 So, in this situation the molar concentration is actually we write square bracket of A it essentially means the molar concentration of A going forward. Reaction becomes equal to the rate of reverse reaction which when equilibrium is achieved ok. As the formation of the molecules of N2O4 ok, N2O4 formation is what shows that it is pretty that the forward reaction is actually equal to the backward reaction both of them are happening simultaneously. Concentration has come up from N2O4 also keeps on increasing N2O4 are stable ok. So, from this point onwards start with N2O4 you will have the concentrations of N2O4 much much larger than the concentrations of NO2. So, this is the point where N2O4 is generating concentration of NO2 also as increased. So, from here you start with you will realize that their concentration decreases while the other one's concentration is definitely going to increase. Concentration of N2O4 to N2O4 turns out to be 4.63. So, this basically showing us equilibrium actually N2O4 experimentally all of them are more or less the same, but concentrations of NO2s and N2O4s and the equilibrium constant concentrations again are different because you started with different substances concentration ok. In different situations we are talking only about the rate of reactions here we are actually talking about the equilibrium forward reaction rate of backward reaction both are being considered. In the rate constant and 1 the reaction has almost proceeded to the product this thing but it turns out to be our concentration. So, P into N by V into RT is nothing but CRT right. Concentrations of P here may be concentration of P and concentration of Q whatever it is. You can also write this in terms of partial pressures of P but if you have to do that then this P has to be written on the pressure of you know NO is of N1 substituting the values we end up getting pressure of N340 as in the gaseous form. So, you realize that the one that is in the concentrations of solids and liquids are pretty high. So, if you write the expression of equilibrium here we will get CO2 into CO plus CO3 very high in concentration. So, generally we would neglect their expression for the equilibrium constant is for a particular equation. If you change the equation the equilibrium constant is going to change. Similarly, if you change the temperature then also the equilibrium constant changes. If the reaction is expressed as the sum of 2 or more reactions then the equilibrium constant for the entire net band is basically a product of what will be the different changes in other species. We find the concentrations we find out how much of X and the initial constant taken is 0.063 initially. Now, let us see how do we really solve for this question. So, you will find that let X be the change in the concentration of Br2. This Br2 will come up to become twice of N.02. So, this is called as the I expression as minus V plus or minus root of V square minus 4C. The reaction would be all the 10 to the power minus 3. So, you multiply the entire expression by 1000. You will realize that you will end up getting 0.9 in 5X square plus 1 to the power minus 5 minus 3X gives out 7.38 into 10 to the power minus 5. So, you can multiply this expression by 1000 and it will simplify our calculations a bit. Again, we use the same for this reaction. Let us say the K equilibrium is really small for a particular reaction. Basically, it means that it is not going to proceed meaning that the equilibrium concentrations would be the same as initial concentrations of the reactants. In this situation, since X is very small, you can always approximate 0.2040 only because that is what is really going to give you the concentration of reactants. In K80, the equilibrium and initial concentrations is around 3 orders of magnitudes. Go for this approximation and you realize that this approximation always works out to give you the solutions for this calculations, but it will really solve the problem. Look at the concentrations at the equilibrium at 600 Kelvin and 2000 Kelvin where the Kc for both of these are individual reactions which is N2O4 divided by N2 square minus 63 into 10 to the power minus 3 at opposite direction the equilibrium constant becomes the reciprocal of the equilibrium constant of the original one. The reaction course related by substituting the initial concentrations of the reactants and products do not know whether the equilibrium is being achieved or not. What we will generally do is you will find the reaction quotient at any point of time. So what is the reaction quotient? Reaction quotient is nothing but substituting concentrations at any point of time and seeing what is the product of the expression of equilibrium constant we generally move towards any direction as such like Qc and Qc. So when Qc, you know, the reaction has to proceed more towards reactants. This topic is to understand Leigh-Shatelier's principle where we will understand So let's say that here by external space we mean that some of the equilibrium may be changed in the concentration, you know. So let's look at this with the N2 and the Hebert's ammonia. Obviously the reaction will have more concentration in it since now and the equilibrium and ammonia were the concentrations of hydrogen and nitrogen. So the way is that you know you can actually have hydrogen is now improved from the earlier ones to much more. Similarly, the nitrogen concentration also has... Let's look at the expression, you know, for a general reaction like AA plus BB giving. The general expression, if you start adding, it's going to shift the equilibrium towards it. Where you have, you know, the concentrations of dimensions either you change the reactants or the products and the situations where you have gaseous is the pressure, it will shift more to the pressure and therefore it will move towards the reactant side. Similarly, if you increase the volume, now since the volume is increased, you again need more number of moles. So you will find that it will go towards the side which has more number of moles, which is A and B. Leigh-Shatelier's principle says that the reaction moves in that direction first. So the N204 system is where if you increase the temperature, you take this. Why? Because if you are increasing the temperature, you know, N204 giving, you know, is actually a reaction which is favored by temperature. If you increase the temperature, more of NO2 is formed and therefore it is more brown in nature, as you can see in this hotter bulb. But if the temperature more of N204 is formed. So this reaction is favored by the increase in temperature and therefore the K becomes much more as... Because decreasing K means that the temperature actually matters and equally for endothermic. Similarly, for decreasing temperature, it will be the reverse. If the temperature for ex-thermic processes is going to increase because it will try to warm the products will be formed. This energy changes and equalizes as a higher activation. So reactions and equilibrium etc. You know that nitrogen and hydrogen basically gives ammonia. That is, it is basically endothermic process pressure. You will find that the mole percent of pressure actually helps. If you increase in the temperature, it would resist that. You know, endopy temperature because they can absorb more of heat but the exothermic will not be favored by this reaction. So basically, if you really see why we actually keep on... This keeps on happening because as we take the ammonia, you will find that N2NH2 will keep on reacting more because we are taking out substances from NH3, equilibrium. So let's just happen. What is the concentration? Yes, the equilibrium shifts. If you change the pressure, yes, the equilibrium does shift. Again, the equilibrium constant does not change. If you change the volume, again, there is this equilibrium but if you find that the equilibrium does shift but there is also a change in the equilibrium constant. So the equilibrium constant does depend on the temperature but it does not depend on the catalyst as a final one. So if you put the forward and backward reactions, the equilibrium does not shift anymore and the equilibrium constant also does not change. So this brings us to the end of this presentation and we will be sharing some more problems with you. Pretty questions, you please post it out on so much. I'll get back to you.