 Hello. Good morning, everyone. Okay. Bharat is there. Shruti. Odik. See, today in this session, we are going to discuss about a few concepts of various chapters. Okay. Which actually helps you in the coming KVUI, as well as JMAIN's exam. Okay. Since it is actually a quick revision class, we'll see some problems also into this. So obviously, we are not going into all those theory and small, small details. Right. See, only we only discuss those concepts, which is helpful and which it requires. Okay. So. Okay. So kid's question. Fine. You see, first of all, last class, I had taken a division class also of chemical equilibrium. Okay. And there I again did not discuss about Lee Chatelier's principle. Okay. So first of all, we'll discuss today. Lee Chatelier's principle. Obviously, there are some 11th class students also there. So. They haven't done this before. Lee Chatelier's principle, but some 12th class students, you must have done this. Right. Okay. So chemical equilibrium. We also discussed a little bit of a unique also buffer solution. We'll discuss in this session. Okay. After this, and then we'll discuss the concept of equivalent mass. And last, we'll see the concept of GeoC because GeoC, we have too much. We cannot cover everything. Obviously into this few concepts of GeoC also will discuss in the last. Okay. So, let's see. Let's start with Lee Chatelier's principle. Okay. So. Lee Chatelier's principle. If you see the definition of this Lee Chatelier's principle is what whenever a system is in equilibrium. Right. Whenever any reaction you have and that reaction is in equilibrium. Okay. So if you. If you change the equilibrium condition. Right. Or the, if the equilibrium condition of any process is subjected to change, then the equilibrium shift into that direction where. In that direction where the effect of change will be minimized. Okay. So what, what I said just now that whenever a system whenever a system in equilibrium system in equilibrium. Is subjected to change. Subjected to change. It's equilibrium condition. It's equilibrium condition. Then the equilibrium shifts in that direction. That direction by which effect of change. Effect of change is minimized. Minimized. Okay. This is what the Lee Chatelier's principle we have. One more thing I tell you 12 classes to students if you have, if you have, since you have done this before. So whatever you feel important, you can write down in your notes. Everything is not important to write. Okay. Second thing for solution for 11 class of students if they write it properly, it will be helpful for them. Okay. Because this particular importantly Chatelier's principle I won't do again in the class. Okay. So better if they write the notes properly. Okay. So now you see what is the meaning of this when I said that equilibrium shift into that direction by which the effect of change will be minimized. Simple example if I take here. Suppose we have a reaction, any reaction we have and in that reaction if I am increasing the concentration of reactant. Right. Example if reactant concentration is increasing reactant concentration is increasing square bracket I'll put here here. Concentration of reactant is increasing then the equilibrium shift or the reaction goes in that direction by which the concentration of reactant will decrease. Right. The two directions are possible either forward or backward. Right. Two directions are possible. Either the reaction goes into forward direction or backward direction. Right. When we increase the concentration. So if the reaction goes in forward direction then the concentration of reactant will decrease. Correct. And if it goes in the backward direction in the concentration of reactant will increase. So that's what Leach-Atelier's principle is. Whatever the change you have done the equilibrium shift in that direction by which the change that you have produced that change will be minimized. Okay. So obviously when you increase the concentration of reactant the reaction proceeds in forward direction. Like this. So this is the winning of Leach-Atelier's principle. Okay. One by one we'll discuss what is the various factors we have into this. Okay. So factors if you see into this that is the effect of change in concentration. Right. The first one if we see the effect of there are four, five different you know cases we have here that we'll discuss. Effect of change in concentration. Change in concentration. And concentration of reactant or product. Reactant or product. Okay. Now you see in this as I said just now I have taken the example. And we can write direct here that if concentration of reactant square bracket I have written it means concentration of reactant. If concentration of reactant increases reaction goes in forward direction. Second case possible if concentration of product or concentration of reactant decreases. Reaction goes in backward direction. Right. Because concentration of reactant is decreasing. So reaction will come into the backward direction so that it will increase the concentration of reactant. Right. Again the second third possibility is what if concentration of product increases. Then again the reaction goes in backward direction to decrease the product of the reaction. And if the concentration of product decreases then the reaction goes in forward direction to increase the product of the reaction. Concentration of product. Right. Okay. Yeah. So this is the fourth possible case. This is very straightforward and simple. Okay. You can understand this easily. Now the next condition we have or the next factor is that is the effect of temperature. What happens if we change the temperature? Effect of temperature. Right. Temperature effect will actually understand by the nature of the reaction. Nature if I say it means whether it is exothermic or endothermic. Right. So first of all the effect of temperature you can understand by this particular what we say. Equation which is one of equation for K2 by K1 is equals to minus del H divided by 2.303 R into 1 by T2 minus 1 by T1. With this you can understand. And one more thing that you probably know that if K increases reaction goes in forward direction. More value of K more the tendency will be in forward direction. If K decreases the reaction goes in backward direction. This we already know. Right. Now here you see. If I am since I like I said the effect of temperature we can understand by the nature of the reaction if it is exothermic or endothermic. Suppose I will take the both condition one by one. First I am taking suppose endothermic reaction we have. Okay suppose the process or the reaction is endothermic. Endothermic. The first one I will do in little bit of no in detail. Similarly you can do for the other cases also. Right. Now you see for endothermic reaction we know del H is greater than 0. This we know already. Okay. We have to see that in endothermic process if we increase the temperature then what effects on K takes place whether K value increases or decreases that you have to understand. Right. So you see endothermic process we have del H greater than 0 as temperature increases. As temperature increases. Which means what this particular reaction says this particular equation says K2 is the equilibrium constant at temperature T2. And K1 is the equilibrium constant at temperature T1. Right. Okay. So if temperature increases which means what I am taking the condition of T2 is greater than T1. This is the condition I am taking. Right. Condition can be anything. We are discussing all possible cases. Right. So as temperature increases means T2 is greater than T1. Now you see if I reverse this because we have to find out this 1 by T2 minus 1 by 2 and you see one thing here what is the objective here we have. See del H is greater than 0. So this del H is positive. Right. This value is positive. Del H value is positive. This is negative. So positive negative negative. If this value is negative or positive that will actually decide whether log K2 by K1 is greater than 0 or less than 0. Right. So since one particular type of reaction will fix the value of del H whether it is positive or negative according to this this term minus del H by 2.303 R will have either positive or negative. So in this case it is negative. You see positive plus minus minus positive negative negative. So in this case it is negative. Now depending on this value of this whether it is greater than 0 or less than 0 we can say greater than 0. Right. So that is what we are going to discuss. So condition I have taken T2 greater than T1 when temperature increases. So what we can write 1 by T2 is less than 1 by T1 and hence 1 by T2 minus 1 by T1 is less than 0. Right. Which means this value is negative. So when this value is negative you see and this is already negative so negative negative positive. So what we can write under this condition log K2 by K1 log K2 by K1 is greater than 0. Right. Which gives you log K2 minus log K1 is greater than 0 and further K2 is greater than K1. This is the conclusion we get here. So what we can say in endothermic reaction as temperature increases equilibrium constant increases. Right. When T2 greater than T1 K2 is also greater than K1. So that is what by effect of temperature we have in endothermic reaction. Okay. Just write down one line over here after this. That in endothermic reaction in endothermic reaction as temperature increases equilibrium constant K increases and when K increases reaction goes in forward direction. Is it clear? Let me know. Okay. Now in endothermic reaction only we can also understand the second possibilities. The first one I took that as temperature increases this is also possibility as temperature decreases. This is also the second possibility we have. In this case the similar way you can conclude T2 less than T1 1 by T2 is greater than 1 by T1 then 1 by T2 minus 1 by T1 is positive. Right. So there if you see as temperature decreases in endothermic reaction K value also decreases and when K value decreases the reaction shift into backward direction. Reaction shift into backward direction. Okay, so this is the two possibilities we have in endothermic reaction. Similar kind of possibilities are also there in case of exothermic reaction when the reaction is exothermic. So exothermic reaction means what del H is less than 0. This is the difference. Now you see when del H is less than 0 you see this one the first page that we have done here you see del H less than 0 now we have. So this is negative and this is already negative. So this negative negative positive. Right. So if T2 1 by T2 minus 1 by T1 is negative log K2 minus K1 is negative if it is positive log K2 minus K1 is positive. Correct. That is what we are going to do next. So you see del H is less than 0 if temperature increases if T increases this means what T2 is greater than T1 so 1 by T2 minus 1 by T1 is less than 0. Okay. This means log K2 log K2 by K1 is again less than 0 and K2 less than K1. This means equilibrium constant decreases the reactions shift into backward direction. So in exothermic reaction as temperature increases backward shift possible. Reverse of this will be true when temperature decreases. So if temperature decreases then we can say K increases and when K increases forward shift will be there forward shift. Okay. So this is the four different cases we have in exothermic endothermic reaction. The third case here we have third case I'll take and that is effect of pressure effect of pressure in this also there are three cases possible three different types of reaction we have. The first one in this when the reaction which has del NG value is equals to 0 which has 0 del NG value. Okay. Which means what the reaction is suppose if I assume the reaction is this A gas plus one general reaction I am taking B gas gives you C gas and D gas. This is the reaction we have. Right. Since del NG is equals to 0 so we have to find out the effect of pressure in this reaction. Okay. So obviously when del NG is 0 so what we can write Kp is equals to Kc the first thing. Value of Kp and Kc will be same. Correct. So if I write down the expression for Kp that will be what? The partial pressure of C into partial pressure of D divided by partial pressure of A into partial pressure of B. Right. Kp we have to write down the partial pressure. If you write down Kc then concentration term we have to use. Right. Now we know this from Darjeel's law that partial pressure partial pressure is equals to mole fraction into total pressure Pt mole fraction into total pressure. The effect of this pressure only we have to assume. Right. The pressure at which the reaction is taking place. The total pressure. Okay. Now you see according to this formula if I write down the expression of Kp that will be what? Partial pressure of C is equals to mole fraction into C into total pressure into mole fraction of D into total pressure divided by mole fraction of A into total pressure mole fraction of B into total pressure. Okay. So you see here all this pressure term, the total pressure we have will get cancelled in the numerator and denominator. This will get cancelled. Okay. And that's why the expression for this will be Xc into Xd divided by Xn to Xb. Okay. So I see in this one Kp is independent of Pt, right? Kp term is independent of of total pressure Pt. So whatever total pressure you take that would affect Kp or this expression. Okay. And that's why we say the reaction in which del Ng is equals to 0 for that particular reaction pressure has no effect in this. Okay. So you can write down for this kind of reaction pressure will not change the equilibrium condition. Is it clear? Okay. So del Ng is equals to 0. Pressure won't have any effect. Next one you see. Now similarly you can write down the all other possibility you can understand. You see the second case I am taking because in these three cases are possible. First one is when del Ng is 0, second one I am taking when del Ng value is greater than 0, right? Del Ng value is greater than 0. For example, suppose if I write A gas gives D gas and C gas. Suppose this we have. Reaction. Okay. So obviously for this reaction del Ng value is greater than 0. Number of gaseous product minus number of gaseous reaction. Now if I write down the expression of Kp, that will be what again partial pressure of B into partial pressure of C divided by partial pressure of A. And that again we can write XB into XC divided by XA into Pt square divided by Pt, right? So obviously this contains what? XB into XC divided by XA into Pt. This is the expression we have here. Okay. See one more thing here. We can also explain the same thing with some theoretical way also. Like when pressure increases, volume decreases, the number of moles per unit of volume increases. Like that also we can also explain, right? But I think this is the mathematical explanation I am giving you that helps you more to understand the concept, right? That's why I am taking this equation part, okay? Now you see since Kp depends on pressure, the expression contains the pressure term here, right? So obviously the pressure will have some effect on these kind of reaction, okay? What kind of effect that we will see now? Now suppose if pressure increases, if Pt increases, then what happens? That we have to understand. Okay? Okay, so here you see if Pt increases, so obviously this Kp has to be constant, right? Because equivalent constant depends only on temperature. It has to maintain its value, constant value, right? So if this value is increasing, right? To make this ratio constant, Xa should also increase, right? So if Pt increases, to make Kp constant, Xa will also increase. Xa increases. And when Xa increases, Xa is the mole fraction of A, right? This mole fraction of A should increase and mole fraction of A will only increase when the number of moles of A increases, right? And that is only possible in the backward direction. Okay? So I hope you understood this. If Pt increases, Xa increases, and this is only possible when backward reaction is favourable. So here we have in this type of reaction where Derange is written as 0. As pressure increases, backward reaction will be favourable. What happens if Pt decreases? Pt decreases? No, Niranjan, we cannot say if Pt decreases, then Kp decreases. See, Kp has to be constant. Okay? Kp depends only on temperatures. So with pressure, this value will not change, right? So we have to maintain the value of Kp. So if this numerator is decreasing, so denominator also decreases. So that will maintain the ratio. That's the point here, right? If Pt decreases, like I wrote over here, if Pt increases, Xa increases. Why this is true? Because this ratio we have to maintain. Pt by Xa ratio we have to maintain. Right? So if Pt decreases, then Xa should also decrease. The mole fraction of A should also decrease. And mole fraction of A will decrease when the reaction goes in forward direction. Right? So in terms of temperature, we were discussing over there that K value decreases or increases. Why we are discussing about K also over there? Because K depends on temperature. Equilibrium constant depends on temperature. That's why we say if temperature increases, K decreases or increases accordingly. Right? But here we are talking about the effect of pressure. And pressure won't change the equilibrium constant. Right? Equilibrium constant depends only on temperature. We have to maintain the value of Kp and that's why we are doing like this. Okay? Understood? Okay, now the same thing we have to do in the last case where the reaction has del Ng value less than 0. Okay? Del Ng value less than 0. Suppose the reaction is this. A gas gives C gas. Del Ng value less than 0. What is the expression for Kp we write? It is partial pressure of C divided by partial pressure of A into partial pressure of B and that further we can write XC by XA into XB into 1 by PT. Now you tell me here, all of you as PT increases and what is the direction of the reaction? As PT decreases what is the direction of the reaction? Both ways. Backward as P decreases forward as P increases and backward as P decreases. Okay? So if PT increases if PT increases obviously to maintain the ratio XA should also XC should also increase. And XC will only increase when the reaction goes into forward direction. So if PT increases forward reaction favorable and if PT decreases backward reaction is favorable. Backward reaction is favorable. Right? These are the two conditions here. Now you see on the base of this temperature and pressure one question has been asked in previous exam that I will write you over here. Suppose we have A gas plus 2B gas gives C gas plus heat. Reaction given. Equilibrium reaction obviously reversible reaction. Okay? Now this reaction the question is for this reaction forward reaction is favorable. Forward reaction will be favored under which condition? The condition is given low pressure low pressure high temperature second one low pressure low temperature third one high pressure high temperature and fourth one high pressure low temperature. Like this the four reactions are given. Tell me the answer of this question. Option D Where is Bharat? Sugit Bharat What is the answer? Ashutosh is also saying D. Hello. So first of all you see this reaction and this question has been asked in previous year. It is very simple. First of all heat is going out. So it is what? It is exothermic reaction since heat is going out. So exothermic reaction means del H is less than 0. So for exothermic reaction when del H is less than 0 this is favored by what? Forward reaction we have so we must have K2 greater than K1 for forward reaction this is also the condition we have to satisfy and this is only possible when 1 by T2 minus 1 by T1 is positive because this is negative and negative is positive. So 1 by T2 minus 1 by T1 is positive. That is only possible when T2 less than T1. So obviously low temperature will be favorable for this reaction so either B and D will be correct A and C are not possible. Now pressure thing we can understand by del NG value what is the del NG value of this reaction? It is less than 0 del NG less than 0. So just now we have done for del NG less than 0 as pressure increases forward reaction favorable. This is the condition we have here. So obviously we have high pressure and low temperature. B is also not possible. Option D is correct. Okay, Bharat has left. Okay, anyways. Option D is correct, right? So this is how we do the questions of all these Lich Atelier's principle and all. There are a few more conditions we have and that we will see quickly. That is addition of catalyst you write down. Addition of catalyst. The second next condition is a catalyst won't affect the equilibrium state. It only increases the rate of forward and backward reaction by decreasing the activation energy. Okay, so write down into this. Catalyst won't affect catalyst won't affect the equilibrium condition. It only increases the rate of forward reaction RF and RB rate of backward reaction helps the reaction helps the reaction to attain the equilibrium state equilibrium state sooner. That is the addition of catalyst. See catalyst we are assuming positive catalyst here, right? If the catalyst is negative, right? Then that will increase the activation energy. This catalyst is positive we are assuming. If the catalyst is negative that will increase the activation energy and decreases the rate of forward and backward reaction and then the equilibrium will attain after long time. Okay, comparatively more time will have to wait into that. If you see the graph over here the graph we have in this case, suppose this is the graph we have for any reaction. Okay, obviously this is for reactant and this is for product. Okay, so this graph, the yellow one is without catalyst. Now the same reaction is taking place with the help of a positive catalyst and the reaction goes like this. It lowers the activation energy that initial and final position will be same, right? But this green one is with positive catalyst and if the catalyst is negative, if the catalyst is negative, that catalyst will increase the activation energy and then the reaction goes like this to increase the activation energy and then comes down to the final value here. Right, so this blue one is with negative catalyst. So both is possible. equilibrium condition will be same. But with positive catalyst it attains with it attains sooner and with negative catalyst it take long time. That is what you see this is the decrease in activation energy with positive catalyst and this is the increase in activation energy. So when this value decreases the rate of forward reaction will be more and rate of backward reaction will also be more. But when this value increases, rate of forward reaction will be will be less. You see, large is the energy difference here. So we have to the rate of forward reaction, backward reaction both will decrease in that case and hence the equilibrium attains after a long time. That's the meaning. Which one? Which one you didn't understand? Pushpin the catalyst effect. See catalyst effect is nothing. First of all, catalyst won't affect the equilibrium condition. Right. Suppose A and B A and B reaction we have right A and B we have some reaction and here is the equilibrium condition we have somewhere in between. Right. So this A this is the forward reaction and this is the backward reaction we have. So after some time this equilibrium will get into this. But if you have catalyst right, so catalyst does not do anything over here. If the catalyst is positive then the rate of forward reaction will increase. RF will increase over here. It only increase the rate of forward reaction and backward reaction. The rate at which A is converting into B with positive catalyst this rate will increase and the rate of B also will increase. Both reaction whether it is forward reaction or backward reaction is going with a faster rate. Right. And that's why we attain the equilibrium sooner. Right. It requires less time to attain the equilibrium state over here. Condition of equilibrium will not change. Neither this initial and final state will change. That will also be same. Right. But initially without catalyst if the equilibrium is attaining after suppose 10 minutes then with positive catalyst this time will only decrease and it will be anything less than 10 minutes. Right. And that's why we say with positive catalyst by which the reaction the equilibrium is attaining that will be comparatively less. Why it happens because the activation energy decreases with positive catalyst. This is the activated complex and activated complex forms here with positive catalyst at a lower energy only. This is the energy axis and this is the progress of reaction. X axis is the progress of reaction. Y axis is the energy axis here. So energy of activated complex will increase. So if this energy difference which was initially from here to here between reactant and the activated complex now with positive catalyst the energy difference is this only. So less energy difference so it can cover it easily and that's why the rate of our reaction is more and reverse is true when the negative catalyst is there. Okay. This is the effect of catalyst we have next and this is important also this question they have asked many times. Addition of catalyst you must keep in mind equilibrium condition will not change but equilibrium attains with the faster rate if catalyst is positive. Now effect of introduction of inert gas effect of introduction of inert gas the last condition we have here for Lee Chatelier's principle and that you write down addition of inert gas this is possible in two different cases again. We can add or introduce this inert gas at constant volume and constant pressure. Okay. So first condition you write down if you are adding this inert gas at constant volume at constant volume. So at constant volume write down there is no change in the concentration of at constant volume there is no change no change in the concentration of reactant or product no change in the concentration of reactant and product and hence and hence inert gas hence there is no effect simply write down okay hence there is no effect at equilibrium equilibrium. Right. Add an inert gas at at constant volume okay. So that will not change the that will not change the equilibrium condition example you see suppose I have taken this reaction A gas plus B gas gives C and D now in this what we write the Kp value is equals to what the Kp value is equals to partial pressure of A into partial pressure of B divided by partial pressure of C into partial pressure of D right. Now in this suppose if you add an inert gas right inert gas we are adding into this whose number of moles is N right N moles of inert gas we are adding right. If this will change the partial pressure of this individual component A B C and D right then Kp value will change if the partial pressure will not change Kp value will also not change right. So if I write down the formula of partial pressure that is suppose A we are calculating partial pressure of A that will be equals to what mole fraction into total pressure mole fraction what we can write it is number of moles of A by total number of moles that is N A plus N B plus N C plus N D plus the number of moles of inert gas N right N2 what is the partial pressure here that will be total number of moles N T RT P volume right this is the partial pressure we have when we added N moles of inert gas this is the inert gas number of moles of inert gas we have. When we haven't added this inert gas what is the partial pressure then it will be PAA RT by V and PV is equals to NRT we are using so this is the partial pressure when there is N T when we have added this inert gas so what happens here this is very clear from this reaction from this equation that N T that you have here this is the total number of moles which includes what which includes inert gas also and what is this value in the denominator this value is also the total number of moles which is nothing but N T so obviously this and this will get cancelled and the partial pressure again you get here that will be N A RT by V which is exactly same that we had earlier when there is no inert gas added right so whenever we add this inert gas the partial the Kp value will not change since the partial pressure is not changing if the volume is constant we are introducing the inert gas at constant volume this is the first case we have second case will take a constant pressure so whenever we say the partial pressure initial and final is same what we conclude from this that at constant volume addition of inert gas won't affect the value of Kp hence it has no effect at constant pressure understood this now the second condition we are taking addition of inert gas at constant pressure sorry I forgot to write here at constant volume this is and this is at constant pressure now in this also at constant pressure this is the last case we are discussing we have three cases possible we will do this quickly once you understood the first one you will understand the other one also see first case here you are taking that is when del Ng is equal to 0 and since the pressure is constant so we have to see the effect on equilibrium when change in volume is there since we are adding inert gas so that will change the volume right total volume will change after the addition of inert gas suppose some simple example I am taking a gas is giving b gas smaller reaction I am taking del Ng is equal to 0 so what is the value of Kc here that will be concentration of b by concentration a right concentration is what number of moles by volume divided by number of moles by volume volume volume gets cancelled Nb by Na you see this Kc is independent of volume term independent of volume hence addition of inert gas addition of inert gas won't affect equilibrium if del Ng is equal to 0 see one thing you have to understand here we are adding inert gas at constant pressure right so pressure is constant that won't change but since we are adding inert gas that will change the total volume so if Kc is independent of volume term then whatever the change in volume that won't affect Kc that's why we are looking at the volume term over here now this one you see if I take the second case into this and that is when del Ng del Ng is greater than 0 suppose the reaction is this a gives b and c okay so for this Kc will be what it will be concentration of b concentration of c divided by concentration of a so this gives you Nb into Nc by Na into 1 by v this is the expression you get if you solve this because concentration is mole per liter mole per liter here we have v square and here we have v that will get cancelled will get v okay now since we are adding inert gas right so volume increases so what happens here as volume increases right as volume increases then what happens this we are increasing Kc value will not change because again it depends on temperature so if this volume is increasing so to maintain the ratio this Nb Nc value should also increase right and that is only possible when the reaction goes in forward direction right so as volume increases moles of product number of moles of product should also increase to maintain the ratio and that is only possible when the reaction goes in forward direction is it clear del Ng less than 0 A gas so for this if you write down Kc that will be number of moles of C by number of moles of A number of moles of B into V so as volume increases that Na Nb Na or Nb should also increase to maintain the ratio and that is only possible when the reaction goes in backward direction is it clear okay so these are the few you know all possible cases we have discussed for leach-atelier principle now we will see some questions which has been asked already on this leach-atelier principle okay so you see the first question on this and the question is the reaction is given and the reaction is N2 plus O2 N2 plus O2 gives 2 NO plus heat this is the reaction given in the above gaseous reversible reaction all these are gases gas gas and gas gaseous reversible reaction if the pressure increases if pressure increases then what is the effect on effect on Kp equilibrium constant what is the answer one more question let me write down N2O4 gas gives 2 NO2 at a fixed temperature fixed temperature the volume of the reaction container is halved for this change for this change which of the following statement option A neither Kp nor alpha changes option B Kp alpha both changes Kp changes alpha constant alpha won't change means Kp constant Kp constant alpha changes what is this possible ok the first one is pretty clear that derangey value is 0 so Kp will not have any effect right so that is correct second one you tell me what is the answer Kp and alpha union saying D is also saying D anyone else yeah it's right you see D option is correct why because you see the temperature is fixed and Kp is what Kp is the equilibrium constant so if the temperature is fixed then Kp won't change so Kp will be constant that is true but since the reaction container is half that will change the degree of dissociation this is degree of dissociation this will change the degree of dissociation degree of dissociation so that's why answer is D one more question you see here and this question was asked in JeAdvance ok so you see this question the last one for this then we will move on to the next topic the reaction is given CaCO3 solid gives you CaO solid plus CO2 gas this is the reaction now in this for this equilibrium the correct statement is whatever rate you may have more than one answer also possible ok the correct statement you have to find out and options I am giving you ok the options are del H that is the enthalpy of the reaction depends on temperature B K is independent of independent of initial amount of CaCO3 C option K depends on pressure of CO2 pressure of CO2 at a given temperature option D del H is independent of independent of catalyst if any if it is present BC in D BC increases or decreases decreases in volume ok one question that reactionary has asked so does alpha increases or decreases with decrease in volume ok so volume decreases means what reactionary volume decreases means concentration increases no right as volume decreases then concentration increases and that will increase the dissociation of alpha right more concentration will be there so more will be the dissociation so that you will attain the equilibrium see like this also you can understand see you can easily understand this if you write down the expression for KC right if you write down the expression for KC for any reaction which has del NG greater than 0 right so del NG greater than 0 will be what can be like this it is concentration of B concentration of C by concentration of K so concentration of B and C and B and C by NA into 1 by B right so if this volume decreases the number of moles of B and C will also decrease right if this volume decreases volume becomes half right so number of moles of B and C will also decrease to maintain this KC so that will also affect the alpha and when you see this when KC decreases ok one thing you can understand from this first of all you see ok I will just we will come to this question we will come to this question in some time ok first you see this that if this what we say volume of the container is half and what is the change on alpha whether it is increases or decreases alpha will change that is true but what is the change whether it is increased whether it increases or decreases right so that you can understand alpha is what it is the number of moles react divided by the initial number of moles right so if the number of moles react is more obviously directly you can do from here if the number of moles react increases that will increase alpha also when you are decreasing the volume let it be don't write this when you are decreasing the volume obviously the concentration of the reactant will increase and when the concentration of the reactant will increase that will increase its association constant also that will increase its tendency to get dissociated also right and that is why the number of moles reacted here will be more and that's why the alpha will be more in that case ok directly you can do from this ok now this question tell me what is the answer what is the answer tell me see why why C is the answer I don't know why all of you are saying C B is one of the option that is correct Pratik why C just a second I am coming I am sorry ok so you see this partial pressure you see the question is what this question was asked in J advance ok multiple choice question so you see first of all this del H depends upon temperature it's correct one equation you must have done in physics also that is Krzysztof's equation right del H value depends upon temperature with this relation see here I will just write down the answer in this question will be A and B this A and this B is correct this is not correct and this is not correct del H suppose del H you have to find out at any temperature T2 and if you know the value of del H at temperature T1 del H is equals to T2 is equals to del H at T1 plus del Cp into T2 minus T1 this we call it as Krzysztof's equation it is probably given in physics ok so first of all the enthalpy change depends upon temperature what temperature you are taking right ok I am coming to that one another and this B is obviously correct right K is independent of initial initial amount of this right so B is correct coming back to this C K depends on partial pressure of CO2 at given temperature K depends only on temperature it won't depends on partial pressure of any of the component correct this means what when you change the temperature according to partial pressure K also changes right but partial pressure has no role into this so K depends only on temperature option C is not correct del H is independent of catalyst del H is independent of catalyst because you see the activation energy for forward and backward both decreases right see this value is also decreasing and this value is also decreasing if the catalyst is positive so that will decrease in such a manner so that the del H will be constant it is not like the rate of forward reaction or the energy difference between the activated components and reactant is decreasing so this energy is also decreasing with product in this so now I will discuss one small concept of buffer solution in ionic equilibrium same thing same thing so we will start buffer solution that will take around 20-25 minutes and then last we will do that GOC because there are many things only specific thing will discuss did you understand this del H depends upon temperature K independent upon initial amount and D option is not correct because both energy is decreasing since the activation energy is decreasing did you understand this now next you see we will we will start next buffer solution ionic equilibrium will have buffer solution so like I know there are many things that you should know before this buffer solution only but since it is important buffer solution we will discuss this buffer solution only into this and then we will see the GOC so let's discuss this what is a buffer solution see if you have any solution any solution you have which has some pH value means either it is acidic or basic so what happens if you add an acid into it we are adding an acid into it so acid gives what H plus into the solution and when H plus you are introducing into the solution so obviously the concentration of H plus is increasing and when the concentration of H plus increases its pH value decreases because we know pH is equals to what minus log of H plus if you add base into it OH minus then its H plus concentration will decrease pH will increase and that gives you basic solution so the point I am trying to make here it is what that whenever you add an acid and base into any solution that will change its acidity and basicity of the solution buffer solutions are those solutions which I thought some problem buffer solutions are those solutions which resist the change in pH which resist the change in pH actually you see like I said here you are adding acid into it that will eventually change the pH of the solution buffer solutions are those solutions which resist the change in pH change in pH on addition of addition of small amount of small amount of acid or base into it acid or base it will be any solution if you have and if you have added small amount of acid and base if the solution resist the change in pH the solution is said to be buffer solution see it resist the change in pH but it does not mean that the pH of the solution will not change the pH will change but the change will be less than the change when the solution is not a buffer solution right means what if you have a buffer solution and one is non buffer solution if you add equal amount of acid into these two solution one is buffer other one is non buffer then the change in pH of non buffer solution will be more than the change in pH of the buffer solution so one thing you should be very clear that it is it does not mean that the change in pH of the buffer solution is zero no it is not zero but the change is very small and this change is obviously lesser than the change when the solution is not a buffer solution is it clear the definition ok now you see here the first thing you should be very clear buffer solution pH changes but that change is not that much now you see the buffer solution there are mainly actually there are three types of buffer solution but mainly we have two types that we will discuss important all these three types we will discuss in the class right but two types which are more important that we are discussing today acidic buffer the first one is acidic buffer the second one is basic buffer and the third one is salt buffer right acidic buffer, basic buffer and salt buffer you have to be very clear because all those 12th class students they have done this inic equilibrium 11th class you haven't done but you don't worry just try to keep these discussion in mind because buffer solution is important if you get some direct question from this in kvpy you can answer it right but again I am telling you one thing there are many other things which we should discuss before this buffer solution but we don't have time for that much ok so I am directly giving you the concept of this so that you can understand right and most of the 12th class students they will agree with me that in inic solution we have many formulas of pH calculation of pH there are many formulas of pH so what formula you have to apply when that you have to understand ok so keeping this in mind let us discuss this and then we will come back to this how do we analyze this particular thing right acidic buffer you see first of all you should know how this acidic buffer forms right so one thing that you have to keep in mind acidic buffer forms by the combination of forms by the combination of solution of weak acid it is the solution of weak acid plus it is salt with it is salt with strong base weak acid and it is salt with strong base right so weak acid the example of weak acid is suppose CH3COOH and it is salt with strong base means what when the CH3COOH combines with NaOH so that gives you CH3COONA right this is the salt of weak acid and strong base plus H2O we get here right plus H2O so whenever you have this combination CH3COOH plus CH3COONA this forms an acidic buffer right whenever you have combination of weak acid and it is salt salt with strong base it forms an acidic buffer okay now the point is you consider this reaction this reaction here this is what this is a neutralization reaction acid and base gives salt plus water right okay now if I give you one thing here you see if I give you one mole of this and one mole of NaOH is this a complete neutralization reaction is this a complete reaction if you take one mole of this yes because one mole of CH3COOH requires one mole of NaOH so when you take one one mole of both compounds so there is no limiting reagent both reactant will react 100% and converts into salt and water yes but it is because you see the reaction is balanced right reaction is balanced one mole of CH3COOH react with one mole of NaOH so this reaction goes towards this reaction goes towards 100% completion CH3COOH won't fully dissociate fine that won't dissociate NaOH there is one more thing here one mole of CH3COOH it is not like we are not considering the dissociation here see so kids we are not CH3COOH is a weak acid correct dissociation dissociation won't be 100% right but that is true that is true this reaction won't be 100% complete that is true but I am just considering the amount here sometimes you know what happens let me finish this then probably you will understand what I am trying to tell you sometimes what happens here see for this solution let me write down one thing over here for this solution you see when the reaction takes place right the pH that you get for acidic buffer the formula of pH that we can always find out but I will give you the formula directly now that will be equals to pKa plus log of log of salt by acid concentration of salt by concentration of acid concentration of salt by concentration of acid this is the formula we have salt means what let me write down this again pKa plus log of this again just give me a second salt is Cs3COO minus and acid is Cs3COOH this is the formula we have right now you see what happens here when this reaction takes place if you know the degree of dissociation of this if it is given sometimes in the theoretical questions what happens theoretical question I am talking about actually this will not dissociate because that is a weak acid right but sometimes in theoretical questions they will ask you whether this solution forms a buffer solution or not right sometimes in the question they will give you that this is an acidic buffer then you have to calculate pH then you know that okay acidic buffer we have then we have to use this formula okay directly you can use but sometimes they will not mention you that the solution forms a buffer solution then you have to first analyze that whether it is a buffer solution or not so suppose what happens but take a hypothetical situation here right if this Cs3COOH reacts completely with NaOH NaOH suppose you have excess amount right suppose this Cs3COOH reacts completely with NaOH then in the solution we have what we have Cs3COOH Na present we have H2O but we don't have this Cs3COOH right and for acidic buffer we require Cs3COOH and Cs3COOH Na both so if the confusion is what like since you got this that it is a weak acid won't dissociate but sometimes what happens they will give you equal number of moles here right and they will not mention in the question that the solution is forming a buffer solution right and most of the time it should end what does they feel okay one one mole we have so one mole reacts with this so this will react completely so there is no more Cs3COOH present in the solution hence the solution won't form a buffer solution but that is not exactly true if I give you suppose two mole of this right and one mole of this right so we know one mole reacts with one mole so this one mole reacts with again one mole of this only so one mole of Cs3COOH left so here we have one mole of Cs3COOH and one mole of Cs3COOH Na forms so we have acidic acid also and we have this salt also present and whenever these combination are there it forms a buffer solution right that is the point I was trying to make it is true that weak acid won't dissociate like some of you have said that yes it will dissociate right it will form 100% utilization is there right but the point is if they have given equal number of moles then also this particular thing forms a buffer solution right if the number of moles is different then clearly you can say like some of you have said okay this will react 100% and form a salt right but if the number of moles are different then you can understand okay some amount of Cs3COOH is there and Cs3COOH Na also forms the solution will be a buffer solution and then we will apply the formula of buffer solution pH of buffer solution right sometimes in the question what happens they will tell you that this Cs3COOH react completely with NaOH excess of NaOH present okay and that makes the dissociation of Cs3COOH possible and there is no more Cs3COOH left in that case you will use the formula of neutralization reaction where the weak acid react is strong base all these 12 classes students you must have done this neutralization reaction first strong acid is strong base weak acid weak base weak acid strong base strong acid weak base like this four possibilities are there that's why you see this particular reaction is reversible in nature also we should write down this sign here because this also forms here but that is what again a different thing we have but whenever you have combination of Cs3COOH and Cs3COOH Na you have to keep this in mind that the solution may form buffer solution and the type of buffer solution is acidic buffer and then you have to use the formula of buffer solution which is nothing but this Cs3COOH minus and Cs3COOH right this is what the concept you have to keep in mind because there are many formulas of pH what formula you have to apply depends on what type of solution we have right so most of the time in the exam they will give you this but they will not mention it is a buffer solution you have to keep this in mind okay whether the solution is forming a buffer solution or not then you have to use the formula okay anyway this is the one type of buffer solution we have acidic buffer which forms by the solution of weak acid and it's solved with strong base this is the formula we have right and Cs3COOH minus concentration of salt and concentration of acid okay at equilibrium now basic buffer you write down the second type of buffer solution here that is basic buffer and one more thing when I take this revision class for any equilibrium then I will again discuss this okay we will discuss it and see some questions on to this also okay so you yeah so you don't worry with this we will again discuss this because we are not discussing now how buffer solution works right how they will neutralize the pH how they will maintain how they try to maintain the pH of the solution that we are not discussing since we don't have time but we will discuss this in the class when we take the revision class of this okay so you don't worry with that okay now next you write down the second type of buffer solution is basic buffer basic buffer this type of solution forms by the combination of it is a solution of weak base plus it's solved with strong acid reverse solved with strong acid an example weak base we have NH4Cl right and it's solved with strong acid suppose we have weak base and it's solved with strong acid suppose the salt of this is strong acid is I am taking suppose NH4 OH sorry NH4 OH I have written it reverse this is weak base and solved with strong acid basically weak base is NH4 OH and strong acid suppose I am taking HCl right this also be a reversible reaction why we will discuss this later on NH4Cl plus H2O so this is the salt of weak acid and strong base weak acid and strong base NH4Cl and NH4 OH is the weak base right so whenever you have this combination NH4Cl plus NH4 OH that also forms a basic buffer right and the formula of basic buffer is what it gives a basic solution so we write here POH POH is equals to PKB plus log of concentration of salt by concentration of base that is the formula we have ok here also we have same thing you have to analyze and see whether this NH4 OH and NH4Cl is present or not in the solution then only we can say it is a basic buffer now one question only we will discuss so that you can you know understand how to solve the question suppose the question is this on this last question we are discussing on this and then we will see some concepts of GOC question you write down you see it is given CS3 COOH in short I am just giving you data CS3 COOH plus NaOH see they will not give you weak acid and it is solved with a strong base that is the point I am trying to make it is acid and base only you have to analyze that in the reaction mixture whether the acid and it is strong base is present or not ok and another thing is given it is 200 ml volume of this is 200 ml and volume of this is 100 ml it is 0.1 molar and it is 0.1 molar NaOH you have to find out the first thing the pH of the solution if 50 ml of 0.1 molar HCl added T ml of 0.1 molar NaOH added in all these three cases you have to find out pH of the solution one more thing Ka value is given Ka value is 2 into 10 to the power minus 5 this is given then if you understand this question this is very conceptual question if you understand this question you can understand buffer solution easily value of what Sukirt Sukirt dissociation of it is not given COHPK that is not required Sukirt here that is what you are having this doubt few things we haven't discussed in in equilibrium that is why you are asking this ok so we will do this ok wait