 Hello guys, good evening. Can you hear me? Yeah, so last class we were discussing about leach-atelier's principle, right? We just started. Yeah, so we had discussed the first case, right? So a quick recap and then we'll continue with this, right? This is the last topic we have in chemical equilibrium and then we'll start with ionic today. Okay, so we had discussed a leach-atelier's principle. We talked about that, you know, what is leach-atelier's principle and what all, you know, factors we have over here. So first thing we had discussed, like what happens with the effect of the concentration of reactant product, okay? So we had discussed that this principle suggests us that under a given set of conditions, what is a equilibrium condition and if you want to disturb that equilibrium state to what direction the equilibrium post-proceed, right? That is what we discussed and we have seen the first effect over here, that the effect of concentration of reactant and product, effect of concentration and we had discussed that we have forward reaction in two conditions, in two conditions, first when you add a reactant molecule, second case when you remove product. Under these two conditions, the reaction proceeds in forward direction, okay? If you talk about backward, possible when you remove reactant, reactant and you add any product molecule. So these are the two factors we had already discussed last class. Now the next factor we have, effect of change in temperature. Write down all of you. The second factor we are starting with effect of change in temperature. See we had discussed that, you know, Lee-Chartelier's principle says that any disturb or any change in equilibrium state, you know, the reaction will shift in that direction by which the effect of the change is minimized, okay? That is what we discussed last class. So here also, what happens when you change temperature? So we have two different types of reaction possible over here. That is we can think of exothermic reaction and we can think of endothermic reaction, okay? Exothermic. So how do we understand over here that what factor we have here, like obviously temperature, but how the reaction will proceed in forward and in backward direction, how do we understand that? So for this we have ventop equation. Ventop equation, all of you must have, you know, we have done this in the last class. Log of K2 by K1 is equals to minus delta H by 2.303R 1 by T2 minus 1 by T1. This is the equation we have, ventop equation. Since the reaction is exothermic, so delta H is what? Delta H is less than zero negative. We need to find out what happens with the equilibrium constant. K increases or decreases. Generally, what happens at an endothermic reaction, exothermic reaction, the temperature decreases since the heat goes out, right? Heat releases in this process. So we know here K1 is the equilibrium constant temperature T1 and K2 is the equilibrium constant at temperature T2, okay? So if temperature decreases, so T2, suppose it is less than T1, if this happens, you can also think it other way, you'll get the opposite result. Generally in exothermic reaction, temperature decreases. So we can write 1 by T2 minus 1 by T1 is what? Greater than zero. From this to this, we can write. This means what? This 1 by T2 minus 1 by T1 is positive. This is positive, this term is positive. Delta H is negative. So negative, negative, positive, positive, positive, positive. It means what? We can say here, under this condition, log of K2 by K1 is greater than zero, which leads to K2 greater than K1. This means what? As temperature decreases, the equilibrium constant increases in exothermic reaction. Yes, write down. In exothermic reaction, copy this down first, then, okay. So write down the note here. In exothermic reaction as temperature decreases, Kc increases. Yeah, one second. As temperature decreases, equilibrium constant increases, okay? If you talk about this as temperature increases, equilibrium constant decreases. So inversely proportional here, in exothermic reaction. Second one. Like in this only, we have next endothermic reaction. Delta H greater than zero. So we have T2, I am assuming, greater than T1. So 1 by T2 minus 1 by T1 is less than zero. Negative. So went off equation, you see. Log of K2 by K1 is equals to minus delta H divided by 2.303R, 1 by T2 minus 1 by T1. Okay. So 1 by T2 minus 1 by T1, you see. Yeah, this is negative. Delta H is positive. Okay. Negative, negative, positive. So what we can write? Log of K2 by K1 is greater than zero. So K2 is greater than K1. So what we can conclude here? We can conclude in, right on, in endothermic reaction. In endothermic reaction, equilibrium constant increases with, with increase in temperature. Yeah, done? Yeah, one second. Okay. So this is the two condition we have for for temperature, right? Depending upon the nature of the reaction, change in temperature, whether it is increasing or decreasing, we can say that equilibrium will shift in forward or backward direction. Okay. Now, next, write down effect of change in pressure. Third one, effect of change in pressure. What happens with change in pressure over here? See, to understand the effect of change in pressure, we have three different conditions over here. Usually what happens usually what happens, you see, as as pressure increases, then volume decreases because it will try to minimize the change in effect, right? So pressure increases, volume will decrease. Volume will decrease means the number of moles per unit volume, number of moles per unit volume increases because volume is decreasing. So the equilibrium will shift in the direction in the direction where where there is lesser number of moles, lesser number of moles. So with increasing pressure, the equilibrium will go towards the lesser number of moles. Let's say, let's take an example and understand this. If you have N2 plus 3H2 gives 2NH3, okay? N2 plus 3NH2, 3H2 gives 2NH3. So in this case, if you increase pressure, right, then on the reactant side, you see there are four number of moles. We have four, three plus one, four. Here we have two only. So as pressure increases, like I said, the equilibrium will shift in the direction where there is lesser number of moles. It means it will be a forward shift over here, right? So like this also you can do. Sometimes you'll get some confusion in this, like what to think and what not to think over here. So we have a mathematical relation also that you can think of here. So let's see, we understand, we discussed that particular way also. So for that, we have three different conditions we need to take. Like the first case where delta NG equals to zero. Means, suppose we have a reaction in which delta NG equals to zero. It is under pressure only we are doing, right? So for this, suppose I'm assuming a reaction, A gas gives B gas. Delta NG equals to zero for this one. So what is Kp for this? Kp is equals to the partial pressure of B by the partial pressure of A, which is mole fraction of B into total pressure, mole fraction of B into total pressure, right? Pt and Pt will get cancelled and we are left with xB by xA, right? So Kp equals to, Kp equals to xB by xA we are getting. You see this equilibrium constant in terms of pressure that is Kp it is independent of the total pressure, independent of Pt, the total pressure because Pt and Pt will get cancelled, okay? That's why we say total pressure has no effect on equilibrium constant and there's no change in equilibrium here. So right down here, equilibrium constant is independent of, equilibrium constant is independent of total pressure. Hence there is no effect, right? No effect of pressure. For those reactions whose delta N is zero, keep that in mind, any doubt? Okay, all of you finished, copied this? Okay, this is one condition where delta NG equals to zero. Suppose we have the another one in which delta NG is not zero. Suppose we have greater than zero, anything can assume like this, okay? So we can write a equation like this, A gas, 2B gas, Kp is equals to the partial pressure of B square divided by the partial pressure of A square, sorry, only A naught square. So this further equals to mole fraction into total pressure square divided by mole fraction of A into total pressure. This is equals to XB by XA into PT. Okay, copy this. Do you have a school on Thursday, guys? So no school tomorrow, Thursday, right? Tomorrow is also free. Okay, so if it is possible, then we can have some classes in the morning. Is it possible? Yeah, I'm talking about that only just a second. Oh, we have some change. Yeah, fine. So all of you have holiday from tomorrow, right? You're free in the morning. Yeah, okay, got it. No problem. For bio, Praku, you're talking about? Okay, chem. Yeah, it's for, I was asking for bio, not a problem. Like, it's not for chem actually, yeah. Okay, Shraddha, see the bio guys, you please understand this. Your syllabus is a bit lacking, okay? We wanted to finish this by January end, but the syllabus is lacking. That's why we planned for four hours, not a problem. We'll have it for three and a half hours, right? But the sir is going a bit fast because we told him to finish by January end. It's very difficult to finish with three and a half or four hours, or three hours in a week. That's why the thing is, anyways, we'll have it for three and a half hours, not a problem. Okay, yeah. Fine, guys, okay, come back. Let's concentrate here. See, we have this equation. Now, try and understand. We have to understand the effect of pressure, PT. Effect of this, total pressure, PT. So as PT increases, see what you need to think over here? We're assuming, suppose pressure is increasing here, but KP is the equilibrium constant. It is independent of pressure, if you know, independent of pressure. So obviously this won't change. This wants to be constant, right? But PT, you are increasing. Means this ratio, you have to maintain a constant value. The ratio should be. If PT increases, to maintain PT, to maintain KP constant, KP as constant, what we need to do? We need to increase the mole fraction of A. Then PT and XA, whatever increase in PT, we have similar amount, XA will increase, and hence the ratio will become constant. XA increases, XA is the mole fraction of A, right? And that is only possible when the reaction goes in backward direction. Under this condition, backward reaction is favorable. Tell me, did you get this? KP is independent of pressure. So there is no change in KP. You are increasing PT, so XA will also increase. So that the ratio becomes constant, right? Similarly, what happens if total pressure decreases as PT decreases? PT decreases means XA will also decrease. It means we have forward direction here. Okay. So this is the effect of pressure. We have been delta NG is greater than zero. One last case we are left with, that is delta NG less than zero. Could you tell me what happens in this case? I want you to try this on your own, delta NG less than zero. Okay. What happens as PT increases? And what happens as PT decreases backward or forward? Could you try this? Just like we did the previous one. Yeah. Tell me what happened as pressure increases forward or backward? Forward and then backward. Okay. How do we do it? Assume a reaction like we did in the previous one. We have supposed 2A gives B, anyone you can assume. 2A gives B. So it's KP would be, in short if I write down, it is XB by XA into 1 by PT, isn't it? So as pressure increases, XB increases, XB increases means forward. And here we have backward. Yes, any doubt in this? So easily you can do this. Now you'll see, we'll take the same reactions that we did when we discussed this particular thing theoretically. N2 plus 3H2 gives 2NH3. So suppose if this reaction comes into the test and they'll ask you what happens if pressure increases, PT increases. What do we do? As PT increases, you can do it two different ways. For this reaction, you'll find out delta NG. Okay, delta NG is less than zero. So we know when delta NG is less than zero, PT increases forward shift. And you can think of, like this also you can think of. So PT increases, we have forward shift. Means more amount of NS3 forms in this reaction. So here we are getting forward. If you look at this method, the theoretical one that we did initially. Here also we were getting the same answer, you see. This one, P increases forward shift. Okay, so you can do it either by this way, number of moles per unit volume increases. Equilibrium will shift in that direction where the number of moles is less. Okay, in that way also you can think, but it is theoretical way. With mathematical, it's not like you have to write down all the steps again and again. You can do it directly in one step if you do some practice. Right? So both way, whatever you feel easy, you can do that. Clear? No doubt. Okay, next slide down. The next factor we have, effect of fourth one, effect of addition of inert gas. Suppose we have a gaseous mixture and in that gaseous mixture, we are adding an inert gas. Then what happens? Yeah. So we have here two conditions. We can add inert gas either at constant volume or at constant pressure. Constant volume or constant pressure. Right? So what happens if we added constant volume and constant pressure, both conditions we are going to see you. See, first one you write down in this, the first one A is at constant volume. Right? So it means we are adding inert gas, but there is no effect in the volume. Okay? So suppose we have a reaction N1A plus N2B gives N3C plus N4D. This is the reaction we have. N1, N2, N3, N4 are the respective number of moles. Yeah? So if you write down, gaseous is supposed to be taken here. If you write down KP, that would be what? Mole fraction of each component, PT will get cancelled. Right? Or if I write down partial pressure of C into partial pressure of D by partial pressure of A into partial pressure of B. Okay? Now if you talk about the partial pressure of A, we'll write down simply the number of moles of A into RT by V, constant volume we have. So V won't change. Right? We can also write this as mole fraction of A into total pressure of PT. Okay? This would be equals to mole fraction of A is N1 divided by N1 plus N2 plus N3 plus N4. Right? And if you have added inert gas, so number of moles of inert gases Ni into PT. PT is what? Again, total number of moles RT by V. So point here you see, if you write down this particular expression PA, whether this inert gas is present or not, right? This will not have any effect on the partial pressure of A because what happens when you add this inert gas at constant volume, right? Then this number of moles, total number of moles here and total number of moles here Nt will get cancelled, isn't it? Well, both are total number of moles only, right? So this expression and this one will get cancelled. So the partial pressure of A, still we are getting N1 RT by V. It is same only, there's no change in this, which was initially present, right? Which was initially was this. And finally, if you add inert gas, then also it is this only PA. So what we can say, if you add an inert gas at constant volume, this won't affect the partial pressure of any of the component. And hence, there's no effect on equilibrium. Addition of inert gas at constant volume has no effect over here. Yes, tell me. All of you understood this? Yeah, so write down the statement. First of all, you copy down this and then the statement to write down. Yeah, statement write down. Addition of inert gas at constant volume. Addition of inert gas at constant volume. At constant volume does not have any effect on equilibrium. Does not have any effect on equilibrium. Since the partial pressure, since the partial pressure is unchanged. Done since the partial pressure is unchanged. There's no change in the partial pressure. Yeah, done. Okay. So this happens at constant volume. There is no any effect. Second one you see at constant pressure, if you add, then what happens? Okay. Now here also we have three different cases. That is a case of Delta NG. A, if Delta NG equals to zero, right? So let's assume a reaction over here. That is A gas gives B. So here if I write down the KC, that is the concentration of B by concentration of A. That is the number of moles of B by volume and number of moles of B by, sorry, A by volume. Volume volume gets cancelled and B by NA. So you see here, this equilibrium constant is independent of volume. Pressure is already constant and volume is not changing. It is independent of volume, right? Okay. So hence the pressure has no effect on this equilibrium constant. And right down here, KC is independent of, there's no volume term here. KC is independent of volume, independent of volume. Therefore, there is no change, yeah? Therefore, there is no change here. So Delta NG equals to zero, pressure does not have any effect. If you have constant pressure, then there's no effect over here, since it is independent of volume itself. Pressure is already constant. Similarly, if you think of Delta NG greater than zero. So it's KC would be, if I write down this directly, NB is squared by NA into 1 by V, isn't it? Is it right? Tell me. So what happens here? You see, as volume increases, similarly you need to think. As volume increases, obviously KC won't change with volume. So it has to maintain constant. Then NB will increase, right? And NB increases when we have forward shift, right? In this case, if volume decreases, then we have backward shift. Just opposite you can think of, yeah? Okay, next slide down, Delta NG less than zero. No, to increase NB, number of modes of B increases, no. So it is possible when this goes in forward direction. There are only B forms, right? Now the third and the last case in this we have, when Delta NG is less than zero. Could you tell me what happens under this condition if volume increases and volume decreases? Yeah, tell me what happens here. Volume increases Delta NG less than zero. So we can think of a reaction, say this one. We have 2A gives B, right? So it's KC would be NB by NA into NA squared into V. So as V increases, as V increases, NA will also increase, that is backward shift and this one is forward. Got it, clear? Understood? So this is the effect of addition of an inert gas at two different conditions. Okay, any one they can ask you an exam, important we have this point. Okay, the last one we have in this, that is addition of a catalyst, write down. Catalyst, write down catalyst does not affect the equilibrium state. Catalyst does not affect the equilibrium state, but it affects the rate of forward and backward reaction. But it affects the rate of forward and backward reaction. See, clearly understand what I said. Catalyst does not affect the equilibrium state, but it affects the rate of forward and backward reaction. Okay, forward and backward reaction. It's two different things. It helps the reaction to attain equilibrium sooner or later. It does not affect the equilibrium state. Okay, see what does it mean? Suppose we have a reaction A and B and we have an equilibrium state somewhere in between, right? So this is the equilibrium state we have, T equilibrium here. Okay, so this equilibrium state won't change in presence of catalyst also, right? What does it mean? That the position of equilibrium is exactly same. It won't change with in presence of catalyst. This means that the time required to achieve this equilibrium state may be less or more. That we can change the time required to achieve the equilibrium state. That we can change, okay, with the help of catalyst. If you talk about catalyst, we have of two types. Generally, when we say catalyst, mainly we have two types of catalyst, that is positive catalyst or negative catalyst. When we say catalyst only, we mean that it is a positive catalyst, that is understood, right? Positive catalyst is the catalyst which enhances the rate of reaction, right down this point here. Positive catalyst is the catalyst which enhances the rate of reaction. Rate of reaction when I say means both forward or backward, understood? Positive catalyst, what I said, it increases the rate of reaction, increases ROR and when I say ROR means rate of forward reaction increases as well as rate of backward reaction also increases. Yes, so that's why you see this A proceeds to the faster rate, B proceeds to the faster rate and achieve the equilibrium state sooner. Negative catalyst, it decreases the rate of reaction. That is the only difference we have. Means rate of forward reaction decreases, rate of backward reaction also decreases. So it takes more time to achieve the equilibrium state. There's one more thing we have in this, one note you write down. First of all, you copy down this, then you write down one note, write down the note, write down the note, write down catalyst. Catalyst affects the rate of reaction. Catalyst affects the rate of reaction depending upon its nature, right? Affects why I'm saying because it may increase or it may decrease depending upon the nature of the catalyst, right? So catalyst affects the rate of reaction without taking part in the reaction, without taking part in the reaction. Next line, a positive catalyst, positive catalyst enhances the rate of reaction. Yes, yes, I, I intentionally I was on mute. There was some drilling work going on, so it was noisy. So I was just writing it down and thought I'll explain a bit later. Yes, fine. Yeah, so we have this. I'll explain this graph. Let me draw this first, okay? Progress of reaction, this is reactant. And this is product. See this white graph that you see here, this is the graph without catalyst, without catalyst. And if you see the activation energy graph over here, this graph, this energy here, this difference is the activation energy for this reaction. This is EA, activation energy for this reaction, okay? So this graph is without catalyst. This graph is with positive catalyst we have. You see what happens when you use positive catalyst, the activation energy decreases by this value, goes to this and then it comes down, product forms. If you think of this graph, this graph is for negative catalyst, negative catalyst. You see the activation energy increases. Originally it was this and now this is this much more, right? So activation energy increases and hence the reaction proceeds with a slower rate. You can also think of this is the barrier we have. You need to jump this barrier in order to form the product, right? So if the height of this barrier is more, it will be difficult to cross this, right? Same thing we have over here. More of this height, slower will be the rate of reaction, correct? That's why if activation energy increases, reaction rates goes down. If it decreases, reaction rates increases, correct? So catalyst you must take care of. Catalyst does not take part in the reaction. It either increases or decreases the rate of reaction depending upon its nature. How? By increasing or decreasing its activation energy. Got it? Clear? In and out in this? Many times I've asked this question in the test that the effect of catalyst, it affects the activation energy, not the initial and final state. If you see, it is same only, right? The product reactant is this for all the three reactions. Product is this for all the three reactions. So product is same, equilibrium state won't change. Like I said, if the equilibrium state is here without catalyst, it will be here only in presence of catalyst. It won't shift here and there. But the rate at which the equilibrium is attaining, that rate will be different depending upon the nature of the catalyst. So this is the four, five different cases we have for to understand the change in the equilibrium state. Now we have two, three more examples in this, like the application of Lee Chatelier's principle. Okay? Application of Lee Chatelier's principle right down. Last two, three things. The first one is the boiling of water. This is actually physical equilibrium where the phase change is involved. Boiling of water. Try to understand this carefully. So H2O liquid converts into H2O gas. That is the boiling of water. So what happens on the boiling point of water if pressure increases or decreases? Right? So you see what happens? As pressure increases, as pressure increases, then what happens volume decreases? Then the equilibrium will shift towards the lesser volume according to Lee Chatelier's principle. Tell me which one has lesser volume, liquid or gas? Which one has lesser volume, liquid or gas? Liquid. Right? Liquid has a lesser volume. Gas will occupy more volume. Right? The volume of liquid is lesser. It means when you increase pressure, this will shift towards the lesser volume. Means backward shift we have. And when you have backward shift, more water forms. More water forms. Now here we need to understand one thing over here. So the condition of temperature is same over here. We are not changing the temperature. No, the condition of temperature is same. Right? And we are increasing the pressure and more amount of liquid is forming. If more amount of liquid is forming, other way can we say that, that vaporization is difficult here? Can we say that? Or boiling is difficult? Yes. If more liquid is forming, we have the same temperature, but more liquid is forming. It means the boiling is difficult now. And boiling will be difficult when the boiling point has been increased. This means what? That as you increase the pressure, we have backward shift, more water forms. It means the boiling point increases. That's why the boiling becomes difficult. It requires more temperature to boil. So the relation we have here, as pressure increases, boiling point of water also increases. Yes. Tell me, end out. I'll explain it. See what is happening here, you see. Pressure increases. We have backward shift. Backward shift means more water forms. Do you have any doubt in this? Pressure increases. Why backward shift? Did you understand till here? Please respond, all of you. Till here, did you understand? It's fine. Here, it's fine. This one also, it's fine. Yeah. Last point, yes. I can understand last one. Last point, you see, what is the condition now? You have liquid and vapor mixture at equilibrium. And you increase pressure, more amount of water starts forming. So if this is going in backward direction, means opposite if you think, then the liquid to gas conversion becomes difficult. Temperature, you are not changing. Temperature is the same only. But liquid is not converting into gas. That means the boiling has become difficult. And boiling has become difficult means what? The boiling point has been increased. Now it requires more temperature to boil under this condition. And that is why it is not getting boiled, but it is forming, it is producing more amount of liquid for you. Or the reaction is going in backward direction. So since the boiling is difficult, hence we can conclude that the boiling point has been increased. Hence water is not converting into gas, but it is happening other way. Clear now? Clear? All of you? Yeah. This is very important. Okay. They have asked these questions in J NEED exams and all. It's very important. This one is boiling point, boiling of water. Now next one you think over here, see over here, that is melting of ice. Melting of ice means we have H2O solid to liquid H2O solid. Yes, yes, yes. So that's what you see, I3. The condition is same. No, we are not changing the temperature. But what we observe, just to increase the pressure, we observed what that more amount of liquid is forming. It means the boiling is difficult. Boiling is difficult. That is only true when the boiling point has been increased. Now H2O gas to H2O liquid, H2O gas we have what? Sorry, H2O solid. It's not gas, it's solid. Okay, ice. H2O solid is ice and this is water we have. Correct. We know ice has less density and less density means more volume. This has low density and low density means less volume. So if you increase the pressure, what happens? If pressure increases, we'll have the volume will decrease. Same thing. Reaction will go in which direction? Forward or backward, could you tell me? Okay, I have written less and here it is more. Ice density is less, no. This is more, tell me. Which direction? Forward or backward? Why backward here? See, pressure increases volume decreases. It means to minimize this change, the equilibrium will shift in that direction where the volume is less, where we have lesser volume. Lesser volume we have towards the right, forward. So hence the reaction will go in forward direction. Correct. No doubt in this. Yeah, no doubt, correct. Forward. Now you see similarly you need to think over here. We have the same temperature condition because temperature we are not changing. We are just increasing the pressure. But what is happening here? We are getting more amount of liquid. Means the ice is converting into liquid. Means liquefaction becomes easier here. Then the melting point has reduced, very good. So we can say the melting point is decreases now. No, that's what. We are increasing the pressure. So you have to minimize this effect, the increase in pressure. And to minimize this only, it will go towards the lesser volume. One step you have to go back to Jita. Yes, clear? Right. So always keep this in mind that as pressure increases, we'll have the melting point also decreases. This is the case of ice and water. Say it is the exceptional case we have. Because generally solids has more density than liquid. Okay, solid has more density than liquid. Okay, but this is the exceptional case we have in case of ice and liquid. What happens with normal solid that also we'll see. Correct, copied. So if you have any other substance, for example, if you talk about the melting of other substance, anyone, suppose we have any element x solid and it converts into s liquid. Solid, we know it has high density. And high density means there's low density. High density means more volume. Oh, less volume, sorry. Less volume. And this has more volume. So as pressure increases, what happens? Volume decrease to minimize the effect. Forward or backward, could you tell me? Forward or backward? Forward or backward here? Backward, very good. We'll have the backward shift. Now in backward shift, what happens? More solid forms, right? More solid forms. And this is possible when the melting point increases or decreases. Increases. Melting point increases. So melting becomes difficult here. Solid to liquid conversion is difficult, that means. Yes, understood all of you. One last example we have in this. If you talk about diamond and graphite, diamond and graphite. So diamond has high density and this has low density. This has low density. High density means less volume, same thing. Here we have more volume. As pressure increases, volume decreases. Means backward shift. Means more diamond forms here. Melting point also increases. This is a normal compound. So we know as pressure increases, melting point increases. As temperature increases, one more thing you must keep this in mind. As temperature increases, this is forward shift. Forward shift, more graphite will form. This actually you need to memorize. This change, diamond to graphite, this change is endothermic in nature. This factual thing, you must keep this in mind. And once you know this, you can understand the effect of temperature also. In endothermic reaction, tell me what happens if temperature increases? Forward or backward shift? We have done it just now. In endothermic reaction, what happens with increasing temperature? Forward shift. Right. So forward shift means more graphite will form. So for diamond and graphite, these two conditions, you must remember pressure and temperature plus it is endothermic. Okay. So this is it for this chapter. Lee Chatelier's principle, the last topic we had in this chapter that we have finished. Now next, we'll start with Ionic equilibrium.