 Yes. Hello. Good evening, guys. Can you hear me? Good evening, sir. Yes. So last class, I think we discussed till first order reaction. Yes. Yeah. So first order we have discussed graph also we have seen. Next, we are going to start with the second order reaction. So write down second order reaction. So I'm assuming any reaction, for example, or what we can say in the second order reaction, the rate of the reaction depends upon two concentration term, or we can also say a square of the any one concentration term, right? Anything we can say like that. Okay. A few reactions or few examples of second order reaction we have. The first one is suponification reaction. This is the reaction which gives salt off an ester. Salt off. When ester reacts with a base, for example, if I write down one second, so if I write down a reaction, CH3 C double bond O O C2 H5 ester reacts with a base. It converts into CS3 C double bond O O and A and C2 H5 OH. This reaction is a second order reaction. Another example we have two NO2 gives two NO plus O2 O3 gives O2. Second order reaction. Okay. With respect to this. Now I am assuming a reaction here and the reaction is a reactant A gives product. Second order reaction we are assuming. So the rate expression would be what? The rate expression for this equals to minus of D concentration of A by DT is equals to the rate constant K into concentration of A square since we have assumed that the reaction is of second order. Now one more thing you try to understand here. I am writing down this expression with respect to this reaction. Okay. If you have reactions like this, for example, if the reaction is given in the question to A gives product, then we'll write the expression according to this reaction. Correct. So don't try to mug up this relation. Try to understand how to write down the rate expression. Yeah, it's first order reaction to O3 plus O2 if you give. I'm just, I have just written over here with respect to O3. It is second order. Overall order is that's what I said. I have it. I let down here with respect to O3. Overall order is one to A gives product. So I'm just trying to, to make you understand that this expression, you don't mug up. Okay. Expression will write down with respect to the reaction given. If you take this reaction, if this is given in the question, then here the expression of rate would be R is equals to half of minus of D concentration of A by DT is equals to K times A is square. Then minus of D concentration of A by DT is equals to 2K concentration of A is square. And then we can solve this by integration. Correct. Must take care of this. Okay. According to the reaction will write down the expression. Fine. So we'll continue with this. So we have the expression which is minus D A by DT minus of D concentration of A by DT K concentration of A is square. So when you integrate this D concentration of A by A is square. It becomes K DT. A naught to 80. A naught is the concentration when time is T initial concentration. When time is T, this becomes 80. Now you need to integrate this and solve the expression you will get K is equals to 1 by T open bracket 1 by concentration of 80 minus 1 by concentration of A naught. This is the expression of K we'll get. So could you go to the previous slide? Thank you, sir. Yeah, Aditya, we can have it tomorrow. Okay. You can call me at the test seat. Sir, tomorrow is okay. Fine, sir. Okay. Cause I have to go to the airport. Okay. Like I have to read. Okay. So I cannot wait. Okay, sir. No problem, sir. You are, you are talking about this thing. You have some doubt regarding the test tomorrow. That's why. No, no, no, not urgent, sir. Like. Tomorrow last. Yeah. Today I cannot, you know, wait because I have to go to the airport. That's the thing. Anyways, so this is the expression we get. Correct. Now, could you tell me what is T half here? At T is equals to T half. The expression we know it becomes half again. So concentration 80 at T half would be a naught by two. We substitute this and we'll find out the expression for T half, like we did in other reactions. So here T half would be one by K. So here more concentration, we take initial concentration. Lesser will be the value of half life. Okay. Because obviously the rate is, if you compare this with, you know, the first order reaction, the date is directly proportional to the concentration. Here rate is directly proportional to the square of the concentration. So obviously the reaction is proceeding with a faster rate. Correct. So half life should be less than that of first order. That's why you see inversely proportional. More value of AU to get lesser will be the half life. Okay. So T half here for second order reaction is inversely proportional to the initial concentration. Look at this graph here. Okay. So the first graph we are looking at rate and concentration. Rate and concentration. We know the relation. The graph goes like this. Because we know R is equals to K times a square. This is the graph we have. Rate and concentration is squared. The graph will be a straight line passing through origin. Relation is this only we have this graph. This graph is for rate and concentration is squared. So we have this. This one will take Y axis as 1 by concentration of reactant. Okay. AT. And this side we have time. So this graph would be if you draw it will also be a straight line like this. This L value would be 1 by A naught. What is the graph of T half and 1 by initial concentration? Y axis is T half. And this is 1 by A naught initial concentration. This would be again a straight line passing through origin. And the slope of this line is what? 1 by K. 1 by K. Okay. So these are the graphs we have for this. All these three order. Okay. We have discussed zero first and second. Now we'll see that expression for any Nth order reaction. Okay. The last graph T half. T half is same, you know the relation of T half and concentration you see. We have done this relation last slide you check. T half is equals to 1 by K into A naught. Correct. T half and 1 by A naught is what? Y is equals to MX. Slope is 1 by K. Next slide down. Kinetics of Nth order reaction. Nth order reaction. Again the same reaction I'm assuming. A gives product. A gives product. Nth order reaction we have. So rate is equals to. Rate is equals to minus D concentration of A by DT. K times concentration of A to the power N. Similarly we can solve this. And when you solve this the expression would be I'll write down the expression here. The expression would be 1 by. 1 minus N. Concentration of A. T to the power 1 minus N. Minus concentration of A. Not to the power. 1 minus N is equals to. Minus of KT. So which further we can write. A at any time T. The power 1 minus N. Minus A naught. To the power 1 minus N. Is equals to. N minus 1. K into T. Because of this expression you see this expression. This one itself. Is not valid. For N is equals to 1. Just for this expression I'm writing it on this step. Okay. Because. This side will have infinity. This is not valid for N is equals to 1. Okay. Even here it's also not valid for N is equals to 1. It's not like the entire expression is not correct for first order. Like you see if you find out T half here. T half for Nth order. First you copy down this. A naught. T half. Is equals to. We can substitute 80 is equals to a naught by two. So T half could be a light on one by N minus one. Into K. 80 is. A naught. By two to the power. One minus N. Minus. A naught to the power one minus N. Okay. Take care of the science. Okay. Plus minus sign. So when you solve this, you will get T half. Is equals to. A naught. To the power one minus N divided by. N minus one into K. Open bracket one by. Two to the power one minus N. Minus one. This is the expression we have. So the point I'm trying to make is for any Nth order reaction. T half. Is directly proportional to. A naught to the power. One minus N. A naught to the power one minus N. Okay. So you'll get very basic simple questions on these expressions that we are getting. I'm not going to, you know, you know, give you any questions on this now because. When we do others, you know, concepts based questions, like temperature and all, then we'll cover all these concepts. They're only right. So here are the basic questions based on the relation of K concentration and time. We are not doing here now. Later on, after discussing few more concepts, we'll cover up those kinds of questions also. Okay. So this is the expression we get and you see T half is this. It is valid for all the value of N here. You see when N is zero, T half is directly proportional to initial concentration. Fine. N is equals to one. It is independent. N is equals to two. It is inversely proportional. Okay. It is valid for all the. I know this thing of value of N. There are few reactions, which we call it as pseudo first order because the entire chapter, the most and the most important order is N is equals to one first order. So first order is the most important. And then we have zero order. Second order. You won't get much, much question, but second order at this expression. There are some reactions. Which, you know, appears to be. Second order or any other other reactions, any higher order reactions, but actually it's not. It's not. It's not. It's not. It's not. It's not. It's not. It's not. It's, it's not. Any higher order reactions, but actually it is first order reaction. These reactions we call it as pseudo first order reaction. Okay. Right now the heading next. Sudo first order reaction. Just need to know a few examples for this. And that is it. Okay. This is due to excess amount of one of the reactant excess amount of one of the reactant, the second order, or higher order mainly second order. Second order reaction confirms to be, confirms to be to be the first order reaction or it is behaving as a first order reaction. This is due to the presence of excess amount first order reaction. Okay. This type of reaction is called pseudo order reaction. Hence called pseudo first order reaction. Here are those few examples you need to know in this. The first one is hydrolysis of ester acidic hydrolysis. Suppose we have CH3 COO C2 H5 present in water and excess of water we have in presence of an acid, it converts into acid and alcohol acid and alcohol. So for this reaction, if you write down the rate expression, the rate expression would be R is equals to K times concentration of ester into concentration of water. But since it is present in excess, so we have seen that we observe that it's a practical thing. We observe that the concentration of this water which is present in excess, it won't affect the rate of the reaction much. Okay, and hence be neglected. So K into this concentration we assume to be constant K into this will give a new rate constant that is K dash into ester. This is the expression of this reaction. And this K dash is called pseudo rate constant pseudo rate constant. You need to know the concentration of water. Yeah, we are just doing the example. Okay, data will be given if they ask you to compare pseudo rate constant, then accordingly the data will be given concentration of water will be given. You can find out rate constant K and then K into H2O is K dash and then you can find out H2O. Simple thing. Just you need to know what is K dash. K dash is K into H2O, concentration of H2O. Second example we have here is inversion of cane sugar, C12H22O11 plus H2O axis converts into C6H12O6, C6H12O6. Just a second guys, I think someone's there. Yeah, C12H22O11 plus H2O axis it gives this. So for this the rate expression would be same thing I'll write on the rate final value directly the rate expression would be R is equals to K dash the pseudo rate constant into the concentration of C12H22O11. Okay, esterification of anhydride is also another example. Examples you have to keep in mind because sometimes they ask theoretical question, which one of these is this thing, pseudo first order reaction like this also there's mostly need. So the reaction here is CH3COCO CS3 plus C2H5OH alcohol we are taking in excess. So this gives two molecules of ester CS3COO C2H5. So the rate expression like we had in the previous one, rate also is K dash into the concentration of anhydride. This is the expression we have. Apart from this the fourth one we have metabolism of hormones in human bodies that is also pseudo first order. Metabolism of hormones sir. Yes. Okay sir. Metabolism of hormones in human bodies. Okay so till now guys we had discussed about the reaction and concentration we were talking about correct. All these first order reaction will have in case of titration also when volumes are involved. We'll have in case of growth of, no, the growth of bacteria the population growth of bacteria we can have that. Like I said in volumetric analysis we can have it, or when the gaseous phase reaction is given. So when the growth of growth of bacteria is there then what we need to do, we call it as first order growth kinetics just a term. So first question how do we get will see that in volumetric analysis at different different point, the volume is given and how do we get the expression of K, like we had in first order in, in terms of volume that we are going to see. And when the gaseous phase reaction is there, and how we can relate pressure with the rate constant K till now we did with concentration. Okay. So now the first order growth kinetics, growth kinetics, it's very simple will have the general expression of first order only, but only one change is there since the population growth we have. So at point with when the time you know goes, then the population increases in the denominator will have more value means the value will increase in the denominator. Like you see, we're talking about the growth of bacteria here. Suppose that time t is equals to zero. We have certain population, that is suppose a we are assuming at time t, it's supposed becomes a plus X, okay, a plus X. This is the difference we have in the other reactions correct. In the previous reactions you'll see with time the concentration is decreasing, correct. That's why we had a minus X, 80 is a minus X, but since this growth population of bacteria. Correct. So it is a plus X because the population is increasing with time. So it's the same thing exactly same thing you can derive the expression, you know, for first order but since we have done the derivation already, I'm not going to do the entire thing again. So here the K value would be, if you recall the expression for first order, it is one by T ln A by 80 correct, A naught by 80. So if I write down in terms of law, nothing much you need to do just this 2.303 by time t log off initial concentration is a and concentration or population at time T is a plus X. So this is the expression we have any doubt in this. So this will be negative. Yes, one negative sign. The next one is in this the second type in this is the volumetric analysis. Okay, we can say volumetric analysis or like in any reaction if some gaseous product is there on the product side only, then we can take the volume of gas also. Okay, for example, you see next write down volume related method volume analysis any or volumetric analysis correct. So in this, you can do it in two different way. One is, you can directly memorize one formula that I'm giving you. For example, you see the expression of K here is 2.303 by T log off V infinity minus V naught divided by V infinity minus VT all that sign and term will be taken care of. There's no negative sign outside. Okay, where V naught is what V naught is the volume given at time t equals to zero. What is the volume at t equals to zero VT is the volume at T is equals to any time t and V infinity is the volume in finite time. Now what you need to do when you go when we are going to use this expression, whatever volume term is not given, you have to take it as zero. I'm giving you this to solve the question, how do we get this will do that I'll just give you one more example, how to write down the volume related term. So here what you have to just keep in mind if you remember this formula, suppose V naught is not given. Okay, then what you will do. The initial volume V naught is not given, not given, then you substitute simply V naught equals to zero in this expression, the expression of K you will get. Suppose if V infinity is not given, V infinity is not given. You substitute V infinity is equals to zero the expression you get with that expression you can find out the, you know, correct. No, because any reactions, it is given at a certain temperature and pressure, right. So volume in that case is directly proportional to number of modes, so volume and number of modes we can directly correlate. Okay, first of all, you know, let me tell you this thing that this expression if you remember, right, directly you can use this and all these value can substitute according to the data given you will get the answer. Okay, but suppose if you are not able to, suppose if you feel like, okay, how do you memorize this expression. So in that case what you need to do. It's a simple one you can easily derive it. See I'll tell you how. We know for a reaction, a reaction takes place, takes place at a given temperature and pressure, given P and T. So any reaction what we can write the volume is directly proportional to the number of moles, because the number of moles and volume we can coordinate. Okay, for example, you see, if I have this reaction. And I'm taking only one gaseous product N205 with gifts NO2 plus O2. Suppose we have only oxygen in the gaseous state here it is given. Okay. Then, how do we write down the volume expression for this reaction. Okay. So simply here you see, we have initial concentration T0 I am assuming. A0 here. And this is zero. This is zero initially, at time T is equals to T. This becomes A0 minus 2x. This becomes 4x and this becomes x by, sorry, x here. This becomes x. Correct. Now when T is equals to T infinity. This becomes zero because all these N205 convert into O2 and N infinity. This becomes zero. This will be 2A0 and this would be A0 by 2. Okay. So if I write down the expression of K in terms of the concentration, it is 2.303 by T log of A0 by A0 minus 2x. Once again. This is in terms of concentration. Now if you know this A0, how this A0 is related to the volume and this how it is related to the volume, we can write down the expression in terms of volume. Correct. So at any time T, at any time T, the amount that is converted into of N205 converted into O2 is this because we have only gaseous product is O2. So x is the amount of A we have at time T. So I'm assuming corresponding to x, we have certain volume. Okay. That is VT. So what I'm writing down here for oxygen gas only at time T is equals to T, the amount of oxygen gas x, this corresponds to some volume and that I am assuming as VT because number of moles and volumes are directly proportional. At time T is equals to T infinity. What is the volume of what is the value of O2? T is equals to infinity. Value of O2 is A0 by 2 in the reaction is written. This equals to or corresponds to V infinity because T infinity we have. So what is A0 from this you see? A0 is directly proportional to 2V infinity. This if you substitute in the previous expression, that would be k is equals to 2.303 by T log of A0 is 2V infinity divided by A0 is again 2V infinity minus 2x is 2VT. So when you solve this you see 2 and 2 will get cancelled. So expression of k is 2.303 by T log of V infinity by V infinity minus VT. This expression you get in terms of quantity. Now you see if you compare the first method that I told you V0 minus V infinity minus V0 by V infinity minus VT and I told you whatever is not given just substituted as 0. So for this data you see, for this data this given data, what is the value of V0 initial volume? It is not given. So V0 should be what? V0 should be 0. If you substitute this V0 0 in this expression that I've given you here, V0 you substitute 0. You see you'll get the same expression. This is 0 V0 by V0 minus T. So either you memorize this if volume things are, it's very clear like what formula you have to use when volume is given then you have to use this volume formula. Okay and this kind of formula we only use when there is only one gaseous product. Okay, use write down use when only one gaseous product is there. Or we can also say in this way that gaseous molecule is present only on the product size. Write down the volume expression we use when the gases are present only on the product side. We assume that also to use this expression. Okay, so bacteria and this volume it's not much important. Okay, but not that difficult also you can understand it easily. Similarly, we can have one more expression here. That is when the compound is optically active. Okay, so for optically active compounds, what things you have to keep it exactly similar to the expression of volume that we have got. Okay, you see this next is to determine the rate constant rate constant of an optically active compound. So here also all data will be given just you need to use the formula and what is the formula formula exactly same like we had in case of volume 2.303 by T log of instead of the infinity. We have r infinity minus r not divided by r infinity minus RT the least important one but he has to formalize this. Same thing are not is the angle of rotation at zero time I'll write down this are not is the angle of rotation at T is equals to zero. RT is the angle of rotation at time T at T is equals to T and r infinity is again the angle of rotation T is equals to infinity. Okay, again the same thing if any value is not given substituted zero will get the same expression then. Okay, the last one in this type is the most important one and that is to determine the write down to remind the rate constant of gaseous reaction gaseous phase reaction pressure will be given PT will be given okay when pressure is given total pressure will be given. This is the point I'm trying to make is you just need to know that how to write down the cake expression. You don't have to worry about the terms like what terms are given and not you can easily find out with the data given in that question that's not a concern you just try to understand and focus your focus should be that how do we write down the expression of cake. Okay, so suppose I'm taking one very simplest example here, the example I'm taking is a reaction we have in which a gas converts into be gas and see gas. All are gaseous reacting gases product it is also possible that only gaseous only one gaseous product will have that is also possible so based on the reaction you can do all these things. So, I am assuming here are initial pressure as P naught. And after sometime the pressure of able decrease because a converts into B and C that is zero that is zero. This is P1 and this is P1. No doubt. So, here you see if I ask you what is the total pressure at time T what is the total pressure at time T tell me the total pressure T is PT and PT is equals to whatever gaseous molecules are there. Okay, try to understand this carefully. Okay, pressure is because of only gaseous molecule. So whatever gaseous molecules are there we have to add the pressure of all gaseous molecules. Okay, means we have to add the pressure of a because a is a gas, we have to add the pressure of B, we have to add the pressure of C, where P a, P b, P c are the partial pressure of a, b and C. Correct. So, here PT is equals to P a is nothing but P naught minus P1, P b is P1 and P c is P1. So, total pressure PT is equals to P naught plus P. Okay, this PT will be given in the question. Okay, PT will be given and suppose the partial pressure you need to find out, dc pressure basically you need to find out or they may also ask that what is the pressure of a at time T. So, you need to find out that what part of a has been converted into P so that the pressure of a would be where it is to P1, P1 and P1 will get cancelled. No, this one you see P1 and P1, alethe manjunath. Correct. Is this correct guys, total pressure? So, the thing is if the question is you need to find out the pressure of a at time T, it means you need to find out P naught minus P1. Pressure of B at time T, you need to find out P1, you need to find out C is also P1. So, because all these are first order we are assuming. So, if I write down the expression for the first order that would be what? That would be K is equals to 2.303 by T log initial pressure divided by the final pressure at time T. This is what we need to do. Could you please go back? Just a second. Pressure of a at time T. So, I'll write down the pressure of a here like this. This P means pressure of a at time T. Just a second I'll go back. Pressure of a at time T. Okay, done. Okay done guys. Can I go to the next slide? Yes. So, this is the expression we have pressure of a at time T. So, P is we know it is P naught minus P1, correct? A is equals to P naught minus P1. Okay. So, K is equals to 2.303 by T log of P naught by P naught minus P1. Because P1 is a partial pressure like of B and C. So, your expression should not be in terms of the partial pressure because P1 is not given in the question. Pt is given total pressure and Pt relation we have already Pt is equals to what we have just now we have calculated Pt is equals to P naught minus P1. P1 is equals to what? P naught minus Pt. This is correct. Sir Pt is P naught plus P1 we have, correct? P1 is Pt minus P naught, yeah. P1 is Pt minus P naught and this is plus we have here. So, this P1 will substitute here. So, K expression would be 2.303 by T log of P naught divided by P1 is this. So, we'll get 2 P naught minus Pt, correct? So, this is the expression we have. So, you need to understand to write down this expression for data you don't have to worry about. Okay. The data will be given whatever point that data is given for that point will apply this condition that point pressure of A at that point B at that point and all. This is the expression of K in terms of pressure. Any doubt on this? We can also do this in one different way because we know pressure and concentration are also directly proportional. Can I go to the next slide? All of you are done. We can also do it in this phase. Suppose the reaction is this, we have A gas converts into B gas and C gas. So, what happens at different, different time they'll give you the concentration A naught 0 0 at time T it is A T 0 0 and they'll ask you to find out the expression in terms of pressure. Simply we know because it is a concentration given or suppose if moles also it is given here then obviously the volume it is understood and you have to consider the volume one liter. If moles is given then assume volume as one liter otherwise concentration is given is fine. So, we know this fact that pressure is directly proportional to concentration. Like volume we were relating your pressure and concentration also we can relate. So, we can say A naught the pressure at time T is equals to 0 concentration is this. This is directly proportional to the initial pressure that is P naught. Okay and one more thing I just this will have certain concentration of B and C that we do not know suppose X X will have here. Okay. And A T concentration of A at any time P that would be equals to what pressure at this point P. So, concentration and pressure you can also relate like this. What is given take volume as one liter sir. That you can assume if concentration is given then that's fine. Yes sir. Okay and then you can write down the again key expression in terms of pressure because it is directly proportional we can do that. Understood? Yes sir. Sometime what happens they'll give you expression like this. I'll give you go to the next page and show you this. Suppose the reaction is again this A gives B plus C all our gaseous product and gaseous reactant and what they'll do they'll give you at time T is equals to 0 its concentration is A naught this is 0 this is 0. At time T is equals to T this becomes A naught minus X this is X and this is X and at time T is equals to infinity this is 0 this is A naught and this is A naught and the total pressure will be given correct. So, at two points suppose the total pressure is like suppose at this point the total pressure is given it is P1 and here the total pressure is given P2. Then you have to because this P1 and P2 is given then you must write the expression of k in terms of P1 and P2 so that we can find out the other things. Correct. So, what we can do here you see P1 we know this P1 is directly proportional to the concentration at this time so we can write A minus X plus X plus X. Which is P1 is directly proportional to A plus X that's the one relation we have. Similarly, P2 is highly proportional to could you tell me P2 is highly proportional to. Way or not. Way or not. Way or not. Correct. This is the second relation. Now what we need to do that you see we know k is equals to what 2.303 by T log of A naught by A naught minus X this is the expression we have. So, A naught is nothing but P2 by 2. So, A naught there is no problem we can substitute P2 by 2 in terms for A naught over here. And A naught minus X we need to find out. So, to find out A naught minus X what we can do. We can subtract these two this minus this so left hand side what we have P2 minus P1 is directly proportional to A naught minus X is this right understood any doubt please respond guys quickly. Correct. Now just we need to substitute it here. So, k is equals to k is equals to 2.303 by T log of A naught is P2 by 2 divided by it is P2 minus P1. Is this right? Let me check the previous expression. Yeah. A naught minus X is this A naught is P2 by 2. Yeah, that's correct. So, this is the expression we get. So, the expression of k in terms of pressure the given pressure is 2.303 by T log of P2 by 2 times P2 minus P1 this is the expression we get. Yeah. So, it's all depends that what is the condition given why I have taken here because we know these at these two condition only the pressure was given P1 and P2. So, I have taken these two condition. If these two condition is given then we can talk about these two condition. Okay, so it's all depends upon what condition is given the question you can relate directly concentration with pressure and then we can write down the expression of k in terms of pressure. Understood. It's completely depends upon that what reaction is given first of all plus what condition is given like if I give you one more you know reaction here just for your practice you just see if the reaction is this. We have a gas gives to be gas see gas here. Okay, and the pressure is given at any time T at any time T the pressure is is PT. Total pressure is PT initial pressure of P is given that is P not at any time to the pressure is given PT you have to find out the expression of K and at T is equals to infinity. All right, I don't just wait. At time T the pressure is P1. This one, this is one question done. Okay, another question is, which I will do this one only find out the expression of K, assume this is PT only so that can avoid total pressure is this at any time T. You need to find out the expression of K, obviously since P naught and PT is given you need to find out this in terms of P naught and PT. Yeah, it's done to P naught by three P naught and PT. Okay. Yeah, right. So what we'll do here, we'll write down this is P naught and initially the pressure of B and C is zero and zero. So if it becomes P naught minus P one, this becomes two P one, and this becomes P one. So total pressure PT here at this point would be P naught minus two P one P naught plus two P one correct. This plus this plus this P naught plus two P one. So P one from this if you find out because the P one is not given. So it is PT minus P naught divided by two. So the expression of a K for this reaction is 2.303 by T log of A naught. Sorry, P naught by PT, P naught by PT that is P naught minus P one, P one if you substitute this, you'll get the answer as K is equals to 2.303 by T. log of two P naught by three P naught minus PT. This is the expression of K for this reaction. So obviously you see the stoichiometry coefficient changes, the expression will also change. Even in the question when you solve this chapter, right, the numericals, you will have this kind of options also like the reaction is given ABCD like this for options will be given. So this is how we'll do it. So whatever we have done growth of bacteria volume and optical activity rotation and all this particular thing the pressure related question is the most important one in this. Okay. See this question. Yeah, done. See I'll do this week. A simple one the reaction is given a to gas gives be gas and half of C to half of C. Okay. Increasing pressure from 100 mm to 120 mm in five minute. Okay. So, we are assuming the initial pressure is 100 mm, right, and it becomes 120 in five minute of this change. Okay. So, at time T is equals to zero. I'm assuming there's only a so P naught, zero and zero if it is not mentioned that B and C also present initially, you don't have to consider that. Right. In general, the assumption is this only that we have only reacted at time T is equals to zero. It becomes P naught minus P one. This is P one and this is P one. Now according to the given data. We have P naught is equals to 100. And the sum of all these that is P naught plus P one is equals to 120. Right. P one P naught we know what is P one P one is 20 mm. So if you see isn't it half P one. Much a half P two correct. We have this. Oh, then everything's in. Oh, I made a mistake. So half P one we have and when we add all these then we'll get P naught plus half of P one is equals to 120. So P naught is hundreds of P naught P one is 40 correct. We need to find out what we need to find out the rate of disappearance of a the rate of disappearance of a is what. That is minus rate of a is minus D by DT of a pressure that we can write on pressure also here. So it would be P naught minus P one by this time required is five minute. Change is P naught minus P one one negatives and we already have minus P naught divided by the time is okay. Better if you write the average rate that is delta of pressure of a divided by delta of T. So you'll get this. So it would be 40 divided by five. So we'll add eight mm per unit is minutes. This is the answer we get. So they're not asking for the rate at five minutes. They're asking for average rate. So they're asking for average rate not the rate at five minutes. Yeah, obviously because rate at five if you need to find out then what we need to do the pressure of a at this time that divided by this. At that point they cannot they cannot other it does not make any sense rate means what over a period of time how the concentration of pressure is changing that is the rate no ratio. So obviously here it is understood that the change that happens from 100 to 120 what is the rate of that change. That's what the rate of disappearance of it. Okay. If they ask you like this rate at P is equals to five minute then and that would be P naught minus P one divided by five. Yes. Any doubt in this one second. This question you see how do we do this first order reaction you need to prove it. This kind of question usually they ask in board is okay. That's why I've taken one. See what you need to do here. If it is first order reaction, then for this data, you'll have the value of rate constant K. So for this data rate constant K, and for the second time again find out for this data rate constant K, this must be equal. If it is first order. If it is coming out to be equal, the reaction is of first order, otherwise it is not. Okay, you don't have to, you know, I mean what point I'm trying to make is this kind of question they won't ask you in a comparative exam because they won't ask you to prove it. They can ask you for the given data they can ask you what is the order of the reaction like this. So it, it, it would be a bit of, you know, a calculator you need to do some calculation into it method is this only for two set of data, if the value of K is coming out to be same, then whatever assumption you have taken that is correct. So this question would be like this, exactly the data is given and accept this line, it is written over here what is the order of the reaction. In that case, you have to assume first, okay, assume the order is zero. Okay, find out K for this data, K for this data if it is equal, means our assumption is correct, not equal then assume order and is equals to one. Okay, find out this same thing, correct. So it will take some time but the method is this only. So you won't get this kind of question in comparative exam but yes in board exam, you may get this. So one question for that. Did you find out what is the value of K you are getting. See I'll do this. Check. See we have this reaction. See how do you do this to try to understand this pressure the total pressure given. Keep that in mind. But mistake you can make. I'll tell you. SO2Cl2 gives SO2 gas plus Cl2 gas. So initially I am assuming it has obviously it is given so 180 is the initial pressure. And since it is not mentioned so I'm assuming only SO2Cl2 is present pressure is this there is no SO2 there is no Cl2. So when the reaction proceeds this SO2Cl2 dissociates X, X and X. This time here the time which is required this time is given. This is a T is equals to zero. And this is a T is equals to 240 second 240 second. And at this time the total pressure is given Pt is equals to and here since all our gases species take care of this. If this is not gas we won't take the contribution of this into total pressure. Okay, small small things but important. Total pressure will be pressure of this plus pressure of this plus pressure of this that would be 180 plus X. What is the value of X could you tell me Pt is given 244 so value of X would be 64 any doubt in this. Why we need this X that's a question right why we need this X because when you write down the expression for K expression for K it would be 2.303 by T is 240 log of initial is 180 minus X we have this X must be required right so this you need to subtract here 64 now from this data you will find out K. And the same thing you have to repeat for the second set of data means again you can start with from here to here means assume 600 second. Again set of data I'll write down the value of X so to CL to gas. So, this is again 180 zero zero, this becomes 180 minus X, X, X, but this time here I'm choosing the time here is 600 and this time is zero. X value would be 180 plus X so 302 minus 180 that would be 122. Okay, so X for this data we are getting 122. So K is 2.303 divided by T log of 180 by 180 minus 122 from this you'll get the value of K. So if the K value is same for the both set of data it means our assumption that is first order of reaction that's correct. The value of K you will get here is 0.0018 second. Is this the right answer you're getting this second inverse. Yes, this is the answer you get any doubt in this. Thank you.