 We get started today. So, we are looking at temperature effects on rate in equilibria in this lecture. Now, temperature effects we tried to put together by looking at the van toff's equation a celebrated equation d l n k by d t is delta h by r t square. Now, k is the equilibrium constant, t is temperature in Kelvin delta h kilo reaction and r is the gas constant. Now, if delta h is not a strong function of temperature in which case we can integrate equation 1 and then get this relationship l n k by k 298 is given by minus of delta h by r multiplied by this term. Now, how good is the assumption that delta h is not a strong function of temperature? In many cases this is not such a bad assumption. The changes in delta h due to temperatures are not very serious. So, it is not such a bad assumption. So, we can use this relationship to find out if you know k at a given temperature you can find out k at any other temperature using this relationship. Delta h is heat of reaction. Now, we also know from our basic thermodynamics that r t l n k is minus of delta g which is the standard heat of free energy of formation. So, this is also another relationship we might use if you have known the standard heat of free energy of formation you can find k at a given temperature. So, that is how k 298 comes k 298 and you can get the value of k at any other temperature using this relationship. Now, let us see what it means from the point of view of our application to various situations. If delta h is positive which means what it is an endothermic reaction. Therefore, what do we expect as temperature rises? Suppose, our temperature is more than 298 what happens to this term? This term is negative and term outside is also negative. So, what it means the whole term is positive implying that l n k by l n 298 is positive. So, k at any other temperature greater than 298 is much greater than k 298 showing that equilibrium constant for an endothermic reaction keeps on increasing with temperature a relationship that we all know from our basic understanding of thermodynamics. Now, if delta h is negative which is an exothermic reaction the opposite is the kind of relationship you will get l n k by l n 298. Now, this becomes negative correct. So, for the exothermic reaction you will find that the right hand side term is negative and therefore, l n k value of k at temperature is greater than 298 is less than value of k at 298 showing that equilibrium constant for an exothermic reaction keeps decreasing with temperature as you increase temperature. Once again something that we all know from our basic thermodynamics with this let us just look at one example to understand what we are trying to say. Let us take an example simple case of a going to b where we say that it is a first order representation for R 1 and R 2 the two forward and backward reactions. Therefore, the rate of formation of b we write it as k 1 c a minus k 2 c b and in terms of conversion is written as k 1 times c a 0 1 minus of x and minus of k 2 c a 0 x. And at equilibrium of course, the R b is 0 therefore, the equilibrium conversion is simply k by k plus 1 something that we also know from our basics. Now, if you make a plot of the equilibrium conversion versus temperature it looks something like this why it looks like for the case this is exothermic is endothermic. So, we can easily understand here as k increases if it is an endothermic reaction k keeps on increasing therefore, the denominator what happens is denominator x c what happens denominator x c keeps on decreasing therefore, x keeps on increasing. So, you will expect that x keeps on increasing with temperature for an endothermic reaction x keeps on decreasing with temperature for an exothermic reaction. So, the equilibrium conversion keeps decreasing for exothermic keeps increasing let us go forward let us go forward. So, let us look at how see for an exothermic reaction as you can see here that the reactants and products products are at a lower energy than reactants. Therefore, e 1 minus of e 2 which is delta h since e 2 is greater than e 1 delta h is negative which is what we know exothermic reactions delta h is negative. For endothermic reaction the products are at a higher energy than reactants therefore, e 1 minus of e 2 is positive because delta h is positive. So, we can use all these relationships to understand our rate function r b what happens in the case of an exothermic reaction and an endothermic reaction let us look at that now. So, if our rate function r b is k 1 c a minus of k 2 c b and then we represent them in terms of the aniline's dependence k 1 0 e to the power of minus e 1 by r t and so on. So, this function which can be put in this form I have just taken k 2 0 outside it can be put in this form where the left hand side the outside the bracket is essentially rate constant k 2 and inside is the effects of conversion and so on. So, if you look at this term for the case of an endothermic reaction what happens to an endothermic reaction in this case for an endothermic reaction this is r b r b equal to for an endothermic reaction delta h is positive. Therefore, this term inside as temperature rises what do we see the term inside for an endothermic reaction delta h is positive as temperature increases what happens to the term inside it increases the temperature. This is the rate constant as temperature rises rate constant increases the temperature. Therefore, you have for the case of an endothermic reaction the term outside the bracket increases the temperature term inside the bracket also increases the temperature showing that the rate of chemical reaction keeps on increasing as you increase temperature this is for an endothermic reaction also stands to reason we should expect that it is endothermic the higher the temperature better it is. So, that is what we see for the case of an endothermic reaction. Suppose, we look at the case of an exothermic reaction the term outside bracket as you increase temperature you find that the rate constant keeps on increasing. But, the term inside the bracket what we find is that as delta h as temperature increases delta h is negative. Therefore, this whole thing is positive therefore, as temperature rises this whole term decreases. So, for exothermic reaction we find the term inside bracket decreases term outside bracket increases showing that there is a maxima or the other words what we are trying to say is that exothermic reversible reactions you see a maxima in reaction rate. You do not see any maxima in reaction rate if it is an endothermic reaction. So, this is what we are trying to say by looking at this rate function. So, suppose you make a plot of suppose we make a plot of rate of chemical reaction this is a rate function R b is this suppose we make a plot of rate of chemical reaction versus temperature. So, what we would expect to see for an exothermic reaction is that it keeps on going up and then comes down going up and coming. So, you will see there is a point at which the reaction rate becomes maximum when you plot this you will see functions like this. On other words this dotted line is the locus of maxima in reaction rate. So, our interest is to find out what is the equation to this locus if we can find out if the rate expressions are very nice we can find the equation to the locus. In some cases you may not be able to find an expression for the equation to the locus in which case we will do it graphically. On other words this curve can be determined either experiment graphically or mathematically in both case. So, our interest in understanding this function is that we would like to have our designs running around this locus of maximum reaction rates because that is when the reaction rates are the highest. So, we would like to see there are systems operate as close to the maxima in reaction rates as possible. Also you notice here this I have shown here graphs of R b versus T and as you can see here as x increases if you look at this function the rate function this is the rate function here. So, what happens as x increases this term decreases correct. So, what do you expect to see why I have shown here I have shown here as x increases we are moving in this direction. What it means as x increases the magnitude of the reaction rate keeps on decreasing which is also understandable because we are approaching equilibrium. So, as x increases these lines go inwards you understand what I am saying. As x increases this term what I am saying is that as x increases these two terms have the effect of these terms that you move inwards because you are moving towards the equilibrium. Therefore, the net reaction rates are smaller and smaller. So, therefore rate is decreasing in this direction as x that is what we are saying. So, our interest is to find out what is the equation to this locus. How do you determine the equation to the locus as suppose something that we have done before is that we simply look at that function and see where the maxima lies. So, we take the first derivative of this del R b with L t at constant x that means we are looking at what happens to R b at constant x. So, we can differentiate this and then set it equal to 0 I term this x as x m that means the equation to the locus of x versus t is given by this equation 1 minus x m by x m is k by k e 1 by e 2. This is for the case of a first order reaction which is reversible. So, if these rate functions are more complicated it will not be such nice functions, but whatever that may be you will be able to determine that function and plot that function. So, we can put it in this form we can put it in this form that the x m is given by k delta I have got delta is given by e 2 notice here e 1 by e 2 is less than 1. Therefore, x m equal to k delta by 1 plus k delta or x m equal to 1 by 1 plus 1 by k delta this whole term is what x m is all about. So, when we make a plot of x e versus t and then x m versus t you notice here that x m always lies below the equilibrium curve. Why is it you can look at this function here you can say that since delta is less than 1 x m will have to lie below the x c curve. So, you have an equilibrium curve you have the locus of maximum reaction rate curve and therefore, you have all the information that is required for you to look at various design alternatives that might be available to you. So, what is it that we have done if you have an exothermic reversible reaction there is something called a maximum reaction rate curve there is something called the equilibrium curve and you have to operate around these regions. And therefore, we look for strategies that will give us an appropriately good design. So, let us say let us say we are looking at a CSTR a CSTR as a possible strategy for our operation. Let us look at one example operating on the locus of maximum reaction rates CSTR is in our mind what is the CSTR? CSTR operates at a given temperature or in other words the advantage is that you can operate at whatever temperature you want it can be here it can be here it can be here whichever temperature at which you want you can operate. Let us say you want to operate at the highest I mean highest conversion let us say you want this like all this is 0.8. Suppose, you want to get 0.8 conversion in one go on other words you want to get 0.8 in one go. Therefore, you will choose this temperature of operation suppose you want a conversion of 0.2 at the maximum reaction rate you will use this temperature of operation. On other words CSTR is a device which permits you to operate at the maximum reaction rate curve. You can choose that point and run your process at that point. Then question is how do you make it happen? You make it happen by recognizing that you have a material balance equation you have an energy balance equation you have a constraint which is defined by the maximum reaction rate curve. You have three equations and depending upon the number of variables which is specified you have to choose the variable which is not specified which is consistent with these three equations. Then you will be able to operate on the maximum reaction rate curve. So, most important point we must recognize is that CSTR is the device which permits you to operate on the maximum reaction rate curve. The second point which is also important is that suppose you want a very very high conversion 0.8 0.9 whatever 0.95 of the equilibria. So, you will be operating at somewhere here which means you will choose this temperature in this temperature can be very low. The reason is exothermic reversible reactions if you want to get higher and higher conversions your equilibrium constant will tell you that you should go to very low temperatures. Therefore, the reactor operates at a very low temperature and therefore, your reaction rates are very low. So, you are looking at a huge device which is very large in size. So, even though you can do it in one stage, but you might find as an equipment which is so large it is not you know satisfactory from an economic point of view. Therefore, you will often have to go through what is called as an optimization to get a proper size for which you will choose stage 1 here, stage 2 here, stage 3 here, stage 4 here or in other words you will operate 1, 2, 3, 4, 5, 6 number of stages. At each stage the temperature of operation is as per the maximum reaction rate curve and for each of these stages the heat load that you must add or remove will be specified by these 3 1, 2, 3. So, having chosen the temperature from here this defines x m this defines conversion sorry residence time this will define the heat load. You can see here if I specify temperature it will tell you what is the conversion at which you should operate. The moment I know conversion I can determine the residence time from here, moment I know residence time I can determine from here the heat load to be added or removed. On other words for every choice of temperature the equations 1, 2 and 3 will specify the conditions under which you should operate your process. So, 1, 2 and 3 tells you how to make sure that you get what you want. We also recognize that an optimal choice is a choice of decreasing temperature exothermic reversible reaction the optimal choice is a choice of decreasing temperatures. And therefore, a choice of decreasing reaction rates and therefore, larger and larger equipment this is what this curve tells you. So, yeah number of CSTR stage 1, stage 2, stage 3, stage 4, stage 5 instead of you can do this go up to 0.8 or 0.9 of the equilibria in 1 stage. But the reaction rates are very low because of the fact that you are operating at such a low temperature. Yes size if you are going to operate with conversion at 0.8 what it means is that the actual concentrations are very low. And to get along this maximum reaction rate curve you will choose a temperature as per this curve. And that will turn out to be a very low temperature compared to higher conversions. So, this temperature being very low reaction rates are very low low temperature low reaction rates. And therefore, the equipment size is a very large instead if you operate a multi stage stage 1 at this temperature stage 2 at this temperature stage T safe. You will find that the net size of the equipment will be much in superior compared to a single stage operating at 0.8. So, what we are trying to say is that a multi stage CSTR design the optimal design will be a design of decreasing temperatures chosen along the maximum reaction rate curve. And the fact that the temperature are decreasing the reaction rates are decreasing therefore, the sizes are necessarily large. Suppose instead we have a multi stage P F R the question is often asked why multi stage P F R when you can as well operate a multi stage CSTR along the lochs of maximum reaction rates. We will try to address this question as we go along. So, once again we have our reaction A goes to B. So, you have our design equation this is for the design equation of a P F R which we have P C P D T D V R 1 minus of R 2 time delta H 1 star Q. Let us for the moment think that it is an adiabatic. So, that we remove this term in these two terms. So, that we have only an adiabatic process. So, this is the material balance F A 0 D X D V is R 1 minus of R 2. There are two equations dividing one by the other. You get a relationship which tells us that the temperature variation this is the temperature variation in our process V C P F A 0 divided by D T D X is minus of delta H. Now, if in this particular case for example, A goes to B the volume change due to reaction is not there, but there could be change in temperature as a result of which volumes may change. Suppose for example, we assume that volume does not change as an example then V becomes V naught therefore, this term becomes V naught by F A 0 is C A 0. Therefore, you will find C A 0 multiplied by delta H this is how I have written it in the next page. See therefore, D T D X becomes this becomes C A 0 times J which is a constant. Therefore, D T D X becomes a constant times C A 0 ok. What we are saying is that if you have a reaction A goes to B you conduct this reaction adiabatically in a P F R. If V equal to V naught is a satisfactory assumption then D T D X becomes a constant which is C A 0 times J what is J? J is heat of reaction divided by volumetric specific heat it is a known quantity. A constant D T D X means what if you make a plot of T versus X it will be a straight line. So, when you make a plot let us just do that when you make a plot of X versus T if you starting a T 0 then this is a straight line it can go up to the equilibrium curve we go it can go up to equilibrium curve. As you are moving along this curve what do you expect the temperature is increasing conversion is increasing at this point the reaction rate has reached a maximum. And as you move from this point towards the equilibrium curve the reaction rates are decreasing. So, on other words from point T 0 as you move towards reaction rates are increasing up to this point and it is decreasing. Now, if you are doing a design of a single stage. So, of course you will find because of the fact that it is a reversible reaction you are able to reach only up to this point. Say sulphur dioxide for example, ammonia for example, instances where the reaction is a reversible therefore, in one stage you are not able to push the reaction to its completion. So, the question is what is it that you do. So, what people do is that they cool this and run another stage they will cool this. So, if you are starting from here they will cool run another stage run another still you are able to get the extent of reaction that you expect. So, the why multi stage the answer to why multi stage is that in one stage we are not able to push the reaction to the end point that you desire. Why you are not able to push it to the end point you desire because the reaction is reversible. And therefore, as you move along the reactor you will at least you will end up in the end point at the equilibrium point therefore, you cannot go forward. And since the reaction rates become very small you will stop somewhere in little earlier than that otherwise it will become un-wheeled to operate. So, you will stop the process slightly before it reaches equilibrium and then you will run the next stage after cooling it. So, the idea of this cooling from one point to another is to give you the ability to go to higher level of conversion. So, this point all these points are points of design decisions decisions that will determine the economics of your process. So, how far you go in one stage how far you cool during the inter stage cooling and all these are decisions coming from economics only. So, you will have to do a number of iterations before you get the design that satisfies your economic criteria is that clear. And of course, criteria is not just economic there are safety criteria there are various other issues that goes into your objective function. But, once the objective function is defined you can definitely take care of all those calculations and come up with the best kind of design. Then another issue that frequently we have encountered is that if our initial temperature is T 0 therefore, let us say this is our T 0 I just draw here this is T 0. Now, you notice here that if your T 0 becomes much higher that means if you are able to preheat your feed to a higher temperature then you are able to I mean achieve a far better extent of reaction because you are able to go forward to the end point. On other words if you can heat the feed to as high a temperature as possible you are able to approach as close let us say you are able to heat it up to this point you are much closer to the maximum reaction rate curve. On other words preheating the feed facilitates you to reach as close to the maximum reaction curve as possible and to that extent you are able to save on the size of your equipment. So, the idea of preheating is this if you start here you will get a much bigger reaction equipment, but if you start here it is a much smaller reaction equipment because you are much closer to the maximum reaction rate curve. This is something you will see if you go to ammonia if you go to sulphur dioxide reactors these are things that is done routinely to provide the appropriate optimization. Now this interstage cooling as you can see these are all interstage cooling which actually permits you to go to higher and higher conversions for a gas phase reaction this interstage cooling is a huge investment. Why is it because gas phase heat transfer coefficients are small and therefore, and if the reaction is at high pressure for example, you are talking about high pressure equipment at the same time very small heat transfer coefficient ammonia is such an example. Yes number of stages might increase, but the overall size of the equipment will come down yes this is the optimization you will do. See you have to do this optimization to find out what is the best I mean too many stages there are problems all those issues are part of your objective function that you allowed to set for yourself. But even more important is that this interstage cooling whether if it is a gas phase reaction it is a it is a huge investment because it is a heat transfer coefficients are very low and therefore, the size of this size of the reaction equipment and size of the cooling equipment and you have to take both into account. So, that your cost optimization is taken care and ammonia and sulphur dioxide these are all instances was all gas phase coefficients corrosive and all those issues associated with problems that you will encounter. So, you have to do both the cooling equipment design as well as the reaction equipment design so that you are able to get the overall objective as per your objective set out. Now, if you if you go to sulphur dioxide they say what is called as DCDA the process that operates around the world to is called DC double contact double absorption. So, what is meant by double contact see you have you have this contact this is contact that means, you have push the reaction to a certain extent. Then you take this gas for absorption that means, you absorb the SO 3 and then you bring the thing back into the process absorb means automatically it cools. So, this cooling and absorption go together and then you do another stage. So, this is contact then this absorption contact absorption. So, they have double contact double absorption this is what is the sulphur dioxide process it is only two stage process where it take it to the end point this is something that you will see with ammonia I mean ammonia nowadays people do what they call as a radial flow you may have come across this radial flow. The advantage is the radial flow if you just let me just to a small calculation for you. So, these are consulate cylinders your catalyst is here. So, you have feed the gases are moving in like this is perforated this is perforated and then it goes out through this also perforated. So, gases are moving inwards now if it is a reaction which in which there is a decrease in volume SO 2 plus half O 2 is SO 3 is a decrease in volume you can see the facilitation. So, the volume decreases the volume of flow volume also decreases. So, the radial flow has certain and by keeping this thickness very small. You can keep this thickness very small by choosing the size of the cylinders and therefore, the amount of reaction the extent of reaction is controlled number one the extent of heat I mean heat release or what is called as temperature rise is also within the limits that you would have specified. So, the important thing about this radial flow is that you are able to regulate all that ensure that the catalyst call it I mean is does not deactivate because of very high temperatures. So, ammonia and sulphur dioxide are good examples in which this kind of issues become serious. Now, this another point we want to look at is we looked at del R B by del T at constant x and we showed plots of R B versus x and the R B versus T and then. Now, what we want to see is that what happens to this function del x del T at constant R B del x del T at constant R B or in other words what we are saying is that we will take this this function R B as some constant value some gamma and see how this function del x del T at constant R B looks like why we want to do this we will see shortly. So, we want just like we have done del R B del T at constant x now we are going to look at del x del T at constant R B. So, to do that what I have done is set up this function once again that means I have taken R B as constant and then it is k 1 c a 0 1 minus x k 2 c a 0 x just set up the rate function in the form that we all know and then I divided throughout by k 1 c a 0. So, that the right hand side looks like this 1 minus of x x divided by k y is it divided by k is k 2 divided by k 1 is k is it all right I have divided throughout by k n c a 0 this is all right what I have done. So, when you do that first term 1 minus of x the second term c a 0 cancels of k 2 by k 1 is written as capital K. Now, I have combined x together it looks like this therefore, the relationship between x and temperature looks like this where the temperature dependence is involved in capital K and small k. So, x for a given reaction rate x is related to that by this equation 3 is it ok. Now, if you differentiate x with respect to T del x del T at constant R B constant R B means constant gamma gamma does not change. So, we want to do del x del T at this one why do you want to do this what is the rational for doing this the rational for doing this is the following what is this term k by k plus 1 I put it in this form I have this term k by k plus 1 is 1 by 1 plus k as temperature increases if it is an exothermic reaction what happens to k it decreases. So, what happens is whole term. So, k by k plus 1 what happens decreases. So, k by k plus 1 decreases as x increases what happens to the first term 1 minus gamma by k 1 c a 0 as you increase temperature k increases and therefore, this whole thing decreases therefore, this whole thing increases. So, what we are saying is that for an exothermic reversible reaction x equal to this term the first term increases its temperature the second term decreases its temperature is this point clear to all of us this function what we have done we have done we just run through this once again we just run through this once again. We want to look at this function R B for the case when R is held constant and we change temperature. So, this is what we are trying to do. So, we want to see what happens to that function that is for which we have said we have divided throughout by k 1 c a 0 and got this function and so on. So, this increases this decreases and therefore, there is a maxima for the value of x. So, x versus t x versus t should show a maxima to find out what that maxima is what we have said is that we want to see how that function looks like let us do this once again. What we have done previously what we did was del R B del t at constant x. So, that means rate of change of rate with respect to temperature keeping x constant what we are now doing is that rate of change of x with respect to t keeping r constant. So, we are looking at the same function in two different ways these contours need not be the same it depends on the function and so on. So, we are looking at the same thing in a different way to see whether there is something more to be learnt at least from the point of view of a design. So, we are down doing not del R B by del t this we have done already we want to see del R we want to keep R B constant and see what happens to this function as temperature changes. So, now we are doing del x del t at constant R B to do that what we have done is the following. So, we have said this is I must have misplaced it why did I put it somewhere else anyway does not matter anyway this is the our function this is the function which is R B equal to gamma this is the function we want to understand and to understand how x versus t changes when gamma is constant. So, I divided throughout by k and c a 0 then our x looks like this where gamma is a constant we want to now do del x del t at constant R B del x del t gamma is constant. So, when you do that this function so this is the function we want to maximize. So, this is the function x equal to 1 minus the gamma k by k plus 1 del x del t at constant R B. So, this is what we want to do now. So, here the rate function k 1 is the function of temperature sorry k 1 is the function of temperature this k is also function of temperature correct. So, we want to differentiate this by parts this is what has done. So, I am differentiating this by parts let me write this here. So, that we remember x equal to 1 minus of gamma k 1 c a 0 into k by k plus 1 this is what we are differentiating. So, that so we are differentiating this. So, you can see here del x del t 0. So, I have got minus k 1. So, it is minus minus cancels of k 1 c a square d k 1 by d t is k 1 d 1 by r t square it is there k by k plus is the first term is first term differentiate the first term. See this minus term has disappeared because k 1 square with the minus sign comes. So, that is ok and then d k 1 by d t is k 1 d 1 by r t square that is fine. So, this term is all right second term first this remains and then multiplied by you have to differentiate this when I differentiate this. So, it become 1 by k plus 1 d k d t is fine that this differentiate the denominator k plus 1 whole square k plus 1 whole square correct with the minus sign d k fine looks all right looks ok. So, equal to 0. So, this how it looks now what I have to do some manipulation what manipulations have I done d k d t what is d k d t I have got here d k d t is d l n k d t is delta h by r t square. So, 1 by k d k d t is. So, I put d k d t is k delta h by r t square this is ok. So, I have replace this d k d t by k by r t square delta h all right. So, this only we have to simplify now let us see how it simplify see this is what we had correct this is what we had. So, this is gamma k 1 u 1 delta h now I am writing it like this 0 gamma k 1 what happened to k 1 k 1 got cancel k 1 c is 0 square k by k plus 1 1 minus of k by k plus 1 whole square k plus 1 whole square is coming from where which one this term what should it be this one this should be c a naught ok. Thank you thank you now it can be rearrange like this I think I have missed it there, but I have got it right here I think. So, what we are saying is that del x del t at constant r b if you set it equal to 0 this relationship is what we get. That means, this relationship is what comes out of that the differentiation and I have done some more manipulations some very elementary algebra is involved. So, what I have done is that I have replace this delta h as e 1 minus. So, this delta h I am putting it as e 1 minus of e 2 this delta h this delta h where are we this delta h I put a e 1 minus of e 2 it simplifies like this anyway. So, the basically what we are saying is that gamma k 1 c a 0 actually turns out something like this. This is the final form in which we get 1 minus of gamma k 1 c a 0 comes out to be e 1 k 1 plus. So, a relationship which is involves some amount of algebra gamma is rate the reaction rate which is held constant. We started with this equation correct x given by 1 minus of gamma k n c a 0 this is basically or this is the basic balance. So, what we are saying is that equation 1 holds as well as if you want maximum del x del t at constant r b this also should hold equation 6 and this is equation 1. So, if you manipulate further. So, you get essentially you end up with this relationship that for this case of del x del t at constant r b equal to 0 del x del t at constant r b equal to 0 you get x equal to k delta by 1 plus k delta. So, after going through this algebra what we have done we started with this rate function r b equal to this and then we said that this can be simplified to x equal to this. Then we did del x del t at constant r b and found that the relationship that satisfies del x del t at constant r b is given by x equal to k delta by 1 plus k delta and this is exactly the same relationship we got for del r del t at constant x. So, what we are saying is del r del t at constant x and then del x del t at constant r b the locus is the same what we are saying the following. Let us just run through the whole thing we started and said this is see del r del t at constant x it is this this we have done earlier and what is the equation del r del t at constant x or x m is given by k delta by this is x m given k delta by 1 plus k delta. That is for the case of del r del t at constant x now what we are saying is that for the case of del x del t at constant r b also give you k delta by 1 plus k delta or in other words we are saying that this locus this locus also applies for del x del t at constant r b. So, this also applies for del x del t at constant r b the locus is the same. So, what we have tried to do is that to show that locus of del r del t at constant x and del x del t at constant r b it is the same some of this is not clearly brought out in many texts what I find and there is a book written in 1965 it is he says in passing that you see it is the same, but some of it is not proved anywhere. So, I thought it is worth sort of going through this effort and proving this it comes nicely actually. So, let us look at the where is our function our function r this is our function r correct r equal to k 1 c a 0 1 minus k 2 c. So, we can plot this function correct this function r can be plotted for different values of r. So, when you plot this see I have just plotted for different values of r. So, what are we getting here? So, for different values of r you are getting your maxima the maxima is coming at that point only the reaction rates for different values of I mean for the assume values of r where the maxima appears and the other one both are coming out to the same locus. So, the point that we are trying to get across here is that if you look at a point somewhere here somewhere here which is what these are the points what is called as d x d t this is our equation here t t minus t 0 we did this a little earlier I would not be able to find it now, but let me write here t minus of t 0 this is for adiabatic p of r is given by c a 0 delta h x divided by c p we have done this a little earlier. So, what is d t d x here d t d x r d x d t is 1 by j c a 0 d t d x is j c a 0 therefore, d x d t is. So, what is that suppose I want to locate that point I can locate that point here you can see this point these are the points at which d x d t is 1 by j c a 0 or d t d x is j c a 0 d t d x is j c a 0 is that clear. So, d t d x actually the highest value that d t d x will takes is j c a 0 on other words plots of x words in the plots of x x t curve the point where d t d x is the highest value have shown by dotted line here you see this is the dotted line showing locus of points on this x t curve x t curve where the rate of change of t versus x is the highest value which is j c a 0 this is shown as a dotted line. So, we have 3 one is the equilibrium curve one is the maximum reaction rate curve you call it del r b del t at constant x or del x del t at constant r b both are the same. And there is a third line which is shown as a dotted line which is corresponding to d t d x equal to j c a 0 now the interesting point about d t d x at j c a 0 is that d t d x means the rate at which temperature changes as conversion changes and this is the point where the process is most sensitive to control. Because here the change is the highest whatever the process may be you are seeing your change as the highest. So, d t d x has the highest value as j c a 0. So, a process control around this point would be most effective at any other point it may not be as effective. So, the region of design is often people sort of draw a point point here which is something like some shall we say 98 percent of equilibrium conversion. So, and say this is the region in which your design must lie your region of design is between this dotted line and a line which is not at the equilibrium curve, but somewhere close to the equilibrium curve. So, some people take it 98 percent of equilibrium some people do less depends upon the objective function that you set for yourself. So, what we have tried to put across here is that looking at the temperature effects on chemical reaction we have defined a region over which you should look at your design and with respect to the objective that you would set for yourself this objective may involve cost may involve safety may involve consumption of water consumption of utilities. So, many see some people look only at cost, but in many cases cost alone may not be criteria in many designs today the most important criteria is water there is not enough water. I mean water may not cost you so much it may cost you some 50 rupees or 40 rupees the fact that it is not there. So, consumption of water is something that has to enter your the objective function the various issues like this which enters as a result of which the design that you will finally, end up with will have all the compromises that is required make sure that the plant actually runs satisfactorily I will stop there.