 You need to find out the number of moles of A to B2 and C2. Yes, I tried doing that. So, they were asking average molar mass. Correct. So, m of A1, m of A into number of moles of A plus m of B into number of moles of B plus C by N in the center. Yes, but it was not working. Is it coming? What is the reaction? A2 gives B2 plus C2. Yeah. So, Kp is given. You find Kc from that. No, this Kp Rd. I mean Kd by N. Alright. And that would be Kp equals to Kc into Rd. N is 1, the time is 1. Okay, I will try this. So, you write on this question and solve this. The reaction is given. A2 gas, try this one, then we will discuss. The molar mass of A2 is given. It is 70 molar mass of B2 is given. It is 49 and molar mass of C2 is 21. The unit is gram per mole. You need to find out the average molar mass in gram per mole. Average molar mass in gram per mole. Sir, Kp was 9 and B was 7. Yeah, Kp is 9 is given. Kp is given. It is 9 and 7 MOSFET expression, 300 Kelvin. 7 MOSFET expression, 300 Kelvin. Try this one. So, this is B carried out with that pressure. No. In this, the equilibrium constant, Kp is 9 in this reaction. That is given. And the reaction is carried out at 7 atmospheric pressure and 300 Kelvin. Equivalent constant, Kp atmospheric is given for the reaction in this question. It is written that Kp for the reaction is 9 at 7 atmospheric and 300 Kelvin. Means, when Kp is 9, the atmospheric pressure is 7 atmospheric. It means that is the equilibrium pressure we have. Means, when you write down suppose, initially we have only 8 to present. Pressure of this is what? P, 0, 0. At equilibrium, what will happen? P minus x, x and x. So, when you add this pressure at equilibrium, what you will get? P plus x. P minus x plus x plus x, P plus x. So, P plus x is 7. That is given. We need to find out x first. What happens? What is x? 3. We need to factorize this x by minus 32. And what is P? Initial pressure? 4. 4. When pressure is 4, P, M is equals to DRT you can write. All of you got x? See, I will just do it now. The initial pressure I am assuming, that is P and this is 0 and 0. So, when it dissociates, it forms P minus x, this forms x and this forms x. The question says when Kp is 9, pressure is 7 at 300. This is the language we have. So, when Kp is 9, pressure is 7 atmospheric. It means Kp when we are saying, it means the equilibrium has to be established. So, equilibrium pressure is 9 at this given. So, this is the equilibrium pressure. So, when you add this, the total pressure Pt is equals to P plus x and this is equals to 9. Now, once you know this x, you can find out P. And what is the relation we have? Initial pressure is P divided by the equilibrium pressure. We know pressure and molecular mass is what, inversely proportional. We have this relation P, M is equals to rho RT or density RT. PV is equals to NRT. N is what, mass by molecular mass and mass by volume is what? Density. Mass by volume is density. So, P into molecular mass goes to DRT. So, once from this what we can write is pressure and mass is inversely proportional. Because density R and T is constant. So, we can write the initial pressure by equilibrium pressure is equals to the equilibrium molecular mass divided by the initial molecular mass. Can you understand? Okay. Okay. And this is because of only A. Initially, we do not have B and C. At time T is equals to 0, we have only A. Right? So, if you know this, you have to find out this. This you already know and equilibrium you know. No, sir, that only you have to find. No, you ask it. At a pressure, it is here. This you know. That is the pressure because of A, no. This is what? This P plus X. No, P is this. This P that we have. Yeah. P equal to P plus X. So, you know this. If you know this X, you know this P. Yeah. So, you have this P. You have this P equilibrium. That is 7. You know the initial mass that is because of only A. So, mass of A will take here and this you have to find out. So, objective is what? Objective is to find out this X when everything is done. Now, for this we will write down the expression of KB. So, KB expression what will happen? P2. X square divided by P minus. C2 divided by A2. That will be X square divided by P minus X. Then you can substitute. So, P. Yeah. 9 minus X. P is 9 minus X. Yeah. Okay. And KB is 9. So, 9 is equals to 9 into 9 minus. Which is 9 or 7? No. P is 7. PT is 7. PT is 7. PT is 7. So, it is 7 minus 2X. Yes. So, it is X square plus 18X minus 63 is equals to. Okay. X equals to 3 you will get from this. Okay. So, this P is equals to what? The initial pressure P is equals to 4. So, the expression will be 4 pi equal to pressure is 7. M equilibrium divided by. 7. This is 70. This way also we can do. It should be 40. Yes. How much it should be 40? Sir, why those marks B2 and C2? That formula also we can find out. M of A into number of moles of A plus M of B into number of moles of B. But that will be more complicated because this is pressure. Yeah. We have to convert this into number of moles. So, better we will use that. Okay. 30 it also. Yes, sir. One more question you see. Equivalent mixture of 2 gases K2 and B2 is taken and is it reselected at 10 to 300. Reaction is given. Kp3 is given. 2. If the initial pressure in the container is too atmospheric. Initial pressure in the container is too atmospheric. The final pressure developed at equilibrium is 2.75 atmospheric in which the equilibrium partial pressure of gas AB was 0.5 atmospheric. Calculate the ratio of Kp2 by Kp1. Calculate the ratio of Kp2 by Kp1. One information is given. Degree of dissociation of B2. Alpha of B2 is more than to that of Kp2. So, initially there is only A2 and B2. Yes. The reaction is the last one. But it takes place by three different steps. First, A2 converts into 2A, B2 into 2B and then the reaction takes place. So, that is still A2 plus B2. Is that A plus B2 or A2 plus B2? A2 gas gives 2A gas. A2 plus B2 gives 2AB. So, the final pressure AB is given. It is 0.5. No, partial pressure of AB. Yes. Partial pressure of AB is more complex. See, A2 gas gives 2A gas. So, suppose initial pressure I was assuming is 1 and it is 0. So, it is 1 minus xn 2x. So, Kp1 is what? So, you can just assume that it is 1. Yes, you can take it as 1. You don't do that. You always assume it is 1. I will discuss in the class also. Nothing is given. You can take it as 1. Why wouldn't like P1 P2 or P1 plus P2 plus 2 in the next? For that, the required data is not given. If you take any random value of pressure, it is very difficult to solve this question. It is more complex. So, nothing is given. You can take it as 1. So, Kp1 is 2x square pi 1 minus x. Similarly, this reaction can be written. But this dissociation constant is different. Correct? So, why are you doing this? So, Kp2 is 2y square pi 1 minus y. Now, I will find out Kp2 by Kp1. So, X or Y relation is required. Since it is simultaneous, there will be some Z which is used over there as well. Z. Like in the A2 plus B2 plus 2AB, I can say that like some Z comes out from A to Z. Exactly. Because AB also dissociate, B also dissociate. So, Z of A comes out, Z of B2 comes out and just comes out. Coming to that point. But point I am trying to make is, Kp2 by Kp1 you have to find out. So, our objective is what? X and Y value. Now, the next thing was Kp3 is given. Kp3 given is obviously this and this to combines and forms this. So, we can write 1 minus X to A2 and Z dissociate. 1 minus Y to A2 and Z dissociate. And this forms. 2 Z is equal to 0.5. 2 Z is equal to 0.5. Because this is our pressure. So, we know the value of Z. What is the pressure of AB? 0.5. 0.5. So, 2Z is equal to 0.5. So, what is the value of Z? 0.25. Okay. Okay. 0.25 over Z. So, what is this? Square N. Total pressure. Here, total pressure is this plus this plus this. So, and you see one more thing. It is a simultaneous equilibrium. So, in the reaction vessel AB will be present, AB will be present and ESR will be present. It means K2, B2, AB, AB. All are present in reaction vessel. So, all these molecules will contribute to the pressure of the equilibrium mixture. So, pressure of A2 is what? 1 minus X1. Pressure of B2 is what? Pressure of A is what? A is 2X. Pressure of B is? Ui. Pressure of AB is? Ui. Happy Diwali. Happy Diwali. Total pressure, what is the value of this? It is my childhood. Diwali is the question. See again. This will solve the problem. This plus this. So, this will get cancelled. 2 plus X plus Y. 2 plus X plus Y. Okay. Yes. 2X minus X, X2 minus Y, Y and Y plus 1, 2. Yes. And this is 2.75. Now, it is getting easier. X plus Y is in relation. X plus Y is what? 0.75. Is this Kp given? Finally, X2 is given. Sir, Kp3 is given. Kp3 is given. Yes, Kp3 is given. 2. See, X and Y will have one relation. And we need to find out the value of X and Y. So, one more relation if you get it to X and Y, then we can solve those equations. Find X and Y and we will substitute it. Okay. A buskel here, we haven't used. Till now, we haven't used this particular. Okay. So, Kp3 is what? 0.25 squared divided by 1 minus X minus Y. 2Z squared divided by 1 minus X minus Z into 1 minus Y minus Z. So, 1 minus Z is 0.75. So, in that denominator, we can write 0.75 minus X 0.75 minus Y. Is that what? 0.25 into 2.5, so it is 0.5. Square. Okay. So, we can substitute Y as 0.75 minus X in that equation. Yeah. And then you just have whatever you have. It will be quadratic in X. Y is 0.75 minus X. Yeah. And then you can solve that. Okay. Equal equation from percent. That also you can do. Y is 0.75 minus X. So, 0.75 will get cancelled. You get X into this and then you can solve this quadratic. You will get X prices. So, solve it. So, also for Kp1 and Kp2, you will have to change that. You will become 1 minus X minus Z, 1 minus Y minus Z. No. For this relation, see what is A2? A2 is 1 minus X for this equation. So, for this equation, it is only this Kp1 for this equation, not for the entire thing. So, for the equilibrium pressure, that is 1 minus X minus Z. 1 minus X minus Z. Yeah. No, no. See, in this reaction, what we are taking? 1 minus X minus Z we should take. Sir, because it is a simultaneous equilibrium pressure. Oh, yes. Yes, correct. So, it should be minus Z also here. So, can I write this 0.75 minus X? Yes. It should be 0.75 minus X. Equilibrium play has always been 1 minus X minus Z. Yes. Okay. So, one thing is what you can substitute this Y as 0.75 minus X and you will get this X value. One other way also, we have to solve this. See, our use the law of multiplying, Kp1 into Kp2. That is what I was thinking actually. Kp1 into Kp2. The denominator will get the same. Okay. Okay. So, what we are doing here? 4 into 416. X square Y. X square Y square divided by 0.75 minus X into 0.75 minus Y. We are just solving this to get X and Y. So, this is what 0.5 is squared by Kp3 and Kp3 is divided by 2. So, we can write the score. 16 X square Y square divided by this entire thing is 0.5 square into 3 or d2. Okay? Okay. Sir, but you don't get anything from this. Yeah. We should get 5. We need Kp1 divided by Kp2. Sir, you can do that. Solve the problem. Solve the problem. We can solve like that. But I have done this. This way also I have done this. You could say that Kp1 into Kp2 is equal to Kp1 divided by Kp2 into Kp2 square. Okay. You solve this quantity there. What are you getting? Check this. Solve this quantity. You know what? One thing is missing in this problem. You know what is that? It's a2 in between. One thing is missing in this question. What mistake we have is there in this question is this should be A of gas and B of gas. Exactly. It makes more sense. Because those reactions don't exactly do it. Yeah. Two of this and two of this. Okay. Correct. Then only it makes sense. Because when you add these two, you'll get this. Like the reaction is proceeding with A2 and B2. So you should get the final reaction with the help of the intermediate reactions. Okay, sir. And then we have this. Then if you add these two, you'll get this. So we can write this into this is equal to Kp2. Yes. Then the solution will be easier. Okay. That's what I was thinking. Kp1 into Kp2. So this you let it be. And then we can write this Kp1 into Kp2 is equal to Kp3. So you add the equation, you still don't get that. We'll get now. So A2 and B2 will still do that. Yeah. So there's no AB in those two reactions. We can't get it. Why? If you add all these three equations, sorry, if you add all these three, you know. So you get A2 plus B2 equals 2AB. Correct. 2A plus 2B we'll get. A2 plus B2 we'll get 2AB. Okay. If this one I let it be like this. A2, B2, A2, AB. So then we can solve it. So we can solve the equation for this. I got the answer. 3 plus 0.5 plus 0.5. Yeah. You solve like that only. I'll just check this out. Okay. Did you solve this quadratic? Yeah. We're not getting one value for X, one value for Y. X and Y can be either 0.5 or 0.5. So the condition is given. Degree of dissociation of B2 is greater than to that of it. So Y is more than X. Yes, so Y is more than X. Accordingly you find out. What is getting Y and X value? So Y is 0.5 and X is 0.2. Yeah. 0.5. So 1 by 2. Yeah. So Y is 1 by 2. X is 1 by 2. X is 1 by 4. X is 1 by 4. If you substitute it here, you'll get Kp1 and Kp2 ratio. Take it. So we'll start organically. Sir. Oh, by starting? How will you see? We just solved this. We haven't discussed this yet. Substituted. Polyporting as it will see. Minus X as Y. Yeah. X to the support. Minus X to the support. Minus X to the support. Minus X to the support. Minus X to the support. Minus X to the support. Minus X to the support. Minus X to the support. Alpha 1, alpha 1. HSO4 again dissociates and gives H plus and Hso4 to minus. See polyprotein acid is my subject. H2SO4 you have. So when it dissociates, it forms HSO4 minus and H plus. So if it is 1, 0, 0. So it's called your 1 minus alpha 1, alpha 1 and alpha 1. Polyprotein acid means what? More than 1 H plus sign it can give. Okay. So when one dissociation gives, it gives you this. HSO4 minus can further dissociate. It gives you SO4 2 minus plus. H plus again. What was this? 1 minus alpha 1. Okay. This was 0. This was 0. Again it dissociates also. 1 minus alpha 1 minus alpha 1. H plus is alpha 1. H plus is alpha 1. Okay. When this dissociates. So you have a 1 minus alpha 1 minus alpha 2 into 1 minus alpha 1. Okay. Then it forms. HSO4 minus 1 minus alpha 1. Alpha is the area of dissociation. It will just be alpha 1. Yeah. Where? HSO4 minus in second. No. First. Yeah. It is alpha 1 correct. It is alpha 1 only. So it is alpha 1. Alpha 1 minus alpha 1 alpha 2. And then it is alpha 1 alpha 2. And it is alpha 1 plus alpha 1 alpha 2. So HSO4 will write on KPL expression. So we will discuss that polyprotic acid. This is a simultaneous equilibrium. Similar to that. Yeah. Two reactions giving the same ions. So in polyprotic acid you can assume that a simultaneous equilibrium and you can write down the expression. We haven't done this. So cutting after. Alpha 1 minus alpha 1 alpha 2 HSO4. Because degree of dissociation is divided by the unison number of moles. We haven't multiplied by the unison. Alpha is out of one what is reacting. Out of one mole. So our C concentration as C alpha will react. See as a. So you can say in terms of alpha if you write down then you can say 10 minus 10 alpha is the concentration which is left. Means what is reacting? 10 alpha is reacting. Because alpha is defined out of one mole. So 10 minus 10 alpha reacting. In other way what you can say. 10 minus 3 moles is reacting. This is the number of moles reacting. In terms of amount. What amount is getting reacted? This is for one mole really. So in terms of alpha we always write concentration into alpha. C minus C alpha. So this we will discuss. I will share the solution of ionic equilibrium of the center model. So you can take the equilibrium. It should be there. You want I can share that. So because the question is much harder than the previous question. Chemical equilibrium. Chemical equilibrium is easier. Yeah it's a good job. But I'll share it.