 Okay, so next we need to understand law of mass action. Remember that expression delta G expression where we have delta G is equals to delta G naught plus 2.303 RT log Q remember that. Yes. Okay, so what did we do there at equilibrium delta G becomes zero and Q becomes Casey. Remember that Q becomes Casey. What is that Casey, what is that Casey equilibrium constant. So what is equilibrium constant here we'll understand equilibrium constant. Okay, so let's see this. So I have a reaction. I'm assuming a reaction here or hypothetical reactions of a b b balanced hypothetical reaction. C, C, and D, D. This is the balanced hypothetical reaction I'm assuming you can have any reaction like this. So this is law of mass action suggest. It says that the chauvin second I'll go back balanced chemical reaction we have. So according to law of mass action, what we say that the rate of forward reaction rate of forward reaction that is RF is directly proportional to the concentration of reactant concentration of the product of reactant to the power that respective stoichiometric coefficient. If you have the book no NCRT or any book if you have you can go through the statement of law of mass action thing is this only that I'm explaining. According to law of mass action rate of forward reaction is directly proportional to the concentration of the reactant species product of the concentration of the reactant species to the power respective stoichiometric coefficient. If I remove this proportionality sign here, then the expression would be RF is equals to KF concentration of a to the power a concentration of B to the power B. Now what is this KF here? What is this KF? Obviously you won't have any idea. Yes, that is what I wanted to hear equilibrium constant, but it is not the equilibrium constant. It is. It is the rate constant for forward reaction rate constant for forward reaction. So this F stands for forward for forward reaction. Rate constant not equilibrium constant. What is the equilibrium constant? We'll see. Wait. Similarly, according to law of mass action, we can also write the rate of backward reaction, backward reaction, that is RF is equals to exactly same. Sorry, RB is equals to KB concentration of C to the power C concentration of D to the power D. We don't have proof for that. It is according to this law, law of mass action. Same thing. We discussed more concentration, more will be the rate. More concentration, more will be the rate. Like I said this thing earlier also, it's not like it is always directly proportional. I said no, that rate may be proportional to the concentration to the power of something. We can have any number over there. So from that you can relate this easily. Yes, whenever I write down this square bracket, a square bracket C means concentration of C. A square bracket represents concentration, correct? Concentration of C, concentration of D. Yes, okay. So this is RF and RB correlation. Tell me the condition of equilibrium. Condition of equilibrium. What is the condition of equilibrium? F is equals to RB. Yeah, that's right. RF equals to RB. So at equilibrium, we can say this RF equals to RB. Let's see what happens. So I'll write down the expression for RF and RB and hence the expression would be KF, concentration of A to the power A, B to the power B, KB, concentration of C to the power C, concentration of D to the power D. So what is KF by KB? What is KF by KB? It is concentration of C to the power C, concentration of D to the power D, concentration of A to the power A, concentration of B to the power B. This KB, I did not write here, same thing. You can also write it down. This KB here is the rate constant, not equilibrium constant. This is the rate constant for backward reaction, backward reaction. Copy this down. And then we have this. So this ratio of KF and KB, this is the equilibrium constant KC. So what is equilibrium constant? Equilibrium constant is the ratio of the rate constant of forward and backward reaction. Yes, what you have to keep in mind that this KC depends only on temperature, depends only on temperature. But this KF, KB rate constant, I'll write down in this slide, rate constant here, it depends temperature and catalyst. Like I said, rate constant is not the concern for us in this chapter. Our concern is equilibrium constant. Rate constant, rate, everything we have in chemical kinetics grade 12. Correct? So don't bother about it, rate constant. But you must understand that the ratio of rate constant of forward and backward reaction is the equilibrium constant KC. And it depends for concentration. If you take concentration here, it is C. If you take pressure here, it is KB. Clear? Am I clear? Concentration, if you write on the partial pressure of C, D, A and B, it becomes KB. And equilibrium constant, whether it is KC or KB, it depends only upon temperature, nothing else. Okay, so for any reaction, the reaction was this. KC, what we get here? Concentration of C to the power C. Concentration of D to the power D. Concentration of A to the power A. Concentration of B to the power B. What is C, D, A and B? C, D, A and B are the product and reactant. So if you remember, if you go back and recall that particular equation, delta G in terms of that K will write down the concentration of product by reactant. Isn't it? I will see it here then. No problem. We'll have it here also. I'll do it here only you see. Just one equation we have. What is this? The enthalpy change, we can relate like this. Delta G is equals to delta G naught plus 2.303 RT log Q. This Q becomes KC at equilibrium and delta G becomes zero. We'll wait one more time. We'll do it after some time. We'll discuss Q also and then we'll do it. Let it be. Just one equation we have. We'll do it a bit later. Did you understand this? No doubt? Okay. Now suppose the reaction is given. We haven't done that. No problem. We'll have this in this chapter also we have. We'll do that. I will do today only. Just give me some time. See if the reaction is given in gaseous state, the reactant and products are present in gaseous state. For gaseous state reaction or for gaseous reaction, we can write the expression of Kp. Expression of Kp. What is Kp? I'll tell you. Suppose the same reaction that we have written there. We have this is gas. A gas plus BB gas gives C, C gas and DT gas. If you have any solid or liquid, you can ignore that. So could you tell me, because the reaction is taking place in a container, so volume is fixed. So for A, what we can write, partial pressure of A into V is equals to NaRT, isn't it? Pv is equals to NaRT. We can apply for gases. If I write down partial pressure of A, it is Na by V into RT. What is Na by V? It is mole per liter. It is concentration. Pa is equals to, we have the concentration of A into RT. It's fine. So what is the concentration of A? We'll have Pa by RT. What is Pa? Pa is the partial pressure of, is it fine? Any doubt in this? Any doubt? Yes, no doubt. So what would be the expression of concentration of B? Is it Pb by RT? Concentration of C. Not this. Concentration of C. It is PC by RT. And concentration of D. It is PD by RT. Yes. Now what we do, we represent all these concentration terms in terms of pressure in the expression of case. You see what happens. So when we substitute all these concentration terms in Kc. So we have Kc equals to, Kc equals to, we have PC to the power C, PD to the power D by Pa to the power A and Pb to the power B into RT to the power, RT to the power A plus D minus C plus D. End out. One more thing you see here. In this one you see, if I find out delta N for this reaction, that is the delta Ng only. So it is the number of gaseous product minus the number of gaseous reactant. Same like we had in the last chapter, delta Ng. To delta N equals to what? We have C plus D minus D plus P. No doubt. So we substitute this delta N over here. Before that substitution, you see this expression. If we have concentration here, C to the power C, D to the power D, A to the power A, B to the power B, then this term was what? This term was Kc. We have pressure in terms instead of concentration here. Like the previous expression was Kc. So when you write down pressure over here, this term becomes what? This term becomes Kp because the pressure is there. Yes, Kp. So what we can write, you see, Kc is equals to Kp. The difference between these two terms is what? The difference is when you write Kc, we'll write the concentration of product by concentration of reactant. When you write Kp, we can write only for gaseous species, pressure of product by pressure of reactant. That is the only difference. Kc Rt to the power what? Minus of delta N. Is it fine? Look at this expression first. Is this fine? Did you understand this expression? No doubt. So further, what we can write? We can write Kp is equals to Kc Rt to the power delta N. So we can write the concentration or relation of Kp and Kc. Equilibrium constant for concentration, equilibrium constant for pressure. We can write this. Very important relation must keep this in mind. Then copy. Solid and liquid, we don't consider this to ignore that. You don't have to write down the concentration or pressure for solid and liquid. Just ignore it. We'll take the concentration as unity. Did you copy? Till here. Now you see in this, they asked many times this question, the relation of Kp and Kc. Okay, the relation of Kp and Kc. So we have three possibilities in this. We have three small, you know, cases we need to understand. We need to take, and then we'll have some questions on this numerical. Okay. So case one, case one, we have, it's possible that delta N equals to zero for some reaction. For example, we have H2 plus H2 gas plus CL2 gas gives two at CL. Suppose I'm having this equation. Right. So this delta N is zero. For this case, Kp is equals to Kc will have case two, when delta N is greater than zero. So we have CaCO3 solid converts into CaO solid plus CO2 gas. So obviously delta N is greater than zero while the value is one. So this Kp is also greater than Kc for this condition. So case three, when delta N is less than zero, for example, N2 gas plus three H2 gas gives two NH3 gas. See delta N is zero. So Kp is less than Kc. Copy this down. Yes, number of moles of gas is correct for all the three. In this case delta N value would be one minus zero. Now you see the last thing here, the unit of Kp and Kc, the unit of Kp and Kc. R is 0.08. Yes, it's pressure and atmospheric. No, R is 0.08. See the relation of Kp is equals to KcRT to the power delta N. Oh, this is not required. Okay. So Kp is what you see the expression we had written was pressure of C to the power C pressure of D to the party pressure of A to the power A pressure of B to the party. Obviously the unit of Kp is what pressure is ATM. Right. So can we write this as atmospheric to the power delta N based on delta N we can have the unit of it. Isn't it. If you write down the expression of Kp, Kc is equals to what concentration C to the power C concentration of B to the party concentration of A to the power A concentration of B to the party. The unit of Kc would be it is more per liter concentration if it is so unit is more or more to the power delta or you can also write more per liter to the party. That is correct. So it's unit is it's dynamic it's changing with the reaction condition, right, or the reaction is torsion metric coefficient. So let's finish this also characteristics and then we can have the numerical problems on this. So write down the heading characteristics of of Kc and Kp. Right down equilibrium constant equilibrium constant constant is independent of of initial concentration, reactant and product equilibrium constant depends only on temperature temperature. Third one, the expression of Kp and Kc is applicable only when the equilibrium is achieved, the reaction is reversed, then the equilibrium constant constant also get reversed. Suppose we have an I'm assuming a hypothetical action. Okay. It will be easier for us to write. Suppose we have a reaction a gives B, P and Q is the distortion metric coefficient. What is the expression for Kc for this reaction concentration of B by concentration of a to the power q to the party. If I reverse this reaction that is QB gives PA if I reverse this reaction Kc dash it is equals to concentration of a to the power p concentration of B to the power q. So if you look at the relation in this two. Kc is equals to one by Kc dash. This means what if you reverse the reaction equilibrium constant also gets reversed. Yes, can we can do that simply you can do this. Now, the next one is suppose we have two reaction. One is a gives B and other one is C gives D is given for this equilibrium constant and K2 for this equilibrium constant. If you add the two reaction, if you add the two reactions, so a plus C gives B plus D. So we need to find out the equilibrium constant of this reaction, the new reaction we have that would be concentration of D into concentration of D a into concentration of C. If you find out this K1 here, K1 would be what I'll write down here. K1 here is it is concentration of D by a. What is K2. It is concentration of D by C. So K1 into K2 is nothing but D by a into D by C. Right, so we'll write Kc is equals to K1 into K2. So what is the conclusion for this. The conclusion is if you add the two reaction, then the equilibrium constant of the new reaction will be the product of the equilibrium constant of the individual reaction which is added. Yes, tell me is there any doubt all the property you must keep this in mind directly when you use this one last thing we have in this. Suppose we have again two reaction a gives B and C gives D K1 K2 K1 is B by a and K2 is D by C. And we are subtracting the two equation, which means we have a minus C gives B minus D. So what is the equilibrium constant of this reaction, could you tell me K1 by K2. Yeah, that's right. Basically you can see here what that this reaction means what you see this reaction means what we can also write this as a plus D gives B plus C. Which can be formed when we add this and this reaction D gives C. So when you reverse this reaction, its equilibrium constant becomes one by two property we have seen, and one and two if you add, you'll get this right. So for this one Casey would be what the product of equilibrium constant of this and this, which is K1 into one by K2 K1 by K2. So one line is what here the property. Once you add that once you subtract the two reaction, its equilibrium constant will get divided like understood. No doubt. So we'll have a break now, and after the break will start questions on this. Correct. So we'll resume at 623. 623. Okay, take a break.