So last class, can you please tell me what all things we have done, have we finished law of mass action, have we finished law of mass action, the factors affecting the rate of a reaction, have we finished that, factors affecting the rate of a reaction, you can drop some message so that I can get to know, I will get to know that who all are there in this session, okay. Anyone if it is there who like did not in the last session of chemical kinetics, you also let me know so that we can you start accordingly, okay, okay, now I can see there was some technical issue, I could not see the messages, okay, okay, so, hi, hi, everyone, how many of you were not there in the last class, have you seen the session that we have done in the last class of chemical kinetics? I think Shreyas was not there, I guess Mathili was not there, Sayuza are you there in the last session? If not, have you, have all of you have gone through the, all of you have gone through the session that we have done in the last class of chemical kinetics, have you finished that? Let me know quickly then we will start, okay, fine, so like for a quick review then just give me five minutes for that, okay, for a quick review then like we have discussed in the last session that rate of any reaction is like suppose if I write down one reaction here, say if I write down one reaction here, suppose A gives B, right, okay, suppose if I write down one reaction that A gives B, okay, and for this if we, okay, if we have to write down the rate of reactant, okay, rate of reactant that will be minus D concentration of reactant, minus D concentration of reactant divided by the time required for that particular change, similarly if I have to write down the rate of product, rate of product that will be plus D concentration of product divided by T that we have done, okay, and in these two expression we don't have to consider or we don't have to concern, get concerned about the stoichiometric coefficient over here, okay, we don't have to do anything with this stoichiometric coefficient, okay, whatever the stoichiometric coefficient we have here, the expression of rate of reactant or product will be same like we have written, suppose this reaction, this rate I have written for this reaction, if I write down 2A gives 3B also then this reactant, this expression, this expression is correct, okay, but the same thing is not true for rate of the reaction, okay, rate of the reaction depends on the stoichiometric coefficient and we have discussed one question of N2 H2 gives N is 3 that I have given you in the last class, okay, so that thing you must keep in mind, now you tell me like all of you have seen the video, have we finished the, have we started the factors affecting rate of reaction, law of mass action have we done, tell me quick, okay, law of mass action we have done, so we were doing the order, I think order also we have done, right, the various order of reaction and the unit of k, unit of k have we finished, unit of rate constant, okay, fine, okay, okay, we also have discussed about elementary and non-elementary reaction, okay, and suppose if I write this that the rate of any reaction r is equals to k concentration of this reactant to the power x, concentration of reactant to the power y and so on, then the order of this particular reaction whatever the reaction I have written, which is represented by small n and that will be given by the sum of the power of the concentration term in the rate expression, right, so this is what the order we have, order, we have also discussed that order can be 0, can be negative, can be positive or any fractional value, anything it could be, okay, anything it can be, negative positive 0 or fractional value, right, this is what the order we have, now you see in this we have also done the, you know, the definition of k that is the rate constant, okay, the definition of k that is the rate constant and rate constant we also call it as specific rate, specific reaction rate, okay, specific reaction rate, okay, so on the basis of this we will see some question, okay, so if I try to define the rate constant here, okay, so that rate constant will be you see, if I take this concentration of a and b as unity over here, right, a is 1 and b is 1, then k is nothing but r, okay and that is why we are calling it as k as specific rate of a reaction, I think this also I have discussed in the last class is basically a reaction, okay, we also call it as rate constant simply, rate constant or we also call it as velocity constant, velocity constant, okay, so k is what, k is nothing but the rate of the reaction when the concentration of all the species involved in the rate law expression is unity, okay, this expression we call it as rate law expression, in short we write it as RLE, right, rate law expression. Now you see if I, on the basis of this if I give you one small question, okay, just for, you know, practice we will see one question and then we will move on, okay, you see for a particular reaction, you write down this for a particular reaction and the reaction is given to a plus b gives product, to a plus b gives product, the order with respect to, order with respect to, to a is given and that is to, right, order with respect to b is also given, that is 1, okay, so you have to find out, first of all you have to write down the rate law expression, what is RLE, write down the rate law expression and what is overall order, tell me quickly, overall order of the reaction, okay, you have to find out the rate law expression and overall order, a to the power 2 and that is 3, so overall order is 3 and rate law expression is this, okay, simple question I have given you, the rate law expression since we know the order with respect to this reactant, the rate law expression R is equals to the rate constant k concentration of a to the power the order with respect to a and concentration of b to the power the order with respect to b, this is the rate law expression we have, first question. Second question, if I have to answer the overall order, so overall order n is equals to what, the sum of the power of the, the concentrate the species involved in the rate law expression, so 2 plus 1 gives you 3, right, now one more question if I give you in this particular thing, what is the effect on rate, okay, the effect on rate of the reaction, you have to find out the effect on, effect on the rate of reaction that is ROR if the concentration of a is doubled, okay, another one you see, effect on rate of the reaction if concentration of a is doubled, what is the effect on rate of reaction if concentration of b is reduced to half, okay, reduced to half, okay, what is the answer of third and fourth, four times and half, okay, you see simply you have to write down, suppose the new rate R dash will be, rate constant will be same, rate constant will not change because it depends only on temperature, okay, that we will see later on in this chapter, okay, R dash is equals to k, concentration of a is doubled, so instead of this a, what we will write, 2 concentration of a square into concentration of b to the power 1, so that will be what, R dash is equals to 4R when you solve this expression and when you equate these two, when we, you know, compare these two, this is the relation we have, means the rate becomes 4 times the initial rate, right, now here if you see similarly R dash is equals to k, concentration of a will be this and concentration of b is half, so we will write concentration of b by 2 to the power 1, here also to the power 1, correct, that gives you R dash is equals to R divided by 2, so the rate of the reaction becomes half of the initial rate, right, half of the initial rate, okay, one more extension in this one, if volume of the container is doubled, right, if volume of the container is doubled, container means what, the container in which the reaction is taking place, the volume of the container is doubled, then what happens on the rate of the reaction, Arian, okay, we have just started Arian, it will remain same, Atme is saying 1 by 8, okay, you see what happens when the volume becomes doubled, when the container volume you have increased by twice, okay, so obviously what happens, the concentration becomes half, right, because volume and concentration are inversely proportional, right, so here what I write here, the, since the volume is doubled here, so the concentration of the reactant A will become, concentration of A becomes A by 2 and concentration of B becomes B by 2, so if I write down the rate law expression, R dash is equals to K, concentration of A by 2 square and concentration of B by 2 to the power 1, here also 1, so you see this becomes what, R by 8, 8, 1 by 8 of the initial rate, right, so obviously the rate becomes what, 1 by 8 of the initial rate, so in this one what we can write down here you see one, just simple expression I am giving you, just formula in terms of formula you can keep in mind, that the rate of the reaction, right, R is the rate of the reaction is inversely proportional to the change in volume, right, inversely proportional to the change in volume, volume becomes double, inversely proportional to the final volume I will write down here, final volume to the power order of the reaction, okay, basically I have written here final volume since I did not have space over here, but you just understand over here, final volume means what, the ratio or the fraction or the number by which the volume has been changed, right, you see here volume of the container is doubled, right, so if you take the ratio of Vf by Vi, that will be 2, correct, so we will take here 2, right, basically we will take here the ratio of final volume by initial volume, so if the volume becomes double, so we will write here 2, not 2 minus 1 that is 1, okay, or if the volume becomes triple then we will write here 3, only 3, 3 to the power the order of the reaction, so you see order of the reaction we have calculated already it is 3, so rate inversely proportional to volume to the power what 3, that is 2 to the power 3 which is nothing but 8 we have and that is what the answer we get into this, okay, so this you can keep in mind as a you know formula, okay, volume change or volume to the power the order of the reaction, okay, this is what the relation of order, the rate and volume we have, okay, this is important one, this helps you sometimes in solving the question, okay, now the another example which is again important that I will give you just now, you see this, okay, so you see the another one if I take the example here and the example is write down this question first, suppose the reaction we have A gives, A gives product P, okay, and it is given that the rate of the reaction, it is given that rate of the reaction becomes half when the concentration of A, when the concentration of A increases by 4 times, when the concentration of A increases by 4 times, determine the order, determine the order with respect to A, this is what the question we have, you have to find out the order with respect to A, 1 by, okay, Laudhitya minus 2, Atmesh minus 1 by 2, anyone else, Suresh, what is the answer? Minus 1 by 2, Mathle, Snigda minus 1, the answer is correct, minus 1 by 2, I think most of you have done so I will just finish this one quickly, suppose the rate expression if I write, suppose the rate expression if I write here, and I assume since ordered with respect to A we have to find out, suppose the rate expression we have R is equals to K, concentration of A to the power N, this N only we have to find out, okay, so this is the first like relation I have written, I have assumed, now you see when the rate becomes half, right, so now this R becomes R by 2, that is equals to K, concentration of A when becomes 4 times, okay, so what we will write, 4 to the power N into concentration of A to the power N, right, now this is equation 1, and this is equation 2, we have to solve these two and get the value of N, so just you divide 1 by 2, you will get the answer, so 2 is equals to, once we take the ratio 1 divided by 2, so 2 is equals to, we will have 1 by 4 to the power N, and when you solve this you will get N is equals to minus 1 by 2, this is the answer we have the order of the reaction, again in terms of concentration of, you know, reactant change in concentration also we have to write, we can write down the formula, and that is the change in rate, change in rate is equals to the change in concentration to the power N, where this N is what, it is the order of the reaction, okay, since the volume it is inversely proportional to volume, so obviously directly proportional to concentration, okay, right, so this is also one of the expression we have into this, okay, in terms of concentration and in terms of volume that we have done already, okay, okay, so I think all of you can do this kind of question, okay, one question or one more question we will do which is based on the three reactant, okay, so one this kind of questions we will discuss, three reactants when it is there, okay, now you see if you have three reactant, okay, the question first of all you write down, the reaction is A plus B plus C gives product, A plus B plus C gives product, okay, on doubling the volume of container, volume of container, volume of container, the rate decreases by eight times, by eight times if A, and when I write down in square bracket it means the concentration of A, so if concentration of A doubles, concentration of B reduces to half, rate becomes, rate becomes four times, this is the one set of information we have, rate becomes four times, another thing is if concentration of C increases by four times, increases by four times, the rate increases increases by 64 times, 64 times, then we have to find out the first thing here, order with respect to A comma B and C, second one, write rate law expression, write rate law expression, tell me the answer in this question, anyone finished, yes what is the answer, anyone, with respect to A it is one, what about B and C, overall the order is three, Atmesh is saying, overall Atmesh minus one, okay, how it is, okay, three, C it is three, okay, now you see the only thing you have to do here, you have to form the equation, right, so first of all suppose we do not know the order of the reaction, so what we will do, we will just write down the rate law expression by assuming the rate with respect to A, B and C, so suppose with respect to A the order is P, with respect to B the order is Q and with respect to C the order is R, right, so overall order is obviously P plus Q plus R, okay, overall order is obviously P plus Q plus R, correct, so now we are using these, you know, condition that is given, now the first condition is what, on doubling the volume of the container, the rate becomes, rate decreases by eight times, okay, so now when you increase the volume of container by twice, right, the volume obviously becomes half and the new rate will be what, 1 by eighth of this, that will write what, that will write R by 8 is equals to K concentration of A to the power P, concentration of B to the power Q, concentration of C to the power R and this whole divided by 2 to the power P plus Q plus R, we will get this only, okay, now if you take the ratio of these two, okay, when we take the ratio of these two, we will get 1 by 8 is equals to 1 by 2 to the power P plus Q plus R, which gives the information about this P plus Q plus 3, sorry, P plus Q plus R will be equals to 3 into this, okay, so this is the first equation we have from the first set of data, P plus Q plus R is equals to, okay, we will see what could be the answer, okay, so first equation is what P plus Q plus R is equals to 3, now second thing you see what, if A doubles and B reduces to half, you see here in the second line they haven't mentioned anything about the reactant C, okay, they haven't mentioned anything about the reactant C, okay, so if they haven't mentioned anything about the third reactant C, so what we consider, we consider the concentration of reactant is constant that is not changing, okay, so what happens here, if I use the second set of condition and it says if A doubles, if concentration of A doubles and B reduces to half, rate becomes 4 times, so we will write what 4R is equals to K, concentration of A doubles, so A to the power P into 2 to the power P, concentration of B half, so B to the power Q divided by 2 to the power Q into concentration of C is constant so we will write this only, again we will take the ratio of this and this which gives you 4, 1 by 4, 1 by 4, 1 by 4 is equals to, we will write 2 to the power Q divided by 2 to the power P, okay, so when you solve this you will get P minus Q is equals to 2, P minus Q is equals to 2, this is the second equation, okay, now the third one is what, you see if concentration of C increases by 4 times, the rate increases 64 times, right, here also you see, they have mentioned only concentration of C which means we have to take A and B constant, right, so we will write what the rate becomes 64R is equals to K concentration of A to the power P, concentration of B to the power Q and C is what increases by 4 times, so 4 to the power R into concentration of C to the power R, okay, so simply when you divide these two we will get 64 is equals to 4 to the power R, right and when you solve this you will get easily the R value is 3, right, R value is 3, so with respect to C the order of the reaction is 3, when I substitute 3 here this gives you the R value is when we substitute 3 here which gives P plus Q, the value of P plus Q is equals to 2, right, with when R is 3 the value of P plus Q is 0 and P is equals to we have 2 plus Q, right, so these two equations we can solve easily, so this and this if you add we will get 2 P is equals to 2 which gives P is equals to 1 and when you solve you will get Q is equals to minus 1, so 3 1 and minus 1, so with respect to A the order of the reaction is 1 with respect to B it is minus 1 with respect to C it is 3, overall order is 3, so most of you have got it right, okay, okay, so this is what we can do when we have 3 or more reactant present, okay, so you do not have to worry about like if that any information about any reactant is not given you have to assume that as constant with it is not changing, right, okay, okay, see all these things we are discussing under the concentration term only, we will see the order of the reaction specific order and then we will see what all relation we get into this, okay, now we have like one important thing here and this information you should have that is what I am giving you, okay, because sometimes in the you know in the exam you should know the order of the reaction basically, okay, you see some reactions I am giving you as in as an example you just write down this and few of this you should keep in mind, okay, because sometimes you may require the information of these reactions, okay, so some examples you write down the first one I have I am writing down here that is 2 N2O5 gives you 4 NO2 plus O2, okay, this reaction is of first order, okay, we will see that also some examples later on but the rate expression for this reaction is R concentration of N2O5 to the power 1, okay, this is the first order reaction and rate law expression is this, okay, must keep in mind, okay, if I write down the second one that is 2O3 gives 3O2, okay, for this the rate law expression is R is equals to K concentration of O3 square concentration of O2 minus 1, minus 1 we have here, okay, so you see in the rate law expression we have product also possible, we have product also present, okay, that is one important thing you should keep in mind, okay, now the second one the last thing next reaction we have third one that is CH3CHO gives CH4 plus CO and the rate expression for this reaction is K concentration of CH3CHO to the power 3 by 2, okay, all these expressions let me tell you are experimental data it based on experiment, right, so you have to keep this in mind sometime it requires while solving the question, very, very important one more example we have 2SO2 plus O2 this reaction takes place in presence of NO catalyst and this gives you 2SO3 the rate expression for this reaction is found to be K concentration of O2 to the power 1 and concentration of NO to the power 2, okay, these are few examples I have given you that you should keep in mind which sometimes help you in the question, okay, this one is very common N2051 first order reaction it is there in the module also you can go through now the important thing here from this four example you can understand you see this O2 is what this O2 is a product, okay, so in the rate law expression it is not necessary that we will have only reactant we can have product also and O is a catalyst over here, right, so this catalyst is also there in rate law expression, so in the rate law expression we can have catalyst also we can have product also and reactant, so we have seen many times that we have reactant, okay, okay, so on the basis of this few important points we have here that is the first one in rate law expression RLE we can have we can have catalyst comma product reactant anything we can have, okay, second point in this reaction intermediate reaction intermediate never appears appears in RLE rate law expression on the basis of this second point there we have questions also in module also we have one question I will show you just now first you write down two three points into this, okay, so these are the few points we have here okay other points are not actually required all of you have center module now yes or no do you have center module okay all of you open your center module I will give you one question which is based on this second point that I have given you, okay, you see this question now page number page number 1107 page number 1107 question number 32 page number 1107 question number 32 tell me the answer what is the answer RIN is saying is two option two or answer is two RIN option two or answer is two okay two anyone else what is the answer anyone else Atmesh, Lahitya, Sayuza, Snigdha, Lalitha, Mathle, Shreyas tell me the answer where is Tapas are you there Tapas Nidheesh are you there Lahitya you also got two okay tell me one more minute I'll give you till then I'll write down the reaction okay now you see how to do this question first of all you see the first reaction is fast okay this is the fast reaction we have the second one is slow and the fourth one is slow this kind of question is very common for your board point board point of view also okay so how do we go how do we approach this kind of question that you see first of all we know in organic chemistry also we have seen that the slowest step is what is the radio demanding step RDS so first of all you have to find out RDS okay what RDS we have okay so RDS we have this so for this reaction only we'll write down the rate expression so rate law expression is what R is equals to K concentration of A into concentration of B2 concentration of A into concentration of B2 right here you don't have to worry about what is the power of this A and B2 will have here because it is given in the question that it is the slowest step so rate of this reaction depends upon what this one and this one only so with respect to that only we'll write down okay because it is the slowest step now you see in this expression what is this A you see this A is nothing but this A is nothing but the intermediate we have reaction intermediate right because if you write down the overall reaction from these three if you add all these three you see this A will get cancelled this B will get cancelled and the overall reaction will be what A2 plus B2 gives to AB okay so when we write down the rate law expression in the rate law expression we can either have A2 or B2 or AB any third or fourth thing we cannot have in the rate law expression because in the rate law expression we can only have reactant or product or any catalyst if you have right these are not AB are not the reactant or product these are intermediate so we have to replace this okay so how do we replace this you see from the first law we can write down K is equals to from the first expression I am writing down K is equals to what we can write you see A into A that is product by reactant okay from law of mass action reaction question sorry reaction is equals to what the concentration of product by concentration of reactant A2 so from this what is the concentration of A that will be root under of K into concentration of A2 this I can substitute here right and that gives you what R is equals to oh this is K so I will write down this as K dash any other constant okay so K into K dash right K into K dash root over of concentration of A2 into concentration of B2 right so you see this becomes what K into root over of K dash into concentration of A2 to the power 1 by 2 concentration of B2 to the power 1 so this is the rate law expression RLE we have okay because you see we have A2 and B2 that should be the overall reaction we have here right so if the question is overall order you have to find out that will be half plus 1 is equals to 3 by 2 with respect to A2 the order is half with respect to B2 the order is 1 yeah so answer is 3 by 2 okay that is option D yeah is it clear so this kind of question is very common for your board point of view also and J also J means J means also they have asked this kind of question in J means okay that also you must keep in mind yeah that is what you have to keep in mind you have to write down the overall reaction and then you will understand that what all reactant or product you can keep into the rate law expression okay or in the rate law expression what all species we should have okay so all those intermediate steps intermediate thing you have to replace okay understood all of you okay now you see the next we are moving towards the next thing here and that we have is kinetics of various order reaction right on the heading kinetics of various order reaction the first one we have that is zero order reaction zero order reaction right on these are the reaction these are the reaction these are the reaction whose rate is not affected by these are the reaction whose rate is not affected by the concentration of reactants these are the reaction whose rate is not affected by the concentration of reactant. Next line, the rate is independent of concentration of reactant. The rate is independent of concentration of reactant. Next line, these are the only reactions, these are the only reactions in which the rate is constant in which the rate is constant and these reactions goes to 100% completion. These reactions goes to 100% completion. Okay, there are few examples of this zero-order reaction that you must keep in mind. Okay, the first example we have in this, the first one is if you have photochemical reaction, photochemical reaction. For example, if I write down H2 plus Cl2 in presence of light H nu gives 2 HCl. This is photochemical reaction because the rate of this reaction depends upon the intensity of light that we are using, intensity of photon not the concentration of H2 and Cl2. This we call it as photochemical reaction. See I have taken H2 and Cl2 here, this is zero-order reaction but the same thing is not true for H2 and I2, write down somewhere, this H2 and I2 gives 2 H I if you write, this is not zero-order reaction. This is second-order reaction, R is equals to KH2 to the power 1 and I2 to the power 1. Okay, this one thing you keep in mind, okay this is second-order reaction. Another one if you write down H2 plus Br2, H2 plus Br2 gives 2 H Br, this is again a complex reaction, 2 H Br and if this rate expression for this one if you write down that will be K concentration of H2, K1 sorry it is, concentration of H2 to the power 1, concentration of Br2 to the power half divided by 1 plus K2, concentration of H Br divided by Br2. See you do not have to worry that much of all these expressions, okay. Just we have to just go through once, okay because sometimes they ask this since it is a zero-order reaction but the same kind of reaction for I2 and Br2 is of second order and here we have some complex reaction. The H Br is very, the formation of H Br from H2 and Br2 is a complex reaction, okay and the rate expression we get something like this, okay. We can derive this also but that is not there in our syllabus so we are just leaving that, okay. So these are the two more thing over here. Now coming back to this, okay. Now coming back to this example thing, photochemical reactions are the example of zero-order, okay. Next we have the example like NH3, okay. The dissociation of NH3, 2 NH3 in case in presence of molybdenum catalyst we have gives N2 plus 3H2. This is also a zero-order reaction. Enzyme catalyzed reaction, enzyme catalyzed. The reaction in which enzyme is a catalyst. Enzyme catalyzed reactions are also zero-order reaction, okay. Okay. Now so these are the few things that you have to keep in mind. All these rate expressions that I have given you this is experimental, experimental. I am repeating this again and again, okay. So don't get confused or don't think about like how do we get this because order is completely an experimental quantity. So we cannot find out this theoretically. These are experimental observations we have and according to that only we have given the rate law expression, okay. Now coming back to the most important point in this all these order of anti-growth reaction we have that is the derivation, okay. So now for that derivation of concentrate consider reaction. See initially in the last class I told you when we were talking about concentration thing and all these things that we are discussing is under concentration only, right. And I told you there that the relation of concentration and time, right, for any reaction is very complex, okay. And I told you this thing also that the concentration is associated with the rate of the reaction in a very complex way, in a very complex manner, okay. And what is that way or manner depends on that depends on the order of the reaction, okay. So for every order of the reaction I didn't take as in, okay. For every order of the reaction we will find out the relation of rate and concentration, okay. Now suppose we are considering a reaction, okay. And the reaction is suppose any reactant A gives product, okay. See this first two, three steps if you are able to understand you can understand the rest of the thing very easily, okay. This is what the key that we are going to do in every kinetics of every reaction, okay. So A gives product and what I assume that at time t is equals to zero the concentration of A is A naught and there is no product so concentration of product is zero. Now at time t is equals to t obviously when the reaction starts A starts converting into B so the concentration of A decreases and at time t suppose the concentration is A t and this is what A naught minus A t, okay. So one thing you must remember here that the A t is what it is the concentration left, concentration left, right. And that we can write A t is equals to A naught minus x and x is what is the concentration reaction, right. So A t is equals to, is equals to we have A naught minus x. So A naught minus A t is equals to what we have. The concentration of product after some time is nothing but x. Is it clear? Is it clear? Tell me. Yeah. So if I write down, if I write down the rate of this reaction, okay, rate of this reaction, rate will be what? Obviously if I write down the rate of the reaction in terms of concentration, what will write? Minus D concentration of A by D t. Yes. And that should be equals to the rate constant K into concentration of A. Is it clear? Did you understand this? All of you, it is very important, okay, because now onwards from here, we will use this particular expression almost everywhere, okay. So if you are able to understand these two, three steps that I have written just now, okay, you can do it on your own, any question if you get. And let me tell you one thing. I am taking a reference of this reaction. It is not always necessary that you will get the same reaction in the exam, okay. So what reaction you are going to get? According to that, you will have to write down this expression here, correct. So it's the best way of this is what you have to understand how to write down this expression, correct. If you try to memorize this, you will end up with the wrong response, okay, because the reaction can be anything. I have taken just a reference, okay. So is it clear till here or anything I am missing over here? Anything I am missing over here? One thing I am missing here, when I write down the rate is equals to minus dA by dt and that should be equals to that should be equals to what K into concentration of A, correct. This is right, but since the order of the reaction is zero, so we should write down the power here and that is zero, clear. And that is how we write down the rate expression, okay. We must have here power and that should be equals to the rate of the reaction, yes or no? So further if I write down since A to the power zero is one, so the rate of the reaction that is minus of D concentration of A by dt is equals to what we write? K, which is a constant, correct. So that is why you see this is what? This is the rate of the reaction, ROR, okay. And since the order is zero, so ROR is equals to K, which is a constant for any reaction at a given temperature, right. And that is why we say that zero order reaction are those reactions whose rate is independent of the concentration, correct. You see minus dA by dt is equals to K, which is a constant, right. Now this expression only we use for the simplification or to find out the concentration and time relation, correct. Now you see here if I take the next slide, right. The expression that we have here is minus of D concentration of A by dt is equals to K and that we can write minus of D concentration of A is equals to K dt, correct. And we can easily integrate this, right. Now and how do we integrate? That depends on you or what concentration you have to find out after what time, right. So this is the integration sign. Now initial constant at time t is equals to zero we are starting, right. So now at time t is equals to zero, the concentration of A is A naught in the given reaction that we have taken, right. And when time t is equals to t, the concentration here suppose we have A t. So when you solve this you will get what? A naught minus A t is equals to K t, right. Or the concentration after time t that will be A naught initial concentration minus K t, initial concentration minus K t. All of you understood this, okay. So in the another way also we can write down like this if you want. The square bracket of A means concentration of A at time t is equals to the initial concentration of A A naught minus K t. You must keep this in mind that this A t, A t are the concentration which is left after time t, right. This is important, concentration which is left after time t, okay. So now you see if I ask you what is the half-life time, half-life time or half-life period that we denote as t half or t t like this, half-life period, okay. So the definition of this you write down, it is a time required, it is a time required for the completion of, it is a time required for the completion of half of the reaction, for the completion of half of the reaction, right. So now if I have to find out t half of this reaction, so can you tell me if I write down this that when t is equals to, t is equals to t half we have, what is the concentration A t is equals to? Is it A naught by 2? Because half of the reaction has been finished, so half reaction is left. So if A naught, so you are getting A by 2k, okay. So these two value you have to substitute here. A t is what? A t is A naught by 2 that is equals to A naught minus k t half and when you find out t half from this expression, you will get A naught by 2k, right, we will get A naught by 2k. So this is the expression of t half for zero order reaction, okay, for zero order reaction. Like this only, we will get the concentration and time relation for t half for first order also, right. We have second order also, we have third order also, but the first two is most important for J point of view and for board also, zero or second order. We, yeah, we did only definition of half life I guess. I told you what is half life and I have given you the definition. We haven't calculated the half life for zero order, okay. Again we will do this for first order also, the expression for half life of first order. Is it clear till here, all of you? Yeah, now next you see some facts I want to tell you for this which helps you in calculation, okay. Now you see the first point here. When t is equals to zero, at time t is equals to zero, what is the concentration we have initially? A naught. At time t is equals to t, what is the concentration we have? A naught minus kt from the expression that we have calculated just now, okay. At time t is equals to 2t. What is the concentration we have now? A naught minus 2kt. At time t is equals to 3t. What is the concentration we have? A naught minus 3kt and so on we can go easily, right. 4t then 4kt like this, correct. So now all these concentration terms are what? The concentration which is left, okay. All these are at, right, concentration which is left. Now you see in equal time interval, right, because the time interval are same. If you try to understand here, the time interval here is t from here to here, from here to here again we have t time interval and again here we have t time interval, right. So at equal time interval, the concentration of the reactant in zero order reaction follows arithmetic progression. You see here, the common difference is what in this? Common difference, cd is kt. What is common difference here? kt. What is common difference here? kt, right. So the reactant of first of zero order reaction follows arithmetic progression. This point you write down. The concentration of the reactant react, the concentration of the reactant react at equal time interval, the concentration of the reactant react in equal time interval follows the pattern of arithmetic progression, follows the pattern of arithmetic progression. Write down next thing. In a given time interval, in a given time interval, equal amount of the reactant reacts, equal amount of the reactant reacts for zero order reaction, for a zero order reaction, okay. What is the meaning? You see, suppose if the reaction starts from 100 concentration we have and from 100 to 90 concentration to change, the time required, suppose we have t minute, right. For the same reaction, whatever the concentration we have initially, right. For the same change 100 to 90 means 10 changes there, like 10 concentration, whatever 10 mole per liter is reacting, right. So whatever the change, whenever the change in concentration is 10, then the time required is t minute only. Meaning of this is what, suppose 100 to 90 takes 10 t minute, then 90 to 80 also if it has to go, it takes t minute only, t minute only. 50 to 40 also, it takes t minute, right. 10 to 0 also, it takes t minute. Why is it so? Because we know the rate is what? The rate is constant. The reaction is going with the constant rate, okay. It has nothing to do with the concentration. Means what? Means what? You see, wherever you start the reaction with 100 initial concentration, 90 initial concentration, 50 or 10, whatever it is. For 10 change in the concentration of reactant, the time required is same because the rate is independent of concentration. Rate is independent of concentration, okay. Now you can understand this particular thing from an example, right. Suppose the time required, the time required is 12 second for 10 percent of a reaction to complete, 10 percent of a reaction to complete, 10 percent of a reaction to complete, right. Then what is the time required? What is the time required for the completion of last 10 percent of the reaction? Assuming the reaction is zero order, okay. Zero order reaction we have. What is the time required for completion of the last 10 percent of the reaction? Is halogenation of alkene a zero order reaction? Can't say lohitia because it depends on what condition we are taking because halogenation takes place in presence of sunlight also and in dark also, right. But halogenation of alkene is not a zero order reaction because it requires intensity of light as well as the concentration of reactant also, right. So all of you are saying 12 second, right. Okay, so answer is correct, right. You can do this mathematically also, right. What we can assume initially like only one calculation you should do into this, okay, so that you can understand this fact that I have given you here, 10 minute, 10 minute clearly. So since 10 percent is there, so initial concentration I assume as 100, right. 10 percent reaction completes means what? 80 is equals to 90 time is 10 second, right. So for zero order what we can write? 80 is equals to a naught minus kT, right. So k is equals to what from this 80 minus, sorry a naught minus 80, a naught minus 80 is 10 divided by T is 12. k is equals to 10 by 12. Now for the other set of data last 10 percent means what? a naught we have, a naught we have 10, right. And 80 we have zero, correct. So what we can write? You see again 80 is equals to a naught, a naught minus kT. k we have already calculated here. k will not change as far as the reaction condition is same, okay. So what is the time required we have here? So T is equals to, we can easily write a naught minus 80 divided by k. So a naught minus 80 is 10 only, k is 10 by 12. You see easily we are getting 12 second, okay. Don't get confused over here like, okay you have taken 10 initial concentration, so 10 percent of 10 is 1. So 80 will be 9 over here, no. Because since we have started the reaction from here, this reaction when it starts it goes from 100 to 90 in 12 second, first 12 second and then 90 to 80 in next 12 second, 80 to 70 like this it is going then 60, then 50, 40, 30 and then 10 and then to 0, right. So for every 10 concentration of the reacting to react, the time required is 12 second only, okay. This is very important, you know, concept or fact we have for zero order reaction. This helps you a lot when you solve the question, okay, must keep in mind, okay. Now the last point for zero order reaction we are going to see is some graphs of it, right. So write down graph of zero order reaction, okay. So let me ask you, what is possible graphs, we have graph rated question is also important for in this kinetics, okay. So the first graph we can draw, right. Suppose this is time, right and this is concentration and concentration of reactant, right. Similarly we can take time and concentration of product, time and concentration of product. Next one, we have t half here and initial concentration A naught, okay and the last one we have that is the rate of the reaction ROR and time, okay.  This is what the relation we have, okay and this forms what? Y is equals to Mx plus C, right with negative slope. So the graph will be a straight line with negative slope, slope of this is minus k, okay. If you have to find out the concentration of product and time, so concentration of product and time we have what? Concentration of product is what? We have written this, no, A t is equals to A naught minus x, right.  So this gives you what? x is the concentration of product, right. So x is equals to k t we have, x is the concentration of product, correct. So x is equals to k t. So concentration of product and time graph will be a straight line passing through origin with the slope is is nothing but k t half and A naught. What is the relation we have? T half is equals to A naught by 2k. This is what the formula, this formula you should must keep in mind, okay and that will be a straight line passing through origin, right. Rate of reaction and time graph, rate of reaction and time graph again will be a straight line, right, parallel to x axis because we know rate of the reaction is independent of time or even concentration, right. So this is the graph we have. Third has the constant slope 1 by 2k, correct, lohitia. Second is positive slope linear from origin, correct lohitia for the first one. Fourth is parallel to x axis, correct. So x axis whether you take time or concentration you will get the same graph, understood all of you, correct, okay. Now you see the second one that is first order reaction, first order reaction. We will do everything as we have done in the zero order, okay. So first of all we will assume the reaction, okay and suppose the reaction is again A Gibbs product, A Gibbs product. So if I write down the rate of this reaction, rate of reaction is equals to minus D concentration of A by DT is equals to k into concentration of A to the power 1 because the order is 1, right. Now we can again integrate this and we can get the expression, right. But before integrating this you try to understand one thing over here. In zero order reaction there is no A present on the right hand side, correct. So if you see this zero order reaction here we do not have any A, correct. So rate of the reaction is constant then, right. And the rate is going with a constant rate. That is why we have got the graph like this straight line graph parallel to x axis. Now with time what happens since the power 1 over here to the concentration of A, the concentration of A decreases with time, correct. And that is why the rate is also decreasing with time, right. Rate is also decreasing with time. So first of all as the reaction proceeds the rate continuously decreases, right. The rate continuously decreases and if you remember for reaction lifetime I have given you the definition that it is a time required for 98% of the completion of reaction, right. Why not 100% because when the reaction completes about 98% completion is there then the rate becomes extremely slow, correct. When the rate becomes extremely slow then we assume or we consider that okay let it be now the reaction is finished we cannot wait for that long for the reaction to finish 100%. Practically there is no point for the you know calculation of reaction beyond this 98% of completion, right. That is why we usually take reaction lifetime is what it is a time required for 98% of the completion of reaction, okay. Whatever I told you just now regarding this 98% of the completion of reaction is not valid for zero order reaction. Can you tell me why? Why it is not valid for zero order reaction? Why it is not valid for zero order reaction? Because there what happens their rate constant is the rate is constant, yeah correct, correct, correct. Because rate of the reaction is constantly respective of the concentration, yeah right. So one very important point we have here that zero order reactions are the only reactions which go towards 100% completion. This question they have asked once in need exam, okay. They have directly asked this question which of these reactions goes towards 100% completion the option was like zero order first order second order like that, okay. So first of all this thing you must keep in mind since the rate of zero order reaction is constant. So the reaction goes towards the 100% completion, right. So if they ask you directly by saying zero order first order you can easily do that. But sometime what happens they'll write down the reaction itself, right. They'll give you H2 plus Cl2 gives to HCl, okay. They'll give you H2 plus Br2 gives to HBr. Like this they'll give you four different reactions and they'll ask you which one of these reactions goes 100% completion, right. So you should know there that what reactions are of zero order or of first order, correct. So few examples I have given you for zero order reaction. Similarly before solving this integration which I am sure that you can do this easily just write down some examples of first order reaction which are important for you to know. Correct. So the first reaction you write down the reaction the decomposition of H2O2 decomposition of H2O2 gives H2O plus half O2, right. Second reaction you see which I have given you already decomposition of N2O5 to N2O4 plus O2. Third one, third one what all problems we have questions related to radioactivity related to radioactivity. That's why you must have done this in physics also N is equals to N0 e to the power something you get, okay. The similar kind of expression we'll get here in zero first order reaction that we'll see after this, okay. Growth and decay of bacteria growth or decay of bacteria. Fifth one inversion of cane sugar, inversion of cane sugar, right inversion of cane sugar example or this you write down hydrolysis of sucrose if you do C12H22O11 plus H2O when you do this hydrolysis or acidic hydrolysis when you do you get C6H12O6 plus C6H12O6. This is we have glucose and this one we have fructose, okay. Glucose and fructose. Glucose is dextro in nature, okay. It is dextrorotatory. This one is levorotatory and sucrose. This is sucrose. It is also dextrorotatory, right. So when you do the acidic hydrolysis of this you will get glucose and fructose but this fructose is present more in amount over here. This is the major product we get here. That's why conversion from dextro to levo we say and this is the what we say. This is the inversion of cane sugar we have, okay. Acidic hydrolysis of ester also follows first order reaction. Acidic hydrolysis of ester. Hydrolysis of ester follows first order reaction, okay. Now coming back to this derivation part here. So from this relation what we can write minus of D concentration of A by concentration of A is equals to K dt. Now we can integrate this. When time t is equals to 0 the concentration is A naught. When time t is equals to t the concentration is A t, right. D A by A is ln A A naught to A t negative is equals to K t and then when I solve this we will get K is equals to or K t is equals to we will write ln A naught by A t, right. So A naught by A t is equals to we get e to the power K t and A t is equals to A naught e to the power minus K t. A t is equals to A naught e to the power minus K t. Can I doubt? Okay. Now the another way to write down one more expression we will use here that we will write down here only. Here you see we can write down this K t is equals to 2.303 log of A naught by A t and then K is equals to 2.303 by t log of A naught by A t is A naught minus X. This is also the expression we have, okay. A t you must keep in mind this is the concentration that is left when X amount of reactant has been reacted, okay. When X amount of reactant has been reacted, right. So A t is always the concentration that is left, okay. I am repeating this again and again because I have seen many times strands have doubt over here. When you are going to solve the question you will get confused that whether you have to substitute here the X which has been reacted, the concentration which is left. So A t is always the concentration which is left, okay. Now one thing you see here which is again important before going into the graph of this, okay. Since I have this expression that is, okay. At time t is equals to 0, the concentration is A naught. At time t is equals to t, the concentration is A naught e to the power minus K t. At time t is equals to 2 t, concentration is A naught e to the power minus 2 K t. t is equals to 3 t, the concentration is A naught e to the power minus 3 K t. So and so on we can write down this, okay. So now you see the ratio of these two is what? Equal time interval we have here you see. Equal time interval of t only we are taking, right. So we are at equal time interval the amount as we react is this, right. This is t, this is also t and this is also t we have, right. Can you tell me what we are getting over this? See if you take the ratio of this that will be e to the power K t or this by this if you take then it will be e to the power minus K t. The ratio of these two gives you again e to the power K t. The ratio of these to the last two we have again gives you e to the power K t, okay. This means we have a common ratio. At equal time interval we have a common ratio. It means the amount that has been reacted in a first order reaction follows a pattern of geometric progression because we have a common ratio into this, okay. Yeah, it is a GP, okay. That is again very important fact we have here, okay and that you write down, okay. The second important point we have here for a first order reaction for a first order reaction equal percent is reacts a given time interval in a given time interval, right. Always keep this in mind for a zero order reaction we have equal amount react, okay and there what I have written that if 10 seconds is the time required for the change in concentration from 90 to 100 to 90 is 10 seconds then whatever the initial concentration you take 35 to 25 also if you have to go the time required is 10 seconds only that is the meaning we have done over there. But here we have here equal amount does not react but equal percentage react. What is the meaning of this? Suppose if 100 to 90 if you have to go 100 to 90 and the time required suppose we have 10 seconds here. So for the 10 percent of the reactant to complete 10 percent reaction the time required is 10 seconds that is what the meaning we have, right. Then what we say from 90 to 10 percent of it that is what 90 10 percent of 90 is what 9. So what is the concentration left here 81 second sorry 81. So for this change 90 to 80 again the time required is what 10 seconds only, right. If you have to go 30 to 30 to 27 again you see 10 percent reactant we have, okay. So again the time required for this will be 10 second, okay. If I write down 5 to 22.5, okay 10 percent conversion we have again the time required is 10 second. So here we are talking about what the percentage of reaction, okay. But here we are talking about the amount of reactant reaction is taking place irrespective of the initial concentration we are taking because obviously the rate is independent of initial concentration but here the rate depends, okay. So as you see as the initial concentration is more, right more amount also react. Here how many what is the amount react 10, right. But 10 becomes 90 the amount reacted is what 9 becomes 30 the amount reacted is 3. You see as the initial concentration is decreasing the amount that reacts in a given time interval that will also less, okay. Did you understand this particular point, okay. So here we have equal percentage but the amount here we are talking about percentage but in zero order we will talk about amount into this, okay. Now can you tell me what is the expression for t half, t half for this reaction, expression for t half. So when t is equals to t half, right. So a naught is equals to what a naught by 2 and with the value of the formula that we have t is equals to we can write what k, sorry, t is equals to 1 by k ln a naught by a t, right. So a t is equals to a naught by 2 we have here. At t is equals to t half, a t is equals to a naught by 2. So when t becomes t half on the left hand side k will be as it is but a t will be what a naught by 2 so it is ln 2. So t half is equals to what we can write ln 2 the value is 0.693 by k. So t half for first order reaction is independent of initial concentration of initial concentration. Now we will see the last thing in this that is some graph we have and you draw this the first graph we are going to see is log of log of a t and time. The second graph we have, second graph we have concentration of reactant and time. Third one we have rate of reaction and concentration of reactant, okay. So this graph if you want to solve we have an expression like k is equals to 2.303 by t log of a naught by a t, okay. So when you solve this expression that we get here log of a t on the left hand side is equals to minus k by 2.303 into t plus log of a naught we get this, okay. So in this if you have, if you draw the graph the graph will be a straight line like this with the slope here that will be minus k by 2.303, correct. Concentration of r and time. We have the relation what a t is equals to a naught e to the power minus k t. This is what the relation we have. The concentration initially it is a naught and with time it is decreasing and exponential decrease we have so it goes like this but it will never touch this x axis because the concentration of reactant never becomes zero because the reaction never gets 100% complete that you must keep in mind, okay. Rate of the reaction and concentration rate is what minus d concentration of a by dt that will be equals to what k into concentration of a. So rate and concentration graph will be what? Will be a straight line passing through origin, right when with slope is is it clear? Who all are there still watching? Okay, loyta is there, stress is there, lolita is there, rn, where is akmesh, where is sayuja, snigda, tapas, stress, where is apurva? Apurva is not there. Apurva I think he went to New Zealand or what? Okay, okay, watch is in the night, right. Okay, fine. Okay, see since we have discussed zero order and first order reaction, okay. So most of the questions now you can solve, okay. You just try this one, okay. Question number 7 and 8 you try, page number 1105, 7 and 8. Question number 7 and 8 you tell me. The seventh one is what? The plot of concentration of reactant versus time for reaction is a straight line with a negative slope. See whenever concentration and time graph is a straight line which is only true for a zero order reaction, right because we know the concentration is the concentration of the reactant is independent of time over there, okay. The first seventh one is the plot of concentration of the reactant versus time for a reaction is a straight line with a negative slope. The reaction follows a rate equation of zero order, first order, second order and third order, okay. So, okay, I will give you this question number 8, okay. Why you do not have this book Atmesh? You are not at home. Now you see question number 8. Those who have the booklet you can refer, okay. I will just give you the data over here. Question number 8. A to B, the reactant given and it is first order n is equals to 1, okay. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour, okay. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. And what is the time taken for the conversion of? You have to find out the time taken for the conversion of 0.9 mole of A to produce 0.675 mole of B. Option is given, 2 hour, 1 hour, 0.5 option D, 0.25 hour. How it is 1 hour? How did you do this? Lahitya, okay. It is same fraction, okay. All of you have got the right answer. But my purpose was not this, like I do not want to, I want you to solve this question, but one of the purpose of this question is this only also. You see, if there are two possible way to find out this question, okay. One is what? You know the, since the first order is given, you know the relation of first order, that is K is equals to 1 by T ln of A0 by A T. This you know already. First set of data you see, this is what? We have A0, right. 0.6 mole of B is forming.  So A0 you have, A T you have and T also it is given, 1 hour. You substitute here, you will get the value of K from this. K you will have some value, right. Now, this K we can use for the another set of data. For another set of data you see, again this A0 is given, right. To produce this mole of B means this minus this is nothing but A T and T you have to find out, K value you can find out from here, right. Yeah, yeah, correct. But that is not actually you have to do here because it requires a lot of calculation. You will get answer from here also, but it requires a lot of calculation. To make it easy what you can do, initially, suppose you have 0.8 mole, right. So 0.8 into 100 if I do, so you will get suppose A T mole to reduce the calculation. And finally, we have 0.6 mole. So this gives you 60 mole of it, right. This is A and this is B. So what is the percentage of A that reacts? That will be A T minus 60 it is 20 divided by A T into 100 that is 25 percent react, right. Why I am calculating this percentage? Because we know for first order reaction equal percentage reacts, right, in a given time interval. And since the time is 1 hour, right and one of the option is 1 hour also here, right. When you do more practice you can visualize you can see all these things in the question. Now you see if 0.92.65, if the percentage is 25 only, it means the time is same because equal percentage reacting equal time. So 0.9 mole means 90 and 0.65, 675 means what? 67.5. So what is the difference we have over here? What is the difference here? 90 minus this gives you what? 0.520, so it is 22.5 divided by 90 into 100, right. When you solve this, this probably gives you 25 percent I guess. See it is 25 percent means from 0.9 to 0.675 conversion this is also 25 percent, right. And 0.8 to 0.6 this is also 25 percent. So if 0.8 to 0.6 requires 1 hour that is why 0.9 to 0.675 also requires 1 hour answer is B option B, correct. So I will give you some homework from this module exercise 2, okay. You try this at home we will discuss this in the next class if you have any doubt, okay. So exercise 2 section A from question number 1, question number 14 you let it be, you do not do that. From question number 1 to question number 35 till 35. All these questions you can solve from question number 1 to question number 35, okay. All these questions you solve at home, right. Now we will move on, okay, question number 1 to question number 35, okay. Now you see the next reaction we have that is second order, okay. In this suppose if I write the reaction and the reaction is again the same thing we are using A gives product at time t is equals to 0 it is A naught and this is 0. So at time t is equals to t it becomes A naught minus x and this is x suppose or A t and this. So what is the rate expression for this? Rate is equals to again we will write minus d concentration of A by dt is equals to k into concentration of A square, right. And when you solve this I will write down the direct expression that we get. Simply you can take this A square here and k t that side then we can integrate this, okay. And you will get k is equals to 1 by t open bracket 1 by A t minus 1 by A naught 1 by A t minus 1 by A naught. This is the expression we have. Further if you want to solve you can solve this also because A t we can again substitute as A naught minus x, correct. So A t is equals to A naught minus x like that if you solve you will get k is equals to 1 by t open bracket A naught minus A naught plus x by A naught into A naught minus x close this and then k is equals to 1 by t x by A naught into A naught minus x. This is also the expression of rate constant k we have, right. What is the expression of t half? Again at t is equals to at t is equals to t half, A t is equals to what we have A naught by 2. These two will substitute over here, k is equals to 1 by t half 1 by t half here from here we are going through, right. From here A t is 2 minus 1 by A naught and that will be k t half is equals to 1 by k into A naught. Means this t half is what it is inversely proportional to A naught, inversely proportional to initial concentration. Now we will see some graph into this rate and concentration graph. How do we draw rate and concentration graph? This is rate and this is concentration, right, k A square. The graph will be like this, right. R is equals to, rate R is equals to k concentration of A square, right. If I take the graph as rate and, rate and concentration square, concentration square that will be a straight line like this, okay. Where the slope is, okay. Copy down this then we will see the other graph also. Are you tired? You want break also what? Are you saying yes, lawyer is saying no. Yeah, actually it does not matter because already 650. Okay, did you understand till here? Okay, fine. We will continue till 730 no issue. Okay, I will just write down the other graph also, okay. So, now you see if I take the next one, concentration of reactant and time graph. The y axis we have, concentration of reactant and time graph, okay. So, you see this concentration of reactant is decreasing with time, okay. So, it decreases like this. It is non-exponential graph you must keep in mind but it is not linear that is obviously true and it is not exponential also, right. So, it is non-exponential graph, concentration of reactant and time, okay. Now, the next graph we have that is, that is suppose we have and 1 by initial concentration a naught that will be a straight line passing through origin, okay. Another graph we have 1 by concentration of reactant and time graph. This graph will be like this where this point is nothing but 1 by a naught simply the formula that we have, okay. So, these are the few graph we have for second order reaction that you must keep in mind. Now, you see if I write down two, three term that we have done just now. For n is equals to 0, the expression for t half is equals to a naught by 2k. For n is equals to 1, the expression for t half is 0.693 by k. For n is equals to 2, the expression for t half is 1 by a naught k, right. 1 by a naught into k, right. So, you see if I take this term from this for nth order, if I write down the expression for t half, for nth order it is directly proportional to the initial concentration a naught to the power 1 minus n, 1 minus n we have here, okay. Initial concentration n minus 1, okay. Because n is equals to 0, we get this expression true. n is equals to 1, this expression true. n is equals to 2, this expression true, right. So, general relation of t half and a naught is this a naught to the power 1 minus n, okay. Which we can also derive from this expression you see. For nth order, for nth order, we can write down the rate of the reaction dA by dt is equals to k concentration of A to the power n, okay. Now, when you again integrate this the same way that we have done, we get the expression as, I will just directly write down the expression here. Concentration of A at any point of time t to the power 1 minus n, sorry, I will write down like this, at to the power 1 minus n minus a naught to the power 1 minus n is equals to n minus 1 into kT. This is the expression we have for nth order reaction, right.  Now, in this one thing is important here, sometimes they will give you this expression in general form. So, whatever the coefficient of n kT here, that gives you n minus 1, okay. So, from that coefficient also you can find out the order of the reaction, okay. That is the one point we have here, but when you write this expression, when you write this expression, since when you solve this minus de-concentration of a by a to the power n is equals to kDT, right. And when you solve this what you will write, a to the power minus n plus 1 by minus n plus 1. So, this n here the n we have should not equals to 1, that is the condition we have with the formula of integration. This formula have been derived from an equation containing, no you see, so what happens is we are considering nth order reaction, right. The point here is what, suppose if you have 3 reactant A plus B plus C, we are not concerned with the number of reactant present here. If the number of reactant is 3, then also the overall order of the reaction can be 1, however you see that. Suppose with respect to A, the order of the reaction is 0, with respect to B the order of the reaction is 2, with respect to C the order of the reaction is minus 1. Overall order is what, plus 1 only, right. So, the thing is the number of reactant which is present in the reaction that has nothing to do with this particular expression. What we are assuming till now you see that the reaction, the expression for 0 order reaction, the expression for first order reaction, for second order and then for nth order reaction, okay. So, overall order we are talking about, it is not true that we have, we have what, only one reactant present. Your question is right because if you have more than one reactant, we have to write down here in terms, suppose in this case, if you take this expression, what you have to write down the rate here? K concentration of A to the power 0 B square and C minus 1, right. And then we have to solve. So, whenever you have this kind of expression, the calculation becomes extremely difficult, okay. So, first thing that you are not going to get those expressions, okay, in the exam, okay. But in the coming classes, we will see how to solve this kind because when we have 0 order reaction, right, 0 order reactions, there are many possibilities that we will discuss later on. If you have suppose A plus B 2 reactant and suppose I am talking about second order reaction. So, what all the possibilities we have? Either with respect to A, the order of the reaction is 0, with respect to B, the order of the reaction is 2, this is one possibility. Another one is what? With respect to A, the order of the reaction is 1, with respect to B, the order of the reaction is 1, this is one possibility. With respect to A, the order of the reaction is 2, with respect to B, the order of the reaction is 0, another possibility is with respect to A, it is minus 1, with respect to B, it is 3. So, like this, we have many possibilities possible, right. But again, if you are having this kind of more than 2, 3, 4 reactant, if you have, then the complex, then the calculation becomes extremely difficult. Finally, you have to do this only, write down this expression, equate this by D A by D T and then you have to solve the integration there, okay. So, you are not going to have those kind of questions, okay, where the calculation is extremely difficult. But as far as this thing is concerned, we will discuss this calculation, okay, like 1, 0 for 2 reactant, we will discuss later on, okay. But you do not worry with 2, 3, 4, 5 reactant if it is there, you will not get those kind of reactions. Basic thing is what? We will just write down this here and then we will solve the integration, right. So, everything here you have to convert in terms of X or A and then you have to integrate that. The only thing is what? You have to use some integration term over there, integration techniques you have to use, technique you have to use to solve those, okay. Okay, just copy down this, we will do this other one. Understood this one? With 2 reactions we will discuss, okay, more complex thing you are not going to get, okay. So, just you try to understand because sometimes for nth order reaction, you will get this kind of expression. So, only thing important here is what? That the coefficient of KT gives you n minus 1 and from that you can find out the order. This expression is not valid for n is equals to 1 because then this formula of like the integration formula is not applicable over there, okay. This is not applicable for 1. This expression is applicable for all value of n, okay. For all value of n, okay. Now you see, we will discuss like when more than one reactant is there, then how do we solve those questions? And that is if more than one reactant is there, that is generally possible in one order also it is possible, but generally we will discuss in case of second order reaction. Right down the heading, second order reaction we have, order reaction with more than one reactant, with more than one reactant, okay. Now you see in this we have two possibilities. Suppose if I take two reactant, A plus B gives product, product, okay. The thing is what? We can take two possibilities here that since we have two reactant, so we can have either both reactant with same initial concentration, A naught, A naught, or we can have different initial concentration, A naught and B naught. Both possibilities we have, okay. So basically we have in this two case possible, case one you write down, concentration of both reactant are same, both reactant are same and order, order is one with respect to A and B, correct. So now you see if the reaction is this, A plus B gives you product, okay, where the initial concentration of A is A naught, B is also A naught and we are getting some product, okay. So now you see in this, as you see the derivation is not so, only integration and mathematics. So rate what we can write in terms of A, it will be minus D concentration of A by DT and in terms of B if I write minus D concentration of B by DT and that will be equals to what? K into concentration of A into concentration of B to the power 1 and 1 since we have assumed like this. You can also ask me, so why not 0 with respect to this and 2 with respect to this, okay. That is also possible, 0 with respect to one other reactant and 2 with respect to the other reactant, but then the derivation will be same like we have done just now. Here we will have 0, here we will have 2 and then we will take this expression, we will solve that. Are you getting this? Yes or no? Yes. So you see this, if I take this as one thing I missed over here, after time T the concentration of A becomes what? A naught minus x and concentration of B becomes again A naught minus x. One thing you must keep in mind, why it is x and x because here we have 1, 1 mole. If you have 1 and 2 then A naught minus x, A naught minus 2x will have over here, okay that is the change. What happens then that also we will discuss, okay, why not you will get that, okay. So if I find out the rate of reaction with respect to A and B, this expression gives you what? It will be equals to we can write minus D by DT of whether we have A or B, we can write down the expression as A naught minus x final minus A naught the initial we have which gives you DX by DT is the rate of the reaction, ROR, okay. So when I take this DX by DT is equals to k, A value is what? A naught minus x.  DX by A naught minus x square is equals to k DT. Now you can integrate this and you will get the answer 0 to T when T is 0, what is the value of X? 0. When T is T, what is the value of X? It is X only, right. Now when you integrate this, you will get the same expression that we have just now we have done, okay. What we will get here?  One more minus sign will be here because it is minus and here we have X and that will be equals to k T. So when you solve this, this minus minus becomes plus and here we I forgot to write down the limit that is 0 to X and when you solve this, you will get k is equals to 1 by T into 1 by A naught minus x minus 1 by A naught. Like this you will get something which is the same expression that we have done just now. How many of you understood this? Understood? Now you see the another possibility, the another case is what? Case 2 where both reactant has different concentration. You write down in detail this thing, right? When both reactant has different concentration. So if you have A plus B gives some product here, initial concentration, suppose this A has A and this is B and this we do not require A minus X and B minus X after some time T. Order with respect to each reactant is 1 only. There is no change into that, okay. Order with respect to each reactant is 1 only. So if I write down again in the similar way, the rate of the reaction, similar way if I write down the rate of the reaction that will be what? Minus D concentration of A by DT is equals to minus D concentration of B by DT is equals to DX by DT is equals to K concentration of A to the power 1, concentration of D to the power 1. And then we solve this since we have two reactant. So we generally take this DX by DT term which is common like here we are doing, right? So we take DX by DT is equals to K, this is A minus X into B minus X. So now the integration becomes DX by A minus X into B minus X is equals to K DT, right? And we have to integrate this and then we will get the expression is 0 to T and this is 0 to X, correct? Now you see, click down. So you see, okay, your doubt is there, no? So you see, now you see. See, we have two reactants here with different concentration, okay? Then the integration becomes a bit difficult now here, correct? Are you there, so you see, correct? So here you see, if we have two reactant with different concentration, initial concentration, the integration becomes a bit difficult. That's why we are, we cannot, obviously your question was right, we cannot generalize, but most of the time you will get that particular expression only because this is not mathematics, okay? You are not going to show your integration skill into the exam, okay? So that's why we don't get these kind of, you know, questions in the exam. But since we are discussing this one, so when you solve this, you have done this integration thing in your maths, so you can use your integration skill to solve this integration term, okay? But what we are going to do into this to solve this A minus X into B minus X and that we can write as 1 by B minus A, open bracket 1 by A minus X minus 1 by B minus X, close this. So just you substitute in terms of this and this and then you integrate separately, okay? You will get the expression, okay? So you can do this, you can do this on your own, okay? I am just giving you the final expression, right? And the final expression will be K is equals to 1 by T or T is equals to 1 by K into B minus A, open bracket ln A into B minus X divided by B into A minus X plus 1. This is the expression you have when we have two reactant with different initial concentration, correct? With a first order equation. Is it clear? You want, you can do this derivation on your own, okay? I did not do this here because I did, I do not have a space here, otherwise I can do this. The only thing is you can substitute 1 by this in terms of this. B minus A, you take this side, that's why we are getting B minus A finally. And then you write down it separately, DX by A minus X integration plus integration of DX by B minus X and right hand side we have KT only, okay? So you can do this, after this substitution, the integration becomes easy, you can solve this, okay? So that is how we solve these kind of questions, okay? So now you see, we have done like 50% of the chapter. I have done this chapter in so detail, okay? We have done every possible derivation. We have done many questions into it because it is important. You will definitely get one questions from this chapter, okay? So you must solve all the questions. Some questions I am going to give you now, okay? We do not have, no, that much time now so that we can start any new thing. But one to small questions we have, which is important as far as the concept is concerned, that we will see and we will see then volume related and pressure related thing, okay? So write down this question first, okay? So the question is N2O5 gives 4 NO2 plus O2 and expression here it is given minus of D concentration of N2O5 by DT is equals to K1 concentration of N2O5 minus of, sorry, plus of D concentration of NO2 by DT is equals to K2 concentration of N2O5 and the last one plus of D concentration of O2 by DT is equals to K3 concentration of N2O5, okay? These are the three data given. You have to establish the relation between K1 and K2 and K3, that is the question, K1, K2 and K3, okay? Now in this, this K1 here, K1 is the rate constant with respect to N2O5, K2 is the rate constant with respect to NO2, K3 is the rate constant with respect to O2, that is the meaning, okay? You have to establish the relation between the rate constant of all these. K2 is equals to 2K1, what about K3? Anyone else? Loitia, what is the answer? Arian, Lalitha, Snigda, Sayuza, what is the answer? Shreyas, what is the answer? 2K3 is equals to K1, K is equals to 2K2 is equals to K1, what did you write? What is that 2.5, Arian? K3 is equals to 0.5K1, yeah, yeah, yeah, correct stress, correct Atmesh also I think, okay, yeah, yeah, correct, correct. So, you see basically for the reaction, if I have to establish the relation between the rate constant of reactant and product, each reactant and product then what we have to do, right? So, if I write down the rate of this reaction ROR is equals to what we can write? 1 by 2 minus of dN2O5 by dt is equals to 1 by 4 plus d concentration of NO2 by dt is equals to plus d concentration of O2 by dt, all these are equals to K, right, all these are equals to KN2O5 to the power 1 where K is the reaction rate constant, okay. And this suppose if I assume as x, rate of the reaction is x suppose or suppose x will not use, we use R only, okay, because rate is R we are using, yeah, yeah, right, so use 2K1 is equals to K2, correct. Now you see minus dN2O5 by dt is equals to this. So, we can write down minus of dN2O5 by dt is K1 into N2O5. So, we can write down from this is what? K1 into concentration of N2O5 is equals to 2R. Similarly, from this we can write K2 into concentration of N2O5 is equals to 4R, K3 into concentration of N2O5 is equals to R, okay. So, if I take the ratio of first 2, right, what we will get? K1 by K2 is equals to 1 by 2 and the last 2 will get K2 by K3 is equals to 4, right. So, from here we can write, from here we can write K1 is equals to K2 by 2 and from here is equals to we can write K2 by 4 is equals to K3. So, if you equate this, we can write from these 2 that K1 divided by 2 is equals to K2 divided by 4 is equals to K3, right and that will be equals to, if I equate R is equals to KN2O5 that will be equals to K where K is the rate constant of the reaction, okay. Now you see this like one thing that you can keep in mind, you can take as a general result that we have here, okay. K1 is what? K1 is the rate constant of N2O5 and its stoichiometric coefficient is 2 and that is why we are getting 2 over here. K2 is the rate constant of NO2, its stoichiometric coefficient and we are getting 4 over here. K3 is the rate constant of O2, 1 stoichiometric coefficient we are getting 1 over here, right. With this, if I generalize this thing in general, if the reaction is AA plus BB gives CC plus DD, right. For this, if I write down the rate constant of the reaction is K, so that will be equals to the rate constant of reactant A divided by its stoichiometric coefficient is equals to rate constant of the reactant B divided by its stoichiometric coefficient is equals to rate constant of the reactant product C divided by its stoichiometric coefficient, rate constant of the product D divided by, this is what we can write down at any point of time and we can find out the relation if it is required. Is it clear? Yes. Understood? Okay. So, we will stop until here only. We do not have time. We cannot start any new thing. Okay. So, we will start from here in the next class. Okay. This chapter, I know it's a quite big chapter. If you try to understand for JEE mains and advanced lever, see we are discussing every possible thing into it. Okay. So, if you understand this properly, you do not require this to do this again. Okay. Only few questions for practice purpose you must do. Okay. So, 1 to 35 you must solve. Okay. Question number. All those questions which are related to order and graphs. Okay. All those questions you can do now. We have done every possible thing into this. Only few questions like you see in the module question number 36, it is sequential reaction. That we haven't done. Question related to activation energy, we haven't done. Okay. That we will do in the next chapter. Okay. Sorry. Next class. Okay. All these graphs related questions are easy, but we will discuss this in the next class. Okay. So, this is it for today. Okay. You must solve these questions. Okay. Otherwise, you won't understand the concept that we have discussed. Okay. And try to do it in a, you know, like the trick that I have given you AP, GP thing, try to use all those concepts while you solve the question. Okay. That helps you in memorizing those things. Okay. Anyway, so we will start from here in the next class. We solve those questions. If you have doubt, you can ask me anytime. Okay. We will start from here in the next class. Okay. Thank you all. Thank you for joining. We will see you next class. Okay.