 Yeah, audible. Shall we start? Okay, so the next we need to understand is activation energy represented by EA, activation energy. Under this we'll talk about or we understand the relation of equilibrium constant temperature. Like you know, in the beginning only we said that equilibrium constant depends upon temperature. How it depends what is the relation that is what we are going to understand. Two graphs I am going to draw over here. This is the energy axis and this is the progress of reaction or time. Progress of reaction or time. This is also energy. Then we have progress of reaction or time. This is the one graph and other graph is this. This is reactant. This is the energy of reactant. And this is the final product. We have the energy of product. Same thing we have here. Reactive molecules when it starts colliding, when the reaction starts, it will form an activated complex. The energy keep on increasing and finally it forms a very highly unstable complex which we call it as activated complex. The energy of this activated complex is called activation energy. It is basically the minimum amount of energy required for any reaction to process. Okay, so reactant to product. This is forward reaction. Right, so this difference in energy that we have here from this reactant to activation energy. This we call it as activation energy EAF for forward reaction. Similarly, this difference if you see here, it is the activation energy for backward reaction when the reaction goes backward direction. And the difference in between the energy of reactant and product is called the enthalpy of reaction delta H. Is it clear? How do we define enthalpy? Enthalpy of reaction is defined as activation energy for forward reaction minus activation energy for backward reaction. Forward minus backward. Okay, so could you tell me for this graph delta H is positive or negative? You see the enthalpy forward reaction is this much and backward reaction is more, right? This is less than zero, isn't it? Delta H is less than zero means the reaction is what? It is exothermic. Hence, this graph is for exothermic reaction which you can easily understand by looking at the graph itself that the energy of product is lesser than to that of reactant. So this difference in energy will come out exothermic. Okay, and same thing will have over here. This is the activated complex. This is the activation energy for backward reaction and this is the activation energy for forward reaction and difference in these two energy is the delta H the enthalpy of the reaction. Delta H clearly you see Eaf minus Eab delta H is greater than zero. So this graph is for endothermic. Any doubt in this? Now, this is the graphical understanding you must have by looking at the graph you can say what reaction it is endothermic or exothermic. That's one thing. Now, according to Arrhenius equation if you see, according to Arrhenius equation, we can write K is equals to A e to the power minus Ea by RT minus Ea by RT. This is the equation we have Arrhenius equation. Derivation of this we do not have. It is based on some observation. We have this relation. What is this K over here? What is this K? This K is the rate constant. It is not equilibrium constant. Because you see for rate constant and equilibrium constant we have the similar notation K only. So must take care of this. Where it is rate constant, where it is equilibrium constant. Let me repeat this thing again. Rate constant we don't have much to understand over here. It is there in chemical kinetics in grade 12. Here we need to deal with only equilibrium constant. Why we are discussing rate constant then? Because you know rate constant, equilibrium constant K is the ratio of rate constant of forward and backward reaction. Hence we need to see this. But you should know where it is rate constant and where it is equilibrium constant. A is a constant here which we call it as Arrhenius constant or Arrhenius factor. A is a constant. We also call it as pre-exponential factor. Mostly we use the second term pre-exponential factor. EA is the activation energy. R is gas constant. T is temperature. EA is the activation energy. R is gas constant. T is temperature. So if I write down here now you listen to me. This is very important and you know don't get confused. Just listen to me carefully. Kf is the rate constant for forward reaction. This A will be constant e to the power minus Eaf. What is Eaf? What is Eaf? Eaf is the activation energy for forward reaction. Yes, activation energy for forward reaction. Similarly, if I write down Kb backward rate constant for backward reaction. A e to the power minus eab by RT. Is it fine? Activation energy for backward reaction. Again keep that in mind. It is rate constant for forward rate constant for backward. Can we take the ratio of 2? Kf by Kb. What is Kf by Kb? Kf by Kb is the equilibrium constant. I have seen that. Kc and that would be equals to e to the power minus Eaf minus eab by RT. Is it fine if I write this? Fine. I will write this as K1. Let me write this as K1. Eaf minus eab is delta H. So K1 is equals to Ea minus delta H, the enthalpy of the reaction by RT. Is it fine? What is ln K1 from here? If I take ln both sides, ln K1 is equals to minus delta H by RT. If I write this as K1, let me take this as T1 over here so that it will be a no confusion. So K1 is the equilibrium constant at temperature T1 that you assume. Is it fine? Did you understand this? Tell me. Till here it is fine. Clear? No doubt. Very simple expression it is. Similarly, can I write down if K2 is the equilibrium constant at temperature T2, then can we write this ln K2 is equals to minus delta H by RT2? Any doubt in this? Can we write this? Agreed? Tell me. Yeah. This is equation 2 and the previous one, I am assuming this as 1. So 2 minus 1, what we will get here? ln K2 minus ln K1 is equals to minus delta H by R, 1 by T2 minus 1 by T1. Any confusion? Any doubt? Yeah. Tell me. Yeah, I will go back once again. Just a second. Yeah. So we got this. So further we can write this as ln K2 by K1 is equals to minus delta H by R, 1 by T2 minus 1 by T1. And further, if I convert this ln into log, the expression would be log of K2 by K1 is equals to minus delta H by 2.303 R, 1 by T2 minus 1 by T1. This is the relation of equilibrium constant and temperature. K1 is the equilibrium constant at temperature T1 and K2 is the equilibrium constant at temperature T2. Yeah, done. ln is equals to 2.303 log. You see, the relation is ln A, we write 2.303 log A. That's why we will write this in the denominator. Yeah. So this is the relation we have. This relation, we call it as or this equation we have. We call it as Vanthoff equation. Vanthoff equation. Okay. Always keep that in mind. K1, K2 here, it is not rate constant. It is equilibrium constant. Delta H is the enthalpy of reaction. Okay. Now, we have three cases possible. All of you have copied? Done. Three cases possible we have here. Depending on that, we can compare K1 and K2. Case one is for the reaction whose delta H equals to zero. When delta H is zero, we can write log of K2 by K1 is equals to zero, which further means K1 is equals to K2. And this means what? For all those reactions whose enthalpy change is zero, then the equilibrium constant is independent of temperature. Right now, equilibrium constant remains same at all temperature. Constant remains same at all temperature under this condition. It is independent of it. Case two, delta H is greater than zero. Means what? Endothermic reaction. When we have endothermic reaction. What we can write? I am assuming as temperature increases. As temperature increases. Means T2 is greater than T1, I am assuming. You can also assume other way. Just opposite results you will get. So, 1 by T2 is what? Less than 1 by T1. Further, 1 by T2 minus 1 by T1 is what? Is less than zero. Okay. Copy this down. I will go to the next page. Now, you look at this went-off equation because we have to conclude from this equation only. So, went-off equation is what? Log of K2 by K1 is equals to minus delta H divided by 2.303 R. 1 by T2 minus 1 by T1. See, you do not have, you have to memorize this equation. How I have memorized this? Once I write K2 on the top, then corresponding to this, the temperature I will write down first over here. If you write down K1 on the top corresponding to this, T1 you should write down here first. K2, T2, K1 denominator, T1 in the last with a negative sign outside. Okay, like this. So, for, you know, just know what condition we take, T2 greater than T1 greater than 0. 1 by T2 minus 1 by T1 is less than zero. So, this is what? This term is negative. Just we need to understand this K2 is increasing or decreasing. That is what we need to understand. So, this entire term is negative. Once again, I will go back. Negative. I have assumed the reaction is endothermic, right? Delta H greater than 0. So, delta H is positive. Yes. So, this is positive. This is negative. One negative sign already we have. So, negative, negative, positive, which means if temperature increases, we can write log K2 by K1 is always greater than 0, isn't it? Right? Which means what? Which further means that K2 is greater than K1. So, what we can conclude here? We can conclude in endothermic reaction as temperature increases, equilibrium constant KC increases. And temperature decreases, KC decreases. That is what we can say. Any doubt here, guys? No. But the relation is exponential, right? We have seen that. Relation is exponential here like logarithmic relation we have. So, log K we have this side and temperature we have here. So, as temperature increases, K increases, the graph goes like this for endothermic. Keep that in mind. Tell me. Any doubt here? Yeah. Now, what is the third condition left? The third condition is we have delta H less than 0 exothermic, right? So, exothermic reaction generally heat releases, right? So, temperature will decrease. So, if T2 temperature decreases, so this means T2 less than T1, right? So, 1 by T2 minus 1 by T1 is greater than 0. Again, you look at this equation log K2 by K1 is equals to minus delta H by 2.303R, 1 by T2 minus 1 by T1. So, this is positive, right? Delta H is negative, negative, negative, positive. It means log of K2 by K1 is greater than 0. This means what K2 is greater than K1. So, what we can conclude here? In exothermic reaction write down, in exothermic reaction, equilibrium constant, in exothermic reaction, equilibrium constant, equilibrium constant increases with decrease in temperature. No, not always, Prakul. The reaction condition may be different. If temperature you are increasing in exothermic reaction, then maybe you can get the result other way. The reaction condition could be anything. At one temperature, we have certain condition. Then with that temperature, you can decrease the NNPs also. So, based on that, we can have this relation. In general, what we say exothermic reaction if temperature decreases, equilibrium constant increases. So, if you look at the graph here, this is time, sorry, this is temperature and this is log K as log K increases at temperature decreases, right? So, the graph goes this way. You will get here a direct formula-based question, okay? K2, K1 relation, formula-based question you will get. You just need to know the Vantop equation. All value you need to substitute. Delta H, if it is given in Zule, then R value also you will take in Zule, right? Depending upon the value of Delta H, you can take the value of R here, gas constant. You may require some log value to solve this kind of, you know, expression. So, you don't have to, you don't have, you don't bother with that, okay? Because log or anti-log value will be given in the question if it is required. But usually what happens like log 2, log 3, log 5, these values you should know, you should memorize. Based on these values, you can calculate, you can find out the log value. If it is not possible to find out the log value with these three, four values, then it will be given in the question. So, you don't have to, you don't bother about that, okay? Let's see some questions on this. We have a reaction N2 plus 3H2 gives 2NH3. 400 degrees Celsius, Kp value is given, that is 1.6 into 10 to the power minus 4 unit you will have over there. What will be the equilibrium constant at 500 degrees Celsius? The enthalpy change in this process is given minus 25.14 kilocalorie unit you must take care of. Then if you're requiring log value, you can take help of calculator, okay? So, what we'll do over here? Temperature we must take in Kelvin, okay? So, we'll write down log of K2 by K1. Suppose K2 we need to find out, K2 by K1 is equals to minus delta H by 2.303R 1 by T2 minus 1 by T1. So, log of K2 by K1, K2 we need to find out, K1 literally will substitute in the last. Delta H is given in calorie, kilocalorie. So, minus 25.14 into 10 to the power 3 divided by what should be the R value we should take here? Can we take it as 2 approximately 1.98, which is 2? No, 0.0821 is atm liter, right, per mole Kelvin. But delta H is calorie, no. So, R value you also take in calorie. That is 1.98, exact value if you see, approximately 2 we can assume. So, T1 minus T2 is what? T1 is 400 minus T2 is minus 100 divided by T1 plus T2, sorry, T1 into T2. 673 in Kelvin into 773 in Kelvin. You need to solve this. Tell me the value you are getting quickly. When you solve this, the answer you will get, K2 is approximately 1.429 into 10 to the power minus 5, atm to the power minus 2, the unit is. So, here particularly in this particular portion, the Vendorff equation, you will have to use the formula directly. They will ask you questions, all other data will be given, right, all other data will be given, one unknown will be there that you need to find out. You look at this question, one last one and then we will see the Lee Chatelier's principle. At equilibrium, one reaction is given, that is N2O4 gas converts into 2 NO2 gas. This is the reaction given. Observed molecular weight of N2O4 is 80 gram per mole, 350 Kelvin. Find out the percentage dissociation and 204 at the same temperature, 350 Kelvin. Percentage dissociation you need to find out. Could you try this? Yes, done, 15%? Yeah. So, you see what is given over here? This 80 gram per mole is the observed molecular mass. So, if you look at this formula of alpha, first of all, the percentage dissociation means degree of dissociation you need to find out, okay? So, alpha, you know, in terms of molecular mass, m minus small m by alpha minus 1 into small m. This is the formula we have. m is the observed molecular mass, that is 80. This m value is 92, N2O4 we have, I'm sorry, it is 92. N2O4 you can calculate, it is 92. So, it is 92 N minus 1, 92 minus 80 divided by N value is 2, 2 minus 1 into 80. It is 12 by 80. Percentage dissociation is what? 12 by 80 into 100. So, that would be 15, 15 percent we get here. Understood? Simply, you have to use the formula that it add questions you'll get over here, okay? So, one question we discuss on this also. Anyways, so, let's start with the last topic for this chapter, that is Lee Chatelier's principle. Lee Chatelier's principle. Write down the statement, whenever a system in equilibrium, whenever a system in equilibrium, yeah, whenever a system in equilibrium is subjected to, is subjected to change in equilibrium condition, change in equilibrium condition, then the equilibrium, the equilibrium shifts to that direction by which the effect of change is minimized, effect of change is minimized. Copy down this? Yeah, see the simple explanation over here. Suppose you have a pendulum at rest, pendulum at rest. When you take this ball at this position and you leave it, what happens? What happens? This will go back, but it won't stop over here. It will continue to this point and then there will be a two-end promotion, right? Yeah, so this is why it is happening. First thing is that, okay? Because at this position, it was static at equilibrium, isn't it? Once you drag this ball over here, this object over here, right? You disturb the equilibrium state, right? And once you leave it, it tries to gain its equilibrium state again. So it goes till here and then again here and gradually it slows down and attains the equilibrium state here, right? The same thing happens with chemical reactions also, right? If you change the equilibrium state or any condition of any reaction, suppose if I give you this example, we have a reaction A gives B at equilibrium, right? At equilibrium. What happens if you add some more amount of reactant into it? What happens then? This, you are disturbing the equilibrium state, right? Once you add A so that concentration is changed now. So equilibrium is disturbed. Again, the reaction will move in such a direction so that the change is minimized because it wants to gain, it wants to be in that state, equilibrium state, right? So once you add A, the reaction starts going into the forward direction so that the change is minimized and the reaction will be at the equilibrium state. Yeah, understood. That is what the Leach-Atelier's principle means. Any kind of change if you do in any equilibrium state, the system itself, system means reaction or whatever it is, the system itself react in such a way that that change is minimized so that the equilibrium state will be maintained. Similarly, here if you think of the other way, if you add more amount of B, then what happens? The reaction will go in backward direction, right? So that the concentration of B decreases and the equilibrium maintains. Since we are adding B, got it? Understood? Similarly, if you change pressure, correct? Then also there will be change in shift in the equilibrium state. Any reaction condition you see, whether you are changing pressure, whether you are changing temperature, whether you are increasing the amount of reactant, decreasing the amount of reactant, increasing product or decreasing product or if you add some inert gas into this, any kind of change if you're trying to make into that equilibrium state, that will affect the equilibrium condition and hence the direction of equilibrium changes to maintain the equilibrium state again, right? That is what the Leach-Atelier's principle means, okay? So there are three, four, five, so five or six factors we have here that we need to understand that under this factor, how the direction of the equilibrium changes, correct? If you look at this first case you see over here, the first case we have effect of concentration, effect of concentration, right? Concentration means anything you can think of reactant or product. So what happens if the concentration of reactant increases? By any means, you are increasing the concentration of reactant. Reaction will move in which direction? Forward, right? To decrease the concentration of reactant, it will move in forward direction, right? If product concentration increases or reactant concentration decreases, first of all, reactant concentration decreases. So obviously this reaction wants to go in backward direction so that the concentration whatever you have removed, that has been what? That has been maintained or by any means you have to maintain that equilibrium state. Maybe the equilibrium status has been changed. You may have some different concentration but to maintain the equilibrium, this thing will happen, got it? So if the reaction concentration decreases, it is in backward direction, backward direction, right? Similarly, if we have product concentration, product concentration increases, you are increasing the concentration of product, again backward reaction, isn't it? If you are decreasing the concentration of product, then forward reaction phase. So whatever change you are making in, according to Lee Tattelier's principle, that change must be minimized. Understood? We have other conditions also here like effect of temperature we have, effect of pressure we have, in that we have different different conditions, right? That we'll see next class. This is the last topic we are discussing, okay? So next class we'll finish this off and then we'll start with Inic equilibrium, okay? I will share a few DPPs with you today, okay? You have to finish all those DPPs. Only Lee Tattelier's principle is left that we'll see next class, right? Any doubt you have? So Lee Tattelier's principle, you should leave it now. We will discuss this next class, other questions you can solve. Where it is forward? Okay, fine. Understood? So like, okay, fine. We'll wind up the session here. Okay, we'll see the next class. Inic equilibrium, we'll start next class after finishing this. We'll take half an hour more and then we'll start Inic equilibrium. Okay guys, yeah. Thank you so much. Yeah, bye.