise got when the reaction was not catalyzed by the solid catalyst. That means it was a normal reaction taking place in a homogeneous medium. So, this is an additional thing that comes in and what do you have here? These are the adsorption constants or probably desorption whatever something to do with the adsorption and desorption. So, the denominator that you are seeing here is because of the adsorption and desorption of the species or the components involved in the reaction because there is a competition for the empty or active sides for the reaction to take place. So, in a way these adsorbed sides are going to or the already occupied sides are going to be unavailable for the reaction to take place. So, this is how we interpret the overall rate equation that we get. But then of course, this may change depending on the assumption that we make which step is the rate controlling step. So, we looked at different steps to arrive at this particular rate equation how to synthesize this rate equation. But that is not enough because once we get this rate equation what is the guarantee that the rate equation is valid. So, in order to know whether the assumption that we have made are correct or not or the equation that we have got is right or not as I told in the last lecture we need to look at the experimental data. We need to generate experimental data in laboratory and see whether the data fits very well in this rate equation. If it does not then we have to go back and make another assumption and come up with different rate equation and see whether the data fits or data falls in line with the rate equation that we have got. So, in this today's lecture what we are going to do is like we are going to look at a simple example and we are going to look at how experimental data can be used to see the authenticity or validity rather of the rate equation that we are talking about. So, for that we are going to consider one example. So, let us go for a real example where like for example let us say for instance we have a dehydrogenation reaction say cyclohexane hydrogenates to give benzene and hydrogen. Now, this is cyclohexane let me call this as C benzene B and of course hydrogen. So, I am going to denote my concentrations with C B and H 2 as I mean I go ahead. So, let us first synthesize the rate law with certain assumption and then we look at a possible experimental data and see how we select finally the rate equation. So, now I do not want to really go through each and every step quite systematically like what we have done before, but quickly come up with the rate equations for three different cases when first the adsorption controls, secondly the reaction controls and the third is the desorption control. So, let me first start with adsorption controlling before that let me write down the mechanism. So, I have C plus S giving C S adsorption then C S gives B S plus hydrogen. Now, it is very important step look at a way I have made an assumption here chemical reaction. What is happening here is the hydrogen is generated, but it is coming out immediately it is not occupying the sides. One can make some different assumption here saying that C S plus S gives B S plus H 2 S that means H is H 2 or hydrogen is on the in the adsorbed condition. Right now I am making an assumption which is quite similar to a layer ideal assumption where one of the components is not in adsorbed state it does not have affinity for the catalyst as far as adsorption is concerned and as and when it is formed it goes to the bulk. So, that is the meaning of this particular step chemical reaction. So, this particular component is always present in bulk there is no adsorbed hydrogen. Now, you may ask me why I have made this assumption right now there is no basis for this I have just made this assumption I am going to go ahead and see whether this is valid or not based on the experiments. If it is not valid then I will come back and make some other assumption. So, all possibilities are there, but I am just trying to tell you the meaning of or significance of this particular step that is chemical reaction and then the desorption that is B S gives B plus S right. Now, you know how to derive the rate equation if one of the steps is rate controlling. So, in the last lecture we looked at simple homogeneous sorry not homogeneous simple isomerization reaction A giving B reversible and we said that chemical reaction was controlling and we looked at the rate equation. Today let us derive the rate equation for adsorption controlling and then quickly write equations for the chemical reaction and desorption and then look at the validity of these equations based on the experimental data. So, first we will spend some time deriving the equation for the case when adsorption controls. So, C plus S giving C S alright. So, rate of adsorption is equal to overall rate that is my assumption rate determining step rate controlling step slowest step is equal to k adsorption right into C C into C S minus k A dash C S C S right. So, this is the rate equation that I have got and depending on the stoichiometric coefficient I will have a sign here like what you do in writing any normal rate equation. So, I am writing a general rate equation here. So, let us go ahead now in this particular equation we have bulk concentration C C, but apart from bulk concentration we have empty sites concentration vacant site concentrations and a concentration of adsorption species and these are the concentration which are unknown and they will come from equilibrium steps. Now, what are the equilibrium steps here the first step is of course the reaction now reaction is not the slowest step like what we did before in this particular case adsorption is slowest step desorption and reaction they are they are equilibrium control or they are at equilibrium instantaneous. So, the reaction is C S giving B S plus hydrogen. So, adsorption controls the overall rate and these concentration which are C S and C C S which are some which are the concentrations we do not know they will be obtained from the equilibrium state and which are the equilibrium steps here the reaction chemical reaction which is now instantaneous earlier in our earlier example we consider reaction to be controlling here adsorption controls. So, reaction is instantaneous which is at equilibrium. So, the reaction and desorption these two steps are going to give me this bulk sorry the adsorption species concentrations and empty site concentrations. So, this is reaction equation C S in equilibrium with B S plus hydrogen. Now, if you write equilibrium equation for this it is K e q capital K e q is equal to C B S C H 2 divided by C C S. Now, we can express C B S in terms of C C S here. So, let me just write equation 1 here I am going to use equation 1 later to get value of C C S. Similarly, I write desorption which is again an instantaneous process where you have B S that is benzene is going to get desorbed. So, this is adsorbed benzene benzene in bulk and empty site. So, again same thing I can write K capital K D is equal to C B C S divided by C B S. So, C B S is equal to capital K D dash C B C S equation 2. Now, using equation 1 and equation 2 equation 1 and equation 2 I am going to substitute for C B S and C C S in the main equation that is obtained by adsorption to be the rate controlling step right. So, what I need is C C S. Now, how do I get C C S from this equation C C S is equal to some constant into C B S C H 2 right and from equation 2 you have C B S is equal to this. So, what it means is C C S is equal to again K E Q dash K D dash C B C H 2 C S. So, I have got the expression for C C S which is to be substituted in the main rate equation the overall rate of the reaction this is the one C C C this is where I need to substitute for it. So, if I do that what I get is R 0 that is overall rate is equal to rate of adsorption is equal to K A C C C S minus K A dash K E Q dash K D dash C B C H 2 C S. Now, in this I do not know only C S rest all our bulk concentrations. By the way now it all depends on what kind of reaction that I am talking about whether it is a gas phase reaction or liquid phase reaction for the gas phase reaction normally the concentration is expressed in terms of partial pressures if for the liquid phase reaction it is the concentration. So, right now I am just expressing all these in terms of concentrations, but if you are talking about gas phase reaction you can convert it to partial pressure. So, P by R T is equal to concentration fine all right. Now, you have this equation further site balance C S plus C C S plus C B S cyclohexane adsorbed benzene adsorbed is equal to C T. Now, C S what is C C S? C C S is equal to and C B S is equal to which we have already derived is equal to C T right. So, C S is equal to C T divided by 1 plus let me write again some constant here C B C H 2 plus K B C B right. So, this is I have just clubbed all this in digital constants in K double dash right. So, I have got expression for C S look at expression slightly different look at this term which was not there before. Now, it is adsorption controlling that is why you have different terms appearing in the denominator try and observe the differences. Now, I have got the expression for C S I will substitute for C S in the main rate equation that I have got. So, R O is equal to K forward. Now, I am just giving you the final form you can because this is all of obtained by clubbing all those constants right this K F which is function of K A and C T and so on. C C that is concentration of cyclohexane minus K F dash or one can say K reverse K backward K B rather C B C H 2 divided by 1 plus right. So, this is the final expression when adsorption controls right adsorption controls adsorption is a rate determining step adsorption is a rate determining step. Now, this is one expression that I have got suppose I do the same exercise, but then now with reaction to be controlling reaction to be controlling then I will get a different rate equation then I will get a different rate equation and you can do this exercise on your own I am going to do it here I will just write the rate equation the form of the rate equation it is going to be R overall is equal to R reaction is equal to sum K into C C minus oh sorry C C minus K dash C B into C hydrogen divided by it is quite similar to what we have done before K C C C plus K B C B this is when the reaction controls. So, look at a difference this was when adsorption is controlling look at a difference the numerator is not changing much, but the denominator there is a difference there is a difference and we have to exploit this difference to know based on the experimental data which one is the right equation, but it is not yet over now I am going to look at the case when desorption controls case when desorption controls R o is equal to R d is equal to. So, desorption controlling R 0 is equal to R d is equal to sum K C C again the numerator look at this numerator is not changing C B into C hydrogen divided by I will have a very peculiar expression here you have C hydrogen plus K dash C C plus K double dash C C and C hydrogen. Now, you can derive this equation on your own I am just writing the final form the equation because I am going to make use of this equation later to see whether this is the right equation based on the experimental data or not. So, this is when desorption controls. So, I have three expressions when these three different steps are controlling steps the first one is when adsorption controls that is this equation then reaction controls this particular equation and when desorption controls this equation. These are three different expressions that I have got now I want to know whether which equation is valid in a real situation. So, how do I know this as I said before I need to do experiments in laboratory. Now, think of a situation you want to do experiments in laboratory for a solid catalyzed reaction you have cyclohexane getting converted to benzene and hydrogen. So, let us say it is a vapor phase reaction I am using any hydrogenation catalyst say nickel adsorbed on alumina whatever say palladium adsorbed on alumina as a support right the polar support and reaction takes place right. Now, how do I do the experiments in laboratory how do I get a rate value for the rate they are different types of reactors which are used in laboratory for solid catalyzed reactions. So, it is quite similar to what we have already learned for simple reactions normal reactions here we have a differential reactor. So, for example, you have a tube in which you have packed the catalyst and a flow is taking place. Now, this particular length is very small what does it mean that means there are no gradients along the length. So, it is like a differential that is why it is called a differential reactor though it is a tubular reactor in that the catalyst volume that I am I am using is very small. So, that the concentration does not change drastically. So, if that particular volume acts like a CSTR where the concentration is uniform the it has an advantage because it is going to get me the rate equation for the particular concentration the concentration change is not much. So, I can assume the concentration to be the feed concentration at which the reaction is taking place anyway that will be clearer later. So, this is a tubular reactor with difference or one can have a stored reactor a continuous reactor one can have a fluidized bed reactor. But of course, typically like such reactors are not so common in laboratory this is the most common reactor used in laboratory one can have. So, if so much you are dealing with liquid phase the reactants and products are in liquid phase where the catalyst is solid in that case again I will have a slurry reactor similar reactor. Now, these are continuous, but I I may have a batch reactor right and, but then I will have a concerned performance equation for the batch reactor and can evaluate the value of rate can evaluate the value of rate. So, this is how I can perform experiment in laboratory for solid catalyzed reaction. Now, in all these reactors I can use solid quite comfortably. In this case in the fluidized bed case we have to make sure that this particular flow rate should be good enough or high enough to fluidize the catalyst. So, all these reactors are going to give you some performance based on which you can calculate the rate the fluidized bed and the CSTR type of slurry reactors are integral reactors. And by differentiating the conversion versus catalyst loading plot we can obtain the instantaneous rates and can calculate the reaction rate constants. Also it should be noted that the differential reactor has a limitation as I said before that the conversion should be as small as possible. So, if the conversion goes beyond a particular limit then probably the differential reactor will not work like a differential reactor and the analysis will be incorrect. Now, let us get back to our example of cyclohexane dehydrogenation to give benzene and hydrogen. So, in this case suppose I am using a tubular reactor here cyclohexane will go inside and a mixture of cyclohexane plus benzene plus hydrogen will come out. You may use some inert also here of course, which inert is to be used and all will look at it because inert may have some effect inert may get adsorbed on the catalyst surface we will look at it separately. But then we can do this experiment here to determine the rate. So, let us say I determine the rate for this particular reactor by changing the concentration of cyclohexane in a field. Now, it is not pure cyclohexane it you have a inert and let us make an assumption that inert does not adsorbed on the surface it is like hydrogen. Hydrogen of course is not inert here it is a product, but like hydrogen there is another inert say argon or say helium or nitrogen for that matter which does not have any affinity for the catalyst. So, what I am doing is in this particular reactor I am passing cyclohexane along with that I have some inert also going inside. So, it gives me an opportunity to play with the concentration of cyclohexane. So, I can do or I can perform experiments at different concentrations of cyclohexane in the field and get a rate for every concentration. So, if I do this experiment if I do this experiment what I get is at different partial pressures or different concentrations of C C what is the rate? What is the rate of the reaction? This is the rate that I observe this is the rate that that is the overall rate that we have derived the rate equation for and this particular rate should fit very well in the rate equation that I am going to get. Now, C C verses are overall are observed let us let us say I am getting a train like this I am getting a train like this this is the experimental data. Now, I need to know whether any of those equations that I have derived falls in line with this experimental data or not. So, what is that nature here? Initially it is a bit linear and then the slope is decreasing and it is getting saturated after certain concentration of cyclohexane in the field which is sufficiently high there is no change in the rate it is almost becoming constant. So, that is the general observation that I have. So, this is the experimental data observed I am going to get back to this I am going to get back let us observe our earlier rate equations that I have derived. Now, I have the equation when adsorption controls what do I see here? Even if I change the concentration in the feed what is the rate? Now, there is a finite value for this whatever is the concentration that I am doing the experiment with. So, there is some finite value here what is the feed concentration of benzene? Zero reaction is taking place at the feed concentration remember what I said the concentration is not changing much here. So, whatever rate that I am calculating here is at the feed concentration. So, suppose the concentration of cyclohexane entering is 50 percent the extent of reaction is not much it is going to come out with say 48, 49 percent concentration of cyclohexane. Some benzene will form some hydrogen will form which will be very small depending on the extent of reaction, but this extent is small. So, that for all practical purposes I can assume that the reaction takes place at a concentration corresponding to the feed concentration. This is a very important assumption and it is valid provided this catalyst length that is length that I am using is relatively small. So, what happens here? C c is some finite value say 50 percent concentration units of course, C b is negligible 0 C h 2 0 C b 0 C h 2 0. What does it mean? R 0 is equal to k f C c and what does it say? C b is equal to it says that the rate is directly proportional to the concentration of cyclohexane. So, what train will I get? I will get a train which is linear with respect to cyclohexane concentration in the feed. So, it is going to be like this C c versus R overall when adsorption controls this is a train that I would get. Is that a train that I have got in the experiment? No, my train was like this. So, what does it mean? It means that the rate equation that I have got by assuming that adsorption to be controlling is not valid here is not valid here. So, let us look at desorption controlling now. Let us look at the rate equation when we get when we assume that the desorption controls the overall rate. So, what is the rate equation that I have got? This is the desorption rate equation. What do I see here? C c is finite C b is 0 I do not have benzene in the feed I do not have hydrogen in the feed this is also 0 this is 0 this is 0 this is finite. So, R 0 or R observed overall is equal to k divided by k dash because C by C c by C c that we get cancelled. So, rate is constant in the desorption controls rate is independent of C c is a feed is pure cyclo pure in the sense there are no products present it will be only cyclohexane in some inert medium. So, if I change C c and observe the rate and in desorption controls what is the train that I expect? I expect that. So, there is no change when desorption controls I hope this is clear as the concentration changes the rate is unaffected because of this because of this or rather this it is a desorption of benzene that is controlling. So, the rate does not depend on cyclohexane concentration at all. So, is that my observation my fixed bed reactor or other that tubular reactor differential reactor that I am using what is the observation there? I am seeing that the rate changes in this fashion and what I should get is this not true or does not fall in line. So, what it means is from the experimental data that I have got neither that sorption is controlling alone nor the desorption is controlling alone. So, let us look at the reaction now or let us look at the expression when the reaction controls the overall performance or reaction is the slowest step. So, what is the expression that we have got the expression that we have got is this when the reaction controls. In this particular case again for the feed when there is no benzene and hydrogen present this term becomes 0 this term becomes 0. This term becomes 0, but then I get rate equation r overall is equal to k c c divided by 1 plus k c c c. Now, this is a very typical rate equation for very small values of c c when c c is very small r 0 is this is very small r 0 is k into c c. For small values of c c r 0 is proportional to c c is proportional to c c and what happens if c c is very large if c c is very large then this becomes very large compared to 1. So, r 0 for c c to be very large r 0 is constant the very typical rate equation that we have got where in the nature would change depending on the concentration of cyclohexane in the field at very low concentrations the rate is proportional to c c and at large concentrations the rate remains constant or it attains saturation it attains saturation. So, how will it look like on the graph of c c versus rate it looks like this c c versus rate overall or observed it is almost constant here it is linear here and there will be some range where it is neither linear nor constant and this is when it is a reaction surface reaction controlling the overall rate and remember the data experimental data that I had shown you before falls in line with what one would get by assuming surface reaction to be controlling. So, it is like you have initially linear and then going to saturation. So, what it means is the rate equation in this particular case for this experimental data is the rate equation when chemical reaction the surface chemical reaction controls overall rate at least the nature is quite similar. Now what we do now later now nature is similar means I will try and fit this particular model or this particular rate equation in the data that I have got and get the values of parameters. Now which are these parameters the parameters are adsorption constants rate constants. So, I get a rate constants I get adsorption constants by fitting this data. So, if you have this particular equation. So, this is something that I obtain for certain values of the parameters. Now which are these parameters k the adsorption constants then k f k backward or k f dash whatever all these are rate constants these two and these are adsorption constants. If the surface reaction controls the overall rate then the k by k dash is related to the overall equilibrium constant of the reaction and that can be determined independently. So, the unknown parameters are only three those are k c k b and small k that is the forward reaction rate constant. So, these will be determined such a way that my data fits well in this particular equation. So, once you do that if the data fits well over a large range of c c then you have got a right rate equation. And this data fitting is normally done by the many methods of regression is something called as least square method where what we do is you have the experimental data you have the experimental data and you have the curve for certain values of parameters k k f k f dash right. And for every point there is some difference we try and minimize this difference. And there is a function called summation of the square of these errors why square because sometimes it is quite possible that this difference is positive at one point this difference is negative at certain point. So, if we just go on adding the differences they may compensate for each other and say that find the overall total error is 0 but that is not correct. So, we have to square it so that the magnitude is always positive. So, that I get a better best fit fine. So, that is nothing but parameter estimation or regression analysis it is called as. So, least square method is the best method to or that is a commonly used method to estimate the parameters. And there are mathematical techniques optimization techniques which are used to calculate these values of parameters. Now, after doing all this does it mean that you have got a right right equation which will work very well. Now, there is one thing that we are not really looked at what is that you have just looked at the effect of cyclohexane concentration in the feed. We are not looked at the effect of benzene concentration we are not looked at the effect of hydrogen concentration. There are some constant associated with benzene adsorption right in the denominator you would be able to determine them only if you have benzene present in the feed. So, after doing all this you have to generate more data by doing some experiments with either hydrogen in the feed benzene in the feed with certain proportion not negligible now or both of them in the feed. So, doing all this you generate as much data as possible and see whether your rate equation fits or rather it falls in line with the experimental data and that is the complete exercise. Now, I have just shown you three different rate equations here and said that fine possibly the reaction controlling mechanism may work, but in real situation things are likely to be more complicated where I need to make certain other assumptions to get different types of rate equations. As I said before now I am assuming a single site adsorption they can be dual site adsorption I am assuming hydrogen to be present in bulk only, but valadium may have affinity for hydrogen and it has. So, I cannot just make that assumption. So, I have to assume that there is another adsorption step that is hydrogen. So, hydrogen is first present in adsorption state and then there is a desorption from not just benzene, but hydrogen as well. So, it will become more and more complicated, but then if you want the right rate equation you have to bear with this complications and finally when you get a right equation then only you can use it for the reactor design. Because ultimate aim is to have a proper reactor design for the given reaction of interest and that will be valid over the range of concentration that I am studying. Before we go ahead to the reactor design let us look at the effect of temperature. Now, I am not talked about temperature at all. Now, let us look at the rate equation that we get when the reaction controls the overall rate. Now, this is the reaction rate. Now, in this rate expression I have several parameters. This is one parameter, this is another parameter and these are adsorption related parameters. Now, how will the temperature affect these parameters? Now, k is the rate constant, k dash is the rate constant for the reversible reaction. Now, these two are likely to be highly sensitive to temperature. Of course, it will have adsorption constant also, but it has the rate constant or these two will have the rate constants as well in them. See, I have got this by clubbing or other putting together many constants. There is the capacity C t also coming in picture, but in that if you remember you have forward rate constant and backward rate constant also in this and those are the constants which are highly sensitive to temperature. That is why these constants are highly sensitive to temperature. What about these constants? These are only based on adsorption. There is no rate constant here. So, adsorption constants relate you to or compare to rate constants. These two are less sensitive to temperature, less sensitive to temperature, but it is quite likely that because of temperature or high temperature, adsorption will get affected. Now, you all know that if I increase the temperature, adsorption is less favored and desorption is favored. So, the adsorption constants will go down and you may make an assumption that at very high temperature, the denominator which is related to adsorption, these two terms are negligible. What happens? If these two terms are negligible, then you have in a denominator just one. So, this boils down to a normal rate equation which is similar to homogeneous catalysis. So, at high temperature you may see such effects. You can even change the temperature and generate data at different temperatures to get the hidden parameters here, which are the hidden parameters activation energy, frequency factor and so on. Right now, the parameters that I was talking about was only like forward rate constant, backward rate constant and adsorption constant, but then if you generate a data at different temperatures, you can get the activation energies also. Something that we have already done at different temperatures, you have 1 by T versus l n k or l n r 0 that is initial rate. You have a plot like this. The slope gives you the activation energy and the intercept gives you frequency factor, where the k that is the rate constant is nothing but A into exponential minus E by or delta E by r T. So, I can get activation energy and frequency factor for both the rate constants forward and backwards. Now, both the examples that we have considered so far, first isomerization and now dehydrogenation, both the examples have considered the reactions to be reversible, but not necessary that the reaction has to be reversible. The reaction can be reversible as well. So, the same expression also the same methodology or procedure can be applied to irreversible reactions and one can obtain the rate equation. You can do that as a homework. Just take an example of A going to B, but now it is not reversible. It is just irreversible reaction. Fine. So, you have a rate equation. After we get a rate equation, what we do? We need to design the reactor. So, I have told you the procedure to get a rate equation. Synthesize the rate expression for certain assumptions and then validate it against the experimental data. This is the entire procedure that I need to perform in laboratory before I design the chemical reactor. And remember, we have not considered the external mass transfer effects. We have not considered the intraparticle diffusion that will come later anyway. But right now I am just assuming that these are negligible and I have the rate equation. Now, how do I design a reactor? Now, the procedure is not much different from what we have done before. What do I do? Earlier I have a CSTR. I have a CSTR. I write a rate equation F A 0 minus F A plus R A V is equal to 0. You have F A 0, F A here. And then in R A, I have minus K C A for irreversible first order reaction A going to B. And by knowing F A 0, for the required conversion, I get a volume. That is my reactor design. Or for the given volume, I calculate the outlet concentration that gives me the conversion. So, this is the exercise that I do. Now, what is going to change for heterogeneous reactions? You have a similar thing except now we will have F A 0 minus F A plus R A. And in most of the cases, instead of V, we have W equal to 0. Now, what is W? Here it is the volume. Here it is the weight of the catalyst. Why? Because the rate now is minus possibly K C A divided by 1 plus K A C A plus K B C B something like this. If the reaction is reversible, you will have one more term here. So, instead of this, I have this. This is one difference. Another difference is instead of volume, I have W here. Why? Because the rate constant here is per unit weight of the catalyst. The rate constant here is per unit volume of the reactor. Here the rate constant is per unit weight of the catalyst. So, this is the only difference. Otherwise, the entire exercise that I have done, what are problems I have solved for homogenous reactor? Similar procedure will be adopted for heterogeneous reaction. Remember, volume is replaced by the weight of the catalyst and the normal rate equation which is probably of the power lot type is replaced by either Langmuir-Wood or Euler area, whatever mechanism depending on whatever assumption I make. It is slightly complicated rate equation, but the methodology or procedure remains same. Thank you.