 Hello guys. Good evening. Can you hear me? So last class we discussed the relation of KP and KC, isn't it? Yeah. So we discussed about the relation of KP and KC. I think most of you have joined. Yes. So, yeah, so the unit of KP and KC also we have discussed. Unit of KP and KC. Okay. So last class we were here. We stopped at this. We discussed law of mass action. And then we'll see the equilibrium constant constant pressure is equals to the equilibrium constant at no in terms of pressure in terms of concentration. RT to the power delta N. Okay. So if you look at the unit of KP and KC here, unit for KP, it is atmospheric because we have pressure to the power delta N. And KC is concentration to the power delta concentration is more per liter to the power delta N. Okay. We also discussed at depending upon the value of delta N, what would be the relation of KP and KC, all these things. Okay. Yeah. Yes, yes, yes. More per liter. Also, you can write more. That is the molarity. That's fine. Not a problem. You can write. Okay. We also say in a concentration, two molars, three molars. So that's fine. Not a problem. Okay. Next you see the characteristics of of KP and KC of KP and KC point wise you write down the first point here equilibrium constant. And when I say equilibrium constant means we're talking about KP KC both. Okay. Equilibrium constant does not depend on depend on the initial concentration, initial concentration and the amount of amount of reactant and product expression of of case KP and KC is only applicable. Equilibrium can write only when the equilibrium is achieved. Third point, the equilibrium constant constant for a reverse reaction or a reverse reaction is reciprocal is reciprocal of the equilibrium constant of the original reaction. Original means the forward one. Copy down these points. Now look at this example for this point. Suppose we have a reaction. H two plus I two gives two H I KC for this reaction is concentration of H I square concentration of H two and I two. Now if you reverse this reaction, and the reaction would be this to H I gives two plus I two. So for this reaction KC dash if you see this would be concentration of H two concentration of I two by H I square. So if you look at the relation between the two, if you compare this, this expression and this expression if you compare, right, what we get here. We get KC is equals to one by KC dash. So if we say when we reverse a reaction, the equilibrium constant also gets reversed. Okay, that's the point gives two H I we know it's equilibrium constant is KC. Now what happens if you multiply this number by this reaction by any number suppose to be multiplying it with two H two plus two I two gives four H I. So what would be a KC dash for this reaction. The KC dash would be H I to the power of four divided by H two. Square and I two. Square, which is nothing but KC is square. Okay, so in this case the new equilibrium constant. Third point, right, three point I've given you the last one was this only equilibrium constant of a reverse reaction is reciprocal of the equilibrium constant the original reaction. Yeah, this is under that only we're discussing. One more example we're discussing on this. Constant KC dash is equals to KC is square. So any reaction if you multiply by any number to have for whatever it is, the new equilibrium constant, right, is equals to the old one. The previous equilibrium constant the original reaction to the power the number with which you have multiplied the equation. Clear. Oh yeah I'm going back once again. Okay so we're done with this. Now, another one you see in this one. Another point in this. Suppose we have two reactions. Right. Two reactions we have suppose a gives B. And C gives D. For this one the equilibrium constant is K1. And for this one the equilibrium constant is K2. If you add the two reaction what happens. Which means we'll get a plus C gives B plus D. Right. So for this one, when we add the two reaction, the new reaction that equilibrium constant KC would be. Concentration of B. Concentration of A into concentration of D. Concentration of A into concentration of C. If you find out the expression of K1. K1 would be B by A. And K2 would be D by C. D by C. Okay, so if you take the ratio here you see B by A is nothing but K1. D by C is nothing but K2. So the new equilibrium constant equals to the product of the reaction that we have added. Then. Okay, similarly what happens if track the two. Okay. So when you subtract the two reaction you see. This would be A minus C. D minus D. Okay. Which we can also write. This equation we can also write. A plus D. Gives. D plus C. Right. So. The KC for this reaction would be. Concentration of B by A. That is K1. Concentration of C. By D. That is one by K2. So KC is equals to what? K1 into one by K2. So it is something like you have some reverse the first reaction, the second reaction and then added with the first one. This two, three properties you must remember plus you also keep this in mind that equilibrium constant depends only upon temperature as if the temperature is constant equilibrium constant won't change. Okay. Yeah. The next thing you see heading right down predicting the the extent of of reaction. Mostly it is a factual thing. Okay. You should know a few things into this. There's two conditions we have nothing much. That you need to keep in mind. Okay. Not that important also. Okay. So for any reaction, what we have in this. For any reaction if KC value. Is greater than 10 to the power three. Right. Generally what happens if KC increases. The tendency will be more in forward direction. The reaction has more tendency to go in forward direction. More value of KC. More forward reaction will be. Okay. From how like how did we get this value that is experimental. We don't have to bother about it. Right. So if KC value is more than 1000. Then what we say obviously the reaction. Moves in forward direction. Towards forward direction. Right. And the position of equilibrium. Position of. Equilibrium lies towards the product. So basically it is shifted towards the product side. Equilibrium position. If KC is less than 10 to the power minus three. The value is extremely small. And the reaction. Goes in. Backward direction. And the position of equilibrium. Towards the reactant. Just opposite. If KC value. Lies in this range. More than 10 to the power minus three. And less than 10 to the power three. And the reaction. Can go. In forward. Or backward direction. Depending on. On the conditions. Okay. Next you write down the application of equilibrium constant. Application of equilibrium constant. Let us assume a reaction first. And the reaction is. A plus B. Gives C plus B. This is the reaction we have. See whether the equilibrium is achieved or not. We can always find out the concentration of product. By the concentration of reactant. This ratio we can always find out. Whether it is equilibrium or not. Okay. AB starts reacting. Converts into product. This ratio we can always find out at any one time. This ratio is called. Reaction quotient. This ratio is called. Reaction quotient. When time T is equals to. Is not equals to T equilibrium. When T is not equals to T equilibrium means that when the equilibrium is not achieved. The ratio is called. The ratio is called reaction. Caution. The ratio is called. Reaction quotient. Okay. What happens at T is equal to T equilibrium. That means when the equilibrium achieved. Right. This Q becomes. Casey. The equilibrium constant. So equilibrium constant. Is nothing but the reaction quotient. And it is also defined as. The ratio of product by reactant as we know. But the only condition is what. This ratio this concentration of ABCD. We'll take at. Or when that equilibrium is achieved. T is equals to T equilibrium. No doubt in this. So basically at T is equals to T equilibrium. K becomes Casey and that is nothing but the equilibrium constant. That's one thing. Now suppose if I represent this reaction on the scale. Where initially we have. The reactant here. A plus B. And then finally we have the product. C plus T. And somewhere in between. We have the equilibrium condition T equilibrium. So at this point you see. At this point the ratio of C and D is said. Casey. Equilibrium constant. And at any other point other than this T equilibrium. At any point. Right. This ratio is what. This ratio is said to be what. Reaction quotient or QC also sometimes we write. This is also. Basically the ratio of product by reactant. At any point other than equilibrium point. It's called reaction quotient. Okay. This is what the meaning we have here. Now when we have the reaction. Equilibrium is achieved. And if you disturb the equilibrium condition, like suppose if you add some reactant into that. You increase the pressure or anything is there. If you disturb the equilibrium, or we can say if the equilibrium is not. Is not achieved. Then we have two things possible here. Suppose at any point of time. The Q is coming out to be here at this point. Right. So reaction always has tendency to go towards equilibrium state. So if this Q is here. Right. Then the condition is what we can write. Then the condition we have here. First condition. Is Q is less than KC. Q is less than KC. Which is somewhere on the left side of this equilibrium constant. Yeah, suppose what I said that reaction always has tendency. To go towards the equilibrium state. So this has tendency to go towards the equilibrium state. Hence forward reaction is possible here. Right. So when Q is less than KC. The reaction will go in forward direction. End out in this second point. If Q is greater than KC. Then we have backward reaction possible. And when Q is equals to KC. That is the case of equilibrium. Now based on this, you try this question. The question is. We have a reaction given. Give C plus D. Assume the total volume is one liter. It is given is one liter. Okay. And at any point of time, the number of moles of A is one more. B is one more. C is one more and D is one more. The question is find the. Equilibrium constant. The equilibrium constant. Once again, find the molar concentration. Sorry. Find the. Concentration of a. Concentration of B. Concentration of C and. Concentration of D. Right. Concentration of ABCD. At equilibrium. This we need to find out if. The KC value is given first question. The KC is one. The KC is a hundred. And. KC is. Point one. But this three different data of KC. You need to find out the equilibrium constant of the concentration of ABCD. Try this. Done. Yes, volume is given. Volume is one liter. Yeah. Volume is the reaction in which the reaction, the vessel in which the reaction is taking place. The volume is one liter for that. Okay. Did you get it? Okay. See what we have to do in this one. See here. Try to understand this. Okay. Okay. This is the reaction we have. Now you see first of all. This is the number of moles given. Right. Volume is one liter given. So this means. Mole is nothing but the concentration here. We can assume. This is volume is one liter. Mole is nothing but the concentration. Okay. Mole is equals to the concentration we can take. First of all, you see. This is the number of moles given. We do not know at what stays this reaction is. At this point of time. We do not have any idea. Whether this reaction will go in forward direction. Or in backward direction. We don't have this info. So how do we know that first of all. Right. So we do not know whether it is the equilibrium state or not. So if you find out the ratio of the product. And reactant. This ratio is what could you tell me? So what is this ratio? No, it's not Casey. That's what I'm telling you. Let's not, you know, go into this data. Try to understand the question first. Okay. Yes, molarity is concentration. But since volume is one. So concentration is more per liter. Right. Volume is one liter. So more is nothing but the concentration you can consider. Understood. So this value you let it be. What is this Casey values given? Let's first try and understand this. This ratio, since it is not mentioned. That this mole is the number of moles at equilibrium. So this ratio we can check. We can say it is the reaction quotient. Isn't it. Now we do not know whether it is at equilibrium or not. This value is at equilibrium or not. This ratio definitely we can say it isn't. It is a reaction quotient. Isn't it. Yes or no. Please. So if we know the value of Casey. And Casey, we compare with this queue because cuto, we can find out easily more is given one, which is nothing but the concentration. So it is one into one divided by one into one, which is one. So if you know the value of Casey, Casey, we can compare with you. And depending upon the relation, it is equal to Q greater than Q or less than Q. We can say whether the reaction is at equilibrium or it will go towards the product side or towards the reactants. Can we say that? Yeah. So for a given data here, ABCD, we got the value of key Q, which is one reaction quotient is one. Now you look at this question. The first question is given. What is KC equals to one? And Q we have already calculated equals to one. This means what Casey equals to Q. This condition means what could you tell me? What do you mean by this condition? This condition means the reaction is at, is at equilibrium. Right. So whatever the concentration given number of moles given, that would be the equilibrium. Concentration of reactant and product. And that is what we need to find out. Did you get this? So we can compare this with Q and KC and we can say whether we have forward or backward reaction and we can do this. Isn't it? First one you understood. So this is the first case we have. Now, the second question we have here is for the second one, the KC value is given. Is it 100 or 0.01? Can you tell me the second value? Let me go back. KC, the second value is given 100. So this value is 100. So Q we already know. So for the given value of Q, we can say KC, sorry, KC is greater than Q. KC is greater than Q. Now, when KC is greater than Q, could you tell me forward or backward reaction under this termination? Forward reaction, okay. So we concluded this thing that we have the forward reaction. Now we see what we'll do. We have the forward reaction. So we know the direction of reaction now. Initially, we did not know. When the question was given, we were not sure that whether the reaction will go in forward or backward direction. Now we understood this. That's what the application of KC. We can compare the value of KC and Q and we can say whether the forward or backward reaction is there. So what happens when the reaction goes in forward direction? You see, we have A plus B and it is given one, one, one, one more, right? Forward reaction means what? The concentration of A and B decreases. So we assume A and B reacts X mole and forms X moles of C and B. One plus X, one plus X. And the equilibrium is achieved. This is what happens to achieve equilibrium. This reaction takes place. Did you understand this? No, it's not. From one, it was one, right, initially. And then A and B is reacting. So obviously here it would be one minus X. Out of one X mole reacted. That's what we are assuming. Complete reaction won't be there, no. So you can assume a number. You can assume here that's Y. Y is equals to one minus X that you can say. But we'll write like this. Out of one mole, X mole reacts. So one minus X left, one minus X left. One plus X forms, one plus X forms. And the reaction maintains the equilibrium condition in this situation, right? So what we can write here, you see, the KC expression would be what? Can we write KC for this reaction would be one plus X by one minus X square? Is it clear? C into D by A into B, one plus X, one plus X, one minus X, one minus X. Okay. This value is given in the question, 100. Could you find out the value of X from this? Find out the value of X. You substitute it here. You'll get the concentration of A, B, C and D. That is that equilibrium concentration. Yes, got it? Tell me the answer here. No, nine you won't get. Nine by 11 is fine. X equals to nine by 11 you will get. So what is the concentration of A, B, C, D at equilibrium? The concentration of A, because this is what we need to find out equals to B at equilibrium. And that would be one minus nine by 11, which is two by 11. Concentration of C into concept is equal to the concentration of D. One plus nine by 11. That would be 20 by 11. This is the answer for your question. Now, could you tell me for the first third one? When Kc is equals to 0.01. Try this one. Yes. What is the value you are getting? Again, you will say what? Q is equals to one we have. That is given in that question. So we have Q is greater than Kc. Under this condition, we have backward direction reaction. Backward reaction possible. So A plus B gives C plus D. One more, one more, one more, one more. Backward reaction is going on, right? So C and D will react. So it is one minus X. One minus X. One plus X. One plus X. Then you substitute the value of Kc. Expression of Kc would be. One minus X by one plus X square. So one minus X by one plus X equals to one by 10. X value would be what? Again, nine by 11. X value is nine by 11. So concentration of A equals to B. Equals to one plus nine by 11. That is 20 by 11. Concentration of C equals to concentration of D. Is equals to two by 11. This is the answer we have. How many of you understood this? Is it clear? No doubt. Yeah. Now you see we have types of equilibrium. Just the definition you need to know here. Types of equilibrium. The first one is homogeneous. Homogeneous equilibrium. This is the equilibrium in which. In which. All. Reactant. And product. All reactant and product are. In the same phase. Miss. We have only one phase. Okay. Solid or liquid or gas. Only one phase. Same phase reaction. For example, you see. Suppose we have S2 gas. Plus I2 gas. Converts into two at I. Gas all our gaseous phase reaction. This equilibrium is homogeneous equilibrium. And just opposite we have the next one. That is heterogeneous. So equilibrium in which more than one phase. More than one phase is present. That is heterogeneous equilibrium. Suppose we have CA CO3. It is a solid. And when it dissociates it forms CAO solid. Plus. CO2 gas. Two different phase we have. So heterogeneous equilibrium. This two types of equilibrium. This definition is required here. Not important. Mostly we'll discuss with. Homogeneous equilibrium. This definition is required over here. Now there is one important term we need to understand. That is. Degree of. Dissociation. Degree of dissociation. That is. Alpha. It is represented by alpha. How do we define degree of dissociation? It is defined as the number of moles. Number of moles reacted. Per unit mole. Per unit mole of the reactant. Mathematically if I write down alpha degree of dissociation equals to. Number of. Number of moles. Reactive. Divided by. The number of moles initially taken. Number of moles initially. Taken. Okay. You see. Yeah. So suppose I have a reaction. I'm assuming a gives B plus C. The reaction is. At time T is equals to zero. The concentration of a I am assuming or number of moles. I'm assuming a zero zero. At time T is equals to. T equilibrium when a starts converting into B and C. X moles of a reacts. X moles of B forms. And X moles of C. So degree of dissociation. Of dissociation. Alpha is equals to. How many moles reacted X. How many moles initially taken a. So alpha is equals to X by a. X equals to what we can write. A times alpha. A times alpha. So if I write down the expression of KC here. For this reaction. It would be. Again, concentration of B. Into C. Divided by K. Volume I'm assuming. As one. I'm assuming this volume. Just to make you understand in the question. Whatever volume is given. You need to take the ratio of mole and volume. So that you can write down the concentration. Because we know. Concentration is mole per liter. Okay. So concentration of B is X. Which is a alpha. So I'll write down here a alpha. Into a alpha. Divided by a into. One minus alpha. So KC equilibrium constant equals to. A alpha is square. By one minus. This is the formula we get. In terms of. Equilibrium. In terms of degree of dissociation. Okay. See this one. The question is. The equilibrium concentration. Of a. BC in a reaction. And what is that reaction. The reaction is three a. Plus B. Gives C plus B. To see I'm sorry. To see plus T. Are. 0. 0. 0 1. 0. 0 0 8. Respectively. Respectively. Calculate the initial concentration. A. And B. Done. Once again, guys. Okay. Anyone got the answer. See how do we do this. One thing you must. Excuse me. One thing you must take care of here. The reaction is. Three a. Plus B. Gives to see plus D. So initial concentration is not given. I am assuming it as a. And this is B. Because that is what we need to find out. We don't know whether it has. Equal concentration or not. I'm assuming it has NB. It's not mentioned that any product is present initially. So C and D will have 0. At time T is equals to. T equilibrium. We can assume. That out of a mole. X mole of a reacts. So how many moles of B reacts. Could you tell me. If X moles of a reacts. Then how many moles of B reacts. X by three. Very good. So X by three moles of B reacts. How many moles of C forms. How many moles of C forms. Two X by three. Two X by three. And this is. X by three isn't it. Right. How do we do this you see. Three moles of a. Reacts with one mole of B. So one more reacts one by three. X reacts with X by three. So out of B X by three has been. Reacted. Similarly what we can say. Three moles of a. Gives two moles of C. So one mole gives. Two by three moles of C. So X moles gives. Two by three into X moles of C. Right. Similarly for D we can write. Okay, but I won't suggest you. That you assume it like this. Okay you can do it like this way. This way also you will get the same answer. Answer is not wrong. But it's better to consider this. I'll tell you what. It's better to consider. Three X instead of X. What I'm telling you. You have assumed X small reacts. I'm telling you. Whatever the coefficient we have. That into X you assume. This amount reacts. Then what is the benefit of this. Instead of X. We have three X. So everywhere. We'll have three X only. You also instead of X. We have. Three times X. You also have three X. Now I wonder if this is what you see. This three and three will get cancelled. Three and three will get cancelled. And three and three will get cancelled. Point I'm trying to make. If you consider this as now you just. This leave it. This you see over here. A moles we have initially. And the amount that reacts. Is three times into X. This is for a. Then for B it would be B minus. It is three. So three X it is one. So X. It is two. So two X. It is one X. This data is easier to write. And easier to calculate also. So I would suggest. That consider data like this. You understand this. Depends upon the social metric option. Anyone need out in this tell me. Because this is what you need to do in this chapter. Okay. This is the only thing which you need to write first. And then you can go with the other steps. To solve new miracles. Okay. So whatever the coefficient is given in balanced equation. That into X you have to assume. Everywhere. Okay. So here what I'm going to do. I'm assuming this as this is the. Number of moles of ABCD. Obviously one. Later you need to assume here. At equilibrium. So what is given in the question. Now you say it's very easy to do. What is given in the question. The question it is mentioned. That at equilibrium concentration of ABC is this. So we have a minus three X equals to what. 0.03. B minus X equals to what. 0.03. What. 0.01. To X equals to what. 0.008. To find out X from this. You always have the data like this so that you can find out X. Because we have three variables know AB and X. Three equations. Right. Find out X. Substitute here. We'll get a and B and that will be your answer. Any doubt in this. Easy. Could you please respond all of you. Tell me the value of A and B. A value. Okay, I am. Oh, you're getting different answers. A job. So. Is this the answer you got. Yeah, I think this is the answer. So you should know how to write down the expression of Casey. By assuming this X or three X whatever. Let me tell you again. If you assume this also it is not wrong. Okay. You can easily you'll get the right answer. From this, whatever you want, you can assume. Okay, whatever you want, you can assume. But this is easier. You don't have to calculate all this. Whatever the question is given three X. X. Two X X. Finish. Yeah. Okay. Now some more, you know, expression will try to find out. Suppose we have a reaction. To a plus three B. Gives four C plus five D. Okay. T is equals to zero. The initial moles is given. It is a B. Zero and zero. This is given. You need to find out the expression of Casey for this. It is a number of moles we have and volume. We are assuming one liter again. Find out the expression of Casey. Assume like that only X or whatever. And try to find out the. Expression. This is again, I'm coming try this. Yes. Okay. So what I'll write down here, I'm assuming. At time T is equals to T equilibrium. At time T is equals to T equilibrium. Two X. Moles of a reacts. So here we have three X. Two gives four. So two X gives again, four X. So two X gives again, four X. Two gives four. So two X gives again, four X. And five X. Nothing you have to think whatever the coefficient with excellent on. Finish. Casey would be equal to. Concentration of C to the power four. So four X to the power four. Concentration of D to the power five. Five X to the power five. Concentration of E to the power two. A minus two X. To the power five. Concentration of E to the power four. Two X. To the power two. And B minus three X. To the power three. This is what the expression we have. So four to the power four into. Five to the power five into X to the power nine. Divided by a minus two X. Square. B minus three X. Q. Okay. If suppose V volume is given. Then the concentration would be. This divided by V. Divided by V, V and V. You need to divide it by V. This question you try. One more. Of nitrogen. Is mixed with. Three more. Of hydrogen. Three more of hydrogen. In a four liter container. In a four liter container. If. Twenty five percent of. And to. Converted into. Into ammonia. By the reaction and two plus. Three H two gifts. Two NH three. Calculate. Casey and its unit. Once again. Copy this. This one. See how do we do this. The reaction is given. And two plus. Three H two gifts to an history. Okay. Are you getting one point six. Anyone. What is the answer you're getting me. Okay. See the question is we have one more love and two. Initially we have taken one more love. And two. And three more love. We don't have any product. So it is zero initially. Right when the reaction proceeds. Twenty five percent of. And to converts into ammonia. So whenever it is given into percent is no. It is alpha actually. Twenty five by hundred. It is point. Two five alpha. It means out of one. Point two five has been reacted. Degree of dissociation. So one minus. Zero point two five. Three minus. Three into zero point two five. And two into zero point two five. And two into zero point two five forms. End out in this. This is point seven five. This is point seven five. Three miles point seven five two point two five. And this is point five zero. So point seven five. I'll write down this in another way. It is the. Value we have three by four. Two point two five and five point. This we have. So this is. This is three by four. We can write. And this one would be. Two point two five by hundred. So we can write this as. Nine by four. Right. And this one would be. One by two. But this is the number of moles. It is not the concentration. So what would be the concentration here. The concentration would be. Three by four into. Four. In the denominator. Mole per liter. Four liter is the volume here it is given you see. In four liter container. So we need to divide this by four. So nine by four into four. One by two into four. Then write down the expression for KC. KC expression would be. And it's three. Concentration is square. And two. It's two concentration. Two concentration Q. Substitute all the values. And as three concentration is one by eight. A square. And two is. Three by four to three by sixteen. And this is. Nine by sixteen. This is the answer. In this term if you write down. You'll get the answer as four to the power five. Divided by three to the power seven something. This is the answer we have. Got it. We'll do some more questions on this. You will understand how to. Find out. All these things. Okay. Apart from this if you see. The formula of. Degree of dissociation alpha. Right. And that is X by a that's initial concentration. Total initial concentration. But if these two things are not given. Then also we can find out alpha. By these formula. This formula just you need to memorize. It is given as. Alpha is equals to. Capital D. Minus a small D. And minus one into D. Where this capital D is the. Vapor density. If Vapor density is given you can directly use this. Vapor density. When there is no dissociation. That is important. Means initial. No dissociation. Initial Vapor density. Vapor density. The small D is the. Observed Vapor density. When the reaction is start after sometime. At equilibrium. Whatever the density given. Observed. Vapor density. Is this D. And is this D. Vapor density. And is the. Total number of products. Keep that in mind. Total number of products. We'll do some questions on this also. You'll understand. If the reaction is this A gives NB. This N we have over here. This N we have over here. Okay. In this only. If you multiply by two. Two D by two small D. N minus one into D. N minus one into two D. Then this two D. It becomes. The molecular mass. Observed molecular mass. Into M. Capital M write down. It is the molecular mass. When there is no dissociation. A small M is the. Observed molecular mass. Observed molecular mass. Small M. Sometimes in terms of molecular mass also the data is given. Can use this formula to find out alpha. Okay. One more. Formula we have in this. The least important one is alpha is equals to. We have. T1 P2. Minus T2 P1. Divided by. T2 P1. Where this T1 and P1. Are the initial constant. Sorry initial temperature and pressure. P2 T2 P1. P2 T2. Is a final temperature and pressure. These. Three formula also you can use. To find out alpha. Copy this down. It is N minus N into M. Multiplied by not divided by. This one is multiplied. See exactly this. The same formula we have. Instead of capital D. Right on capital M. Small D.