 Yeah, okay. So the next topic we are going to start is collision theory of reaction. See, according to this concept, for, you know, the completion of any reaction, or for any reaction to proceed. We basically have two, you know, hurdles we can say or two barriers we can say, which every reaction has to fulfill, otherwise the reaction is not possible. Okay. So we know when a and b are reacting together, right a and b are reacting together and it forms a product. So the hurdles or the barrier that we are talking about is you know, when a and b reacts, there are collision between a and b. Correct. So we have a molecules and bb molecules, both a and b molecules, they collide. Suppose this is a and this is b. Both collides with each other. And if this collision is effective, okay, this gives a collision. This collision gives you the product. And we have the sufficient energy. Okay. Now this collision, if you talk about here, collision, we have two types. Here also we have collision. One is effective collision. And another one is non, non effective collision, non effective collision. What is effective and non effective collision we'll discuss. When the two molecules collides, we know at the surface of the molecules we have certain active surface area, correct. So to provide the product, we must have active surface area into contact when the molecules collides. Okay. That is one of the criteria, not the only criteria. That is one of the criteria, but important criteria without this also the reaction is not possible. So we are going to discuss about the two barriers here. Both barriers are equally important in order to give the product, right. If one barrier is not there, one condition is not satisfied, product won't form. Okay. So basically all reaction we know reactant molecules collides and it give product. All those collisions, which gives product are called effective collisions. Okay. We can have n number of collision, but it is not necessary that all the collisions will give you the product. So, so the collisions this gives you the product here that collision we call it as effective collision. In effective collision, we get product. If the collision is non effective, then the product won't form. No product. Now for a collision to be effective, we have that two barriers that must be satisfied. Okay. The two barriers that we have we call it as the first one is orientation barrier, orientation barrier, energy barrier, orientation barrier and energy barrier. If both condition are satisfied, then only the collision is effective and product forms. Okay. So before going into this, just one few small term will see that we call it as collision frequency. So first you write down the heading here, collision frequency. It is the number of collision, it is the number of collision takes place per unit time per unit volume of the reaction mixture represented by Z and don't get confused with compressibility factor. Okay. Number of collision takes place per unit time per unit volume of the reaction mixture reaction mixture. Next slide down all collision does not bring all collision does not bring a chemical change chemical change means formation of product does not bring a chemical change. Next, the collision which produces which produces a chemical change a chemical change is called is called effective collision right and those who does not bring a chemical change non effective college. And there are two barriers, barriers which are need to be cleared, which are need to be cleared in order to, in order to provide effective collision. Okay. Now the first point to barriers we have both are important. Okay, it's like only one barrier is there only one barrier is satisfied other one is not satisfied then also the reaction the collision won't be effective. So both are equally important. You'll get theoretical questions on this with statement is right or wrong like that. Okay, the first barrier that we have right down. It is orientation barrier orientation barrier means what the molecules which collides the reacting molecules right down reacting molecules must have a proper orientation. I said the active surface area must involve in the collision right so reacting surface area must have proper sorry reacting molecules what I am writing it down. The molecules must have proper orientation orientation without proper orientation. The reaction is very difficult to proceed because the active surface area is not taking part in the in the collision right orientation is very important. See if two molecules are colliding if two molecules are colliding. So they will have an intermediate where a bond is about to break BB bond is about to break at AB bond is about to fall. And then eventually what happens a and BB bond breaks will get AB and AB. This is the product we get provided the other barrier other condition is satisfied. If the other condition is also satisfied here this intermediate forms and then it's convert back into the reactant. This cannot be the final product, never. This will form either it will convert into the product, final product, or it will convert back into the reactant. This won't be there because it is activated complex it has the maximum energy highly unstable. So we can say it is activated complexity, the peak of the graph. The next one we have here energy barrier right on to this. If you look at this graph, similar kind of graph we have discussed in Maxwell distribution curve if you remember. This side we have fraction of molecule, the y axis molecules, and this side we have energy axis fraction of molecule energy right. So the graph goes like this. So all these molecules we have. Okay, but it's not like all the molecules. Obviously it represents what the fraction of molecules. Just a second. One second guys. Yeah, so. So all these fraction of molecules this represents what this represents the energy of the various fractions like the fraction is this, you know, the energy is this and like this we have energy is this and so on. So all these energies, we should have a minimum amount of energy, like energy for a college to be effective. It's not like this fraction of molecule has certain energy this fraction of molecule a certain energy, and all these molecules will provide, you know, the effective the minimum amount of energy that is required for the college and to be effective that energy we call it as threshold energy. Okay, suppose I'm assuming that threshold energy is this. This is the ET we have ET stands for T stands for threshold energy. Right. So corresponding to this energy, the fraction that we have. You see this. This is a fraction of molecule, this energy. So, all those molecules which has energy below this threshold energy will not provide effective collagen collagen will be there. So those collisions are non effective collagen. Okay, so all molecules all fraction of molecules, which are towards the right of this right side of this threshold energy provides the effective colleges, these are the fraction of molecules, which provides effective collagen right the fraction of molecules, which provides effective collagen, provided the fact that orientation barrier is there right that barrier is already, you know, is satisfied effective collagen. Okay, this energy ET right on here it is a threshold energy threshold energy. It is a combination of threshold energy. First of all, the threshold energy ET right down it is the minimum amount of energy. It is the minimum amount of energy of energy, which colliding molecule, which colliding molecule must possess to make chemical reaction possible, make chemical reaction possible. Right, this is threshold energy. Next slide and write down all these, you know, things you will get theoretical questions. Okay, that's why I'm writing down each and every point here. So that you also write it down. Okay, a collision between high energy molecule, a collision between high energy molecule forms forms and unstable unstable molecular cluster called activated complex, pivoted complex, right this highly unstable it is written already, hence the life span of these complexes very small. And further these complex turn to this further this complex activated complex breaks into into either either the either the reactants or a new substance, a new substance called product. So either this activated complex will convert back into the reactant, or it converts into the product. Okay, the energy associated with activated complex is called activation energy right on the last point, the energy associated with activated complex is called activation energy. Copy this. The last thing is the graph we need to see for this to grasp we have first will discuss the first one. And similarly, you can understand the second one also. Okay, so this y axis is the energy x axis is the progress of reaction progress of reaction. So the graph would be like this. We have reactant and with reactant the molecule collides energy increases and then it comes down. Right. This you see this peak that we have here is the activated complex. No, we don't have any concluding facts for that triple. We don't have any energy that the purpose is that the energy provides the cluster the unstable cluster that you have because if these two molecules collides, whether they have the same energy or different energy less or more. But when collision takes place it should form this. If this does not form no matter what the energy we have no or high equal or not equal that does not make any sense. The reason is the collides and they should provide this minimum amount of energy, so that it converts into the product. Yes, we can have any combination in that. Okay. So this point the peak that you have is called activated complex. This is reactant. This is product. Okay. And this, this energy, the energy of activated complex. This energy gap. This energy gap from the product to activated complex. This energy gap from reactant to activated complex. Okay. This we call it as activation energy for forward reaction. Reactant to activated complex. This is activation energy for backward reaction. And this gap that you have this energy gap. Wait, from the reactant to product this energy gap that we have. This energy gap is called the enthalpy of the reaction delta H or delta E the definition of delta H or delta E the mathematical relation we have it is defined as the activation energy for forward reaction minus the activation energy for backward reaction. Obviously if you look at this difference here activation energy for backward reaction is more than this delta H or delta E is coming out to be negative and hence the reaction is exothermic in nature. This graph is for exothermic reaction. Okay, obviously the reactant energy is more than the product energy. So this amount of energy has been released. Generally we call it as one more small thing here you see this generally we call it as activation energy. Correct. Forward to backward activation energy is this plus the potential energy of the reactant molecule. Right, this is a plus the potential energy of the reactant molecule. This sum is the threshold energy it is coming out to be this one. So there's a relation and on this relation also they ask question in the exam. The relation is here I'll draw one more graph here. I'll write on this only way. The relation is threshold energy energy is equals to the potential energy of reactant plus the activation energy. If you solve question any book, you'll get one question based on this relation threshold energy is close to this and this will give you activation energy is close to this minus this this kind of question you will get. Okay, threshold energy one point you must remember it is independent of temperature independent of temperature and fixed for a reaction. It won't change fixed for a reaction. That's why if you recall this fact that we had discussed in grade 11 also. If you increase the temperature and this graph if you draw at higher temperature I'm just drawing here to make you understand the graph if you draw at higher temperature the curve becomes flat like this it goes and goes like this. Remember this or just a second goes like this and it comes. Okay, you see the threshold energy is same only it is not changing threshold energy is still this. But the graph shift towards the right so more number of molecules we have which can provide effective college and that's why at higher temperature, you know, the rate of reaction increases because the collision frequency has been increased. Right, so ET the threshold energy, it won't change with temperature whatever temperature you have it will be saying, and it is fixed for a reaction. Understood this point any doubt. Yes, sir. Now, this you can also drop for endothermic the graph will be exactly opposite right. If you look at this graph I'm just drawing it for endothermic the energy of product will be more than to that of reactant. Okay, so the graph of endothermic reaction, all the excess are same so it will go like this reactant. This is the activation energy backward. This difference is the delta H and delta H is positive reaction is it's clear. Done. One minute. Done. Yes. When got finished. Finished. I'm waiting for you only. Okay, so this is it we are done with this chapter almost. Okay, but like for exams like J or need, especially J. Okay, these are the two three types of reactions and some calculation we are going to do now. Okay, this most probably they won't ask you in need. Okay, but for Jay it is important. The derivation is quite big. Okay, you cannot do this derivation in the examination hall, because of the you know, because the space is not that much. Or nowadays they're providing sheet but obviously it will take time. If the integration is not tough, you can easily do this is just an application of, you know, integration and all can easily do this. But if you can memorize the final formula, then it will save a lot of time in the exam. Okay, so here we are going to derive each and everything we are going to derive, but I would suggest if you can memorize this, that will be, you know, better. So the first reaction we have calculations we are going to do on this sequential reaction panel reaction and reversible reaction. Okay, so first of all, we are going to discuss sequence sequence reaction sequence reaction means what one reactant is giving a product and that product is again forming one more product, like for example you see, we have a reaction gives B and B is again converting into C. This kind of reaction is there. Rate constant I'm assuming for the first step it is K1 and then it is K2. Okay. So what we are going to find out in this each and every step you look at properly otherwise you'll get confused. And T is equals to zero. The initial concentration is a not. This is zero, and this is zero. At time T is equals to T. We have a certain concentration of a at time T of B it is BT at time T and see it is CT. One more thing if I write simply like this a it is a concentration of a at any time T. If I write this or this both are same thing you take care of. Similarly for B and C also. So we need to find out in this, what is this concentration at any time T, what is the concentration of B at any time T, what is the concentration of C at any time T. This is the three things we need to find out and apart from this, we need to find out also what is the P max concentration what is the maximum concentration of B. At what time and at what time. Concentration of B becomes B max. Did you understand. Right, what are the things we need to find out at BT CT, plus at what time since a is converting into be correct. So B is forming and the moment B forms, it starts converting into C also. So we'll definitely have a point where the be concentration is maximum and beyond that the concentration of be cannot be and it will it converts into C. Correct. One by one we are going to discuss this. Did you copy this. One second. See first of all, and all these are first order reaction we are assuming correct first order reaction. So rate of disappearance of a rate of a what is that. It is minus of D concentration of a by dt minus of D concentration of a by dt is equals to the rate constant K one into concentration of a to the power one. Right. So from this we can find out the concentration of a at any time T that is a not e to the power. This is minus K one. Any doubt respond all of you please. What about this ATV got because a not will be given ATV. Now, for B you see what we need to do a rate of formation of be I'm writing it down. What I'm writing rate of formation of be of be that would be what the rate at which is disappearing. Rate of disappearance of a bracket close rate of disappearance of a is nothing but the rate of formation of be but at the same time be is also converting into a so rate of disappearance of a minus rate of disappearance of the is it fine. Yes, net rate of formation of be is this one, the rate at which be is forming minus the rate at which be is disappearing. Yes. Correct. So, the rate of formation of be is equals to what we can write minus D by dt of be is equals to. Plus of D by dt of be is equals to rate of disappearance of a is minus K1 into a minus minus K2 into be it should be plus no, because minus D by dt is K1 a correct minus D by dt is K1 a minus D by dt is K2 be any doubt. Tell me. So, we can write this as Dp by dt is equals to K1 into a minus K2 into this it's all about mathematics, there's nothing convinced into this which we can understand. Right, just need to solve this differential equation that is it. Okay, so what we can do here you see D. Sorry. We have D concentration of be by dt plus K2 concentration of be is equals to K1 and concentration of a we have a not e to the power minus K1 t correct. Yes. Now we need to solve this, what we'll do to solve this I'll just multiply here e to the power k2 t here and decide also e to the power k2 t. Now, could you tell me what is this entire term on the left hand side differential equation. Can I write this entire term as D by dt of concentration of be into e to the power. K2 t, can I write this be you into be by DX. Yes, yes. So we get this, and on the light and see, maybe you find this derivation is complex because there are so many terms, but the result that you get known it's very, you know, small and beautiful result, you can easily memorize that expression. Okay, so maybe the, you know, the derivation is there are so many terms and we are finding it complex now, but the result is very simple and clean. Okay, this side what we get k1 a not e to the power what we can write e to the power k2 minus k1 into t. Now you just did you take it that side and differentiate and integrate it. So we can write next D concentration of be e to the power k2 t is equals to k1 a not e to the power k2 minus k1 into t dt, and then we'll integrate it. Copy this down. None. Can I move on. Yes, sir. So once we finish this class. Okay, at least after that last you have your dinner. And after that, just sit for 1520 minutes, half an hour, and do this derivation on your own ones. Right. Okay, after this you see, when you integrate this, we get on the left hand side it is concentration of be e to the power. k2 t is equals to we get k2 minus k1 into dt we have no, so we'll multiply by this so we'll have a k1 a not we already have. Okay, and the integration of this would be e to the power k2 minus k1 into t divided by k2 minus k1 plus see, is it fine. Yes. Yes, sir. If you get the C we can substitute C and we can find out the concentration of B. Yes, no. So what is the condition we can apply to find out C condition is what at time t is equals to zero. What is the concentration of be. Yes, there's no be present initially. We'll substitute it here. So we'll get what C is equals to C is equals to k1 by. This side it becomes zero so k1 by k1 minus k2. Can I write down this directly because if it comes this side will have a negative sign I'll take it inside. Into a not is it fine. Now we'll substitute this C here and find out be so be into concentration of be into e to the power k2 t is equals to. k1 by k2 minus k1 into a not e to the power k2 minus k1 into t plus C is k1 by k1 minus k2 into a not. Okay, here we have k2 minus k1 k1 minus k2. So what I'll do, I'll take this common k1 by k2 minus k1 a not. Take this is common it becomes e to the power k2 minus k1 into t minus one fine are you getting it all of you. Yes. Okay, now what I'll do I'll take this e to the power k2 t because be we need to find out anyways we have to take this this site in a denominator. This e to the power k2 t I'll divide it here. This you keep it as it is this you divided this site. And then you open this bracket also e to the power k2 t into e to the power minus k1 t you will get an expression over there. That's I'm going to do now you see concentration of B is equals to k1 by k2 minus k1 into a not. And inside the bracket we have e to the power k2 t into e to the power minus k1 t minus one, sorry, minus one divided by e to the power k2 t bracket close. Now we'll divide this e to the power k2 t in both the terms. So what we get k1 by k2 minus k1 a not open bracket e to the power minus k1 t minus e to the power minus k2 t. So concentration of be here is equals to k1 by k2 minus k1 a not open bracket see k1 here in the numerator. So here you start with k1 and then you write on k2. The result will be symmetrical you can memorize this. This is the concentration of be at any time t everything will be given in the question you will you solve some advance level assignment you will get questions on this and you will realize once you know the final answer final formula, you can do these questions in seconds. Otherwise you have to derive all this and mind my words, this kind of derivation in the exam and the pressure you will definitely do wrong somewhere. You never know if like it's then the slavers they can ask you this question. Right. So this is beauty. Now, what is important what kind of question that they have asked I'll tell you in the last but let's finished it first right. So this is BT we have already calculated a see we need to find out so see easy can find out by mass balance okay by mass balance you see what we can do because is is the only one which converting into B and C. Any time we can balance the mass by mass balance what we can write that concentration of a initial that is a not is equals to the concentration of a B and see at any time because from a only it is B and C it is forming right this relation is true at all the time a concentration you know a not e to the power minus k1 t be just now you have calculated and a not is a not only. So if you subtract this a not minus a minus B you'll get C. So I'm just giving you this expression here final expression of C. And this they haven't asked. The concentration of C the point is a you don't have to memorize you have to memorize only be once you know be if they ask you see you can subtract you can find out see also the concentration of see if you solve this method I told you a not open bracket. One minus e to the power minus k1 t comes from this a and then be it would be k1 by k2 minus k1. And since I start with k1. So we'll write down first e to the power minus k1 t minus e to the power minus k2 t. So this is the concentration of C at any time. What next we need to find out be max max and time at what time the concentration of B becomes maximum okay be max you can simply apply the mathematical you know thing and you can find out the concentration of B max C the concentration of B is this if you differentiate this and equate to zero will find out the time at which B becomes maximum yes you know. And rate of formation of B from a equal to disappearance for C no sir. What. So will the ability max when rate of formation of B equal to disappearance of B. For that condition we can find out no that we cannot say depends. Okay. Because it depends upon ABC it's a general expression no. So obviously we can say the rate of B will be maximum, you know, when rate of a to B is more and B to C is less. That is the thing we can say but that you know won't lead you anywhere it's anywhere correct. We need to find out the time at which B becomes maximum so what we are going to do, we are going to differentiate this and equate it to zero. We'll see what we are getting here. So expression of B. We have expression of BC will start with K1 so K1 by K2 minus K1. A naught we have open bracket e to the power minus K1 t minus e to the power minus K2 t. Okay. Now when you differentiate this you'll get a very beautiful expression here. Okay. Don't worry that we'll get a complex expression very simple expression very simple expression you'll get. All these things are constant so I'll write on this as it is K1 by K2 minus K1 a naught. And this when you differentiate you will get minus K1 e to the power minus K1 into t plus K2 e to the power minus K2 into t. No doubt into this and when you equate this to zero what we can write K2 e to the power minus K2 t is equals to K1 e to the power minus K1 t. So we'll have K2 by K1 is equals to e to the power K2 minus K1 into t. Now look at this expression. If you take Allen both side, Allen of K2 by K1 is equals to K2 minus K1 into t. The time at which B becomes maximum is Allen K2 by K1 divided by K2 minus K1. This question they have asked many times in the exam. And out the time at which the concentration of B becomes maximum you see nothing is required K1 K2 will be given you just need to substitute and end out the answer in two seconds. Understood this. Yes, Allen K2 minus Allen K1 by K2 minus. This is the expression we have. Now again when you substitute this T there in the concentration of B, you'll get the maximum B. Yes. Yeah. Yeah, I'll take five more minutes will finish it off. You see calculation of B max. So B max equals to we have K1 by K2 minus K1 a not open bracket e to the power minus K1 t. So K1 t is Allen K2 by K1 divided by K2 minus K1 minus e to the power minus K2 Allen K2 by K1 divided by K2 minus K1. This is also looking like very complex but again, you'll get a very simple expression in the last. Okay. Now you see if I write down this expression as K1 by K2 minus without did you get did you get the expression. Can I write this as K2 by K1 to the power minus K1 by K2 minus K1. Is it fine. Yes, quickly. Or this would be K2 by K1 minus K2 by K2 minus K1. Can we write this. Yes, sir. Yes, this, this, this will go up and e to the power L and e something so that expression will have here. No doubt in this all of you. Correct. Now what you do is just take the LCM and this you solve this, this expression will get cancelled. K1 by K2 minus K1 a not when you take this in the denominator and you multiply this here. So can I write this as K2 by K1 to the power to the power K2 minus K1 by K2 minus K1 minus one. And the entire thing is divided by K2 by K1 to the power. K2 by K2 minus K1 bracket close. Is this fine. Yes, sir. Yes. Brother is fine. Triple on it. Yes, sir. No doubt. Yes, sir. Okay. It's simple. It's not tough. Expression looks like a bit tough, but it's not that tough. You cancel it. Take care. Now when you take this LCM here so it becomes that this numerator you see. It becomes K2 minus K1 by K1 correct. So the expression here would be K1 by K2 minus K1 a not and this numerator would be K2 minus K1 by K1. And this expression we can write K2 by K1 in the numerator whole to the power minus of K2 by K2 minus K1 fine. I have taken this in the numerator so negative sign here this LCM you take so K2 minus K1 by K1 is this correct. No doubt. Right. This K1 this K1 will get cancelled K2 minus K1 K2 minus K1 will get cancelled so the expression of B max. The final expression is, we have K2 divided by K1. Hold to the power minus of K2 by K2 minus K1 a not. This is the final expression for B max. And this is it. No, when you, when you, you know, memorize this result one more thing you have to take care of that the reaction was this A to B and B to C. So it is K2 and this is K1. So B max so B current constant we are taking here K2 by K1 K2 by K2 minus K1. With respect to this only you memorize all the results at your time. Right. So we are doing this. So, in this, they have asked this B max to find out and the time at which the concentration of B becomes maximum. Okay, so concentration of B if you don't remember fine, let it be, but that is the only one thing that you have to keep in mind. Okay, but B max and T you must remember. T also T also you see we have got the value of T. We have K2 by K1 so K1 K2 is what K1 K2 K2 is for B to C K1 is for this with respect to this only all the relation you remember. Okay, fine. So this is it one last thing this one last minute. Okay, I just finished this one last minute the graph of this okay because graph based question also they have asked once. Okay, so graph of this is we have concentration of a concentration of a decreases with time. Okay, so a will keep on decreasing. So this is what we have a there was this question which graph is not correct. So this is concentration a not we have. So it decreases, and it goes like this. And C concentration is increasing continuously. See was initially zero C increases like this, it won't go beyond a not so take care of that. See goes beyond this. Okay, be will be bought at one point be becomes maximum. So the graph of be would be like this it could be like this and then it comes. So remember, none of these graph will go beyond this maximum value of a that is a not based on this they have asked question ones. Okay, so because a not is the maximum concentration with a not only B and C is for me correct. So the graph cannot exceed or no go beyond this line. Okay, the blue one is for B. Yellow is for C white is for a continuously decreases and the point at which, you know, the concentration of B becomes maximum. This T is. This is the time axis we have this T is equals to an K2 by K1 divided by K2 minus K1 corresponding to this point. Here this one is the B max we have. This is it for this any doubt in this. Okay, so today after the dinner, you just go through it once the entire derivation. I would suggest you should do it on your own. Once you do it, you will memorize it. Correct. Like this, we have two more derivation. One is for parallel reaction and other one is for reversible reaction so next class will finish this. What are the chapters we are left with after this. DF. Yes. Okay, fine, fine, fine. We'll see next class will decide something else. Thank you, sir. Thank you, sir.