 Okay. Just a second. Okay, so last class you tell me, any one of you Naman, Amogh, Ramchiran, what is the sequential we have finished, right? So we have done the calculation of the concentration of A concentration of B at any point of time concentration of C, what is the maximum concentration of B and what at, by what time or after what time the concentration of B becomes maximum, correct? Graph also we have done, right? Graph, we didn't do. Graph where B is the maximum, right? Yeah, yeah, yeah, concentration time graph, we have done that, right? Okay, so you see we are moving forward into this, this is probably the last session of kinetics, okay. So we'll probably finish this particular thing in one hour, one and a half hour, then we'll start, we'll see some questions and then we'll start solution probably, okay? So the next thing that we are going to start here in that we have is parallel first order reaction, parallel first order reaction, okay? So since it is a parallel reaction, so we have one reactant and that reactant will convert into two different products, correct? Like this you see, if you have one reactant suppose A and this may convert into B and this may convert into C, right? Suppose the rate constant for the first reaction A to B is K1 and this is K2, okay? Now you see I have taken a very simple reaction here, like to make you understand, very simple example we are taking. But in the question that you get, there you may have here some coefficient also, like it is 2A gives 3B or 5C, like this, correct? So if some coefficient, stoichiometry coefficient is there, then your expression will change, like whatever expression we are going to get in this after derivation, okay? And we'll do the derivation in the similar manner that we have done earlier, right? In sequential reaction, okay? So that derivation and the final product and the final expression will change, right? The expression will change, little bit change will be there, if the stoichiometry coefficient is not same, right? So whatever we are going to get here, that is not the formula, okay? It is just an expression. But with the help of that, how do you write down the expression for any reaction with any stoichiometry coefficient? That also we are going to understand, because every time we are not going to derive these things again and again in the examination hall, we don't have that much time, okay? So first, we'll solve these questions on the basis of this very simple example that we have taken A gives B and A gives C, correct? Now you tell me one thing, and this is the very important step we have, and that step is to write down the rate law expression, okay? Or the rate expression, correct? So if I ask you, if I write down the rate of appearance of B, what will be that? Rate of B. If you are able to write this expression, then you can do that. After this, after this, we have only mathematics, concept of mathematics you have to use, correct? So rate of B, if you write, so that will be what? Con, D concentration of B by dt, yes or no? And this will be equals to, you see, this A gives B, right? So D B by dt from this reaction, what we can write? K1 into concentration of A, right? Similarly, if I write down rate of C, rate of C, that will be what? D concentration of C by dt is equals to what we write? K2 concentration of A. Clear, Naman? If you want, you can write down plus here, plus here also, because both are product, correct? Rate of disappearance of A if I ask you, that is a little bit, you know, tricky, the expression that you have, because A is converting into B, and A, at the same time, A is also converting into C, two parallel reaction we have, right? So if I write down the rate of A, that will be what? Rate of A is equals to what we can write? First you write down, let me write down this one, minus D A by dt, correct? Because A is the reactant, right? And that will be equals to what? Rate at which B is appearing, right? Plus the rate at which C is appearing, correct? Because A is converting into B and C simultaneously, right? One after the other, right? So the rate of appearance of B plus rate of appearance of C is equals to what we can write? Rate of disappearance of A? Yes or no? Correct? So that, what we can write? Minus D A by dt is equals to plus D B by dt, concentration of B by dt. And again, we can write what? Concentration of B by dt. And D B by dt is what? K1A, D C by dt is what? K2A. So if I substitute these two here, what we get? Minus D concentration of A by dt left-hand side is equals to K1 plus K2 into A. See, I will, maybe in between I will skip some of the steps, okay? So if you are not able to understand, just you can ask me directly, okay? Because we have done 2, 3 derivation already. Similar pattern we are going to, right? Now, can you integrate this and solve this differential equation? Yeah? So what we can write the expression here? Minus D concentration of A by concentration of A is equals to what we can write? K1 plus K2 constant into dt, right? So if I integrate this, if I integrate this 0 to t and when the time is 0, what is the concentration of A? It is A naught, right? A, A t. And when you differentiate this, what you will get? Minus sign, suppose if I take right-hand side, okay? And we get what? ln of A naught minus, so minus sign will take here only. So can I write this ln of A naught by A t is equals to K1 plus K2 into t minus, sorry, okay? Can I take, can I have this? Minus or plus? Just check this. I think it is plus, we will get here. That's okay. Plus we will get here, no? Okay. So what is A naught by A t? That will be e to the power K1 plus K2 into t, right? And what is A t? It is A naught e to the power minus K1 plus K2 into t. Similar kind of expression we have already done in which reaction. Can you tell me first order? What is the expression we have in first order? A t is equals to A naught e to the power minus K t. So if you write down here, in terms of, since we have first order reaction, the pattern is same, but the only difference is what? There we have rate constant K. And instead of K what we have here? K1 plus K2 is the rate constant. So whatever the rate constant we have here, K1 and K2, you have to add these two to find out the rate constant of this reaction, overall reaction. Correct? Yes or no? Okay. So we got A t. Now how do we find out B t? For B t, can we use this expression? Just see here, just see here. The first equation you see, right? This is A is nothing but what? It is A t only. Yes or no? A is the concentration at any point of time t. Correct? What is A? A naught is the initial concentration. Whether you write A, it is a concentration at any point of time. Correct? A t is what? It is concentration at time t. So these two are same only, no? Same time we are taking though. Okay? Yeah. So now you see, now you see this is A t and this is also A t you can write. A t we can substitute like A naught into e to the power minus K1 plus K2 into t. Correct? So what we will write? d B by d t is equals to K naught into all these terms. Yes or no? That is what the next derivation we will write. So you also write it down. I will rub this off. All of you have copied this down, right? Okay. So now you see d B by d t we write. That is d concentration of d by d t is equals to K1 into what we can write? A instead of A what we can write? A naught e to the power minus open bracket K1 plus K2 into t. Right? So this d B we can write what? d B is equals to d concentration of B is equals to K1 A naught e to the power minus K1 plus K2 into t d t. Right? Now we can integrate this, right? Zero to some concentration and zero to t. If this time is zero then concentration of B is what? Zero initially. When time t is t then the concentration of B is what? d t. Right? So when you solve this we get what? Concentration of B at any point of time t is equals to K1 by K2 minus here. Right? Into A naught open bracket e to the power minus K1 plus K2 into t minus one. Is it clear? You solve you will get this only. When you put limit e to the power zero will be one. Okay? So the expression of B I will write down the final formula here. Expression of B d t that will be K1 divided by K1 plus K2 into initial concentration of A that is A naught open bracket one minus minus I have taken inside. Okay? One minus e to the power minus K1 plus K2 into t. Is it? Right? Now on the basis of this can you tell me what is the concentration of C? We don't derive this. Okay? Just you tell me look at the expression and you tell me what is the difference we have in both the expression. Instead of K1 we will have K2 that is the only difference we get into this. Right? So formula of C is what? K2 by K1 plus K2 A naught open bracket one minus e to the power minus K1 plus K2 into t bracket close. Right? So this is the expression. Right? And you don't remember this as B t is equals to K1 plus K2. You try to understand this that this numerator will have the rate constant which represents the equation because the equation what we have? The equation is this A gives B. See what you have to keep in mind that you try to understand. This is K1 and this is K2. Correct? If I write down K2 over here then here we will get K2. Are you getting me what I am trying to make you understand? Right? So it is not K1 always whatever here the rate constant we have in which the reaction A gives B that rate constant comes over here in the numerator. Correct? Which you can understand easily but you have to keep this in mind. Don't memorize this. Okay? Bt is equals to K1 by K2 into this. Don't memorize this. Okay? Because anything you can have in the exam. Okay? Now one more expression we will try to find out here. Okay? Because Bt, Ct all these terms you have to find out. Okay? I will see here. If I write down this, if I ask you this one, concentration of A that decompose what is that expression? Can I write down this the initial concentration? Right? Minus the concentration at time t. What is the meaning of AT? Yeah. What is the meaning of AT? It is a concentration which is at time t which is not, this is not the concentration which reacts. Okay? I am telling you this thing again and again. AT is the concentration of A at time t. Okay? A0 minus AT when you write that concentration has been reacted during that time interval. Correct? Yeah. So AT formula we have already. So we can write what A0 minus AT is what A0 into 1 minus e to the power minus K1 plus K2 into t bracket close. Okay? So when I take this A0 common, 1 minus 1 will get cancelled. Yeah. Right? No. Oh, one thing I am making a mistake here. This will be A0 into e to the power minus K1 plus K2 into t. Correct? AT is equals to A0 into e to the power minus K1 plus K2 into t. That is what we have derived. Yeah. Yeah. So when you take this A0 common, we get what? 1 minus e to the power minus of K1 plus K2, K1 plus K2 into t. Okay? Now you see what is this we have? What is this all term? A0 into this. What is this? What is this? This is AD compose. Yes or no? The same thing we have here. No? So this formula you also keep in mind AD compose is equals to this. Write down in a square box. Okay? AD compose is equals to this. Okay? Sometimes they ask you this also. The concentration of the reactant which decomposes. So what we can write you see here? BT is equals to what we have? BT is equals to in terms of AD compose, we can write K1 divided by K1 plus K2 into AD compose. Yes or no? Similarly, concentration of C at any point of time t is equals to what we can write? K2 divided by K1 plus K2 into AD compose. Same thing just we are simplifying it. Right? In terms of AD compose. All this formula you write down in a square box. Correct? Now the last thing here is I ask you to find out the ratio of the concentration of product at any point of time. Right? What is the product we have? B and t? The concentration of B at any point of time t divided by concentration of C at any point of time t is equals to what? The ratio of the rate constant. Everything gets cancelled. You can check. Right? So if they ask you and this is important also, you don't have to calculate that much. Okay? That's why we are calculating over here. Okay? Don't try to derive this in the exam. Try to memorize this. Okay? Try to go through the expression and when you write down this expression 2, 3, 4 times, you will memorize the expression. Okay? Whenever you have to find out BT by CT, just you take the ratio of K1 and K2 where K1 is the rate constant of this reaction and K2 is the rate constant of this reaction. Is it clear? Okay. Now you see the next thing that we have discussed initially. Like you see, reaction I have taken A gives B, A gives C. Correct? But the reaction can be anything. No. Reaction can be this also. 3A gives 2B and 5C. This is also possible. Means we can have any number. Possible? Okay? So now we are not going to derive this. Complete. We are not going to derive this. Okay? Little bit of thing we will do and then you will understand how to write down the formula directly from this. Okay? Now you see if I write down the rate of B, rate of B what we have? Can I write down this? 1 by 2 plus D concentration of D by DT. Correct? And that will be equals to K1 into concentration of A. Yes? And rate of C will be 1 by 5 plus D concentration of C by DT is equals to K2 into concentration of A. Correct? And then if you write down this rate of A that will be what? 1 by 3 minus D concentration of A by DT is equals to what we can write? What we can write you tell me? This rate of appearance of B plus rate of appearance of C. Right? Rate of appearance of A plus C and this is equals to this only so directly I am writing down this. K1 plus K2 into concentration of A. Clear? Now when you rearrange this, D A by concentration of A if you write down here and right hand side what we have? 3 into K1 plus K2 into DT. Now when you solve this you will get the expression like this. You tell me yes or no. A T is equals to A0 e to the power minus 3 K1 plus K2 into T. This is what you get? Because if you compare this expression and the expression which we get here in that equation the first equation will not have 3 over here only. That is the only difference. Right? We did not have 3 over there. Right? Where this 3 comes from? From this, the stoichiometry coefficient. Because we are talking about A here so whatever the stoichiometry coefficient of A we have here that comes over here. Correct? And which comes in the final expression. Correct? Similarly you can do this for DBN for BNC also. Correct? BNC. Last expression did we calculate and T half for that reaction? The last one? You find out T half for that reaction somewhere in between there only. Tell me what is the expression you are getting? What is that T half expression for first order? Yes. What is that T half for first order? 0.693 by K. By K in the first order. Right? When we have one reaction. So since instead of K we have K1 plus K2 here that is the difference we have. Right? So in T half expression also instead of K we will write down K1 plus K2. Correct? So that is the formula we have in that equation. Now here you see instead of K what you are getting over here? 3 K1 plus K2. Correct? 3 K1 plus K2. So if you find out T half for this reaction T half will be what? Nn2 divided by instead of K1 plus K2 what we have? 3 into K1 plus K2. You can find out this also you can solve you will get this only. Fine? Is it clear? So whatever constant coefficient of A we have here that we have to take here with K1 plus K2 that will be your expression for A. Is it correct? Now you see from this expression if I write down dB by dt first expression you see from this if I write down dB by dt what we will write you see plus dB by dt plus dB by dt is equals to what we write? 2 K1 into A is what? e to the power minus 3 K1 plus K2 into T instead of K1 what we have here? 2 K1 which is the coefficient of here we have. See this 2 is coming over here. Correct? So what is the expression of dB we have over there? In this case the concentration of B at any point of time T will be instead of K1 we will write what? 2 K1 plus K2 we have in the denominator instead of K1 plus K2 what we have to write? 3 K1 plus K2. Okay? 3 K1 plus K2 and the expression will be what? A naught or A decompose we can write down here or we can also write down A naught into 1 minus e to the power minus 3 K1 plus K2 into T. This term is nothing but the A decompose. A naught into this is what? The concentration of A that decompose. Correct? Similarly can you tell me the expression for C concentration of C at any point of time? 5 K2 by 3 K1 plus K2 into A decompose. Correct? Tell me this BT by CT what is BT by CT? 2 K1 by 5 K2. Correct? 2 K1 by 5 K2. Okay? This is it you have to find out in this parallel reaction. Right? These formula you have to keep in mind. I don't think the formula or the expression is that much tough. If you can look at the expression properly you can write it down in the exam. Okay? You don't have to derive this again and again. But since I have done the derivation suppose you got stuck also somewhere or if you are not getting the formula then you can derive this quickly. So you must practice this two or three times. Okay? So that if you got stuck if you are not getting the formula in the exam you can derive this quickly. Okay? So which is not that tough? The derivation is not that tough right? Yeah. So only integration you have to keep in mind. Okay? Okay. Let me check one question based on this. Okay? You see one question I'll give you on this this thing only. Okay? You see how the like the expression becomes or the question becomes so easy when you know these formula that we have done just now. Okay? You see I'll write down I can I rub this off? Yeah? Okay. You see the question we have. It is given in the centre module also similar kind of question I am not sure with the data but yeah it is given. So you can go through those questions you can solve. Okay? You see two reactions A gives B and A gives C. This is K1 and this is K2 and while you are solving these questions don't try to look at the final formula or expression that we have derived just now. You try to write down on your own. Okay? So you see K1 value is given. The value of K1 is 1.26 into 10 to the power minus 4 second inverse. K2 is equals to 3.8 into 10 to the power minus 5 second inverse. Okay? You have to find out the percentage distribution of B and C. Percentage distribution of B and C. Done? Means the reaction is complete. Done or not? You are doing calculations? Yes sir. Why so? You see what is the formula of B? B formula is what? See what you have to find out you have to find out the composition know percentage distribution of B and C. Means out of 100 you have to find out. So what we will write? K1 by K1 plus K2 into A decompose. So this is same in both the expression know. So you find out this and that gives you the answer. You see what we have? What is K1? K1 plus K2 will be what? That will be 0.38 from here. 0.38, 0.1 plus 2, 6 that will be 1.64. So K1 plus K2 is what? Yeah, it is close. 77 percent 76.83. 1.64 into 10 to the power minus 4. Right? So if I write down this B here, so K1 is 1.26 into 10 to the power minus 4 divided by 1.64 into 10 to the power minus 4. Right? Since this thing is common only, so when you find out the percentage composition or ratio if you take B, T and C this will get cancelled only. Correct? So percentage composition will be this only. You will get this as 0.7683. That will be 76.8 percent. And similarly you find out concentration of T that will be K2. That also you can do. K2 K1 plus K2 into A decompose. Right? So basically this term and this term you do not have to find out. Just you find out this. Since you are taking out of 100 only, these are common. So you can eliminate this, not required. So all these things you will understand when you solve the question. Otherwise you will know everything. Then also you are not able to solve the question because of some calculation thing. Okay? Okay. So now we are moving ahead. See the next reaction we have and that is right down the kinetics of reversible first order reaction. The last question also if you see, there is nothing like first order or second order or whatever the order it is. It is not mentioned over there. Right? But whatever formula we have used in the last question, those formulas are applicable only for what? First order reaction. Okay. So that you must keep in mind. Most of the time you will not get any other order in this kind of question because the calculation becomes so difficult. Okay. So you will get first order only. Okay. But yeah, that's what. But you see the K value is given over there and unit is second inverse. That implies the order is first only into that. Okay. So you should note down like you should note all these things. Okay. You should ask this question to yourself only because you have only reaction. Now what is the order? Because those formulas were derived only for first order. So you should ask this question to yourself. Okay. What is the order of the reaction? Since order is not mentioned, but K unit is given. So from there it is confirmed that the reaction is of first order. Correct? Okay. Now coming back to this kinetics of reversible order reaction. A, suppose I am assuming A gives B and this is a reversible reaction. K1 is the rate constant of forward reaction. K2 is the rate constant of backward reaction. Okay. Now in this again, we will do the same thing. You see, if I write down the rate of A, the rate of A for forward reaction we are assuming. What is this? Minus D concentration of A by dt is equals to what we write? K1 into concentration of A, rate of B we write down, rate of B and that is for backward reaction. And when I say backward reaction, it means B becomes what? Reactant over there. Yes or no? That will write what again? Minus D concentration of B by dt is equals to K2 concentration. D. Yeah. Okay. Now since it is a reversible reaction we have. So after some time, the rate of forward reaction is equal to rate of backward reaction. That will be the case of equilibrium. Okay. So write down when equilibrium is achieved, write down this thing. Okay. When equilibrium is achieved then we can write what? Rate of forward reaction is equals to the rate of backward reaction. The rate of backward reaction. And that will write what? K1 into concentration of A is equals to K2 into concentration of B. Fine. But the concentration of A and B here will be at equilibrium. Correct? These are the concentration at equilibrium. Right? If they ask you what is the ratio of the B and A at equilibrium, answer will be what? K1 by K2. This is the first thing you have to keep in mind. Okay. B by A is forward by backward. K1 by K2. Correct? Now this expression we will use later on also. Okay. So you see the next thing we have. Can we write this at any point of time at any time? We can write what? A0 is equals to concentration of A at any time plus concentration of B, conservation. Right? Because from A only we are getting this concentration of A and concentration of B. Right? And this is applicable at equilibrium also. Correct? So we can also write A0 is equals to concentration of A equilibrium plus concentration of B equilibrium. Correct? No. Can we substitute this B equilibrium in terms of A equilibrium? Yeah. So you see here, if I write on B equilibrium from here, that will be B equilibrium will be A0 minus A equilibrium. Okay. B equilibrium we can write what? K1 by K2 if I am not wrong, you let me know. Like if I am not, if I am wrong, you let me know. K1 by K2 into A equilibrium. Right? From the previous formula that we have done. Into A0 minus A equilibrium. What is A equilibrium from here? Concentration of A at equilibrium will be K2 by K2 by K1 plus K2 into A0. Okay. Now K1 by K2 what we can write? It is the rate of rate constant of forward reaction and rate constant of backward reaction. This is nothing but equilibrium constant Kc. Right? So further we can write the concentration of A at equilibrium is equals to A0 divided by Kc plus 1. Is it correct? This is also the formula and this is also the formula here. Both you can keep in mind. Both are actually the same thing. Okay. Now if you know A equilibrium, you can find out B equilibrium also. They are substitute here. Right? So B equilibrium is equal to what we have. Concentration of B at equilibrium is equals to K1 by K2 into A equilibrium. Right? From the first relation. A equilibrium is this only. So we will write what? This is nothing but Kc. Right? Kc into A0 divided by Kc plus 1. Correct? So all these formula you write down in the square bracket. Okay. Done. Okay. I will just stop this off. Okay. Now you see if I write down the rate of the reaction, that will be what? Minus D concentration of A by dt and that will be equals to what? Rate at which it is going in the forward direction minus the rate at which it is coming going into the backward direction. Correct? Forward minus backward. That is the rate of the reaction, net rate of the reaction. Okay? Now you have to simplify this only. From here only we will go. K1 into concentration of A and B at any point of time, we can always write as concentration of A0 initially minus concentration of A. Correct? Because A0 is equals to what? A plus B at any point of time. Right? So when you multiply this, you see this minus minus plus K2A. A we can take common. You will get what? K1 plus K2. Ramcharan, how are you there? Okay. K2 A0. Correct? Yeah. So K2 A0 what we can write in terms of A equilibrium. What is the formula of A equilibrium? You see. K2 by K1 plus K2 into A0. Yes? So what is K2 into A0? K1 plus K2. Okay. So that is what we have. So can I write down this minus dA by dt is equals to K1 plus K2 into A minus A equilibrium. Correct? Now you can solve this integration again. The same thing. So we can write minus D concentration of A by concentration of A minus A equilibrium is equals to K1 plus K2 into dt. Right? A equilibrium will be constant. Yes or no? Because it is the concentration at equilibrium. So that will not change. Okay? So you have to integrate this and this. What is the limit we will substitute? At t is equals to 0. What is the concentration we have initially? A0. At t is equals to t. What is the concentration? A or At also you can write. Correct? At you are writing better. Okay. Now when you solve this, this left-hand side becomes what? ln concentration of A minus A equilibrium, open bracket. And here we have to substitute the limit. It is from A0 to At and this will be equals to K1 plus K2 into t. Here we have minus sign also. Right? So what we will get here? If you take this minus sign inside, the limit will get reversed. Correct? So the expression we get here is, you know, something like this ln A0 minus A equilibrium divided by At minus A equilibrium is equals to K1 plus K2 into t. You see this expression is also same of first order. In first order, we have only A0 by At. Here we have minus of A equilibrium, minus of A equilibrium in numerator and denominator. Correct? So from here, if I write down A0 minus A equilibrium divided by At minus A equilibrium is equals to what we have? E to the power K1 plus K2 into t. Right? So this At we have to find out. Correct? This is what our objective is. What is our objective? See, in all this derivation, we are finding out concentration at any point of time. So this At we have to find out. Correct? A0, in terms of A0 and A equilibrium are also constant only. Correct? So only we have to, what we have to do? We have to simplify this. Okay? Did you complete till here? Yes? I will see. If I reverse that, I have to find out At. Okay? So At I will write down on the numerator. Okay? So concentration of A at any point of time t minus concentration of A equilibrium divided by concentration A0 minus A equilibrium is equals to what we can write? E to the power minus plus K2 into t. Right? So At minus A equilibrium is what? You have to multiply this. Okay? This becomes what? A0 E to the power minus K1 plus K2 into t minus A equilibrium E to the power minus K1 plus K2 into t. Okay? Now A equilibrium we can substitute. Yes? See A equilibrium here also we have and here also we have. So we can take this, this side. Right? And take common because we have to, our objective is what? We have to find out this At. See we are just simplifying this. You can do it on your own way also. At is equals to what we have? A0 E to the power minus K1 plus K2 into t. And then if I take plus A equilibrium common, okay? So what we will get next? 1 minus E to the power minus K1 plus K2 into t. Now A equilibrium what we can write? We can write down this in terms of A0 also. We have calculated that already. Correct? And what is that expression? After this I will give you the expression. Okay? After this you can simplify. A0 E to the power minus K1 plus K2 into t plus A equilibrium is what? K2, K2 by K1 plus K0, K1 plus K2. In terms of Kc also you can write down. Into A0 open bracket 1 minus E to the power minus K1 plus K2 into t. Correct? Yes. So now when you solve this you just multiply with this then take A0 common from all the expression. Okay? You will get some expression here and that final expression of AT I will just give you. Okay? You can do this on your own. Okay? Only two, three steps we have. I don't have that much. I am not writing down. Okay? And the final expression you get here that will be initial concentration A0 divided by Kc plus 1 open bracket Kc into E to the power, E to the power minus K1 plus K2, K1 plus K2 into t plus 1. This is the formula you get. If you want you can do this on your own. Final expression I have given you. The only thing is what you see. Kc is equals to what we have. Kc is equals to K1 by K2. Only this thing you have to keep in mind. And then you see everything is there. A0 is there. A0 is here. K1, K2, t. K1, K2, t is here plus 1. You can take common. You will get it easily. If you analyze this also, you see A0 if I take common A0 by Kc plus 1. Just two steps you have. We can write down. You will understand. Okay? You have to get this expression. Only thing you have to take this thing common. You will get this thing automatically. Similarly, if you have to find out concentration of B, can I rub this off? I will give the idea how to find out the concentration of B. Concentration of B at any point time t is equals to, we can write what A0 minus A, correct? A0 expression we have already done just now. That we can substitute here. Right? Where is Naman? Okay, are you getting this? A0 expression just we have done just now. You have to substitute here. And then we can solve this for BT. A0, A0 you can take common. And the final expression of BT you will get here. Concentration of B at any point of time. That will be Kc by Kc plus 1. See, all these expressions look so similar. Okay? 1 minus e to the power minus K1 plus K2 into t close bracket. This is what the expression of BT. So this is it for this derivation. Okay? So to understand this, what all things we have done? We have done sequential reaction last time. And today we have done parallel and reversible reaction. Okay? One question you see on the basis of this reversible reaction. Whatever formula we have calculated, enough formula we have calculated. Okay? You do not have to, you know, worry about what kind of question you will get. If you have reversible reaction or sequential or parallel, these formulas you keep in mind you can apply and you will get the answer. It is not here. Okay? It is not here. It is not here. Okay? I will give you the questions. Okay? I have to check. Anyways, anyways. So this is what the thing that we have to discuss. Okay? Some questions we will see. General questions of this chapter only. And then we will do the last part of it. That is reaction mechanism. Okay? That takes around 10, 15, 20 minutes, not more than that. Okay? So this question you write down. What is the activation energy for a reaction? What is the activation energy for a reaction if its rate doubles when the temperature raised from, what is the activation energy for a reaction if it rates, if it rate doubles when the temperature raised from 20 degrees Celsius to 35 degrees Celsius, 2235. Okay? R value you take as 8.314. Tell me the answer. Let me know. Does it require? Think on this one. What happened? Temperatures must be in Kelvin. See, what have you taken K1 and K2? You see, we have this formula log of K2 by K1, this formula log of K2 by K1 is equals to minus EA by 2.303R open bracket, 1 by T2 minus 1 by T1. Correct? This is the formula we have. Now you see the raise, rise in temperature at 2035 degrees Celsius. That the rate constant increases to double or it becomes twice or twice for every 10 degree rise in temperature. Right? So this log K2 by K1, we have to find out, we have to write this log K2 by K1 as log 2 because it is however it is the difference is 15 degrees Celsius but we consider as this only because this is not the, you know, for around 10 degrees Celsius rise in temperature. Okay? So they should give actually 30 only over here but for 20, 12, 13, 14, 15 also we can take twice or thrice. Generally we will take two times only over here. So K2 by K1 becomes log K2. Now EA by 2.303R, all the value you substitute, R is 8.314, 1 by T2 minus T1. T1 minus T2, what is the temperature given? No. Difference is 15 only. So the numerator will get 15 and denominator will get what? 293 into 308. Some noise is there. Check. Tell me now. Some noise is there. I'm not getting you. Reconnect. I think reconnect. You reconnect. You reconnect. You reconnect. Approximately, huh? You can take depending on the option. Depending on the option you can take. Okay. And then you can solve. So answer is around 34, I think 34.7 kilojoule. Per mole. Okay. So when you solve this, then let me tell you this chapter, this kind of, especially the question related to Arrhenius equation. Okay. This requires some calculation. Okay. You will have some calculation into it. So when you solve the question, you try to take approximation according to the option given in the question. Okay. When the options are very close, don't take that approximation. Correct? One more question you see. So this is the catch we have in this question. Nothing was there in the question. Only this on almost around 10 degree Celsius rising temperature we have. So K2 by K1 becomes 2. Okay. That you must keep in mind. Okay. Now this question you see. The integrated rate equation is given. RT is equals to log C0 minus log CT. RT is equals to log C0 minus log CT. The straight line graph is obtained by plotting the straight line graph obtained by plotting time versus CT. Oh, it's simple only. I thought something was there. Time versus CT. 1 by T versus CT. 1 by T versus 1 by CT. What is the answer? Tell me quick. You have to find in which of these four options will get the straight line graph. Correct? Let it be this question was rubbish. Okay. Now one more question you write down. Consider an endothermic reaction X gives Y. This is an endothermic reaction. This one is I think probably better. Endothermic reaction X gives Y with the activation energy EB and EF with the activation energy EB and EF for backward and forward reaction respectively. So it must be reversible for backward and forward reaction and forward reaction respectively. In general, forward reaction respectively. Then we have four options given. We have to find out which one is correct. There is no relation no relation between EB and EF. Second one we have EB is equals to EF. Third one we have EB greater than EF. Fourth one EB less than EF. Yes what is the answer? Theoretically you have done. See whenever endothermic exothermic is mentioned no you write down del H first of all. Del H is greater than 0 for endothermic and what is the change in enthalpy? I have given you this formula activation energy for forward reaction minus activation energy for backward reaction. Did I? This formula I have given you. Have we finished that temperature term activation energy, Arrhenius equation? We have finished Arrhenius equation no. There we have done the graph. You see that graph first. Endothermic exothermic graph. This is also I have given there only. Enthalpy change. Yeah enthalpy change is equals to activation energy for forward reaction minus activation energy for backward reaction. Keep this in mind. Now you see enthalpy change if we have so del H is greater than 0. So this term is greater than 0. So forward is more than backward. D-option is correct. So you see it is very straight forward but you have to keep this in mind that what you have to apply. One more question. Last one we see then we will do the mechanism part. See two reactions are given. A gives B and this is K1 rate constant. A gives C. It is K2 rate constant. For this reaction activation energy is EA1 and for this one activation energy is EA2. The relation which is given over here that is EA2 is equals to 2 EA1. This is given. EA2 is equals to 2 EA1. Then you have to write down the relation of K1 and K2. Option I will just give you. First option is K2 is equals to K1 e to the power e to the power EA1 by RT. First option is this. Second one K2 is equals to K1 e to the power EA2 by RT. Third one K1 is equals to A K2 e to the power EA1 by RT. EA1 by RT. And last option we have K2 is equals to 2 sorry K1 not K2. A1 is equals to 2 K2 e to the power EA2 by RT. Option number? Ammo what happened? What is your answer? How did you do? A1 no K1 is equals to K1 is equals to A1 not A because for other one it is A2. Both will be different because two different reactions we have now. Correct correct correct correct. Option C is correct Ammo. You see you have to do this. You see K1 is equals to A1 e to the power minus EA1 by RT. And K2 is equals to A2 e to the power minus EA2 by RT. When you divide these two because in every option you see we have either K1 by K2 or K2 by K1. Correct ratio we all have in all options. So we will divide these two K1 by K2 is equals to we get what? A1 by A2 and e to the power e to the power EA2 minus EA1 EA1 by RT. Correct. So we will have this ratio becomes again A. Right. So K1 is equals to K2A e to the power EA1 by RT. That is what we have the option. Right. Option C is correct. Okay. Now you see the last thing of this chapter we are going to discuss that is reaction mechanism. Write down. Correlation theory we haven't done. Finish. No no no it's different. It's different. Reaction mechanism. Write down into this. This is the sequence of steps. This is the sequence of steps. Sequence of steps by which the reactants are converted into product. Next line. All these series of steps all these series of steps are based on observed rate of reaction. Observed rate of reaction. Okay. Example you see suppose if I write down this reaction H2 plus I2 gives 2 HI. Okay. This reaction goes like this. I2 goes into 2I first. Fast step it is. Organic chemistry we have done now. The various 2-3 step reaction. This one is fast. This one is slow. Formation of carbocation is a slow step and it reacts very fast. So that mechanism all those mechanism the observation of the mechanism and idea of the reaction mechanism we get from this particular concept only. At which reaction which step is fast which step is slow. Correct. So how do we know that one particular step is fast or slow that we get from this only? How you see? Those are most of more of the thing it is related to this of like experiment only but little bit of question few questions we'll get and that we'll try to understand from this that how to solve the questions. H2 then reacts with one of the IE from H2I. This is again the fastest step we have. And at last this H2I reacts with IE and it forms 2HI. Okay. And when you add all these okay this is the slow step the last one. When you add all these what you'll get H2 plus I2 gives 2HI overall reaction. Okay. Now for this if I write down the rate of this reaction will be what? R is equal to since it is slow step. So we'll write from this only okay. K1 concentration of H2I into concentration of IE right. This is slow step no. So this is what this is the rate determining step RDS. Correct. And with RDS only we write down the rate law expression RLE. Yes. One thing you have to keep in mind here that whenever you write down RLE in this you do not have to write you don't you should not write this intermediate product into the rate law expression. Okay. H2I is what? H2I is the intermediate product yes or no? No. It forms what are the intermediate product? It forms during the course of reaction and destroys at the end of the reaction. Correct. So here you write down this expression first and then here you mention it it is the intermediate product. Intermediate product write down the definition of this what is what is this intermediate product. These are the substance or compound these are the substance or compound which obtained which obtained during the course of reaction during the course of reaction and get destroyed and get destroyed at the end of the reaction get destroyed at the end of the reaction. Okay. It is highly unstable it is it is highly unstable one more important point to write down. We cannot write intermediate product in the rate law expression. We cannot write intermediate product in the rate law expression. Okay. Now you see rate law expression we always write with the step which is the rate determining step. Correct. So whatever the rate determining step we have which we get from experimental data. Okay. So this is the RDS lowest step with this will write down the rate law expression and that will write the constant into the concentration of reactant in that reaction H2I and I what is this H2I? H2I is what? H2I is the intermediate product. So we cannot write this H2I in this rate law expression. So we have to replace this H2I. Okay. And how do we replace you see that from this expression can we write down the constant of this expression K2 will be concentration of product that is H2I divided by the concentration of reactant which is H2 into I. Okay. Now H2I is equals to what we can write concentration of H2I that will be K2 into concentration of H2 into I. Right. Now I is again in the intermediate product. Okay. This I we can replace by this first reaction for first reaction if I write down K3 that will be what I square product by reactant I2. So I square will be what or I will be K3 root over I2 power 1 by 2. Yes. So now this I we can substitute here. So H2I becomes what? K2 H2 and I is root over of K3 I2 to the power 1 by 2. Okay. This we can substitute. This we can substitute in the rate law expression that is this. So R is equals to what we get K1 K2 root K3 into H2 into H2 into I2 into I something we are missing somewhere. I is also an intermediate product now. You see I is also intermediate. So that we have to substitute like this. So we get what? Do we get this K1 K2 K3 H2 I2. Do we get this? Yes or no. Tell me all of you. So this K1 K2 K3 finally will write a constant that is K. Okay. So rate law expression for this equation is what? R is equals to K concentration of H2 concentration of I2. This is the final rate law expression we have. Okay. Now how they like form the question from this. Okay. See here they can give you it is K1 it is K2 and K3 and they will give you the final rate law expression like this. Then they will give you what relation of this K K1 K2 K3. One option they will give you this K is equals to K1 K2 K3. Another option they will give you K is equals to K1 into K2 divided by K3. Like this they form four options. Do you understand this? First option, second option. So if so what you have to do? You have to write and they will mention this fast slow fast whatever it is. So with the slowest step write down the rate law expression. If any intermediate is there that intermediate you replace with the help of this K1 and K2. Right. Whatever equation in that equation the intermediate step you have you write down product by reaction and you keep on replacing the intermediate product. In the final rate law expression you must keep in mind that you must have the reactant okay which is there in the overall reaction H2 and I2. You see we are finally H2 and I2 only. Understand this? Can you do one question? One question based on this only exactly similar kind of question. Yeah. So you see one equation I am giving you the second example we are doing. Okay. The reaction is this NO plus BR2. This we have K1 and it gives NOBR2. This reaction is fast. Next we have this NOBR2 plus NO. This is K2 gives 2 NOBR. This reaction is slow. Right. You have to find out. You have to find out. Just a second you have to find out the rate law expression finally. Okay rate law expression you have to find out. Overall reaction is what? If I write down the overall reaction we add these two. So we will write 2 NO plus BR2 K gives what? NOBR. Overall reaction is this. Okay. So with the help of this data you have to find out the relation of K K1 and K2. What is the answer? Can I do this? Okay. What is the slowest step? Second one, this one? So what is the rate? What is the relation of K, K1, K2 that you tell me? NO square and BR2 is correct. K is equals to K1 into K2. So R is equals to what we can write? K2 concentration of NOBR2 into concentration of NO. Right. Okay. What is this NOBR2? It is an intermediate product. You see. It is intermediate. No. First this reaction forms NOBR2. This again reacts with NO forms to NOBR. Okay. So final reaction is this and in this reaction there is no NOBR2 present. It means it is an intermediate product. So this we have to we cannot write this in rate law expression. Correct. So NOBR2 if I replace from this so K1 is equals to what we can write? Concentration of NOBR2 divided by concentration of NO into concentration of BR2. Right. So NOBR2 is equals to this into this that we will substitute here. So that will be K2 into K1 concentration of NO concentration of BR2 into NO we already have. Okay. This becomes what? R is equals to K1 K2 concentration of NO square concentration of BR2. This is what the expression we have and from this if you write down the rate law expression that will be what? R is equals to KNO square into BR2. So if I rate these two K is equals to what we have? Is it clear? Correct. So this kind of very easy questions you will get. Okay. You will understand what you have to do. Now you see the last part of this. See we have done this reaction. Okay. So any reaction which takes place in more than one step. Okay. So basically if you draw the diagram. Okay. So just write down the heading. We have this is the last thing of this chapter. Energy diagram a multi-step reaction. Okay. In a multi-step reaction. So now you see if I draw the diagram here. Okay. This y axis you write down it is the energy and this side we have progress of reaction. Okay. You see the energy diagram can be anything. If you have multi-step reaction. Okay. So we will have what? We will have multi-step multi like number of peak also we will have accordingly. Suppose the reaction peak is like this. Wait I will just draw it properly. Get this. Then suppose we will get this. Then suppose we will get like this. Okay. This what we have? This is reactant and this is the final product we have. Now you see in this. This from here to here. This is suppose intermediate one. This is intermediate two and this is the final product. So from here from this reactant to this intermediate one we have step one. We have written no two three steps are there. Correct. One step reaction proceeds gives you the intermediate one and now this intermediate again reacts gives another intermediate. This is step two and finally the reaction completes in step three. So the number of steps we have is equals to the number of peak. First information is this. If this energy diagram is given the number of peak is nothing but the number of steps. Correct. What all information we can retrieve from this diagram that we are going to understand. First of all you write down this thing. Number of peak is equals to number of steps. First thing is this. Correct. Yeah. Overall the reaction is endothermic or exothermic. First of all you tell me this. Step one. Step one endothermic or exothermic. Endothermic because you see reaction starts from this state and it is going over here. So this is the higher energy state. Intermediate is that higher energy state? Yes or no. Right. So first step is what? Endothermic and for this del H value is what? Greater than zero. What about step two? Exothermic. For this del H value is what? Less than zero. What about step three? Again exothermic. Del H value is what? Less than zero. Overall reaction is what? Exothermic. For overall if I ask you then you have to see this reactant and product. Since the product is at lower energy state. So overall the reaction is exothermic del H is less than zero. This is another information we can get. Now another thing. Okay. Step one. If I ask you about the activation energy. Right. So what is the activation energy of the first step? This energy difference. Right. So this we write it as EA1 and this we write it as EA2 and this we write it as EA3. Correct. Yes. Which one has maximum value from this diagram? This diagram. EA1 has maximum and then we have EA2 and then we have EA3 suppose. Correct. EA2 suppose it is more than EA. From this diagram only we are talking about. Okay. Okay. Now you see when EA is maximum then what is if I ask you what is the order of rate constant? Order of K according to EA. What is activation energy? K is equals to A e to the power minus EA by RT. So when EA is maximum, K is minimum. Right. So what we can write? Order of K will be what? K1 is minimum. Then we have K2 and then we have K3. Right. Minimum value of K is what? Minimum rate. If K is minimum, the rate is also minimum and for this the rate is maximum. More K value, more will be the rate of reaction. Right. Okay. So now if this kind of diagram if it is given and you have to answer the question that which one is the fastest reaction? Fastest reaction means what? Maximum rate. Maximum rate means what? Maximum K value. Maximum K value means what? Minimum activation energy. So this reaction step 3 will be the will have the maximum rate. Step 1 will have the minimum rate or minimum rate or what we can say? Slowest step or what we can say next? RDS. Rate. Data mining step. All this information you can get. Yes or no? Okay. So this kind of graphical question you also get. So what all data we can retrieve? According to the graph, according to the energy of intermediate reactant and product, we can easily understand that which step is endothermic which step is exothermic. Okay. Overall reaction nature if you have to understand, just you have to check the energy of reactant and the energy of product. Correct. If the energy of product is more than the energy of reactant then the overall reaction is endothermic. Right. Number of peak is equals to the number of, number of peak is equals to the number of steps involved. Higher value of activation energy. Minimum value of rate constant. Okay. Minimum value of K means what? Minimum is the rate. Minimum rate means what? Slowest step. Slowest step means what? Rate. Data mining step. Is it clear? Okay. So this is it for this chapter. Okay. So we have the thing in this particular chapter. Now you can solve all the questions of center module. Okay. You can go through and you can solve all the questions. Understood? Any doubt you have in this? No? Okay. So we will take a break. Okay. 15 minutes break we will take. It's 630 already. So we will start 640 or 5 or 650. Correct. Okay. So we will start solution then. Okay. Okay. Yeah. Hello. Hello. Just give me a second. Okay. So next we are going to start another chapter and that is a solution. Okay. So second last chapter we have a physical chemistry. After this we are left with only what we say. Surface chemistry. Surface chemistry is not that important. This one is again. Amines. What? We haven't done amines. In that regard we haven't done amines. So I forgot. Okay. I will do. We will do. I thought I have finished. Anyways. We will do that also. Kormangala we have done no amines. Just a second. Okay. This solution chapter is it in the school is it done? No. Kinetics. They have done kinetics in the school? No sir. No. Okay. If you are doing solutions. Right. Okay. So what all things you have started already? Can you see the screen? You can see the screen. Right. Okay. So I say as far as this J is concerned this chapter is important. Okay. And you will have definitely one questions from this. Okay. So if I ask you first thing. Okay. That what is the difference between a solution and a mixture? Okay. If I ask you what is a solution or what is a mixture? Do we have any difference in these two terms? Solution is a mixture. Yeah. Right. Solution is a mixture but mixture is not a solution. Correct. Mixer are actually mixture are of generally two types. Okay. Don't write anything now. Just listen to me. Mixer are generally of two types. One is what? One is homogeneous mixture and other one is heterogeneous mixture. Any example of homogeneous mixture? What is homogeneous mixture? Definite composition. I suppose if I take water and in water if I mix sugar, does it form a homogeneous mixture? Yes. If you mix salt, if you mix sand, no, because sand gets settled down. Okay. It won't form a mixture. It won't form a homogeneous mixture actually. Okay. So homogeneous mixture are those mixtures in which the composition and properties are same throughout. Okay. So you write down this definition, homogeneous mixture. Here you write down homogeneous mixture are those mixture in which the composition and properties are same throughout. Composition and properties are same throughout. Okay. Example you can write on water plus sugar, water plus salt. Okay. What is heterogeneous mixture? Heterogeneous mixture in which the concentration and properties are not same throughout the mixture. Okay. So write down this definition of heterogeneous mixture also. Heterogeneous mixture are those mixture in which the composition and properties are not same. Example, example you write down water plus sand, water plus oil. See what happens when you take water and when you put sand into it. Okay. Why this sand particles get settled down in this? What is the reason of that? Suppose this is the water we have. Any reason for this sand particles to settle down to get settled into this? Huh? If you mix sand into it. See actually what happens whenever you have any mixture. Okay. Mixture you must have heard about emulsion, suspension and all. Right. This kind of solution. Okay. So what is the difference of solution? What is the difference of solution and suspension, emulsion? These are the questions we have one genuine question because all these things are mixture. Suspension is a mixture. Right. Emulsion is a mixture. Solution is also a mixture. Okay. So what happens when you put sand into this water molecule? Okay. Because of the size of the sand particles, right? Because of the size of sand particles, it experience more force of attraction due to gravity. Right? It more, it experience more force of attraction due to gravity. That's why it starts settled down into the bottom of the, like, you know, the jar that you have got the vessel. Understood? Like this, it forms a suspension, not a solution or even, yeah, not a solution. Right. So this thing we'll again discuss in surface chemistry also different kind of mixture, how the suspension forms and all. So major thing is what if you have, if you have large size of the substance, because of the mass, that particular object will experience more gravitational pull, right or wrong? Correct. So because of gravitational pull only, it starts settling down into the bottom of the vessel. Okay. And it won't form a homogeneous mixture. So for solution, we have a definite range of size of the solute that we are using, right? In that range only, the, if you see, you try to understand this thing. If the size of the particle is large, right, then more will be the gravitational attraction has less tendency to form a solution because it has more tendency to get settled down. Clear? Right. So when solution is forming, so size is one of the factor, size of the solute particle is one of the factor we have in this case. Okay. Lessor size if you have, then force of attraction due to gravity will be less. And then it has tendency to float into the solution. Right. It has more tendency to float into the solution. And in that way, it may form a solution. Okay. So the point here is what this suspension and five of the particles will discuss in surface chemistry again. Okay. But here what we have, as I told you that mixture are of two types, that is homogeneous mixture and heterogeneous mixture out of this solutions are what solutions are homogeneous mixture. Right. You must have seen water plus sugar solution, water plus salt solution. Can you observe sugar particle into that solution? It is completely dissolved. Right. We cannot see sugar crystal into that solution. Correct. When it is completely dissolved, then it forms a homogeneous mixture. Homogeneous mixture is a solution. Right. Heterogeneous mixture is not a solution. Right. So what we can say from this solution is nothing but, sorry, mixture is, solution is nothing but a homogeneous mixture. Correct. So all solution is a mixture, but vice versa is not true. Correct. All solution is a mixture. Right. Whether it is homogeneous heterogeneous, that is another thing. But all kind of solution is a mixture, but all mixture are not a solution. Is it clear? Because it is only homogeneous mixture. Okay. So first of all, you write down the definition of the solution. Right. Write down. It is a homogeneous mixture. It is a homogeneous mixture of two or more. It is a homogeneous mixture of two or more non-reacting substance, non-reacting substance whose composition varies within, varies within certain limits. Okay. Now, see, solution also we have two types. Okay. Ideal solution and non-ideal solution. Okay. You see, actually what happens, little bit of idea I will give you here, but this ideal and non-ideal thing we will discuss later on also in this chapter. Okay. Suppose you have a solvent here. Okay. First of all, we will discuss this. See, solution has what? Solution has two components and what are those components? Solute and solvent. Okay. So we will come to this ideal and non-ideal part later on. Okay. What is the definition of solute and solvent? Can you tell me the definition of solute in lesser amount? Generally, this is the definition it is given in the book. Right. The solute is a part of the solution or one of the component of the solution which is present in lesser amount. Solvent is what? Solvent is the part of the solution again or one of the component of the solution which is present in larger amount. Correct. This is the assumption we take. Okay. Write down the solute, the definition of solute. It is a component of solution. It is a component of solution which is present in lesser amount in the solution. It is a component of solution which is present in lesser amount in the solution. Solvent, you write down. The component of the solution which is present in larger amount. Larger amount. Larger amount by mass also we write sometimes. Okay. But larger amount will be the solvent. Okay. Now, if I take this example, if I take this example, the first example, if I write down this 60 percent of ethanol solution, ethanol solution. Can you tell me the solute and solvent in this? What is solute and what is solvent? Oh, suppose it is the first one. I will write down your 20 percent, not 60. If nothing is mentioned regarding the solvent, right, then we assume water as a solvent. Correct. So 20 percent and generally we consider this by mass. So 20 gram of ethanol present in present in what 100 gram of solution. Right. So 20 gram of ethanol we have and 80 gram of water present into that. Okay. That's where the solvent is water. If I write down now the first case, if I take 60 percent of aqueous solution of ethanol, what is the meaning of this aqueous solution? Now when aqueous is mentioned, it means it is clear that the solvent is water. Okay. Now 60 percent of ethanol we have. So what is the solute in this? Now solute is water. Right. Solvent is what? Solvent is ethanol. Right. So now what happens? Third case, if I take 50 percent of aqueous solution of sugar, 50 percent of aqueous solution of sugar. Now you tell me what is solute, what is solvent? Why? Here we have 50-50 percent. Anything can be anything. Okay. Anything can be anything you can say when both are liquid. Okay. Then you take anything. But see one like when I, when we say that larger amount is solvent, lesser amount is solute. This is just an assumption. It is not a definition. Correct. So one more way we have by which we define this solute and solvent and that is I feel is a more appropriate way. Okay. And that way is what? Solvent is that component of the solution, which is present in the same state as the solution is present. Means state of solvent and solution must be same. Correct. So now when you say 50 percent of aqueous solution of sugar, sugar exists in which state? Solid. Aqueous solution means solvent is, correct, solvent is water. So when you put sugar into water, it forms a solution and what is the state of the solution? Liquid. So liquid is the state of the solution. So the component which exists in liquid state, that will be the solvent. Is it clear? This is also one of the way that we explain what is solute and solvent. Okay. That is also you must keep in mind. So in this case, the solute, if you have to take 50 percent of aqueous solution of sugar, solute will be sugar, solvent will be water. Correct. If you take 50 percent of ethanol, anything can be anything you can take. Correct. Now, if you mix two liquid, okay, two liquid if you are mixing, it is not always necessary that they always form a homogeneous mixture. What did I say? If you are mixing two liquid, it looks like a homogeneous mixture, but it is not necessary. Correct. Write down this thing. If we mix two liquid, they may or may not form homogeneous mixture. They may or may not form homogeneous mixture depending upon, depending upon the nature of the liquid. Did you solve that kinemical kinetics question, symptom module? What about you, Naman? Which one, sir? Uh, symptom module. Have you started or you are like doing? Correct. What about you, Ramchiran? Okay. Okay. So now that, that, that you can do now, I think. Okay. So what is, do you feel any difficulty or you understood the concept properly? Like, do you have to? Most of the questions were direct. Okay. So maybe, maybe you have got the concept properly. Then you can do this. Anyway, so if you are able to solve this fine, otherwise you can refer the solution also. Or if you have doubt in solution also, then you can ask me. Okay. But do solve the question much. If you have any other book also, just go through the questions because practice is required. You must have seen in most of the questions, you know everything that this formula we have to apply like this, the data we have to put, then you stuck where you stuck somewhere in calculation part. Okay. So you solve all the questions. You will definitely get one question from this format is too much. Okay. So don't miss this particular chapter in, even in solution also, there are very less chance that you will have wrong answer in this question because this chapter is mathematical completely. If you see everything is mathematics, only calculation you need to require and concept and interesting also. I don't feel any, there are very less thing to memorize in this chapter, chemical kinetics. Correct. So solve the questions properly. Okay. Anyways, so depending on the nature of the liquid, they may or may not form homogeneous mixture. Correct. So now there are three types of liquid we have. Okay. So first type you write down. First type, we have completely miscible liquid. Completely miscible means what? Completely mixed. They mix which the liquid which mix properly. Okay. And the definition is not important. So what is the example we can take into it? Polar liquids dissolving polar, like dissolves like you know, right? When both liquids are polar, when both liquids are polar, they form what? Completely miscible liquid. Example, ethanol plus water can we write? C2H5OH plus H2O, both are polar. When both liquid are non-polar also we can take, like dissolves like either both are polar or both are non-polar. Correct. Example is what? We can take benzene plus cyclohexane also we can take. Example is not important just you write down. Okay. Second one you write down. Completely miscible we have done. Next you write down. In this one more just line you mentioned. These type of liquid mix properly to form a homogeneous mixture. These liquids mix properly and forms homogeneous mixture. Okay. First is completely miscible, then we can have completely immiscible also possible. Right. Completely immiscible liquid. Okay. Write down. This kind of liquid forms when one liquid is when we have polar plus non-polar combination. Right. They do not dissolve at all. Polar plus non-polar. Example you write down like the C2H5OH plus cyclohexane benzene plus water. Correct. Third one you write down. Third one. Partially miscible. Partially miscible. Partially miscible means what? They do not completely dissolve and it like the nature is in between. Completely miscible and completely miscible liquid. We also call it as partially immiscible liquid also. Right. Write down to this. When two liquids are not exactly similar. When two liquids are not exactly similar in nature. When two liquids are not exactly similar in nature. And also not completely dissimilar. And also not completely dissimilar. Example you write down ether plus water, phenol plus water. Okay. Now how many components are possible in a solution? Multiple solutions. Right. So yeah. So basically on the basis of next write down. On the basis of the number of components present. The solutions are generally classified. The solutions are generally classified into three categories. Into three categories. First one is binary solution. Number of components. How many components we have into this? Number of components are two into this. Example anything you can write. H2O plus sugar. Two components we should have into this. Second one. Ternary solution. Number of components are what? Three. Example what we can write. H2O, sugar and milk. I am making tea now. Okay. The third one we have. That is quaternary solution. Okay. Quaternary solution. Number of components are what? Four. In this example we have what? H2O, sugar, milk and then we can put tea leaves into this. Okay. So these are on the basis. We can have many other things possible. Just give me a minute. Okay. I am getting caught. Yeah. Hello. So we can have any number of components possible. Okay. But generally we classified into three categories. Okay. And in all these three only binary solution is there in our syllabus. Okay. So these two we are not going to discuss. Correct. Okay. Now you see like I said that solution has two components. Main two components we have. That is solvent and solute. Right. Solvent and solute. We are looking at the classification of solution. Okay. Solute. Now this solvent can be what? Solvent can be solid also. Can be liquid also and can be gas also. Right. Solute or what? Solute can also be solid. Can also be liquid and can also be gas. Correct. So on the basis of the state of this component solvent and solute again solution are classified into three different categories. Like you know what is the meaning of this? See if you take solvent as solid. Right. Solute can be solid, liquid or gas. Three types of solution. Yes. If the solvent is liquid again solid can be solid, liquid or gas. Nine type of six type of solution. And again if solvent is gas, then solid can be solid, liquid and gas. Understood. Okay. So next next line you write down on the basis of on the basis of the state of solute and solvent on the basis of the state of solute and solvent. The solution are classified into nine different categories. Okay. So now first thing you write down in this. If solvent we are taking as solid. Okay. In this case, solute can be, solute can be what? Anything, any of the three states of matter that we have? Solute can be solid, can be liquid or can be gas. Okay. Solid plus solid. Any example? Alloys. Okay. Solid plus liquid. Any example? HD in gold. Water of crystallization you can write down. Okay. Examples are not important, but just write down. Solid plus gas. You must have heard the, we use palladium for the storage of hydrogen gas. Right. We can store hydrogen at the ports of palladium. Okay. So H2 in PD is the example of this. Okay. Now this solvent can be what? This solvent can be liquid also. When the solvent is liquid, then again, solid can be, solute can be solid, liquid or gas. Solute can be solid. It can be liquid. It can be gas. So liquid plus solid. We have many examples. Right. Water plus sugar you can write down. Many examples we have. Okay. Liquid, liquid solution we have what? Alcohol plus water. Okay. Liquid and gas. Oxygen, H2O. Right. And this is the reason of the living of this aquatic animals. Right. Okay. So these are the three types. Now, the one more we have and that is when the solvent is, when the solvent is gas. When the solvent is gas, solute can be again solid, liquid or gas. So what I said? What is the doubt? Huh? Possible. What is air? Air is a mixture of gas. No. That we can consider as homogeneous mixture. Okay. So, solute can be solid, can be liquid and can be gas. Example of this you see. Gas plus solid solution. Gas plus solid solutions are what? Dust particle in air. Correct. Dust particle in air. Liquid is what? Moisture. Yes. Moisture in air. What is gas? Air itself a mixture of gas. Okay. So there are nine different types of solution possible. Yeah. Now you see, you must have seen that when you park your bike or car anywhere after sometimes like two, three, four hours, you will see some dust particle is there on the seats of the bike or on the mirror or a glass of the car. Right. Yes. So how this, where from where this dust particles comes from? From atmosphere. Right. Right. So basically what happens, these dust particles are settled down, are settling down with time. Correct. So when settling is there due to gravity, it means the solution is not homogeneous. Sorry, mixture is not homogeneous. Yes. Due to gravity, what happened in a solution of water plus sand due to gravity settling, the sand particles are settled down and that's why the solution is not homogeneous. Okay. The same thing happened over here also. Dust particle in air and moisture in air also. You must have seen in winter season that early morning, if you see there are some moisture on the leaves, tree leaves or on the grass. Right. That also comes from atmosphere only. Right. Condensation. So because of that also the solution or mixture may not be homogeneous. Okay. Sometimes what happens, these two, these two does not form a homogeneous mixture. Okay. And they, and we, you know, in NCRT it is written that there are nine different types of solution possible, but in this gravity settling is possible. Right. So because of that, what happens at some point, it may be homogeneous, may not be homogeneous also. Right. So true solution, if we say, right, but in CVC, you don't write. Okay. In J, you also it is, in J, all and all it is important. If you talk about true solution, true solution, we must have homogeneous mixture. Right. Then we have only seven different types of solution possible. We do not consider these two. Right. Because they may form, at some point of time, they may form heterogeneous mixture due to gravity settling and all. Right. Write down these two points. Due to gravity settling, due to gravity settling property, I'm talking about these two. Okay. So write down the arrow over here and then you write down these two processes. Sorry. Due to gravity settling property, this solution can be homogeneous at some point, can be homogeneous at some point and it can be heterogeneous also. And that is why we do not consider these two as a true solution. But one thing also I'm telling you, in NCRT it is clearly written that there are total nine types of solution possible. Correct. In the school, you don't write seven types of solution we have. Okay. You write nine type only. I have just given you one more idea over here. Right. In this, gravity settling possible. Okay. So this is the basic introduction or idea we have got from this, all this discussion. Okay. Now the another thing, here it is important in this chapter, which you have already done in the previous chapters like mole concept and cathesis state and all. And that is concentration term. Write down next concentration term. Okay. Concentration term we'll discuss in two different groups. Okay. And that is suppose group one. Okay. Two different groups I have classified this. Group one, which is based on volume or volume related term means volume related concentration term. Four different concentration terms possible we have. That is percentage by volume. The first concentration term we have. Okay. Second one, percentage mass by volume. Volume. Third one, molarity. Fourth one, normality. All these are volume based concentration term we have. Okay. Second group we have group two and this is mass related concentration term. Mass related. First one, we have percentage by mass. Second one, we have mole fraction. Third one, molality. And fourth one, we have parts per million that is PPM. Correct? Parts per million that is PPM. Yes. So now you see if I say percentage by volume, what it means? Volume of solute divided by volume of solution. Yes or no? We always write solute by solution in all this. Only one concentration term we have that is molality where we use the mass of solvent in the denominator if you remember. Yes. Correct? All other concentration term, numerator we always write solute and denominator we always write solution except this term which is molality where we have solvent. Percentage by volume is what? Volume of solute divided by volume of solution into 100. Correct? Similarly, mass by volume is what? Mass of solute divided by volume of solution into 100. Correct? Right? Molarity is what? Number of moles of solute by volume of solution in liter. All this definition you know? Yes sir. Okay? Normality is what? Number of equivalent by volume of solution in liter. See all these concentration term is required in this chapter. Okay? And most important is molality. Molality is most important. Molarity is also important over here. But you should know all these concentration term. Okay? Similarly, percentage by mass means what? Weight of solute by weight of solvent weight of solution into 100 since we have mass only. So numerator also will write mass. Denominator also will write mass. Numerator what mass will take? Mass of solute. Denominator what mass will take? Mass of solution into 100. Mole fraction you know already. Molality is what? Number of moles of solute by mass of solvent in kg. Right? So all these are concentration term we have. Now if I ask you one thing that suppose this like the concentration term of group 1 and group 2 if you have to relate, if you have to establish the relation between these two, any concentration term of group 1 and any concentration term of group 2, what is required in this? What is the term we need? What is the value we need into this? Mass and volume. How do you relate mass and volume? Density. So if density is given, you can relate this to concentration term of the two group easily. We will see some questions into this also. One question we have. Can you establish? I will give you one formula here which is required important also in this chapter. Percentage weight by volume. Percentage weight by volume is equals to, we will write percentage weight by weight into density of solution. Can you establish this relation? Can you understand this? How? Yes sir. Yeah, so percentage weight will get cancelled both side actually. Right? And density on the numerator. So from the right hand side if the weight will go to the left hand side then we get what? Weight by volume only. That is nothing but density. So this is very one very important formula we have that you write down. It helps you somewhere in the question. Percentage weight by volume is equals to percentage weight by weight into density of solution. Do you know the relation of molality and molarity? Molality is equals to, you haven't done this relation. On this they have asked question price in J e. 1000 into m, m is molarity by 1000 into d minus molarity into x. I will write down like this. And molarity into molar mass of solute. Capital M you may write down here. Capital M is the molarity. D is the density of solution. D is the density of solution. And this is the molar mass of solute. One question you write down on this. An aqua solution contains, I will write down the next page. An aqua solution contains 28 percent weight by weight of KOH. Density of solution is given. That is 1.25 gram per ml. You have to find out the mol fraction of KOH. You have to find out percentage weight by volume, molarity and the fourth one is molality. Tell me the answer. Can you do this? I don't know the answer. Done. See to solve the question of this chapter you have to be very comfortable with these things. Concentration related problem, mol fraction, concentration, molarity, molality, conversion, all these things you have to be very comfortable in all this. Mol fraction is 1 by 9, correct. Ramcharan, correct. Weight by volume is 35 percent. Molarity 5, no I think not 5. We will solve some questions on the concentration term because you have to, first of all you have to be comfortable with this. Then only because everywhere you require mol fraction, molality and all. 0.6 I don't think so. No, it's not 0.6. You see, why you are getting this wrong, I don't understand. First of all, directly can I write this? Percentage W by V is equals to percentage W by W into density of the solution. W by W is given. What is that 28 into 1.25 and you can multiply this, correct? One thing you also see, 1.25 we can write this as 5 by 4 also. So, you see this you can multiply direct also, you will get the answer. 1.25 we can write at 5 by 4. So, it is not multiplied. This we can write like this also, that is the meaning here. Now, when you write 5 by 4, your calculation will be a bit easy, right? No, 35 gram, 4 into 7 and 7 into 5 is 35. So, 35 gram of KoH in 100 ml of solution. Is it right? Can I say this? Because it is weighted by volume. Yes. So, what is the solute in this? KoH only. Right. So, if I see just see here, 35 gram present in 100 ml. So, can I say 350 gram present in 1000 ml, right? So, if I take volume as 1 liter, correct, which is 1000 ml, what is the mass of the solute? It is 350 gram, correct? So, can I find out the number of moles from this? Yes. That will be 355 divided by molar mass of KoH is what? 56. So, what is molarity now? 350 divided by 56. We will get molarity from this. Is it clear? Okay. Now, you see one thing. We can also derive one formula of molarity from this. What did I say? That this, how do we get 350 and 1000 ml? If you multiply this weight by volume ratio by 10, okay? Right, no? So, if I write the formula of molarity like this, suppose percent is weight by volume into 10, which is nothing but 350 by 1000. Yes or no? Right? And this, when you write this, this means what? It is 350 gram divided by 1000, correct? 1000 is what? Volume, 1 liter, it's fine. But in the numerator, we required what? We required number of moles. And number of moles is what? This divided by molar mass. This molar mass you have to include. This represents, percent is weight by volume into 10, represents what? 350 by 1000. If I divide this by molar mass of solute, does this give you molarity? Is it? Ramchiland, understood? This formula, did you get this? Did you get this formula? Tell me. Otherwise, you can do by the normal formula that we have. We have done already like that. You see this? Yes. We have done already like this, okay? But this is one of the formula which may be useful at some point in this chapter, okay? Percent is weight by volume into 10 divided by molar mass of solute is nothing but molarity, correct? Right? Now, the next question we have what? Mole fraction also we will find out. Next question, you see, 28 percent weight by weight solution, what is the meaning of it? Right? Means if 28 gram of KOH we have, right? Then what is the mass of solvent? Gram of H2O, nothing is given, so we will assume H2O, correct? What is the number of moles of KOH? It is 28 by 56, yes or no? 1 by 2. What? What is the mole of H2O? 72 by 18, that is 4. So we can find out the mole fraction of KOH and that will be what? 1 by 2 into divided by 1 by 2 plus 4. I am doing this in short. I hope you all understand this. There is nothing in this 1 by 9, correct? If I have to find out the mole fraction of H2O, which is not there in the question, what we will do? 1 minus 1 by 9, that will be 8 by 9, right? How do we find out molality? Now, everything is easy. You see, molality is what? Number of moles of solute, which is 1 by 2, correct? Mass of solvent is what? 72 gram, correct? So what we can write? 1 by 2 by 72 into 100. This is the molality, 1000, sorry. Is it clear? So all these conversion concentration term is very important in this chapter. That is why I am giving you some questions so that you will be comfortable with all these, okay? And the formula that I have given you, you must remember those formula. One more formula I will give you now. Can I rub this off? Okay. Now you see, molality formula we already know. One more formula we have of molality and that is mole fraction of solute divided by mole fraction of solvent into 1000 divided by moled mass of solvent. This is also one of the formula we have. Can you figure it out how this formula we get? Try to do this. How do we get this formula? Molality is what? It is the moles of solute. Moles of solute by mass of solvent. And if you take this in gram, we will get here 1000. This 1000 is because of this gram only, correct? If I divide, numerator by total number of moles, suppose capital N we have, I am just trying to derive this, okay? Here also we have to write down this capital N, okay? So what is moles of solute by total number of moles? It is mole fraction of solute, right? Divided by mass of solvent means what? It is the number of moles of solvent into molar mass of solvent and this divided by total number of moles we already have into 1000. You see what we get? You get this? Yes or no? It is also useful, okay? Mole fraction of solute by mole fraction of solvent. Have you seen this formula before? Okay. You see this is the same thing actually. Number of moles you divide by number of moles you will get this, okay? See how it is very easy because if you know the solute, okay? Solute if you know, then you have the molar mass of that, right? And if you have information of any one of this solvent or solute, if you know the mole fraction of solute, 1 minus that will be the mole fraction of solvent, right? And if you know solvent also, you can find out this, okay? Suppose in this what we required? You required only the mass of solvent. Suppose solvent you know what is the solvent and mass solvent you know, right? So with mass and molar mass since the solvent is mentioned, so you know the molar mass of that also. You can find out the number of moles of solvent, correct? Right? And with that number of moles of solvent, you can find out mole fraction of solute also because you will have the data, correct? And then you can find out mole fraction of solute. This formula it is useful sometimes, okay? So you must keep this in mind. One more question we see, this is the last for today, okay? Right? On how much water should be added, how much water should be added in 500 ml of, how much water should be added in 500 ml of semi molar HCl solution, semi, semi 500 ml of semi molar, semi molar HCl solution to make it Desi molar. Will you understand the question? Semi, SEMI, semi circular like that. Semi molar HCl solution to make it Desi molar. That is what you have to, why not? See how much, how much water should be added? See you have, you have a solution of HCl, right? Which is 500 ml of volume and it is semi molar. Semi molar means what? Molarity is M by 2. Ramcharan, semi molar means molarity is M by 2. So Desi molar means what? Molarity is M by 10, correct? Yes, what is the answer? Tell me. 2 liter is correct. Yes, 2 liter is correct. You see, what is the, see initial volume is what? V1 is 500 ml, right? And molarity, first we have, suppose M by 2, right? M1. Volume, this you have to find out and M2 is given, it is M by 10. So what we use? M1 V1 is equals to M2 V2, into V2, right? So we will get what? V2 is equals to 2500 ml. So 500 we already have, final solution is, volume is this. So what is the volume we have added? 2500 minus 500, so 2 liter, 200, 2000 ml or 2 liter. Semi molar means M by 2, molarity. Desi molar means M by 10, molarity, correct? Okay, so we will stop here only. Okay, we will wind up the class here only. Okay, we will start from here to the next class, okay? And you solve all the question of chemical candidates, right? Okay, chalo, bye-bye. Thank you, thank you.